# ATAR Notes: Forum

## VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TrueTears on November 26, 2011, 10:07:42 pm

Title: VCE Specialist 3/4 Question Thread!
Post by: TrueTears on November 26, 2011, 10:07:42 pm

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Original post.
similar to the methods one Methods [3/4] Summer Holidays Question Thread! post away your questions from your summer holidays self-studying, everyone can discuss and benefit! I'll try answer as much questions as possible too ^^
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 27, 2011, 12:00:02 pm
Let w = 2cis(θ) and z = w + 1/w
Show that z lies on the ellipse with equation (x^2)/25 + (y^2)/9 = 1/4
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on November 27, 2011, 12:28:30 pm
z = w + 1/w = 2cist + 1/(2cist) = 2cist + 1/(2(cost + isint)) = 2cist + (cost - i sint)/(2) = 5/2 cost + 3/2 i sint
parameters:
x = 5/2cost
y =3/2 sint
convert into cartesian form and you get the equation of the ellipse.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on November 27, 2011, 08:47:43 pm
what do you mean by complicated derivatives? the method of 'antideriving through derivatives' is called integration by recognition, which is just a more abstract form of integration by parts. not sure if this answers your question, but try wikipedia-ing integration by parts.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 27, 2011, 08:48:49 pm

heys you know the spesh derivatives that are complicated, it can done be through 'anitderiving through derivatives right'?
is it true that it is possible to antiderive through complicated derivatives? is it way of vce level?

if its possible can any of you guys show me an example of it?

I'm not sure what you mean by anti-deriving through derivatives...
Perhaps you mean is it possible to find an anti-derivative when given a similar derivative? If that's your question, then the answer is yes. Here is an example:

Given that dy/dx = cos(2x), d/dx (sin(2x)) = 2cos(2x) and that x = pi/4 when y = 5, solve for y.
So first you convert the integral of cos(2x) into 1/2 * Integral of 2cos(2x)
Then you use the information they gave you to say:
1/2 * Integral of 2cos(2x) = 1/2 * sin(2x) + C
Now you use the initial conditions to solve for C:
y = 1/2 * sin(2x) + C
5 = 1/2 * sin(pi/2) + C
5 = 1/2 * 1 + C
C = 5 - 1/2
C = 9/2
y = 1/2 * sin(2x) + 9/2

That is called a "differential equation" if you're interested.

what do you mean by complicated derivatives? the method of 'antideriving through derivatives' is called integration by recognition, which is just a more abstract form of integration by parts. not sure if this answers your question, but try wikipedia-ing integration by parts.

Ye know too much for a year 10 student!
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 27, 2011, 09:11:07 pm
Someone please tell me where I went wrong in this question!

Given that z = -3/2 sinθ + 5/2 i cosθ, show that |z - 2i| + |z + 2i| = 5

z - 2i = (-3/2 sinθ) + (5/2 cosθ - 2)i
|z - 2i| = sqrt((-3/2 sinθ))^2 + (5/2 cosθ - 2)^2)
|z - 2i| = sqrt(9/4 sin^2(θ) + 25/4 cos^2(θ) - 10cos(θ) + 4)
|z - 2i| = sqrt(9/4(sin^2(θ) + cos^2(θ)) + 16/4 cos^2(θ) - 10cos(θ) + 4)
|z - 2i| = sqrt(9/4 + 4cos^2(θ) - 10cos(θ) + 16/4)
|z - 2i| = sqrt(4cos^2(θ) - 10cos(θ) + 25/4)
|z - 2i| = sqrt((2cos(θ) - 5/2)^2)
|z - 2i| = 2cos(θ) - 5/2
Similar process for |z + 2i|:
z + 2i = (-3/2 sinθ) + (5/2 cosθ + 2)i
|z + 2i| = sqrt(9/4 sin^2(θ) + 25/4 cos^2(θ) + 10cos(θ) + 4)
|z + 2i| = sqrt(9/4 + 16/4cos^2(θ) + 10cos(θ) + 16/4)
|z + 2i| = sqrt(4cos^2(θ) + 10cos(θ) + 25/4)
|z + 2i| = 2cos(θ) + 5/2
|z - 2i| + |z + 2i| = 2cos(θ) - 5/2 + 2cos(θ) + 5/2
|z - 2i| + |z + 2i| = 4cos(θ)

Where did I go wrong? I was supposed to get the answer 5 but instead I got 4cos(θ) :(
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on November 27, 2011, 09:42:39 pm
|z - 2i| = sqrt((2cos(θ) - 5/2)^2)
|z - 2i| = 2cos(θ) - 5/2
here's the error. cosθ =< 1, which means  2cosθ - 5/2 =< -1/2
so sqrt((2cos(θ) - 5/2)^2) = abs(2cos(θ) - 5/2) = 5/2 - 2cosθ
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 27, 2011, 09:51:46 pm
I never knew you had to do that...
What if it was sqrt((cosθ - 1/2)^2) ?
How would you know whether to take the positive or negative solution?
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on November 27, 2011, 09:55:27 pm
oh absolute values!

if it were ur alternative case, then itd be cos(t) <=1 hence cos(t) -1/2 <= -1/2
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on November 27, 2011, 09:56:32 pm
I never knew you had to do that...
What if it was sqrt((cosθ - 1/2)^2) ?
How would you know whether to take the positive or negative solution?

depends on the restrictions of θ. if none are specified, then you'll have two solutions.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 27, 2011, 10:00:12 pm
That makes sense... actually I think I remember a similar problem when taking an integral which ended up with an absolute value and I was unsure whether to write ln(5 - x) or ln(x - 5) so I had to use the initial conditions to see which solution it was.
This is basically the same concept.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on November 27, 2011, 10:00:23 pm

heys you know the spesh derivatives that are complicated, it can done be through 'anitderiving through derivatives right'?
is it true that it is possible to antiderive through complicated derivatives? is it way of vce level?

if its possible can any of you guys show me an example of it?
yup exactly what brightsky said, this is just integration by parts, you don't need to know this for spesh either, however it's not too hard to learn, it's just derived from the product rule, wikipedia is nice like brightsky said, however just for heck of it, i'll provide an example

basically to derive it, check wiki, you simply just antiderive the product rule, then after rearranging this is the result:

$\int u\, dv=uv-\int v\, du\$

basically u and v are functions, ill show u an example and this will make more asense, note that here im stressing more about the application of this "formula" rather than going through rigorous details etc

eg $\int x\sin(x) dx$

first we need to guess which function is u and which is dv, ie what i'm doing is basically "equating equations" (again im not stressing formality, simply showing you the mechanics) so what im doing is this:

we have the integration by parts rule  $\int u\ dv$ then we got our "equation" that we need to integrate, ie $\int x\sin(x) dx$

so its like saying $\int u\ dv = \int x\sin(x) dx$

so we have to guess, what is u and what is dv, ie, lets guess u = x and dv = sin(x)dx

yes if you're wondering, there's another guess we cudda taken, ie, u = sin(x) and dv = x dx [as you will see only one "pair" would work]

so if we let u = x and dv = sin(x) dx

then du = dx and v = -cos(x)

then all we gotta do is sub this into the formula! $\int u\, dv=uv-\int v\, du\$

so it becomes $\int x sin(x) dx = x(-cos(x))-\int -cos(x) dx$

but we know how to do $\int -cos(x) dx$

and so the rest is trivial

you go to try the other "guess" and you will see that the 2nd integral is not something we can integrate easily, so picking the first guess is better.

hopefully that makes a tiny bit more sense, note integration by parts just takes practise, there's no set method, you gain experience as you do more, so lets say

$\int \log_e(x) dx$ [note this is actually $\int 1 \cdot \log_e(x) dx$]

here u can either do let $u = \log_e(x)$ and dv = 1dx or u = 1 and $dv = \log_e(x) dx$ but it is very clear why the 2nd guess doesnt work!

Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on November 27, 2011, 10:12:51 pm
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on November 27, 2011, 10:19:26 pm
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
heh, no need to worry, just do some self study and this stuff will be easy ^^
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on November 27, 2011, 10:25:04 pm
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
heh, no need to worry, just do some self study and this stuff will be easy ^^
From the textbook? I don't think I'm getting that till late Jan next year.. because I'm going overseas to Japan!! I'm thinking of taking some material over there though, are there any online places where I can learn the basics?
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on November 27, 2011, 10:26:07 pm
ill send u the ebook? pm me ur email brah :D
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on November 27, 2011, 10:32:46 pm
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
heh, no need to worry, just do some self study and this stuff will be easy ^^
Good times will come, young padawan.
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on December 05, 2011, 07:08:09 pm
Illustrate this on an Argand diagram:

$|z-6|-|z+6|=3$

Can someone explain why there's only HALF a hyperbola?

I've done this so far:

$|z-6|-|z+6|=3$

let $z=x+yi$

$|x+yi-6|=3+|x+yi+6|$

$\sqrt{(x-6)^{2}+y^{2}}=3+\sqrt{(x+6)^{2}+y^{2}}$

$(x-6)^{2}+y^{2}=9+6\sqrt{(x+6)^{2}+y^{2}}+(x+6)^{2}+y^{2}$

$x^{2}-12x+36+y^{2}=9+6\sqrt{(x+6)^{2}+y^{2}}+ x^2+12x+36+y^{2}$

$-24x-9=\sqrt{36((x+6)^{2}+y^{2}}$

$576x^{2}+432x+81=36((x+6)^{2}+y^{2})$

$576x^{2}+432x+81=36x^{2}+432x+1296+36y^{2}$

$540x^{2}-36y^{2}=1215$

$\frac{4x^{2}}{9}-\frac{4y^{2}}{135}=1$

Asymptotes: $y= \pm\sqrt{15}x$

Answers only show right half of the graph.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 05, 2011, 07:26:08 pm
i'll give you a hint, it lies in this step

$-24x-9=\sqrt{36((x+6)^{2}+y^{2}}$

$-(24x+9) = \sqrt{\text{positive number}}$

now think about the restrictions on x ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on December 05, 2011, 07:38:38 pm
ill send u the ebook? pm me ur email brah :D
do you have ebook of heinemann spesh book? ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 05, 2011, 07:47:01 pm
ill send u the ebook? pm me ur email brah :D
do you have ebook of heinemann spesh book? ;)
nope :(
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 05, 2011, 07:56:35 pm
Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
=6( 2tan x/1-tan^2 x)
= 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1

ehhh
i probs made some huge and obvious and rather silly mistake:p

Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 05, 2011, 08:02:03 pm
3+tan^2 x = 1
root3 +tan x = 1

lol u cant do that
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 05, 2011, 08:04:51 pm
yeahh thats right haha,
hence 'im stuck' :p

I know i shouldnt have even gone there..., so do you think it was something to do with not converting tanx to sin/cos ?
or just some arithmetic?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 05, 2011, 08:05:32 pm
Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
=6( 2tan x/1-tan^2 x)
= 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1

ehhh
i probs made some huge and obvious and rather silly mistake:p

Use the tan double angle formula then see where you can get to.
tan(2x)=(2tan(x))/(1-tan^2(x))

Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 05, 2011, 08:07:39 pm
Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
=6( 2tan x/1-tan^2 x)
= 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1

Use the tan double angle formula then see where you can get to.
tan(2x)=(2tan(x))/(1-tan^2(x))

:\ i suck haha
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 05, 2011, 08:13:11 pm
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, $x=\pi$ or $x=2\pi$  or no solution

So x=0, $\pi$, $2\pi$

EDIT: forgot sols
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 05, 2011, 08:14:35 pm
x also equals pi and 2pi from b^3's working
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 05, 2011, 08:16:16 pm
x also equals pi and 2pi from b^3's working
Yeh I'm a bit rusty, I clicked post, then realised and my com froze.

Moderator action: removed real name, sorry for the inconvenience
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 05, 2011, 08:18:27 pm
yeah other mistake was

4 tan x(1-tan^2 x) = 12 tan x

cancelling the tan killed off solns
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 05, 2011, 08:19:03 pm
yeahh thats right haha,
hence 'im stuck' :p

I know i shouldnt have even gone there..., so do you think it was something to do with not converting tanx to sin/cos ?
or just some arithmetic?

Cancelling is a big nono ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 05, 2011, 08:21:29 pm
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, $x=\pi$ or $x=2\pi$  or no solution

So x=0, $\pi$, $2\pi$

EDIT: forgot sols

legend! thanks bro!

just ah... wouldnt:
-8-4tan2(x)=0
-4tan2(x)=0 +8
tan2(x)= 8 /-4
tan2(x)=-2 ?

so yeah no solution because u cant take the sq.root of a negative number anyways!

And thanks alot guys, will remember this for next time! :D
+1

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 05, 2011, 08:24:32 pm
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, $x=\pi$ or $x=2\pi$  or no solution

So x=0, $\pi$, $2\pi$

EDIT: forgot sols

legend! thanks bro!

just ah... wouldnt:
-8-4tan2(x)=0
-4tan2(x)=0 +8
tan2(x)= 8 /-4
tan2(x)=-2 ?

so yeah no solution because u cant take the sq.root of a negative number anyways!

And thanks alot guys, will remember this for next time! :D
+1

YES! I did it in my head, and didn't type it, I've lost it already :(. Stupid holidays :|
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 05, 2011, 08:26:27 pm
ok sweet, atleast im getting something right ahaha... yaya.

thanks again guyss
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 06, 2011, 07:43:41 pm
Was wondering what other people's approach to this question is :

Use vector methods to prove that the midpoint of the hypotenuse of a right-angled triangle
is equidistant from all vertices.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 06, 2011, 07:48:53 pm
let one side (adjacent to the right angle) be denoted by vector b, other side (the other side which is adjacent to the right angle) be denoted vector a, the hypotenuse is given by a-b, half of this vector is given by 1/2(a-b), from the right angle vertex to the midpoint; this vector is given by b+1/2(a-b) = 1/2(a+b)

compute the magnitude, the magnitude are also the same hence equidistant.
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on December 06, 2011, 10:10:07 pm
TT, $\frac{1}{2}(a-b)+b=\frac{1}{2}(a+b)$. Your attempt doesn't use the fact that it's a right angle triangle so that alerted me.

What you should now do is compute the length:

$||\frac{1}{2}(a+b)||^2=\frac{1}{2}(a+b).\frac{1}{2}(a+b)=\frac{1}{4}(a.a+2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Likewise we can show that $||\frac{1}{2}(a-b)||^2=\frac{1}{4}(a.a-2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Hence the magnitudes are indeed the same (notice this is false for certain triangles that are not right angled e.g: an iscosceles triangle with two very long equal sides and one very short side, this is a good way to check your proof, does it use all the assumptions?)
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 06, 2011, 10:28:54 pm
TT, $\frac{1}{2}(a-b)+b=\frac{1}{2}(a+b)$. Your attempt doesn't use the fact that it's a right angle triangle so that alerted me.

What you should now do is compute the length:

$||\frac{1}{2}(a+b)||^2=\frac{1}{2}(a+b).\frac{1}{2}(a+b)=\frac{1}{4}(a.a+2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Likewise we can show that $||\frac{1}{2}(a-b)||^2=\frac{1}{4}(a.a-2a.b+b.b)=\frac{1}{4}(a.a+b.b)$

Hence the magnitudes are indeed the same (notice this is false for certain triangles that are not right angled e.g: an iscosceles triangle with two very long equal sides and one very short side, this is a good way to check your proof, does it use all the assumptions?)
thats right, i got lazy and didnt end up computing the lengths lol but yeah in my head using a.b = 0 was quite obvious from the start ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 06, 2011, 11:00:35 pm
I don't get why the hypotenuse is a-b?
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 06, 2011, 11:41:24 pm
I don't get why the hypotenuse is a-b?

It's just a matter of 'how' you draw the triangle. I think the triangle is supposed to use the positive x, y axes, with the hypotenuse slanting down from the positive y axis to the positive x axis. Here I'm only saying x and y axes for illustrative purposes.

edit: actually a better visualisation: if vector b goes from origin to point B, and vector a goes from origin to point A, and AOB is the right angle, then the hypotenuse is from B to A, which is given by a-b. Draw it!
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 06, 2011, 11:57:47 pm
Easier to see why the hypotenuse = a - b, if you imagine the vertex corresponding to the right angle as the origin, O.
Therefore, the other two vertices will be represented by their position vectors,

OA = a
OB = b

And the hypotenuse, AB becomes,

BO + OA =
(-OB) + OA =
-b + a =
a - b

Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 07, 2011, 12:16:33 am
Thanks dc302,  making sense :)

Thanks to you too argonaut! Just checking, if you're going by the same diagram as dc302,  you've said ab = a-b,  but isn't that ba?  Ab would be - a + b wouldn't it?

Which leads me to asking,  the hypotenuse could be a-b or b-a yes?
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 07, 2011, 12:20:24 am
Which leads me to asking,  the hypotenuse could be a-b or b-a yes?
yeh doesnt matter, but then the vector expression for the vector starting at the right angle vertex to the midpoint would also change, the final result is the same
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 07, 2011, 12:25:31 am
Which leads me to asking,  the hypotenuse could be a-b or b-a yes?

Indeed it can be either.
But for the purposes of the (correct) solution posted by kamil, in order to find the relevant magnitudes, you end up forming the dot product,

(a - b) . (a - b)

which is equivalent to

(b - a) . (b - a)

Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 07, 2011, 07:34:33 am
Alright then!  Thanks again :)  and woops forgot to thank you before TT!
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on December 08, 2011, 11:26:08 am
Can someone please explain what this question means?
"Prove that the median to the base of an isosceles triangle is perpendicular to the base".

What is the "median to the base"?
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 08, 2011, 11:37:40 am
Can someone please explain what this question means?
"Prove that the median to the base of an isosceles triangle is perpendicular to the base".

What is the "median to the base"?

http://en.wikipedia.org/wiki/Median_(geometry) There's a nice picture too.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 09, 2011, 09:32:16 am
Hey, guys i really got stuck in this question :/, i tried yesterday night and this morning

$\int_{-1}^{1}\frac{-1}{4-(x-1)^2}dx$
$I=\int_{-1}^{1}\frac{1}{(x-1)^2-4}dx$
$\int_{-1}^{1}\frac{1}{(x-3)(x+1)}dx$

$=\frac{1}{4}\int_{-1}^{1}\frac{1}{x-3}-\frac{1}}{x+1}dx$
$=\frac{1}{4}\ [log_e\vert\frac{x-3}{x+1}\vert]_{-1}^{1}dx$
$=\frac{1}{4}(log_e(1)-log_e(\frac{4}{0})$ this is where i got stuck in, please help
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 09, 2011, 10:17:36 am
Question must be wrong. What does the answer say?
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 11, 2011, 10:16:51 pm
This question has been puzzling me for a while :/.

Part of the graph with equation $y=(x^2-1)\sqrt{x+1})$  is shown below.

Find the area that is bounded by the curve and the x-axis. Give your answer in the form
$\frac{a\sqrt{b}}{c}$, where a, b and c are the integers.

Do you know how to do linear substitution?

edit: hint: let u = x+1
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 11, 2011, 10:19:31 pm

This question has been puzzling me for a while :/.

Have you tried the substitution u=x+1 ?

.
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 11, 2011, 10:25:12 pm
Yep, so you would sub in u=x+1 and expand it out. You should get some fractional powers.

The a,b,c refers to your answer. Your answer will be in this form. For example, if I found the area to be 2sqrt(5)/3, then a=2, b=5 and c=3. Of course, these combination of numbers should be in simplest form.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 11, 2011, 10:32:31 pm
- int^(1)_(-1) (x^2 - 1)sqrt(x+1) dx
let u = sqrt(x+1), du = 1/(2sqrt(x+1)) dx
so the integral becomes:
- int^(sqrt(2))_(0) (2u^2*u^2(u^2-2)) dx
= -2 int^(sqrt(2))_(0) (u^6 - 2u^4) dx
= -2 [u^7/7 - 2u^5/5]^(sqrt(2))_(0)
= -2 (8sqrt(2)/7 - 2*(4sqrt(2))/5)
= -16sqrt(2)/7 + 16sqrt(2)/5
= (-80sqrt(2) + 112sqrt(2))/35
= 32sqrt(2)/35

hopefully didn't make any careless mistakes..
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 11, 2011, 11:27:35 pm
Thanks Guys, I will put this up incase for someone who wants to tackle a question like this later on :).

$-\int_{-1}^{1}(x^2-1)\sqrt{x+1} dx$
$u=x+1$
$x=u-1$
$\frac{du}{dx}=1$
$x^2-1=(u-1)^2-1$
$=u^2-2u$

$x=-1, u=0$
$x=1, u=2$

$=-\int_{0}^{2}(u^2-2u)\sqrt{u})du$
$=-\int_{0}^{2}(u^2\frac{1}{2}-2u^1\frac{1}{2})du$
$=-\left [ \frac{2u^\frac{7}{2}}{7}-\frac{4u^\frac{5}{2}}{5} \right ]_{0}^{2}$

$=-\frac{2\cdot 2^\frac{7}{2}}{7}+\frac{4\cdot 2^\frac{5}{2}}{5}-0$
$=-\frac{2^\frac{9}{2}}{7}+\frac{2^\frac{9}{2}}{5}$
$2^\frac{9}{2}(-\frac{1}{7}+\frac{1}{5})$=
$=\frac{2^\frac{11}{2}}{35}$
$=\frac{32\sqrt{2}}{35}$

Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 11, 2011, 11:47:40 pm
Perfect.

(PS. Just small thing about the limits of integration. Get into the early habit of not writing,

x=-1, u=0
x=1, u=2

but

x=1, u=2
x=-1, u=0

so that you dont have to do mental handstands to make sure that they are the right way up.
Yes, I know it sounds like weird advice ... but there you have it :)   )
Title: Re: Specialist 3/4 Question Thread!
Post by: mr.politiks on December 13, 2011, 04:27:54 pm
Hey, guys i really got stuck in this question :/, i tried yesterday night and this morning

$\int_{-1}^{1}\frac{-1}{4-(x-1)^2}dx$
$I=\int_{-1}^{1}\frac{1}{(x-1)^2-4}dx$
$\int_{-1}^{1}\frac{1}{(x-3)(x+1)}dx$

$=\frac{1}{4}\int_{-1}^{1}\frac{1}{x-3}-\frac{1}}{x+1}dx$
$=\frac{1}{4}\ [log_e\vert\frac{x-3}{x+1}\vert]_{-1}^{1}dx$
$=\frac{1}{4}(log_e(1)-log_e(\frac{4}{0})$ this is where i got stuck in, please help

The integrand is not defined for x=-1 (vertical asympotote), hence definite integral with x=-1 is undefined ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 14, 2011, 04:11:54 pm
COMPLEX NUMBERS

If z=x+yi find the values of x and y such that (z-1)/(z+1)=z+2

Pretty simple question you would think, I know the process in how to do it but I'm just stuck in the algebra.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 14, 2011, 04:31:27 pm
z - 1 = (z+1)(z+2) = z^2 + 3z + 2
z^2 + 2z + 3 = 0
z^2 + 2z + 1 -1 + 3 =0
(z + 1)^2 = 2i^2
z + 1 = +- sqrt(2) i
z = -1 +- sqrt(2) i
hence x = -1 and y = +- sqrt(2)
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 15, 2011, 07:27:21 pm
Makes sense, thanks.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 16, 2011, 09:35:17 pm
(http://imageshack.us/photo/my-images/528/spec2.jpg/)

http://imageshack.us/photo/my-images/528/spec2.jpg

Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 16, 2011, 09:37:28 pm
(http://imageshack.us/photo/my-images/528/spec2.jpg/)

on what? lol
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 16, 2011, 09:38:41 pm
(http://imageshack.us/photo/my-images/528/spec2.jpg/)

on what? lol

If you click modify you can copy the image link and see it. Don't know why it didn't come up thought.

EDIT: Or just quote it and copy the link.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 16, 2011, 09:43:20 pm
edited - dont know why it didnt come up! links up in the prev. post!

here it is anyways: http://imageshack.us/photo/my-images/528/spec2.jpg
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 16, 2011, 10:05:38 pm
Chp7, MQ, pg 314,

Differential Equation question.

I got a bit stuck on it for a while

Q9. Verify that $y=e^\sin(2x)$ satisfies the differential equation.
$y{}''+(4sin(2x))\cdot y=(2cos(2x))\cdot y{}'$

When a question asks you to verify something, what they mean is they want you to show that the 2 statements agree with each other. In this case, you just need to substitute y=e^sin2x into the differential equation and show that LHS = RHS.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 16, 2011, 10:08:28 pm
(http://imageshack.us/photo/my-images/528/spec2.jpg/)

http://imageshack.us/photo/my-images/528/spec2.jpg

since they want you to prove something look at the required expression for clues:

|c|^2, that reminds you of c.c, so try c.c and then use the fact that a+b = -c to expand and simplify ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 16, 2011, 10:23:20 pm
If you do that, you get: c^2 = (a+b)^2 yea?
so, c^2 = a^2 + 2ab + b^2
tf, c^2 = a^2 +2(abcosB) + b^2

so you end up with c^2 = a^2 +b^2 + 2abcosB

should be negative before the 2abcos B :s
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 16, 2011, 10:28:01 pm
no watch out, c = -(a+b)

so c.c = -(a+b).-(a+b)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 16, 2011, 10:42:53 pm
no watch out, c = -(a+b)

so c.c = -(a+b).-(a+b)

then wouldnt it end up as:   c^(2)=-a^(2)-2ab-b^(2)  ,
so c^(2)=-a^(2)-b^(2)-2(abcosB)
?
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 16, 2011, 10:46:45 pm
you cant square vectors, there is no such thing as c^2 if c is a vector.

you have to dot product the expressions
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on December 16, 2011, 10:50:00 pm
No the algebra is correct, the only confusion arises from what $B$ is.

Let's call the angle between between the vectors $a$ and $b$ by $\theta$

Then $B=\pi - \theta$ (draw the diagram)

Thus $cos(\theta)=cos(\pi - B)=-cos(B)$ hence the negative sign.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 16, 2011, 10:54:04 pm
No the algebra is correct, the only confusion arises from what $B$ is.

Let's call the angle between between the vectors $a$ and $b$ by $\theta$

Then $B=\pi - \theta$ (draw the diagram)

Thus $cos(\theta)=cos(\pi - B)=-cos(B)$ hence the negative sign.
if he writes c^2 and the expressions on a sac or exam, marks will definitely be taken off
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on December 16, 2011, 10:56:44 pm
Bad notation yes but that's not my point, my point is that the lack of a negative sign didn't come from some incorrect algebra but rather getting the angle mixed up.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 16, 2011, 11:00:13 pm
i know, just clearing things up - which is why i tempted him to redo his workings again using the correct notation ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 17, 2011, 11:31:30 am
okay so in terms of notation, i shouldve written:

|c|^2 = |a|^2 +2ab + |b|^2
i.e c.c = a.a = 2a.b + b.b ?

you cant square vectors, there is no such thing as c^2 if c is a vector.

you have to dot product the expressions

okay :)
but arithmetically, that'd mean you end up with : |c|^2 = |a|^2 +|b|^2 + 2a.bcosB

No the algebra is correct, the only confusion arises from what $B$ is.

Let's call the angle between between the vectors $a$ and $b$ by $\theta$

Then $B=\pi - \theta$ (draw the diagram)

Thus $cos(\theta)=cos(\pi - B)=-cos(B)$ hence the negative sign.

I drew the diagram. but i didnt introduce theta. i just used B haha..
can you explain why theta was needed to be introduced into this? i dont get that bit
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on December 17, 2011, 12:19:44 pm
$B$ is the angle in the actual triangle. In the formula $a.b=|a||b|cos\theta$ the angle $\theta$ is the angle between the vectors $a$ and $b$.

See diagram for which angle is which.
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 17, 2011, 03:24:34 pm
Do you have to use vectors? There's another (probably less confusing) proof with just plain geometry.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 17, 2011, 03:34:13 pm
I guess not since the question never stated use vectors! but it is vector proofs so i dunno :p
but its made sense now, i just didnt think you'd need to go that deep into it haha, but ehhh its spesh!
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 17, 2011, 05:21:02 pm
not sure if everything's been cleared up yet, but essentially what the guys have been trying to say is that in the formula: a.b = |a||b|cos(theta), the 'theta' is the angle made by the two vectors when you put their 'tails' together. so basically, the 'theta' in this case isn't angle B (since then you are taking the angle between the 'head' of vector a and the 'tail' of vector b), but the supplementary angle (picture moving vector a so that it's tail coincides with the tail of vector b. recall that you can 'move' normal vectors around and it will still remain the same vector). hence a.b = |a||b|cos(pi - theta) = -|a||b|cos(theta), with theta = angle B.

also, the geometric proof that dc302 alluded to is worth probing in order to consolidate your understanding of the cosine rule. (hint: try drawing the altitude of the triangle and do some trig off that). if i'm not mistaken, i think this particular proof is in your textbook.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on December 18, 2011, 11:49:44 am
thanks, - yeah i figured out that I had to make them tail to tail for dot product to make sense :p
then i realized it was simply pi-theta as in... straight line- angle on one side.. therefore angle on other side = pi - theta.

which book ? this is off the heinneman
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 18, 2011, 04:12:42 pm
COMPLEX NUMBERS

If z=x+yi, determine the values of x and y such that z=sqrt(3+4i)
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 18, 2011, 04:25:22 pm
let sqrt(3+4i) = a + bi, for a,b E R
3+4i = a^2 + 2abi - b^2
3+4i = (a^2 - b^2) + (2ab) i

a^2 - b^2 = 3..(1)
2ab = 4 --> b = 2/a
sub into (1):
a^2 - (2/a)^2 = 3
a^2 - 4/a^2 = 3
a^4 - 4 = 3a^2
a^4 - 3a^2 - 4 = 0
(a^2 - 4)(a^2 + 1) = 0
since we are only looking for solutions in R,
a^2 - 4 = 0
(a-2)(a+2) = 0
a = 2 or a = -2
b = 1 or b = -1
so we have two solutions:
a = 2, b = 1
or
a = -2, b=-1
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 18, 2011, 04:44:05 pm
That does correspond with the answers in the book, thanks a lot.
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 22, 2011, 10:59:45 am
COMPLEX NUMBERS

If z^n=(1+i)^n, determine the smallest value of n = element of N, so that z^n is equal to:

a) (sqrt(2))^n
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 22, 2011, 11:30:09 am
$z^{n}=(1+i)$

let $w= 1+i$

$\vert w \vert=\sqrt{2}$
$\theta =\tan^-^1(1)$
$=\frac{\pi}{4}$
$w=\sqrt{2} cis(\frac{\pi}{4})$
$z^{n}=(\sqrt{2}cis(\frac{\pi}{4})$
$=(\sqrt{2})^{n}cis(\frac{n\pi}{4})$
$=(\sqrt{2})^{n}\left (\cos(\frac{n\pi}{4}+\sin(\frac{n\pi}4)i \right$

a) $z^{n}=\sqrt{2}^{n}$ if $cos \left ( \frac{n\pi}{4} \right )=1$

and $sin\left ( \frac{n\pi}{4} \right )=0$

$n=0,8,...$

Smallest $n\epsilon$ $N$ is $n=8$

hoped i helped im a bit rusty with complex numbers
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 22, 2011, 11:36:49 am
Yes that helped, I was confused because I thought the answer was 0 but then I saw that the answer had to be a positive whole number so the arg(z) had to be 2pi and hence n=8. Thanks.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 22, 2011, 11:40:49 am
This outflow/inflow question is troublesome for me :/.

Q9. A tank containing 600 litres of water has a sugar solution of concentration 0.1 kg/L pumped in at a rate of 10 L/min. if the mixture is kept uniform and is pumped out at a rate of 8 L/min, set up a differential equation for the amount of sugar, x kg, at any time t. (Do not attempt to solve).

Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on December 22, 2011, 11:49:46 am
It should be dx/dt=0.1*10-x*8/(600+2t)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 22, 2011, 11:50:24 am
For these questions where the flow in is not equal to the flow out, it is good to use this formula.

$\frac{dx}{dt}=rate_{in}*concentration_{in}-Rate_{out}*\frac{x}{V_{0}+(Rate_{in}-Rate_{out})*t}$

$Initial \, Volume \, V_{0}=600L$
$Concnetration_{in}=0.1kg/L$
$Rate_{in}=10L/min$
$Rate_{out}=8L/min$
$\frac{dx}{dt}=10*0.1-8*\frac{x}{600+(10-8)*t}$
$\frac{dx}{dt}=1-\frac{8x}{600+2t}$
$\frac{dx}{dt}=1-\frac{4x}{300+t}$

Hope that helps.

EDIT: Explaining the equation
The first part of the equation is giving how much of "x" is following in per second.
The second part of the equation is giving how much of "x" is following out per second.
You can see this by "rate"*"concentration"
In the second half of the equation, the concentration is  $\frac{x}{V_{0}+(Rate_{in}-Rate_{out})*t}$, think of it as c=mass/volume, mass is x and the volume is changing as t changes.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 22, 2011, 11:51:36 am
It should be dx/dt=0.1*10-x*8/(600+2t)
For these questions where the flow in is not equal to the flow out, it is good to use this formula.
$\frac{dx}{dt}=rate_{in}*concentration_{in}-Rate_{out}*\frac{x}{V_{0}+(Rate_{in}-Rate_{out})*t}$
$Initial \, Volume \, V_{0}=600L$
$Concnetration_{in}=0.1kg/L$
$Rate_{in}=10L/min$
$Rate_{out}=8L/min$
$\frac{dx}{dt}=10*0.1-8*\frac{x}{600+(10-8)*t}$
$\frac{dx}{dt}=1-\frac{8x}{600+2t}$
$\frac{dx}{dt}=1-\frac{4x}{100+t}$

Hope that helps.

Thanks guys :).
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 22, 2011, 11:55:58 am
I added a quick explanation of where the equation comes from (in an edit) if you wanted it.
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 29, 2011, 04:42:38 pm
If P(z) is a polynomial of degree 4 with all of its coefficients real with ai, bi (a,b = R) as two of its zeros, then the term that doesn't contain z is:

A. ab
B. a-b
C. a+b
D. a^3b^3
E. a^2b^2
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 29, 2011, 05:02:04 pm
$\pm ai \pm bi$

$P(z)=(z+ai)(z-ai)(z+bi)(z-bi)$
$=(z^2+a^2)(z^2+b^2)$
$=z^4+(a^2+b^2)z^2+a^2 \cdot b^2$

$=a^2 \cdot b^2$

E) $a^2 \cdot b^2$

EDIT: sorry man latex doesn't work, don't know whats really going on with the latex, and solution doesn't look to clear without latex.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 30, 2011, 12:05:24 pm
Kinematics - Rectilinear Motion.

An object is dropped from rest so that the acceleration due to air resistance is 0.2v where is the speed of the object. The acceleration due to gravity is g m/s^2

a) show that $\frac{dv}{dt}=g-0.2v$

b) Find $v$ in terms of $t$

c) Find the $\lim_{t\rightarrow \infty } V(t)$, that is, the maximum veloctiy

d) Find the distance fallen after 5 seconds (to the nearest metre)
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on December 30, 2011, 12:20:35 pm
Lol @ "acceleration due to air resistance" whose phrase is this? I'm sure the better way of saying it is the air resistance is $0.2vm$ upwards where $m$ is the mass of the object. In that case the total force is Gravity - Air Resistance = mg-0.2vm Hence the acceleration is by Newton's ith law $\frac{mg-0.2vm}{m}=g-0.2v$. (note downwards is taken as positive direction)

b) just use the "fact" that $\frac{dt}{dv}=\frac{1}{\frac{dv}{dt}}=\frac{1}{g-0.2v}$ and if I remember my calculus correctly (I havn't done that in a while, too busy with maths) you get something with log.
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 30, 2011, 12:26:12 pm
^ Of course calculus isn't maths :P
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 30, 2011, 12:28:36 pm
Lol @ "acceleration due to air resistance" whose phrase is this? I'm sure the better way of saying it is the air resistance is $0.2vm$ upwards where $m$ is the mass of the object. In that case the total force is Gravity - Air Resistance = mg-0.2vm Hence the acceleration is by Newton's ith law $\frac{mg-0.2vm}{m}=g-0.2v$. (note downwards is taken as positive direction)

b) just use the "fact" that $\frac{dt}{dv}=\frac{1}{\frac{dv}{dt}}=\frac{1}{g-0.2v}$ and if I remember my calculus correctly (I havn't done that in a while, too busy with maths) you get something with log.
Sorry its meant to be acceleration due to gravity, I'm sorry for not checking it :P twice.

Anyways Thank you very much Kamil :).
Title: Re: Specialist 3/4 Question Thread!
Post by: samad on December 30, 2011, 01:21:58 pm
Kinematics - Rectilinear Motion.

An object is dropped from rest so that the acceleration due to air resistance is 0.2v where is the speed of the object. The acceleration due to gravity is g m/s^2

a) show that $\frac{dv}{dt}=g-0.2v$

b) Find $v$ in terms of $t$

c) Find the $\lim_{t\rightarrow \infty } V(t)$, that is, the maximum veloctiy

d) Find the distance fallen after 5 seconds (to the nearest metre)

I think the shortcut to finding terminal velocity is calculating v when acceleration is zero (this can be conceptualised by looking at the velocity- time graph for a falling object):(http://www.gcsescience.com/Velocity-Time-Graph-Rocket.gif)
when terminal velocity is reached, the gradient, and hence the acceleration, is zero.

so: a=g-0.2v
let a=0
g-0.2v=0
therefore terminal velocity=v=g/0.2

check if this is right using calculus  :)
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on December 30, 2011, 06:07:38 pm
This outflow/inflow question is troublesome for me :/.

Q9. A tank containing 600 litres of water has a sugar solution of concentration 0.1 kg/L pumped in at a rate of 10 L/min. if the mixture is kept uniform and is pumped out at a rate of 8 L/min, set up a differential equation for the amount of sugar, x kg, at any time t. (Do not attempt to solve).

b^3 has provided an explanation already, but I'll add mine because I have a slightly different way of doing it ;D

dx/dt = rate of change of amount with respect to time (kg/min)

Volume at time t = V(t) = 600 + 10t - 8t = 600 + 2t

Rate of change of x (in)= Concentration (kg/L) x Flowing In Rate (L/min) = 0.1 x 10 = 1 kg/min

Rate of change of x (out) = Concentration (kg/L) x Flowing Out Rate (L/min) = x/v(t) * 8 = (8x) / (600 + 2t) = (4x) / (300 + t)

Now you find dx/dt = Rate of change of x (in) - Rate of change of x (out) = $\frac{dx}{dt}=1-\frac{4x}{300+t}$

Generally I use this method to do it because it doesn't need formulae (I hate formulae =.= it's more to remember to be honest) - you kind of use the units to guide you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 30, 2011, 06:21:04 pm
This outflow/inflow question is troublesome for me :/.

Q9. A tank containing 600 litres of water has a sugar solution of concentration 0.1 kg/L pumped in at a rate of 10 L/min. if the mixture is kept uniform and is pumped out at a rate of 8 L/min, set up a differential equation for the amount of sugar, x kg, at any time t. (Do not attempt to solve).

b^3 has provided an explanation already, but I'll add mine because I have a slightly different way of doing it ;D

dx/dt = rate of change of amount with respect to time (kg/min)

Volume at time t = V(t) = 600 + 10t - 8t = 600 + 2t

Rate of change of x (in)= Concentration (kg/L) x Flowing In Rate (L/min) = 0.1 x 10 = 1 kg/min

Rate of change of x (out) = Concentration (kg/L) x Flowing Out Rate (L/min) = x/v(t) * 8 = (8x) / (600 + 2t) = (4x) / (300 + t)

Now you find dx/dt = Rate of change of x (in) - Rate of change of x (out) = $\frac{dx}{dt}=1-\frac{4x}{300+t}$

Generally I use this method to do it because it doesn't need formulae (I hate formulae =.= it's more to remember to be honest) - you kind of use the units to guide you :)
It's still basically the same way, you've worked out the inflow of x per second and the outflow of x per second
EDIT: Explaining the equation
The first part of the equation is giving how much of "x" is following in per second.
The second part of the equation is giving how much of "x" is following out per second.
Its the same thing, just that you do $\frac{dx_{in}}{dt}$ and $\frac{dx_{out}}{dt}$, this is the way that I did it all the time when there wasn't a changing volume, as it is easier to just remember flowin-flowout.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 30, 2011, 07:01:26 pm
Thanks Paulsterio :), I'd stick to more of b^3 workings. I finished differential equations and im almost finishing off kinematics and im heading to vector function/vector calculus.

Thanks again Paulsterio, would look back again soon :).
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 30, 2011, 07:47:56 pm
Yeah what b^3 and paulsterio did were the same thing if you ask me :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 30, 2011, 07:49:24 pm
Q(z) = z^3-3iz^2-3z+4+i

Determine the values of a=C, b=R if Q(z) is of the form Q(z) = (z-a)^3+b
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on December 30, 2011, 07:54:09 pm
Expand (z-a)^3+b out to get:

z^3 - 3z^2a + 3za^2 - a^3 + b

Equate coefficients.

a = i,b = 4
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on December 30, 2011, 08:33:11 pm
Question c of Q17 Maths Quest right :P?

c) $Q(z)=z^{3}-3i^{2}-3z+4+i$
$Q(z)=(z-a)^{3}+b$
$=z^{3}-3z^{2}+3a^{2}z-a^{3}+b$

Equating with $Q(z)$

$a=i$
$-a^{3}+b=4+i$
$-i^{3}+b=4+i$
$i+b=4+i$
$i+b=4+i$
$b=4$
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on December 30, 2011, 11:14:35 pm
Thanks to you both and yes Q17c Maths Quest :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on January 01, 2012, 12:44:07 pm
Given that cis 3θ = (cos^3 θ − 3 cos θ sin^2 θ) + i (3 cos^2 θ sin θ − sin^3 θ), it follows that cos 3θ
is equal to:
A cos^3θ
B 4 cos θ − 3 cos^3 θ
C 3 cos^2 θ − sin^3 θ
D 4 cos^3 θ − 3 cos θ
E −(2 cos^3 θ + 3 cos θ)

Ans. =D

Could someone plz go through this one?!
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on January 01, 2012, 01:03:46 pm
cos(3x) = cos^3(x) - 3cos(x)sin^2(x)
= cos^3(x) - 3cos(x)(1-cos^2(x))
= cos^3(x) - 3cos(x) + 3cos^3(x)
= 4 cos^3(x) - 3cos(x)
so D.
Title: Re: Specialist 3/4 Question Thread!
Post by: tony3272 on January 01, 2012, 01:07:01 pm
$cis(3\theta) = cos(3\theta)+isin(3\theta)$

So you only need the cosine component to do this question

$\therefore cos(3\theta)=cos^3(\theta)-3cos(\theta)(1-cos^2(\theta))$

$\therefore cos(3\theta)=4cos^3(\theta)-3cos(\theta)$  once you have simplified it down
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on January 01, 2012, 01:09:45 pm
silly me, i even wrote it out and still didn't realise it was 'sin^2' !!!!

thanks haha ;D
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 01:45:45 pm
This is such an easy question, yet I keep getting the wrong answer :(

The radius of a spherical ball is $2.5m$ and its volume is increasing at a rate of $0.1m^{3}/min$. At what rate is the radius increasing with respect to time?

$V = \frac{4}{3}\pi r^{3}$
$\frac{dV}{dr} = 4\pi r^{2}$
$\frac{dV}{dt} = 0.1$
$\frac{dr}{dt} = \frac{dV}{dt}/\frac{dV}{dr}$
$\frac{dr}{dt} = \frac{0.1}{4\pi r^{2}}$
$\frac{dr}{dt} = \frac{1}{40\pi r^{2}}$

But I got the answer wrong. I suspect that my flaw might have been in the $\frac{dV}{dt}$ part, but I'm not completely sure.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 02:04:09 pm
$\frac{dr}{dt} = 0.00127m/min$ (approximation, not exact value).
Someone please tell me where I went wrong.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 01, 2012, 02:11:18 pm
Plug in the value of r (2.5m) as this is the radius at the point in time that you are looking for. Then you get dr/dt=0.00127m/min.

Also dont forget to put the \ in front of the "pi" in your LaTeX code (also leave a space between the \pi and the r) so that you get $\pi r$.
Title: Re: Specialist 3/4 Question Thread!
Post by: butene on January 01, 2012, 02:16:49 pm
This is such an easy question, yet I keep getting the wrong answer :(

The radius of a spherical ball is $2.5m$ and its volume is increasing at a rate of $0.1m^{3}/min$. At what rate is the radius increasing with respect to time?

$V = \frac{4}{3}pir^{3}$
$\frac{dV}{dr} = 4pir^{2}$
$\frac{dV}{dt} = 0.1$
$\frac{dr}{dt} = \frac{dV}{dt}/\frac{dV}{dr}$
$\frac{dr}{dt} = \frac{0.1}{4pir^{2}}$
$\frac{dr}{dt} = \frac{1}{40pir^{2}}$

But I got the answer wrong. I suspect that my flaw might have been in the $\frac{dV}{dt}$ part, but I'm not completely sure.

dr/dt = 0.1/(4pir^2)
= 0.1/(4pi(2.5)^2) = 0.00127m/min

sorry i cant use latex LOL
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on January 01, 2012, 02:18:15 pm
Plug in the value of r (2.5m) as this is the radius at the point in time that you are looking for. Then you get dr/dt=0.00127m/min.

Also dont forget to put the \ in front of the "pi" in your LaTeX code (also leave a space between the \pi and the r) so that you get $\pi r$.
+1.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 02:24:59 pm
Plug in the value of r (2.5m) as this is the radius at the point in time that you are looking for. Then you get dr/dt=0.00127m/min.

Also dont forget to put the \ in front of the "pi" in your LaTeX code (also leave a space between the \pi and the r) so that you get $\pi r$.

Oh so I got it all right I just didn't plug in the value.
Thank you!
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 02:37:45 pm
So if I then wanted to work out the rate of change of surface area to time, would I say:
$A = 4\pi r^{2}$
$\frac{dA}{dr} = 8\pi r$
$\frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt}$
$\frac{dA}{dt} = \frac{8\pi r}{40\pi r^{2}}$
$\frac{dA}{dt} = \frac{1}{5r}$
$When$ $r = 2.5$
$\frac{dA}{dt} = \frac{2}{25}$
$\frac{dA}{dt} = 0.08m^{2}/min$
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 01, 2012, 03:27:24 pm
Looks good to me.
Title: Re: Specialist 3/4 Question Thread!
Post by: jaydee on January 01, 2012, 07:13:00 pm
can someone please explain linear independence and dependence? the maths quest book uses a=mb+nc whereas the essential book uses c=ma+nb. Im confused :(
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on January 01, 2012, 07:39:25 pm
i'm a bit iffy about the geometric significance of linear dependence/independence myself. however, all you really need to know is that the vectors a, b and c are linearly dependent if there exists non-zero constants p, q, r such that pa + qb + rc = 0 (the letters don't matter, all you need to know is the idea). if a set of vectors aren't linearly dependent, then they're linearly independent. the a = mb + mc, etc. are merely 'shortcuts' that you can take.

the explanation below given by wikipedia represents my current understanding of linear dependence:

"A geographic example may help to clarify the concept of linear independence. A person describing the location of a certain place might say, "It is 5 miles north and 6 miles east of here." This is sufficient information to describe the location, because the geographic coordinate system may be considered as a 2-dimensional vector space (ignoring altitude). The person might add, "The place is 7.81 miles northeast of here." Although this last statement is true, it is not necessary.

In this example the "5 miles north" vector and the "6 miles east" vector are linearly independent. That is to say, the north vector cannot be described in terms of the east vector, and vice versa. The third "7.81 miles northeast" vector is a linear combination of the other two vectors, and it makes the set of vectors linearly dependent, that is, one of the three vectors is unnecessary.

Also note that if altitude is not ignored, it becomes necessary to add a third vector to the linearly independent set. In general, n linearly independent vectors are required to describe any location in n-dimensional space."
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on January 01, 2012, 08:03:08 pm
if you guys wanna really understand linear independence etc formally, check out elementary linear algebra by anton, uploaded here http://ifile.it/0k9e8i/__Elementary_Linear_Algebra_with_Applications_9_edition.l_72x2j2n5nx4tx41.pdf

pg 370, section 5.3 a whole section on linear independence/dependence etc
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 08:15:32 pm
The equation of a curve C is $x^{3} + xy + 2y^{3} = k$ where k is a constant.
$\frac{dy}{dx} = \frac{-3x^{2} - y}{6y^{2} + x}$
C does have a tangent parallel to the y axis.
Show that the y coordinate at the point of contact satisfies $216y^{6} + 4y^{3} + k = 0$

I was thinking that since it is parallel to the y axis, it makes a vertical line, so I could make the substitution let dy/dx = infinity.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on January 01, 2012, 08:19:50 pm
The equation of a curve C is $x^{3} + xy + 2y^{3} = k$ where k is a constant.
$\frac{dy}{dx} = \frac{-3x^{2} - y}{6y^{2} + x}$
C does have a tangent parallel to the y axis.
Show that the y coordinate at the point of contact satisfies $216y^{6} + 4y^{3} + k = 0$
a few hints:

implicit differentiate.

from then you can easily manipulate to satisfied the required expression
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 08:22:55 pm
I think I get it. Can I let the denominator equal 0?
So 6y^2 + x = 0.
But I haven't got any y^6 terms...
Wait I think I know:
x = -6y^2
Now substitute into the original equation:
x^3 + xy + 2y^3 = k
(-6y^2)^3 + (-6y^2)y + 2y^3 = k
-216y^6 - 6y^3 + 2y^3 = k
-216y^6 - 4y^3 = k
-216y^6 - 4y^3 - k = 0
216y^6 + 4y^3 + k = 0

Got it. Thanks TrueTears and Paulsterio :)
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 01, 2012, 08:26:45 pm
notice that it says at the point of contact.

x = -6y^2 from what you had

sub that back into the eqn for the curve, you get the answer
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 01, 2012, 08:29:09 pm
$\frac{d}{dx}(x^{3})+\frac{d}{dx}(xy)+\frac{d}{dx}(2y^{3})=\frac{d}{dx}(k)$
$3x^{2}+x*\frac{dy}{dx}+y(1)+6y^{2}\frac{dy}{dx}=0$
$\frac{dy}{dx}(6y^{2}+x)=-y-3x^{2}$
$\frac{dy}{dx}=\frac{-y-3x^{2}}{6y^{2}+x}$
Now for the gradient to be parallel to the y-axis the gradient will be undefined, i.e. the bottom of the fraction will be equal to 0.
$6y^{2}+x=0$
$x=-6y^{2}$
Sub this back into the original equation.
$(-6y^{2})^{3}+(-6y^{2}y)+2y^{3}=k$
$-216y^{6}-6y^{3}+2y^{3}=k$
$216y^{6}+4y^{3}+k=0$

EDIT: Beaten, dam latex is slower.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 01, 2012, 08:31:44 pm
EDIT: Beaten, dam latex is slower.

1) - You made an error in the last line with y^6

2) - You always seem to get beaten cause of extravagant formatting/LATEX :P
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 01, 2012, 08:33:02 pm
EDIT: Beaten, dam latex is slower.

1) - You made an error in the last line with y^6

2) - You always seem to get beaten cause of extravagant formatting/LATEX :P
Thanks Paul, and not always beaten, its just there was a lot of fractions in that one, too much {} and so on. It's not always slower, just depends on when you pick up the question. Plus LaTeX looks nicer :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 08:33:55 pm
Hence show that k≤(1/54)
I have no idea how to do this part either...
Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 01, 2012, 08:39:04 pm
let y^3 = Y, then 216Y^2+4Y+k=0, this must have one or two solutions for the case when there are two points of vertical tangents
So the discriminant must be greater than 0, do this and you should get k<=1/54
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 01, 2012, 09:02:25 pm
damn, that wasn't too tough, it just shows how much I've gotten rusty just two months after exams! damn!!
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 01, 2012, 09:05:38 pm
wow 2 months? I forget everything after 1 week after uni exams.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 01, 2012, 09:09:05 pm
wow 2 months? I forget everything after 1 week after uni exams.

hmm, I guess Year 12 is a little harder to forget ;P
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 01, 2012, 09:15:46 pm
Find the possible values of k in the case where x = -6 is a tangent to C.

I spent so long on this question and I give up because I keep going around in circles and I don't know what to do :(
If it was any equation that didn't lead to a vertical line with infinite gradient, then I could've solved it so much quicker.

I think I know!
After literally 30 minutes of thinking, the idea occured to me:
Rather than letting dy/dx = infinity, I can simply let dx/dy = 0.
Actually I don't know if that helps me with this specific question :( It would've helped me in the last one though.

OMFG yes I finally found the answer after almost an hour!
Let dx/dy = 0
(6y^2 + x) / (-3x^2 - y) = 0
6y^2 + x = 0
y^2 = -x/6
y = +or- sqrt(-x/6)
when x = -6
y = 1 or y = -1
If y = 1:
216 + 4 + k = 0 so k = -220
If y = -1:
216 - 4 + k = 0 so k = -212
So k = -220 or k = -212
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 02, 2012, 10:25:15 pm
Find an antiderivative of cos(x) / (sin(x))^3

My first method was:
cos(x) / (sin(x))^3 = cot(x)*csc^2(x)
Let u = cot(x) and du/dx = -csc^2(x)
= Integral of -u du
= -1/2 u^2 + C
= -1/2 cot^2(x) + C

My second method was:
Let u = sin(x), du/dx = cos(x)
= Integral of u^(-3) du
= -1/2 u^(-2) + C
= -1/2 sin^(-2) + C
= -1 / (2*sin^2(x)) + C

-1/2 cot^2(x) is not the same as -1 / (2*sin^2(x)).

Why do I get a different answer???
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 02, 2012, 10:29:55 pm
I think they might actually be the same, sub something like pi/6 in and see if it comes to the same answer, if it does, try and manipulate it with theorems...it's got me stumped!
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 02, 2012, 10:31:50 pm
yep, confirmed it, they are both right, both give cos(x) / (sin(x))^3 after differentiation :P
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 02, 2012, 10:33:53 pm
Remember that 1/sin^2(x) is csc^2(x)
Now also note that 1+cot^2(x)=csc^2(x)

Now what you have are translations of each other along the y axis, as the +C is not the same in both cases,
i.e. -1/2cot^2(x)=-1/2*(csc^2(x)-1)+C1
=-1/(2sin^2(x))+1/2+C1
=-1/(2sin^2*x) +C2 where C2=1/2+C1

I.e. it is just a translation on the y-axis, as you get two different C's but they are both correct.

Hope that helps and I hope there aren't any mistakes in there (it is likely, I'm pretty tired right now)
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 02, 2012, 10:34:22 pm
firstly the 2nd answer should be -1/2*csc^2(x), anyways heres a hint, 1+cot^2(x)=csc^2(x)
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 02, 2012, 10:39:26 pm
I get it now. They are the same graph (and same gradient) but a different translation.
Thank you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 03, 2012, 01:17:36 pm
$\underset{\sim}{j}$

Code: [Select]
\underset{\sim}{j}
e.g. $\underset{\sim}{i}+\underset{\sim}{j}+\underset{\sim}{k}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 03, 2012, 01:30:38 pm
Do we need to know integration by parts?
Should I learn it just incase?
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on January 03, 2012, 01:33:52 pm
Do we need to know integration by parts?
Should I learn it just incase?
I heard integration by parts is useful for complicated derivatives via 'product rule' right? I think it would a good idea to learn it.
Don't know if VCAA examiners would accept that process of integration though :P.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on January 03, 2012, 01:48:19 pm
Do we need to know integration by parts?
Should I learn it just incase?

You can learn it to check answers, but can't use it in the exams :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 03, 2012, 01:48:50 pm
I heard integration by parts is useful for complicated derivatives via 'product rule' right? I think it would a good idea to learn it.
Don't know if VCAA examiners would accept that process of integration though :P.

That's unlucky :( I've never really found derivatives complicated, it's integration that sometimes messes with my brain. Speaking of which:
Show that $\int\frac{1}{a^2 - x^2}dx$ for a>0 = $\frac{1}{2a}ln(\frac{|x + a|}{|a - x|}) + C$
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on January 03, 2012, 02:05:57 pm
Can't you let $x=a\sin(\theta)$ and work from there? Things should cancel out I think.

I'm really out of it lol
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on January 03, 2012, 02:17:09 pm
Do we need to know integration by parts?
Should I learn it just incase?
a quick tutorial if u ceebs wiki-ing it Re: Specialist Summer Holidays Question Thread!
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 03, 2012, 02:23:49 pm
Do we need to know integration by parts?
Should I learn it just incase?
a quick tutorial if u ceebs wiki-ing it Re: Specialist Summer Holidays Question Thread!

That was very helpful and unlike Wikipedia, I can actually understand it :) Thank you!

I pm'd the last problem to Thushan. Hopefully he can solve it.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on January 03, 2012, 02:32:59 pm
$\int\frac{1}{a^2 - x^2}dx, \ a>0$
$Let \ x=a\sin(\theta)$
$\int\frac{1}{a^2 - a^2\sin^2(\theta)} a\cos(\theta) d\theta$
$\frac{1}{a}\int{\frac{cos(\theta)}{cos^2(\theta)}}d\theta$
$\frac{1}{a}\int{\frac{1}{cos(\theta)}}d\theta$

Yeah, you know how to do that :)

(do it the long way, and not multiply by the identity btw)

Title: Re: Specialist 3/4 Question Thread!
Post by: samad on January 03, 2012, 02:33:15 pm
Hey, that question involves partial fractions:
express the integral:
$\int\frac{1}{a^2-x^2}dx\\= \int\frac{1}{(a-x)(a+x)}dx$
Now:
$\frac{1}{(a-x)(a+x)}=\frac{1}{2a(a-x)}+\frac{1}{2a(a+x)}$
Using standard partial fractions techniques
now the integral is:
$\frac{1}{2a}\int\frac{1}{(a-x)}+\frac{1}{(a+x)}dx$
$=\frac{1}{2a}(-\ln|a-x|+\ln|a+x|) + C$
$=\frac{1}{2a}\ln\frac{|x+a|}{|a-x|} + C$ as required
pi's appraoch very valid as well but not on course

Moderator action: removed real name, sorry for the inconvenience
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on January 03, 2012, 02:34:57 pm
lol, I was so off :D
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 03, 2012, 02:35:47 pm
pi, your way would have worked reasonably, except it's extremely long :P

Moderator action: removed real name, sorry for the inconvenience
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 03, 2012, 02:36:08 pm
$\int\frac{1}{a^2 - x^2}dx, \ a>0$
$Let \ x=a\sin(\theta)$
$\int\frac{1}{a^2 - a^2\sin^2(\theta)} a\cos(\theta) d\theta$
$\frac{1}{a}\int{\frac{cos(\theta)}{cos^2(\theta)}}d\theta$
$\frac{1}{a}\int{\frac{1}{cos(\theta)}}d\theta$

Yeah, you know how to do that :)

(do it the long way, and not multiply by the identity btw)

Since when do you know how to do that in year 12?

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Title: Re: Specialist 3/4 Question Thread!
Post by: pi on January 03, 2012, 02:38:55 pm
Since when do you know how to do that in year 12?

Our SACs... :(

pi, your way would have worked reasonably, except it's extremely long :P

Yeah, I see that, samad's way was much better, should've seen that (I'm really out of it lol)

Moderator action: removed real name, sorry for the inconvenience
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 03, 2012, 02:39:25 pm
Hey, that question involves partial fractions:
express the integral:
$\int\frac{1}{a^2-x^2}dx\\= \int\frac{1}{(a-x)(a+x)}dx$
Now:
$\frac{1}{(a-x)(a+x)}=\frac{1}{2a(a-x)}+\frac{1}{2a(a+x)}$
Using standard partial fractions techniques
now the integral is:
$\frac{1}{2a}\int\frac{1}{(a-x)}+\frac{1}{(a+x)}dx$
$=\frac{1}{2a}(-\ln|a-x|+\ln|a+x|) + C$
$=\frac{1}{2a}\ln\frac{|x+a|}{|a-x|} + C$ as required
pi's appraoch very valid as well but not on course

Thank you! You're a legend :)

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Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 03, 2012, 02:40:51 pm
Since when do you know how to do that in year 12?

Our SACs... :(

pi, your way would have worked reasonably, except it's extremely long :P

Yeah, I see that, samad's way was much better, should've seen that (I'm really out of it lol)

Ah I see, I didn't know cause I don't remember ever seeing that (1/cosx) in vce :P

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Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 03, 2012, 02:43:22 pm
Yeah, I see that, samad's way was much better, should've seen that (I'm really out of it lol)

LOL! same here, I can't believe I didn't recognise the difference of squares when I first looked at it either!

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Title: Re: Specialist 3/4 Question Thread!
Post by: pi on January 03, 2012, 02:45:00 pm
Yeah, I see that, samad's way was much better, should've seen that (I'm really out of it lol)

LOL! same here, I can't believe I didn't recognise the difference of squares when I first looked at it either!

Yeah same! Can't believe I saw the trig way first... That's pretty damn slow.

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Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 03, 2012, 02:46:26 pm
Hmm, the log should have given it away though, like you always get logs with partial fractions, I guess this is what happens when you don't do this stuff for 2 months, I wonder if I'll have anything left in the tank at the end of 2012, I was really looking forward to doing the 2012 spesh exam (A)!! :D

Btw, Year 12s, Tip: - If you see logs, think partial fractions
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 03, 2012, 02:52:11 pm
Btw, Year 12s, Tip: - If you see logs, think partial fractions

Thanks. I'll remember that.
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on January 05, 2012, 02:16:23 pm
Find all z satisfying:

z^8+1=0

Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on January 05, 2012, 02:31:25 pm
$z^{8}+1=0$
$z^{8}=-1$
$r=1, \theta=\pi$
let w= $r \cdot cis(\pi)$
$z^{8}=cis(\pi)$
$z=(cis(\pi)^{\frac{1}{8}}$
$=cis(\frac{\pi}{8}+\frac{2k\pi}{8})$
$k=0, z=cis(\frac{\pi}{8})$
$k=1, z=cis(\frac{3\pi}{8})$
$k=2, z=cis(\frac{5\pi}{8})$
$k=3, z=cis(\frac{7\pi}{8})$
$k=4, z=cis(\frac{-7\pi}{8})$
$k=5, z=cis(\frac{-5\pi}{8})$
$k=6, z=cis(\frac{-3\pi}{8})$
$k=7, z=cis(\frac{-\pi}{8})$

EDIT: beaten :/
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 05, 2012, 03:35:56 pm
I don't really know why you have introduced a 'w' that you never used, but in any case, you should have z^8 = cis(pi + 2kpi) from the beginning, not 3 lines later as this is inconsistent.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 06, 2012, 10:51:19 am
Btw, if you intend to write k = 4, you'll have to say that z = cis(9pi/8) = cis(-7pi/8) otherwise your answer wouldn't make much sense

Alternatively, you can let k = 0, 1, 2, 3, -1, -2, -3, -4 instead
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 07, 2012, 02:51:54 pm
Is there supposed to be a picture attached?
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on January 07, 2012, 05:48:43 pm
you guys are so ahead! takes me forever to do one exercise  :o
anyway i don't understand part b of the last question in exercise 1E (essentials, if anyone has done it yet):
10. a)Find an expression for the infinite geometric sum 1 + sintheta + sin^2*theta + ...
b)Find the values of theta for which the infinite geometric sum is 2.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on January 07, 2012, 05:49:18 pm
just equate the expression from part a) to 2 and solve for theta
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on January 07, 2012, 05:55:33 pm
just equate the expression from part a) to 2 and solve for theta
yar i did but i got stuck in the last part....
how do you go from sintheta = 1/2 -----> answer: theta = pi/6 + 2kpie, 5pi/6 + 2kpi, k = intergers
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on January 07, 2012, 05:58:59 pm
just equate the expression from part a) to 2 and solve for theta
yar i did but i got stuck in the last part....
how do you go from sintheta = 1/2 -----> answer: theta = pi/6 + 2kpie, 5pi/6 + 2kpi, k = intergers

oh it's just general solutions, Re: Mathematical Methods Guides and Tips i used the exact same equation as an example lol
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on January 07, 2012, 06:18:39 pm
just equate the expression from part a) to 2 and solve for theta
yar i did but i got stuck in the last part....
how do you go from sintheta = 1/2 -----> answer: theta = pi/6 + 2kpie, 5pi/6 + 2kpi, k = intergers

oh it's just general solutions, Re: Mathematical Methods Guides and Tips i used the exact same equation as an example lol
OHH kewl thanks. why haven't i ever been taught about this, is this part of the 1/2 methods book - maybe we skipped the exercise on it :'(
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 07, 2012, 11:45:40 pm
Is there supposed to be a picture attached?

was that at me? LOL?! i'm lost!

Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 07, 2012, 11:55:49 pm
Is there supposed to be a picture attached?

was that at me? LOL?! i'm lost!

Nah there was a post but it got deleted.
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on January 13, 2012, 04:38:40 pm
Express (2sqrt(3)+2i)/(1-3i) in polar form.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on January 13, 2012, 05:24:11 pm
Express (2sqrt(3)+2i)/(1-3i) in polar form.
$\frac{2 \cdot \sqrt{3}+2i}{1-\sqrt{3}i}$
$=\frac{4 \cdot cis\left ( \frac{\pi}{6} \right )}{2 \cdot cis\left ( \frac{-\pi}{3} \right )}$
$=2 \cdot cis\left(\frac{\pi}{6 }+ \frac{\pi}{3}\right)$
$=2 \cdot cis\left (\frac{\pi}{2} \right )$

Hoped I helped :).
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on January 13, 2012, 10:28:36 pm
hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on January 13, 2012, 10:47:27 pm
hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

Are you talking about whats inside the square root? if there was -3 in $\sqrt{-3}$, than it would result as $-3 \cdot i$, because we know that as general rule that $\sqrt {-1} =i$, so if we expand out $\sqrt{3i}=\sqrt{3} \cdot \sqrt{-1 }= \sqrt{3} \cdot i$.

Don't worry to much about it though, just skim through chapter on complex numbers you will understand a lot if you revisit some of the sub topics of complex numbers.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on January 13, 2012, 11:49:26 pm
hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

Are you talking about whats inside the square root? if there was -3 in $\sqrt{-3}$, than it would result as $-3 \cdot i$, because we know that as general rule that $\sqrt {-1} =i$, so if we expand out $\sqrt{3i}=\sqrt{3} \cdot \sqrt{-1 }= \sqrt{3} \cdot i$.

Don't worry to much about it though, just skim through chapter on complex numbers you will understand a lot if you revisit some of the sub topics of complex numbers.

I think what dominator was asking was,  in the denominator,  its : 1-3i?  You've used 1-root3 I
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on January 14, 2012, 09:51:30 am
hey thanks but the denominator, isn't the imaginary part -3 instead of sqrt(3) or did I miss some rule when going through the chapter? Otherwise the root3 would make the question a whole lot understandable and easier.

Are you talking about whats inside the square root? if there was -3 in $\sqrt{-3}$, than it would result as $-3 \cdot i$, because we know that as general rule that $\sqrt {-1} =i$, so if we expand out $\sqrt{3i}=\sqrt{3} \cdot \sqrt{-1 }= \sqrt{3} \cdot i$.

Don't worry to much about it though, just skim through chapter on complex numbers you will understand a lot if you revisit some of the sub topics of complex numbers.

I think what dominator was asking was,  in the denominator,  its : 1-3i?  You've used 1-root3 I

Yeah that's what I meant
Title: Re: Specialist 3/4 Question Thread!
Post by: Ravit on January 14, 2012, 07:11:14 pm
need help!

Find the point P on the line x-6y=11 such that the vector OP is parallel to the vector 3i+j
Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 14, 2012, 07:20:08 pm
let P=(x,y)=(x,1/6(x-11)) from the equation of the line, then OP=xi+1/6(x-11)j

now OP=k(3i+j) since they are parallel, so xi+1/6(x-11)j=3ki+jk, then equate the i, j components, solve for x by eliminating k, then sub back into the line equation to get the y value.

you should get P=(-11, -11/3) i think...if i did it right...
Title: Re: Specialist 3/4 Question Thread!
Post by: Ravit on January 14, 2012, 07:23:35 pm
thank you very much :D
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 20, 2012, 06:30:13 pm
I hate these extremely long worded questions...

"A tank initially contains 400 litres of water in which are dissolved 10 kg of salt. A salt solution of concentration 0.2 kg/L is poured into the tank at a rate of 2 litres per minute; the mixtury, which is kept uniform by stirring, flows out at a rate of 2 litres per minute.
If the mass of salt in the tank after t minutes is x kg, set up and solve the differential equation for x in terms of t."

My thoughts are that every minute, 0.4kg of salt is being added and 1/200 times the current concentration is being removed from the tank. How would I express this?

Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 20, 2012, 06:44:11 pm
let Q=amount of salt at time t.

Then dQ(in)dt=0.2 kg/L * 2L/min = 0.4 kg/min . That should make sense, I'm putting in 2L per minute and if the solution has 0.2kg per litre, then there are 0.4 kg/min going in.

Now coming out, the concentration is not constant, it is Q/V where Q is the amount and V is the volume, which is actually just 400L since there is 2L/min going in and out.
So dQ(out)/dt=Q/400 kg/L * 2L/min = Q/200 kg/min.

so overall, dQ/dt=dQ(in)/dt - dQ(out)/dt = 2/5 - Q/200. now you should be able to solve this, put it on a common denominator and flip both sides etc.

then you use the initial amount of salt, 10kg to solve for the constant of integration

hope this helped :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 20, 2012, 06:59:44 pm
Thanks, but it gets harder:

"If, instead, the mixture flows out at 1 litre per minute, set up, but do not solve, the differential equation for the mass of salt in the tank."

let Q=amount of salt at time t.
Now coming out, the concentration is not constant, it is Q/V where Q is the amount and V is the volume

I don't understand this part...
I get that if the volume is constantly changing, then the percent of input/output in relation to the volume will not be a constant value, but how do I work it out, and what does Q/V actually mean?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 20, 2012, 07:05:42 pm
Thanks, but it gets harder:

"If, instead, the mixture flows out at 1 litre per minute, set up, but do not solve, the differential equation for the mass of salt in the tank."
Do the same thing as above for finding the equation but for the flow out part, instead of using $\frac{Q}{400}*2$, the bottom part is your volume will change as time changes. So initally it was 400 and the volume us increasing at $V_{in}-V_{out}=2-1=1$L/min.
So your volume will be 400+(1)t
i.e. $\frac{dx_{out}}{dt}=\frac{x}{400+t}*1=\frac{x}{400+t}$

Then do the rest, i.e. dxin/dt-dxout/dt
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 20, 2012, 07:13:33 pm
Do the same thing as above for finding the equation but for the flow out part, instead of using $\frac{Q}{400}*2$, the bottom part is your volume will change as time changes. So initally it was 400 and the volume us increasing at $V_{in}-V_{out}=2-1=1$L/min.
So your volume will be 400+(1)t
i.e. $\frac{dx_{out}}{dt}=\frac{x}{400+t}*1=\frac{x}{400+t}$

Then do the rest, i.e. dxin/dt-dxout/dt

Thank you. I sort of get this, but not fully...
Why is it (400 + 1t) as the denominator?
What sort of situation would make it (400 + 2t) or (400 + 5t)?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 20, 2012, 07:36:09 pm
This is from a couple of pages back in this thread
For these questions where the flow in is not equal to the flow out, it is good to use this formula.

$\frac{dx}{dt}=rate_{in}*concentration_{in}-Rate_{out}*\frac{x}{V_{0}+(Rate_{in}-Rate_{out})*t}$

$Initial \, Volume \, V_{0}=600L$
$Concnetration_{in}=0.1kg/L$
$Rate_{in}=10L/min$
$Rate_{out}=8L/min$
$\frac{dx}{dt}=10*0.1-8*\frac{x}{600+(10-8)*t}$
$\frac{dx}{dt}=1-\frac{8x}{600+2t}$
$\frac{dx}{dt}=1-\frac{4x}{300+t}$

Hope that helps.

EDIT: Explaining the equation
The first part of the equation is giving how much of "x" is following in per second.
The second part of the equation is giving how much of "x" is following out per second.
You can see this by "rate"*"concentration"
In the second half of the equation, the concentration is  $\frac{x}{V_{0}+(Rate_{in}-Rate_{out})*t}$, think of it as c=mass/volume, mass is x and the volume is changing as t changes.
So basically if you do chem you know that c=n/v (in M) or c=m/v
So that is what we are doing, the flow in of x is  equal to m=cv.
now the conentration of x in we know as well as the volume in.
The flow out of x is m=c*v flow out, but we dont know the concentration or volume of the solution
The volume out will be give by inital volume +(flow rate in - flow rate out)*t, as we have the intial volume and every minute we add so much to it (flow rate in - flow rate out).
Now the concentration out will be c=m/v, so x represents the amouhnt of x (i..e. m) and v is the changing volume.
so from that we have the concentration of the solution as $\frac{x}{V_{0}+(Rate_{in}-Rate_{out})*t}$
Now we times that by the volume out to get the flow out of x.
Then $\frac{dx}{dt}=\frac{dx_{in}}{dt}-\frac{dx_{out}}{dt}$

I think I may have over complicated a simple explanation but maybe someone else can come along and give a more concise explanation. Whenever you get to these types of questions , work out the flow rate of x in and the flow rate of x out, and minus the second from the first. Hope it helps though.
Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 20, 2012, 09:31:07 pm
if anyone is interested, the general solution to these problems is as follows:

you have a tank with volume, $V_0$ initially with $Q_0$ kg of salt present. You pump a solution of salt of concentration $C_{in}$ in at a rate of $\frac{dV_{in}}{dt}$. The tank is continuously stirred and at any time $t$ there is $Q$kg of salt present. The mixture is pumped out at a rate of $\frac{dV_{out}}{dt}$.

Now you differential equation is:

$\frac{dQ}{dt}=\frac{dV_{in}}{dt}C_{in}-\frac{dV_{out}}{dt}\frac{Q}{V_0+t \left (\dfrac{dV_{in}}{dt}-\dfrac{dV_{out}}{dt} \right )}=V_{in} C_{in} - V_{out} \frac{Q}{V_0+t\Delta V}$

the second one is just an abbreviated form of the first one, so i don't have too many fractions everywhere...

Now, to get the general solution we use the Integrating Factor.

Basically, let $I(t)=e^{\int \frac{V_{out}}{V_0+t \Delta V} dt} = \left ( V_0+ t \Delta V \right )^{\frac{V_{out}}{\Delta V}}$

then using the formula on that wiki page, $Q=\frac{1}{I(t)} \int V_{in} C_{in} \left (V_0+t \Delta V \right ) ^{\frac{V_{out}}{\Delta V}} dt$

then crank the handle to get:

$Q=\frac{V_{in} C_{in} (V_0+t\Delta V )}{V_{out} + \Delta V} + \frac{k}{( V_0 + t\Delta V )^{\frac{V_{out}}{\Delta V}}}$

where $k=\left ( Q_0 - \frac{V_{in} C_{in} V_0}{\Delta V (\frac{V_{out}}{\Delta V} + 1)} \right ) V_0^{\frac{V_{out}}{\Delta V}}$

pretty sure all of that is right, if anyone wants to check for any errors that'd be awesome.

This should only be used to check answers, the methods used are out of the VCE scope and they will never ask you to solve one of these without prompting the solution first and you showing that it is indeed a solution. also its pretty cool i reckon haha :)

and of course it reduces to the general solution when $\Delta V=0$ which is the ones you have to know how to solve, actually it might be a bit more complex given the $1/\Delta V$ in the exponent, but using some identities it should reduce haha
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 20, 2012, 09:36:52 pm
^ more like this won't be used at all. They will at most ask you to plug the solution back into the DE to check that it's right. Good job anyway.
Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 20, 2012, 09:53:02 pm
haha yea, i just remember back in vce i wondered how you would solve the ones with changing volumes, or if you could at all, now that im in uni and know i may as well show other spec kids if they're wondering, also it can never hurt to learn a bit more maths :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 21, 2012, 05:10:02 pm
I have another differential equation I can't solve :(

"A conical tank has a radius length at the top equal to its height. Water, initially with a depth of 25cm, leaks out through a hole in the bottom of the tank at a rate of 5*sqrt(h) cm^3 /min where the depth is h cm at time t minutes.
Construct a differential equation expressing dh/dt as a function of h, and solve it."

I have to use weird rules like dh/dt = dh/dr * dr/dV * dV/dt for example (just an example, not necessarily this exact formula).
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 21, 2012, 09:54:45 pm
Nvm I found the answer to the last question.
Now I'm doing Euler's Method and I don't fully understand it...

If dy/dx = cos(x), given that y-sub-0 = y(0) = 1, find y-sub-3 using h = 0.1
Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 22, 2012, 12:22:50 am
so you use the formula $y_{n+1} = y_n + h\cdot y'_n$ where $h=0.1, y' = \cos (x)$

using $y_0=1,y'_0=\cos (0) = 1$ then $y_1=1+0.1\cdot 1 = 1.1$

now using $y_1 = 1.1, y'_1 = \cos (0.1)$ then $y_2 = 1.1 + 0.1\cos (0.1)$

now using $y_2 = 1.1 + 0.1\cdot (0.1), y'_2 = \cos (0.2)$ then finally, $y_3 = 1.1 + 0.1\cos (0.1) + 0.1\cos (0.2) \approx 1.297507074$

this seems correct as when you solve the DE analytically, you get $y=1+\sin (x)$ and then $y_3=y(0.3)=1+\sin (0.3) \approx 1.295520207$, so we are pretty accurate using eulers method and haven't made any serious blunders :)

these problems you just have to avoid silly mistakes so do a few until you're confident with the formula and how it works.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 25, 2012, 03:38:50 pm
Thanks moekamo.
Here's another one:

dy/dx = sqrt(x), y-sub-0 = y(1) = 0, h = 0.01, find y-sub-4.
Does that mean I have to start at y(1) then go to y(1.01) then y(1.02)?
So would y'-sub-0 = y'(1) = sqrt(1) = 1? Is that right?
Title: Re: Specialist 3/4 Question Thread!
Post by: moekamo on January 25, 2012, 11:33:28 pm
yep, so $y_1 = y(1.01) = 0+ 0.01 \times 1 = 0.01$ then continue like that until you get to $y_4=y(1.04)$
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on January 27, 2012, 10:47:17 pm
Just wondering if we need to know Derivates of inverse circular functions?
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 27, 2012, 11:20:07 pm
They're on the formula sheet, so you won't have to memorise them, but in all honesty, you might as well :P
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 27, 2012, 11:31:49 pm
They're on the formula sheet, so you won't have to memorise them, but in all honesty, you might as well :P

If it's on the formula sheet then you should know it--is how I went by :)
Title: Re: Specialist 3/4 Question Thread!
Post by: acinod on January 28, 2012, 12:04:29 am
Just wondering if we need to know Derivates of inverse circular functions?

Most people in my class remembered it as they worked through the year.
Even during the exam I still remember I had to look it up in the formula sheet. It's kind of annoying and you're better off learning it but don't worry if you don't.
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on January 28, 2012, 12:59:17 pm
hey guys, i need help with a problem about complex numbers. I'll type the question out :D.. show working out if possible, or tips on how to solve this kind of equation.

Solve to find x and y in the following.
a. (x+1) + (y-1)i = 2+3i
c. (2x+i) + (3-2yi) = x+3i

Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on January 28, 2012, 01:17:15 pm
ohhh thanks! i get it now, haha
could you please do this one, im kind of stuck on it D:

(2x-3i)+(-3+2y)i = y-xi
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on January 28, 2012, 01:24:23 pm
2x-3i-3i+2yi=y-xi
2x-6i+2yi=y-xi
2x+2i(y-3)=y-xi

2x=y [1]
2(y-3)=-x [2]

solve simultaneously.

correct me if i'm wrong xD
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on January 28, 2012, 01:26:54 pm
2x=y [1]
2(y-3)=-x [2]

sub [1] into [2]

2(2x-3)=-x
4x-6=-x
5x=6
x=6/5

sub x=6/5 into [1]
y=2x
y=12/5
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on January 28, 2012, 01:27:36 pm
yeep, thats the answer thanks heaps trinh :D
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 29, 2012, 10:50:20 pm
v(t) = 4(1 - e^(-0.5t))
A particle travels for 30 seconds starting at t = 0. What distance does it cover?

I know area = definite integral under curve, but I have the velocity curve not the position curve :( I was thinking of maybe integrating it twice. Once as an indefinite integral to find x(t) and then once to find the area under the curve from 0 to 30. Would that be right?
Also, since distance and displacement are not the same thing, do I have to find when the velocity gets less than 0?
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 29, 2012, 11:01:00 pm
Distance  is the total area under the curve so graph it and see if it goes to the negative, if it does split it into two integrals, if not then just integrate it.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 29, 2012, 11:23:05 pm
Oh I get it. I was overcomplicating it. All I had to do was x(30) - x(0), or the definite integral from 0 to 30 of the velocity function.

Thank you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on January 29, 2012, 11:45:33 pm

Oh I get it. I was overcomplicating it. All I had to do was x(30) - x(0), or the definite integral from 0 to 30 of the velocity function.

Thank you :)

Yes you were overcomplicating it, and now you are oversimplifying it.
xZero also pointed out that you must check that v(t) has no horizontal axis intercept for t in [0,30] ... which happens to be the case here.
Only then can you state with certainty that,

distance travelled = x(30) - x(0)
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 30, 2012, 01:02:51 pm
Ugh I'm so confused! The question is:

"A particle of mass 2kg hangs from a fixed point, O, by a light inextensible string of length 2.5m. It is pulled aside a horizontal distance of 2m by a force P N inclined at an angle of 75 degrees with the downward vertical, and rests in equilibrium. Find P and the tension of the string."

I thought that the length of the string didn't matter and the only thing that mattered was the force of the string. I tried making a triangle of forces but I didn't have enough information to solve it. Someone please tell me how the length of the string affects the forces.
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 30, 2012, 01:09:21 pm
the length of the string gives you the angle of the tension force, there should be 3 forces in total and you need to use simultaneous equations to solve for tension and P force
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 30, 2012, 01:24:14 pm
I thought the forces gave you the forces? After all, it wouldn't matter if the string was 1cm long or 1km long, there would still be the same force applied to it, right?

None of the previous problems I did even mentioned how long the string was. They just said the force and the angle and it was a simple matter of resolving forces. But I don't know what to do when I have forces, angles and lengths. How are they related?
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 30, 2012, 01:44:50 pm
Heres a shitty diagram i drew on mspaint, see if you can connect the dots
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 30, 2012, 04:46:13 pm
I'm not really sure why you drew that diagram that way...
Title: Re: Specialist 3/4 Question Thread!
Post by: dc302 on January 30, 2012, 05:13:53 pm
I'm not really sure why you drew that diagram that way...

The purpose of the lengths is to find the angle, which will ultimately help find the force.
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 30, 2012, 05:43:18 pm
I'm not really sure why you drew that diagram that way...
I never use triangle of forces, i simply drew the question in a 2d diagram and analyze the forces (mind you the triangle of forces dont work in engineering and physics, its all about free body diagram). Also as dc302 said, length of the rope gives you the angle of the tension force, the rope can be 1km or 1m as long as the ratio between opposite and adjacent side remains constant.
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on January 30, 2012, 07:25:31 pm
Can someone go through the question please? I haven't done problems like that for a while and I'm a bit rusty...

"A particle of mass 2kg hangs from a fixed point, O, by a light inextensible string of length 2.5m. It is pulled aside a horizontal distance of 2m by a force P N inclined at an angle of 75 degrees with the downward vertical, and rests in equilibrium. Find P and the tension of the string."
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on January 30, 2012, 07:45:44 pm
Can someone go through the question please? I haven't done problems like that for a while and I'm a bit rusty...

"A particle of mass 2kg hangs from a fixed point, O, by a light inextensible string of length 2.5m. It is pulled aside a horizontal distance of 2m by a force P N inclined at an angle of 75 degrees with the downward vertical, and rests in equilibrium. Find P and the tension of the string."

Draw a diagram like xZero has represented, I think he has the 2kg particle missing in the diagram, to make that easier I would have placed the 2kg particle on dot that connects between the vertical force 2.5cm and resultant force which is unknown. Than resolve it as trig, since they are asking us to the find the tension of 'P', Use something formulas that are resolving vertical and horizontal forces that are like $w_x=m \cdot \cos(\theta)$ and $w_y=m \cdot \sin(\theta)$. Also small thing since its a 'fixed point' the particle wouldn't swing and sway around in that incline angle.
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 30, 2012, 08:19:27 pm
Yeh something like that, the black line is the roof, red lines are the length and blue lines are the forces, at the centre of them blue lines is the weight. Sorry not use to drawing the objects in, i usually just represent everything with a dot. What you do is resolve them vertically and horizontally. For vertical you have Tsin(51.xx)=2*9.8 + Psin(75). Do the same thing horizontally and punch them into your cas and solve them simultaneously
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on January 30, 2012, 08:34:16 pm
Shouldn't the hypoteneuse in xZero's diagram be 2.5m? Hence the vertical component of the triangle would be 1.5m...
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on January 30, 2012, 08:47:11 pm
bleh my bad, well as long as the the image conveyed the idea its all good
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on January 31, 2012, 12:29:18 pm
The purpose of the lengths is to find the angle, which will ultimately help find the force.

Ahh, this makes sense. Thanks!

But the diagram you drew looks a bit different to the "solution", which was this:
http://img848.imageshack.us/img848/4896/mathsproblem12.png
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on January 31, 2012, 11:39:41 pm
there's more than one way to draw it, depending on how you arrange it
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 01, 2012, 06:29:11 pm
Hopefully this isn't a noob question.

Could someone do this equation for me, I get stuck when I square root everything, the book doesn't explain how they got a certain co-efficient.

|z+1|+|z-1|=4, turn to the cartesian equation form.

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on February 01, 2012, 07:01:48 pm
let z=x+yi
|(x+1)+yi)| + |(x-1)+yi| = 4
sqrt((x+1)^2 + y^2) + sqrt((x-1)^2 + y^2) = 4
sqrt(x^2 + 2x + 1 + y^2) + sqrt(x^2 - 2x + 1 + y^2) = 4
sqrt(x^2 + 2x + 1 + y^2) = 4 - sqrt(x^2 - 2x + 1 + y^2)
x^2 + 2x + 1 + y^2 = 16 - 8sqrt(x^2 - 2x + 1 + y^2) + x^2 - 2x + 1 + y^2
4x = 16 - 8sqrt(x^2 - 2x + 1 + y^2)
16 - 4x = 8sqrt(x^2 - 2x + 1 + y^2)
256 - 128x + 16x^2 = 64x^2 - 128x + 64 + 64y^2
48x^2 + 64y^2 = 192
3x^2 + 4y^2 = 12
x^2/4 + y^2/3 = 1
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on February 02, 2012, 04:19:55 pm
The vector resolute of i + 2j - 3k in the direction perpendicular to vector 2i - j - 2k is given by:
A. 1/3 (-i + 8j - 5k)
B. -2/3 (2i - j - k)
C. 1/3 (7i + 4j - 13k)
D. 1/9 (19i + 28j - 7k)
E. -i + 3j - k

The main problem I have with this question is that I don't know how to find a vector that is perpendicular to a 3D vector. If it was 2D, I would just let the dot product equal 0. But this is 3D. Wouldn't there be heaps of different possibilities, since angles are only defined in 2 dimensions?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 02, 2012, 07:13:34 pm
Use vector resolutes (refer to the attached image).
u is the vector resolute of a in the direction of b that is parallel to b
w is the vecotr resolute of a in the direction of b that is perpendicular to b

Parallel component $\underset{\sim}{u}=\frac{\underset{\sim}{a}.\underset{\sim}{b}}{\underset{\sim}{b}.\underset{\sim}{b}}\underset{\sim}{b}$
Perpendicular Component $\underset{\sim}{w}=a-\frac{\underset{\sim}{a}.\underset{\sim}{b}}{\underset{\sim}{b}.\underset{\sim}{b}}\underset{\sim}{b}$

So for this question let
a=i+2j-3k
b=2i-j-2k

So the perpendicular component is
$\underset{\sim}{w}=(\underset{\sim}{i}+2\underset{\sim}{j}-3\underset{\sim}{k})-\frac{(1*2)+(2*-1)+(-3*-2)}{2^{2}+(-1)^{2}+(-2)^{2}}(2\underset{\sim}{i}-\underset{\sim}{j}-2\underset{\sim}{k})$
$\underset{\sim}{w}=(\underset{\sim}{i}+2\underset{\sim}{j}-3\underset{\sim}{k})-\frac{2}{3}(2\underset{\sim}{i}-\underset{\sim}{j}-2\underset{\sim}{k})$
$\underset{\sim}{w}=(\underset{\sim}{i}+2\underset{\sim}{j}-3\underset{\sim}{k})-(\frac{4}{3}\underset{\sim}{i}-\frac{2}{3}\underset{\sim}{j}-\frac{4}{3}\underset{\sim}{k})$
$\underset{\sim}{w}=-\frac{1}{3}\underset{\sim}{i}+\frac{8}{3}\underset{\sim}{j}-\frac{5}{3}\underset{\sim}{k}$
$\underset{\sim}{w}=\frac{1}{3}(-\underset{\sim}{i}+8\underset{\sim}{j}-5\underset{\sim}{k})$

(Hopefully I haven't made mistake somewhere, or swapped the vectors around, I'm a bit rusty :P)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on February 03, 2012, 12:01:19 pm
(refer to the attached image).

We can't :P
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 03, 2012, 05:29:41 pm
(refer to the attached image).

We can't :P
Sorry guys, mustn't have attached it, it's attached in the original post now.

Moderator action: removed real name, sorry for the inconvenience
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 03:06:47 pm
If w= -(1/3) - ((square root 3)/3)i, express w^(-4) in polar form :D
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 04, 2012, 03:15:53 pm
Just this question for me:

Find the tangent to the curve x=2-3y^2, at point x=-1.

I saw the worked solution to it but it's still very confusing.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 04, 2012, 03:17:57 pm
find dx/dy, flip it to get dy/dx, sub in x = -1 to find gradient, then just use general equation of a linear line
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 03:21:37 pm

If w= -(1/3) - ((square root 3)/3)i, express w^(-4) in polar form :D

ugh, i wish i knew latex, but this will have to be sufficient :)

(http://i42.tinypic.com/mkesr4.jpg)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 04, 2012, 03:28:04 pm
If w= -(1/3) - ((square root 3)/3)i, express w^(-4) in polar form :D
$tan(\theta)=\frac{(-\frac{\sqrt{3}}{3})}{(-\frac{1}{3})}=\sqrt{3}$
$\theta=\frac{-2\pi}{3}$ (base angle is $\frac{\pi}{3}$ but is in the 3rd quadrant
$r=\sqrt{(\frac{-1}{3})^{2}+(\frac{-\sqrt{3}}{3})^{2}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$
$w=\frac{2}{3}cis(\frac{-2\pi}{3})$
$w^{4}=(\frac{2}{3})^{4}cis(4*\frac{-2\pi}{3})$
$w^{4}=\frac{16}{81}cis(\frac{-8\pi}{3})$
$w^{4}=\frac{16}{81}cis(\frac{-8\pi}{3}+2\pi)$
$w^{4}=\frac{16}{81}cis(\frac{-2\pi}{3})$

Now we want $w^{-4}$ so we flip the r and the angle will become postive since $cis(-\theta)=\frac{1}{cis(\theta)}$

So we end up with $w^{-4}=\frac{81}{16}cis(\frac{2\pi}{3})$

Beaten but there is your LaTeX Paul.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 03:30:34 pm
Find the tangent to the curve x=2-3y^2, at point x=-1.

(Again, sorry for no LATEX :P)

(http://i39.tinypic.com/530ojm.jpg)
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 03:33:00 pm
Ah i seee! Thanks paulsterio + b^3 :)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 04, 2012, 03:33:43 pm
Find the tangent to the curve x=2-3y^2, at point x=-1.

(Again, sorry for no LATEX :P)

(http://i39.tinypic.com/530ojm.jpg)
But it feels and looks sooo much better! :P

EDIT: THAT WASN'T MEANT TO BE A PUN........
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 03:37:31 pm
But it feels and looks sooo much better! :P

EDIT: THAT WASN'T MEANT TO BE A PUN........

It's Microsoft Word Equation Editor, or something like that, basically it's like LaTeX, except you have like drag-and-drop templates - I don't know syntax D:
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 04, 2012, 03:45:07 pm
Hey, thanks for the explanation.

Heres another question.

Find the argument for 1-(i x square root of 3), in [0, 2 pi].

Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 03:57:43 pm
Find the argument for 1-(i x square root of 3), in [0, 2 pi].

Edit: Whoops, sorry you wanted [0, 2pi]

(http://i43.tinypic.com/21nlb89.jpg)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 04, 2012, 04:08:06 pm
Hey, thanks Paul.

Just this one last question on Complex numbers. (Yay!)

Simplify 4cis(2pi/3)/32cis(-pi/3).

(Also: How do you get that Microsoft Word Equation Editor?)
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on February 04, 2012, 04:14:06 pm
Hey, thanks Paul.

Just this one last question on Complex numbers. (Yay!)

Simplify 4cis(2pi/3)/32cis(-pi/3).

(Also: How do you get that Microsoft Word Equation Editor?)

4cis(2pi/3)/32cis(-pi/3)

4/32 cis(2pi/3 -(-pi/3))
=1/8 cis (2pi/3+pi/3)
=1/8 cis (3pi/3)
=1/8(cis pi)
=-1/8

I hope that's right...
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 04:55:54 pm
(Also: How do you get that Microsoft Word Equation Editor?)

(http://i41.tinypic.com/2608i7o.jpg)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on February 04, 2012, 05:15:44 pm
Install Microsoft Maths too (its legally free!): http://www.microsoft.com/education/ww/products/Pages/mathematics-4.0.aspx :)
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 04, 2012, 05:17:58 pm
Use texnic centre + Miktex!
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 05:50:13 pm
@ is the theta sign: Consider the complex number [email protected] + [email protected], where -pi < @ < pi. A) expand z^3 algebraically using the binomial theorem or otherwise. B) use DeMoivre's theorem to expand z^3 c)Hence, show that [email protected] = 3cos^[email protected]@ - sin^[email protected] and [email protected] = cos^[email protected] - [email protected]^[email protected]
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 04, 2012, 05:56:03 pm
z^3= ([email protected][email protected])^3

use this 'formula': http://en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem

using demoivre's theorem means just apply this 'formula' http://en.wikipedia.org/wiki/De_Moivre%27s_formula

then c) is just manipulation, equate the expression in a) and b), then solve for [email protected]
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 06:07:59 pm
Thanks :) but i'm still a bit lost with part c) :o. can you show me how you get the answer :D
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 04, 2012, 06:09:51 pm
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 06:16:18 pm
Is it a) z^3= cos^[email protected] - isin^[email protected] and b) z = [email protected] so z^3 = [email protected] + [email protected] :/
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 06:20:26 pm
O i think my part a is wrong :(  EDIT: is a) z^3 = cos^[email protected] + 3icos^[email protected]@- [email protected]^[email protected] - isin^[email protected]
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 04, 2012, 06:22:43 pm
i dono, cant be bothered doing the workings, however working from what you have, you should try equate real and imaginary coefficients, then just manipulate to have [email protected] on one side and simplify the mess on the other side of the equation

EDIT: so yeah, do the same thing, group the reals and imaginary components, then just equate.
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 09:44:37 pm
Thanks TT, equating coeeficients at the end got me to the right answer :) anywho, i have another q.  : Show that |z + w|^2 = |z|^2 + 2Re(z(conjugate of w)) + |w|^2
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 04, 2012, 09:49:23 pm
i assume z and w are both an element of the complex field?

well just let z = a+bi and w = c+di, sub them in and expand and do your algebra, then do the same to the RHS and show that you get the same expression :)
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 11:08:36 pm

anywho, i have another q.  : Show that |z + w|^2 = |z|^2 + 2Re(z(conjugate of w)) + |w|^2

Here's an alternative to TrueTears' way, his way leads to some really tough algebra, so I thought about it and it seems this works out more nicely :)

(http://i40.tinypic.com/x5eag4.jpg)
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on February 04, 2012, 11:21:56 pm
Not sure if this is valid...

$Prove |z + w|^2 = |z|^2 + 2Re(z*\frac{1}{w}) + |w|^2$

let $z= a+bi$
let $w=c+di$
conjugate w = $\frac{1}{w}= c-di$

$z+w= a+c+(b+d)i$

$(a+c)^2+(b+d)^2 = a^2+b^2+2Re((a+bi)(c-di))+c^2+d^2$

$a^2+2ac+c^2+b^2+2bd+d^2=a^2+b^2+2Re(ac-adi+bci+bd)+c^2+d^2$

$a^2+2ac+c^2+b^2+2bd+d^2=a^2+b^2+2(ac+bd)+c^2+d^2$

$a^2+b^2+c^2+d^2+2(ac+bd)=a^2+b^2+c^2+d^2+2(ac+bd)$

Proven.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 04, 2012, 11:25:53 pm
It's valid, except you probably need to make it look a little better.
When proving, you usually don't work with both sides of an equation, you work with one side (e.g. RHS) and prove that it equals the LHS.

Btw, good work, that method is even faster than mine!
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 11:35:20 pm
Isn't monkeywantsabananas way the same as wat truetears was talking about? :) and thanks paulsterio, made me look at it a different way :D
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 04, 2012, 11:46:02 pm
Actually i have another question (part b of the previous) : Hence, show that |z + w| is less or equal to |z| + |w| . This is called the triangle equality.  :D
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on February 05, 2012, 12:01:30 am
Isn't monkeywantsabananas way the same as wat truetears was talking about? :) and thanks paulsterio, made me look at it a different way :D

Yeah it's the one TT had... Paul did it another way which I thought was more complex than what TT had so I decided to write it out so you can compare the different ways and choose for yourself.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 05, 2012, 12:11:52 am
Yeah, I didn't see it the way you saw it, so when I worked it through using TT's method I ended up with A LOT of a, b, c and d's because I expanded the |z+w|^2 side, which I probably shouldn't have done :P

But yeah, TT's method is the faster method in this case, sorry TT, you're too good :P
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 05, 2012, 12:14:23 am
Actually i have another question (part b of the previous) : Hence, show that |z + w| is less or equal to |z| + |w| . This is called the triangle equality.  :D
ahh the famous triangle inequality, there are so many different ways to prove it, here's one of my faves that i tend to use (uses the inner product and the famous CS inequality)

$||x+y||^2 = = ||x||^2+++||y||^2 \le ||x||^2+2||+||y||^2 \le ||x||^2 + 2||x||||y||+||y||^2 = (||x||+||y||)^2$ and hence the result.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 05, 2012, 12:16:36 am
Lol, is there a way of doing it that doesn't involve using Vector Spaces? :P
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 05, 2012, 12:17:04 am
yeah ofcoz heaps lol there's a cool geometric proof too
Title: Re: Specialist 3/4 Question Thread!
Post by: HERculina on February 05, 2012, 12:19:11 am
Lol i don't really understand what u just did :P can you do a way that extends from part a of the question
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 07, 2012, 06:39:43 pm
Decide whether this is linear dependent/independent.

(http://i1054.photobucket.com/albums/s481/BondJames5/3x1matrice.png).

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: nubs on February 07, 2012, 06:59:40 pm
A group of vectors are always linear dependent if la + mb = c if my memory is correct (it's been a while)
(there is an exception to this I think, maybe you need to ensure that they're not parallel - someone else can confirm this, but your textbook should tell you anyway)
l and m are just random constants btw

so yeah you will get:

4li + lj + 3k + 2mi - mj + 3mk = -4i + 2j + 6k

Then you equate all the vectors in the direction i, j and k

so for the ones with direction i:
4l + 2m = -4
2l + m = -4
j:
l - m = 2
k:
3l + 3m = 6
l + m = 2

So there we have 3 equations

Try and solve for l and m by choosing two of the equations

Say you picked the first two equations, once you get the solutions for l and m, sub them into the third equation to see if it actually equals 6

from observation, it looks like we won't find solutions that will satisfy all 3 of the equations simultaneously, so therefore, they are independent. If we did find solutions that satisfied all 3 of them, then the set of vectors would be linearly dependent

as you can tell, this is quite rushed, so I may have made some silly little errors or a massive one somewhere
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 07, 2012, 07:12:16 pm
From observation, they are independent. You can work it through, but once you have experience, you can look at the sizes of the numbers as well as their signs and have a feel for whether it's dependent or independent.
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on February 07, 2012, 07:20:24 pm
If I'm right, another way is to take those three vectors and put them into a 3x3 vector and find its determinant (using CAS lol).

If the determinant=0, then the set of vectors is linearly dependent; vice versa..

So using the vectors you've given, you can either enter each vector columnwise or rowwise; you'll get the same number...

With the vectors you've given, the determinant=-72

Therefore, the set of vectors are linearly independent.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 07, 2012, 07:51:45 pm
trinh, i wish i knew that! D:

damn it!! i should have thought of that! it's genius!
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on February 07, 2012, 09:34:42 pm
If I'm right, another way is to take those three vectors and put them into a 3x3 vector and find its determinant (using CAS lol).

If the determinant=0, then the set of vectors is linearly dependent; vice versa..

So using the vectors you've given, you can either enter each vector columnwise or rowwise; you'll get the same number...

With the vectors you've given, the determinant=-72

Therefore, the set of vectors are linearly independent.
Wow, so simple >.> Never thought of that LOL.

//feels like dumbass.
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on February 09, 2012, 08:16:50 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on February 09, 2012, 08:25:51 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. $-i$ is represented by a point 1 unit vertically downwards)
so, $z^3 = cis\frac{-\pi}{2}$
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on February 09, 2012, 08:29:37 pm
$z^3=-i$
$z^3=cis(-\pi/2)$
$z=[cis(-\pi/2+2k\pi)]^{1/3}$, k∈Z
$z=cis(-\pi/6+2k\pi/3)$, k∈Z

now sub in values of k

Thus, $z=cis(-5\pi/6), z=cis(-\pi/6), z=cis(\pi/2)$

Hope this is right :s

EDIT: beaten as usual ><
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on February 09, 2012, 08:38:29 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. $-i$ is represented by a point 1 unit vertically downwards)
so, $z^3 = cis\frac{-\pi}{2}$
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

It's just I had a lesson today with a new tutor and she had this totally different method to the book (with the 2kpi etc, letting k=1, k=2) and it confused me. She made sin(theta)=0 and cos(theta)=1, and then had like 0, 2pi, 4pi etc. Still got the right answers but it was weird and not sure if better.

edit:

to make it more clear, if z^3 = 8i was the thing, she'd make r=2 and then sin(theta)=1 and then cos(theta)=0. I kinda understand what she's doing with the expansion of cis(theta) and then equating real and imaginary sides, but I don't entirely get what was done from there.
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on February 09, 2012, 08:49:53 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. $-i$ is represented by a point 1 unit vertically downwards)
so, $z^3 = cis\frac{-\pi}{2}$
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

It's just I had a lesson today with a new tutor and she had this totally different method to the book (with the 2kpi etc, letting k=1, k=2) and it confused me. She made sin(theta)=0 and cos(theta)=1, and then had like 0, 2pi, 4pi etc. Still got the right answers but it was weird and not sure if better.

edit:

to make it more clear, if z^3 = 8i was the thing, she'd make r=2 and then sin(theta)=1 and then cos(theta)=0. I kinda understand what she's doing with the expansion of cis(theta) and then equating real and imaginary sides, but I don't entirely get what was done from there.

I suggest you use the method that is taught in the book and make sure that method is clear to you, then have a go at your tutor's method. If you don't understand your tutor's method at all, don't let it concern you too much, you won't be at a disadvantage by only knowing one method, but you will be at an advantage by knowing 2 methods.
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on February 09, 2012, 08:53:18 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. $-i$ is represented by a point 1 unit vertically downwards)
so, $z^3 = cis\frac{-\pi}{2}$
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

It's just I had a lesson today with a new tutor and she had this totally different method to the book (with the 2kpi etc, letting k=1, k=2) and it confused me. She made sin(theta)=0 and cos(theta)=1, and then had like 0, 2pi, 4pi etc. Still got the right answers but it was weird and not sure if better.

edit:

to make it more clear, if z^3 = 8i was the thing, she'd make r=2 and then sin(theta)=1 and then cos(theta)=0. I kinda understand what she's doing with the expansion of cis(theta) and then equating real and imaginary sides, but I don't entirely get what was done from there.

I suggest you use the method that is taught in the book and make sure that method is clear to you, then have a go at your tutor's method. If you don't understand your tutor's method at all, don't let it concern you too much, you won't be at a disadvantage by only knowing one method, but you will be at an advantage by knowing 2 methods.

Yeah not my tutor's method I should have said, just a friend who did pretty decently in Spesh, but thank you very much! She's pretty bad at explaining things hence my hope someone here would know what to do haha.
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on February 09, 2012, 09:16:48 pm
Ok, I don't know if I'm helping here, but for the example you showed:

$z^3=8i$, where $z=r*cis(\theta)$
$[r*cis(\theta)]^3=8cis(\pi/2)$
$r^3cis(3\theta)=8cis(\pi/2)$ => $r=2$
$cis(3\theta)=cis(\pi/2)$
$cos(3\theta)+i*sin(3\theta)=cos(\pi/2)+i*sin(\pi/2)$
$cos(3\theta)+i*sin(3\theta)=0+i$

Equating coefficients,

$cos(3\theta)=0$ AND $sin(3\theta)=1$, where $-\pi<\theta\le\pi$, and thus $-3\pi<3\theta\le3\pi$

So solving $cos(3\theta)=0$, $\theta=-5\pi/6, -\pi/2, -\pi/6, \pi/6, \pi/2, 5\pi/6$
and solving $sin(3\theta)=1$, $\theta=-\pi/2, \pi/6, 5\pi/6$

now we pick out the common solutions, which are of course $\theta=-\pi/2, \pi/6, 5\pi/6$

and then sub back into $z=2cis(\theta)$

If this indeed what you were talking about, I'd still stick to the classic $2k\pi$ method because it doesn't involve trig (which I found deadly sometimes)
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on February 09, 2012, 09:20:15 pm
Ok, I don't know if I'm helping here, but for the example you showed:

$z^3=8i$, where $z=r*cis(\theta)$
$[r*cis(\theta)]^3=8cis(\pi/2)$
$r^3cis(3\theta)=8cis(\pi/2)$ => $r=2$
$cis(3\theta)=cis(\pi/2)$
$cos(3\theta)+i*sin(3\theta)=cos(\pi/2)+i*sin(\pi/2)$
$cos(3\theta)+i*sin(3\theta)=0+i$

Equating coefficients,

$cos(3\theta)=0$ AND $sin(3\theta)=1$, where $-\pi<\theta\le\pi$, and thus $-3\pi<3\theta\le3\pi$

So solving $cos(3\theta)=0$, $\theta=-5\pi/6, -\pi/2, -\pi/6, \pi/6, \pi/2, 5\pi/6$
and solving $sin(3\theta)=1$, $\theta=-\pi/2, \pi/6, 5\pi/6$

now we pick out the common solutions, which are of course $\theta=-\pi/2, \pi/6, 5\pi/6$

and then sub back into $z=2cis(\theta)$

If this indeed what you were talking about, I'd still stick to the classic $2k\pi$ method because it doesn't involve trig (which I found deadly sometimes)

That's it, yeh 2kpi seems way better, ty!
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on February 09, 2012, 10:18:45 pm
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 09, 2012, 10:20:10 pm
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

Yeh they were quite hard last year.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 09, 2012, 10:22:31 pm
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

although bora in your dp makes me quite happy hor..

the gif's from the mah boy mv right? lol

watch it [email protected][email protected][email protected]# http://www.youtube.com/watch?v=X6XXia5B2Wg
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on February 09, 2012, 10:43:40 pm
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

although bora in your dp makes me quite happy hor..

the gif's from the mah boy mv right? lol

watch it [email protected][email protected][email protected]# http://www.youtube.com/watch?v=X6XXia5B2Wg
I miss you, bebe. =(
How long did it take you past students to 'master' Vectors?
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on February 09, 2012, 10:45:46 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. $-i$ is represented by a point 1 unit vertically downwards)
so, $z^3 = cis\frac{-\pi}{2}$
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

just wondering what the actual answer is ?!
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on February 09, 2012, 11:33:48 pm
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. $-i$ is represented by a point 1 unit vertically downwards)
so, $z^3 = cis\frac{-\pi}{2}$
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

just wondering what the actual answer is ?!

$z^3 = cis(\frac{-\pi}{2} + 2k\pi)$
$z = 1^\frac{1}{3} cis(\frac{\frac{-\pi}{2}}{3} + \frac{2k\pi}{3})$
$z = cis(\frac{-\pi}{6} + \frac{2k\pi}{3})$
$z = cis(\frac{-\pi}{6}), cis(\frac{\pi}{2}), cis(\frac{-5\pi}{6})$
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 10, 2012, 06:09:35 pm
Okay.

This is to find the volume when rotated about the y-axis.

(http://i1054.photobucket.com/albums/s481/BondJames5/volume.png)

I get stuck at the last line. I can anti-differentiate but when I look at the answers I am wrong.

Could someone complete the last line and explain why it is?

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on February 10, 2012, 06:13:44 pm
Not sure about the rest, but in the 2nd last line it should say (y^4 + 2y^2 + 1) instead of (y^4 + 2y + 1).
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 11, 2012, 01:04:12 pm
Dis-regard my last question.

Just this one.

Express sin(3x), in terms of x.

I used the general formula to get this: sin(x)cos(2x)+cos(x)sin(2x). I am probably wrong here but could someone show me how it's done?

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on February 11, 2012, 01:15:34 pm
now apply the same thing to cos(2x) and sin(2x)  (which turns out to be the double angle formula)
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 11, 2012, 01:19:36 pm
$sin(3x) = sin(2x+x) = sin(x)cos(2x)+cos(x)sin(2x)$

$= sin(x)(cos^2(x) - sin^2(x))+cos(x)(2sin(x)cos(x))$

$= sin(x)cos^2(x) - sin^3(x)+2sin(x)cos^2(x)$

$= 3sin(x)cos^2(x) - sin^3(x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on February 11, 2012, 06:11:40 pm
Express sin(3x), in terms of x. --> it's already in terms of x?
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 11, 2012, 06:17:31 pm
lol maybe he meant in terms of sinx
Title: Re: Specialist 3/4 Question Thread!
Post by: iirene on February 12, 2012, 03:01:51 pm
I request some help please :)
Not having much luck with this...

Simplify the following expression:
cos(4x)sin(4x)

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 12, 2012, 03:13:43 pm
$sin(2x)=2\sin(x)\cos(x)$ (double angle formula)
Which means
$sin(8x)=2\sin(4x)\cos(4x)$
$\frac{1}{2}sin(8x)=\sin(4x)\cos(4x)$
$\therefore \sin(4x)\cos(4x)=\frac{1}{2}sin(8x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on February 12, 2012, 03:17:54 pm
Referring to the double angle formulae, $sin(2x)=2sin(x)cos(x)$

More generally, $sin(2ax)=2sin(ax)cos(ax)$

So... for your question $cos(4x)sin(4x)$, it can be seen that

$cos(4x)sin(4x)=\frac{1}{2}[2sin(4x)cos(4x)]$

Just for the sake of clarity, referring back to the general formula $sin(2ax)=2sin(ax)cos(ax)$, it is evident that for our specific question, $'a'=4$, and thus $'2a'=8$

Thus $2cos(4x)sin(4x)=sin(8x)$

and $cos(4x)sin(4x)=\frac{1}{2}sin(8x)$

Edit: beaten
Title: Re: Specialist 3/4 Question Thread!
Post by: iirene on February 12, 2012, 04:14:40 pm
Thank you! :) Was stuck on that since last night.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 12, 2012, 06:44:06 pm
Ugh, This question is not very well explained.

Find the asymptotes for y-1/x^2 = x^2.

I moved the -1/x^2 over, to make it +1/x^2 and then sketched the +1/x^2 and the -x^2 on the same axis but with this book, it just wants me to find the asymptotes WITHOUT sketching.

Could someone explain how?

Thanks. :D

Note: Not really relevant but does anyone know of an online calculator? Similar to the CAS? My CAS calculator was nicked. :(
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on February 12, 2012, 06:52:46 pm
y=x^2 + 1/x^2
you notice that when x --> +-infinity, y=x^2, so you have an oblique asymptote there. then we have another vertical asymptote at x = 0 for obvious reasons.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on February 12, 2012, 07:28:47 pm
Thanks for that.

Just this last spesh question for today. :(

Volume of a solid cylinder is 128pi cm^3.
Show that the total surface are, A cm^2, is A=2pir^2 + 256pi/r, where r>0.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 12, 2012, 07:32:01 pm
$V = 2pi. r^2 h = 128$

Solve for h

$h = \frac{128}{2pi.r^2}$

Now sub into A = 2pi.r(r+h)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 12, 2012, 07:32:50 pm
$Volume=\pi r^{2}h$
$\pi r^{2}h=128 \pi$
$h=\frac{128}{r^{2}}$
$SA=2\pi r^{2}+2 \pi rh$
$SA=2\pi r^{2}+2 \pi r\frac{128}{r^{2}}$
$SA=2\pi r^{2}+\frac{256\pi}{r}$
and r>0 as the radius can't be negative or 0 :)

EDIT: Half beaten, working above
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 12, 2012, 07:45:24 pm
Thanks b^3, i know how to write pi in LaTeX now ;D

Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on February 13, 2012, 02:40:07 am
Thanks b^3, i know how to write pi in LaTeX now ;D

Too bad you don't know the correct formula for the volume of a cylinder...

Kidding! (Don't take that the wrong way ;))
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on February 13, 2012, 04:04:25 pm
Thanks b^3, i know how to write pi in LaTeX now ;D

Too bad you don't know the correct formula for the volume of a cylinder...

Kidding! (Don't take that the wrong way ;))
LOL!
Title: Re: Specialist 3/4 Question Thread!
Post by: ahh.liz on February 13, 2012, 09:02:55 pm
Hello Specialist Mathematics friends, it would seem that I require some mathematical assistance. How would one simplify the following expression?

cos(3x)cos(x) + sin(3x)sin(x)

Thank you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 13, 2012, 09:04:39 pm
think about compound angle formula :P

$\sin(3x \pm x)$ or $\cos(3x \pm x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: ahh.liz on February 13, 2012, 09:09:30 pm
OOh! Haha, greatly appreciated! That's my homework done - time for some Restaurant City! :DDD
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 14, 2012, 06:45:52 pm
Too bad you don't know the correct formula for the volume of a cylinder...

Kidding! (Don't take that the wrong way ;))

Title: Re: Specialist 3/4 Question Thread!
Post by: pascalleung10 on February 14, 2012, 10:03:17 pm
Could someone help me with this question? (:

Find 'x' for 4sinxo= secxo, x E [0,2pi]

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on February 14, 2012, 10:15:43 pm
\begin{aligned}4\sin(x) & =\frac{1}{\cos(x)}
\\ 4\sin(x^{\circ})\cos(x^{\circ}) & =1
\\ 2\sin(2x^{\circ}) & =1
\\ \sin(2x^{\circ}) & =\frac{1}{2}
\end{aligned}

So $0\leq x\leq2\pi$
$0\leq2x\leq4\pi$
$2x^{\circ}=30^{\circ},\:180^{\circ}-30^{\circ},\:360^{\circ}+30^{\circ},540^{\circ}-30^{\circ}$
$2x^{\circ}=30^{\circ},\:150^{\circ},\:390^{\circ},\:510^{\circ}$
$x^{\circ}=15^{\circ},\:75^{\circ},\:195^{\circ},\:255^{\circ}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on February 15, 2012, 09:58:16 pm
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on February 15, 2012, 10:48:25 pm
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Converting to polar form we have: $z = \sqrt{2}cis\left(\frac{-\pi}{4}\right)$
Then, using De Moivre's theorem: $z^{-3} = \sqrt{2}^{-3}cis\left(\frac{-\pi}{4}\times{-3}\right)$
$z^{-3} = \frac{\sqrt{2}}{4}cis\left(\frac{3\pi}{4}\right)$
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on February 15, 2012, 11:06:12 pm
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Converting to polar form we have: $z = \sqrt{2}cis\left(\frac{-\pi}{4}\right)$
Then, using De Moivre's theorem: $z^{-3} = \sqrt{2}^{-3}cis\left(\frac{-\pi}{4}\times{-3}\right)$
$z^{-3} = \frac{\sqrt{2}}{4}cis\left(\frac{3\pi}{4}\right)$

$r = \sqrt{1^{2}+(-1)^{2}} = 2$,

Fourth quadrant, $\theta = 0 - \tan \frac {-1}{1}$, $\theta = \frac {- \pi}{4}$

$z=r \cdot cis (\theta)$, Sub it in.

Than you get, $z= \sqrt{2} \cdot cis (\frac {-\pi}{4})$.

Using De Moivre's Theorem, $z^{-3}= \sqrt {2} \cdot cis (\frac{-\pi}{4} \cdot -3)$.

Results as, $z^{-3}= \frac {\sqrt{2}}{4} \cdot cis (\frac {3 \cdot \pi}{4})$

Don't rush the questions your having trouble with take it slow and steady you might get an answer out of it.

This is not a sac.

Title: Re: Specialist 3/4 Question Thread!
Post by: motocross53 on February 15, 2012, 11:19:44 pm
could someone please show me how to do this,
4.4 Q5a of essential book:

show that sin(theta)+cos(theta)i=cis(pi/2-theta)

thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on February 15, 2012, 11:22:00 pm
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Converting to polar form we have: $z = \sqrt{2}cis\left(\frac{-\pi}{4}\right)$
Then, using De Moivre's theorem: $z^{-3} = \sqrt{2}^{-3}cis\left(\frac{-\pi}{4}\times{-3}\right)$
$z^{-3} = \frac{\sqrt{2}}{4}cis\left(\frac{3\pi}{4}\right)$

$r = \sqrt{1^{2}+(-1)^{2}} = 2$,

Fourth quadrant, $\theta = 0 - \tan \frac {-1}{1}$, $\theta = \frac {- \pi}{4}$

$z=r \cdot cis (\theta)$, Sub it in.

Than you get, $z= \sqrt{2} \cdot cis (\frac {-\pi}{4})$.

Using De Moivre's Theorem, $z^{-3}= \sqrt {2} \cdot cis (\frac{-\pi}{4} \cdot -3)$.

Results as, $z^{-3}= \frac {\sqrt{2}}{4} \cdot cis (\frac {3 \cdot \pi}{4})$

Don't rush the questions your having trouble with take it slow and steady you might get an answer out of it.

This is not a sac.

Thanks fletch-j and abd123, but the part I got stuck on was after all that... Like I have no idea what to do with the 3pi on 4 when I expand it to try to get it back to Cartesian form... :s
Title: Re: Specialist 3/4 Question Thread!
Post by: tony3272 on February 15, 2012, 11:25:46 pm
could someone please show me how to do this,
4.4 Q5a of essential book:

show that sin(theta)+cos(theta)i=cis(pi/2-theta)

thanks

$cis(\frac{\pi}{2}-\theta)=cos(\frac{\pi}{2}-\theta)+i sin(\frac{\pi}{2}-\theta)$

$=sin(\theta)+ i cos(\theta)$ as required
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on February 15, 2012, 11:26:30 pm
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Hmmm question:
Find z^(-3) if z= 1 - i
(I've tried this twice and too tired to reatttempt it, would be grateful if someone could show me how to solve this.)
Converting to polar form we have: $z = \sqrt{2}cis\left(\frac{-\pi}{4}\right)$
Then, using De Moivre's theorem: $z^{-3} = \sqrt{2}^{-3}cis\left(\frac{-\pi}{4}\times{-3}\right)$
$z^{-3} = \frac{\sqrt{2}}{4}cis\left(\frac{3\pi}{4}\right)$

$r = \sqrt{1^{2}+(-1)^{2}} = 2$,

Fourth quadrant, $\theta = 0 - \tan \frac {-1}{1}$, $\theta = \frac {- \pi}{4}$

$z=r \cdot cis (\theta)$, Sub it in.

Than you get, $z= \sqrt{2} \cdot cis (\frac {-\pi}{4})$.

Using De Moivre's Theorem, $z^{-3}= \sqrt {2} \cdot cis (\frac{-\pi}{4} \cdot -3)$.

Results as, $z^{-3}= \frac {\sqrt{2}}{4} \cdot cis (\frac {3 \cdot \pi}{4})$

Don't rush the questions your having trouble with take it slow and steady you might get an answer out of it.

This is not a sac.

Thanks fletch-j and abd123, but the part I got stuck on was after all that... Like I have no idea what to do with the 3pi on 4 when I expand it to try to get it back to Cartesian form... :s

Just remember that 3pi/4 = pi-pi/4 = pi/4. So it's just root2/4 x cos(pi/4) + root2/4 x isin(pi/4) .

edit: that is, sin(pi-x)=sin(x) and cos(pi-x)=cos(-x). What I said above is just applicable in the context of sin/cos/tan etc :P.
Title: Re: Specialist 3/4 Question Thread!
Post by: motocross53 on February 15, 2012, 11:43:27 pm
could someone please show me how to do this,
4.4 Q5a of essential book:

show that sin(theta)+cos(theta)i=cis(pi/2-theta)

thanks

$cis(\frac{\pi}{2}-\theta)=cos(\frac{\pi}{2}-\theta)+i sin(\frac{\pi}{2}-\theta)$

$=sin(\theta)+ i cos(\theta)$ as required

sorry mate,
i still dont really following you,
could you please give a longer explanation
Title: Re: Specialist 3/4 Question Thread!
Post by: nubs on February 16, 2012, 12:28:28 am
$cis(\frac{\pi}{2}-\theta)=cos(\frac{\pi}{2}-\theta)+i sin(\frac{\pi}{2}-\theta)$

$=sin(\theta)+ i cos(\theta)$ as required

To do a 'show that' question, you need to pick a side, and adjust it and manipulate it so that it equals the other side. so pick the left hand side, or right hand side (whichever you think is easier to manipulate and work on). Tony picked the RHS, which he then manipulated to = the LHS, so he did this:
RHS = $cis(\frac{\pi}{2}-\theta)$ (He then proceed to expand this)

Remember:
$cis(\theta)=cos(\theta)+i sin(\theta)$

So $cis(\frac{\pi}{2}-\theta) = cos(\frac{\pi}{2}-\theta)+i sin(\frac{\pi}{2}-\theta)$

Also remember that, $cos(\frac{\pi}{2}-\theta) = sin(\theta)$ and $sin(\frac{\pi}{2}-\theta) = cos(\theta)$

So, $cos(\frac{\pi}{2}-\theta)+i sin(\frac{\pi}{2}-\theta) = sin(\theta)+ i cos(\theta)$ = LHS
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on February 16, 2012, 10:11:13 pm
Complex number question: (From VCAA 2006 Exam 1)

Solve the quadratic equation $z^2+2z-\sqrt{3}i$, expressing your answers in exact cartesian form.
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on February 17, 2012, 07:06:38 am
Complex number question: (From VCAA 2006 Exam 1)

Solve the quadratic equation $z^2+2z-\sqrt{3}i$, expressing your answers in exact cartesian form.

$z^{2}+2z- \sqrt{3 } \cdot i$

$z^{2}+2z +1- \sqrt{3} \cdot i -1$

$(z+1)^{2}-(1 + \sqrt {3 })\cdot i=0$

$(z+1)^{2}=1+ \sqrt{3} \cdot i$

In polar form: $(z+1)^{2}= 2 \cdot cis ( \frac{\pi}{3}+2k\pi)$

Hence, $z+1= \sqrt {2} \cdot cis ( \frac{\pi}{3}+2k\pi)$

Let $k=0$$(z+1)^{2}= \sqrt{2} \cdot cis ( \frac{\pi}{6})= \frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}$

Therefore, $z= (\frac{-\sqrt{6}-2}{2})+\frac{\sqrt{2}}{2} \cdot i$

let $k=-1$

$z+1=\sqrt {2}\cdot cis({\frac{\pi}{6}-\pi})=\sqrt{2} \cdot cis (\frac{-5\pi}{6})$.

Thus, $z=\frac{(-\sqrt {6-2})}{2}-\frac{\sqrt{2}}{2}i$.

You can also use $z=a+bi$, Where a, b holds r.

Hoped I Helped :).
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on February 19, 2012, 03:48:06 pm
Must be a gap in my knowledge here, but from
$(z+1)^2=2\cdot cis(\frac{\pi}{3}+2k\pi)$
and when you square rooted doesn't the argument change?
I'm confused because when you subbed in k=0, you got $\sqrt{2 }\cdot cis\frac{\pi}{6}$
but wouldn't it be $\sqrt{2 }\cdot cis\frac{\pi}{3}$  because  $2\cdot 0\cdot \pi=0$
If that makes sense haha
Also I was looking through my textbook and couldn't find any examples of this type of question, is this assumed after all knowledge learnt?
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on February 19, 2012, 03:58:24 pm
Must be a gap in my knowledge here, but from
$(z+1)^2=2\cdot cis(\frac{\pi}{3}+2k\pi)$
and when you square rooted doesn't the argument change?
I'm confused because when you subbed in k=0, you got $\sqrt{2 }\cdot cis\frac{\pi}{6}$
but wouldn't it be $\sqrt{2 }\cdot cis\frac{\pi}{3}$  because  $2\cdot 0\cdot \pi=0$
If that makes sense haha
Also I was looking through my textbook and couldn't find any examples of this type of question, is this assumed after all knowledge learnt?
When he wrote $(z+1)^2=2\cdot cis(\frac{\pi}{3}+2k\pi)$
the next line should have been $z+1=\sqrt{2}\cdot cis(\frac{\pi}{6}+k\pi)$

Hope that clears it up. Also, this is a VCAA past exam question, so it would be good to make sure you understand it fully.
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on February 19, 2012, 04:09:56 pm
Ah I see, opposite of De Moivre's theorem.
Thanks man :)
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on February 24, 2012, 05:23:22 pm
Why is this invalid?

$\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}).(\underset{\sim}{a}-\underset{\sim}{b}})}$ = $\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}})^{2}$ = $(\underset{\sim}{a}-\underset{\sim}{b})$
Title: Re: Specialist 3/4 Question Thread!
Post by: nubs on February 24, 2012, 05:56:18 pm
Let a = 2i+j-k
Let b = i+j+k

Using these values, find ((a-b).(a-b))^(1/2)

Now find (a-b)

What does that tell you?

Remember that a.a doesn't equal a^2, it equals |a|^2

Similarly, (a-b).(a-b) doesn't equal (a-b)^2, it equals |a-b|^2
So ((a-b).(a-b))^(1/2) = |a-b|
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on February 25, 2012, 07:13:34 pm
$\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}).(\underset{\sim}{a}-\underset{\sim}{b}})}$ = $\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}})^{2}$ = $(\underset{\sim}{a}-\underset{\sim}{b})$

Nirbaan explained it, but in simple terms,$\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}).(\underset{\sim}{a}-\underset{\sim}{b}})}$ DOES NOT EQUAL $\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}})^{2}$

in fact, $\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}).(\underset{\sim}{a}-\underset{\sim}{b}})}$ = $|a-b|$
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on February 25, 2012, 07:34:41 pm
Why is this invalid?

$\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}).(\underset{\sim}{a}-\underset{\sim}{b}})}$ = $\sqrt{(\underset{\sim}{a}-\underset{\sim}{b}})^{2}$ = $(\underset{\sim}{a}-\underset{\sim}{b})$
Well firstly consider this statement $(\underset{\sim}{a}-\underset{\sim}{b}})^{2}$

Is "squaring" defined for a vector? There is no distinct "multiplication" operation defined for a vector in your usual sense.
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on February 26, 2012, 12:02:44 am
Ah ok thanks!

Title: Re: Specialist 3/4 Question Thread!
Post by: powerrrr on February 26, 2012, 12:04:49 pm
Hi, I'm having a bit of trouble with De Moivre's...

If W = 54rt2 cis (3pi/4), find solutions to cuberootW

I only got one solution which is 3rt2cis pi/4

Can someone help please? There's 2 more..
Title: Re: Specialist 3/4 Question Thread!
Post by: trinh on February 26, 2012, 01:52:20 pm
$w=54\sqrt2cis(\frac{3\pi}{4})$
$\sqrt[3]w=w^\frac{1}{3}=[54\sqrt2cis(\frac{3\pi}{4}+2k\pi)]^\frac{1}{3}$, k ∈ Z
$w^\frac{1}{3}=3\sqrt2cis(\frac{\pi}{4}+\frac{2k\pi}{3})$, k ∈ Z

Sub in k=-1, 0 and 1 to get:
$w^\frac{1}{3}=3\sqrt2cis(\frac{-5\pi}{12}), 3\sqrt2cis(\frac{11\pi}{12}), 3\sqrt2cis(\frac{\pi}{4})$

You pretty much just forgot to add on the $2k\pi$ in step 2, hence missing out on the other two solutions
Title: Re: Specialist 3/4 Question Thread!
Post by: claireb on March 04, 2012, 07:09:55 pm
Having a mental block and can't seem to work out these questions (tech free) :(

edit:

thanks fletch-j :)

- If sin(x)=4/5 for x E [0,pi/2] and tan(y)=5/12 for y E [pi,3pi/2], find:

- If tan(x)=2 and x E [0,pi/2], use trigonometric identities to find the exact values for:

- Simplify the expression tan(x/2)/(1-tan^2(x/2))
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on March 04, 2012, 07:45:26 pm
Having a mental block and can't seem to work out these questions (tech free) :(
The first two I can simplify partially but not fully to get the stated answers, the last two I'm just lost so any insight on how to solve them would be much appreciated.

First one:

$\frac{cos^2(x)}{sin(x)}+sin(x)$             Take sin(x) out as a factor
$=sin(x)\left(\frac{cos^2(x)}{sin^2(x)}+1\right)$
$=sin(x)(cot^2(x)+1)$                          $cot^2(x)+1 = csc^2(x)$

$=sin(x)\cdot csc^2(x)$
$=csc(x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on March 04, 2012, 07:50:25 pm

$tan\left(\frac{\pi}{2}-y\right)$
$=\frac{sin\left(\frac{\pi}{2}-y\right)}{cos\left(\frac{\pi}{2}-y\right)}$
$=\frac{cos(y)}{sin(y)}$
$=cot(y)$
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on March 04, 2012, 08:24:21 pm
Having a mental block and can't seem to work out these questions (tech free) :(
The first two I can simplify partially but not fully to get the stated answers, the last two I'm just lost so any insight on how to solve them would be much appreciated.

First one:

$\frac{cos^2(x)}{sin(x)}+sin(x)$             Take sin(x) out as a factor
$=sin(x)\left(\frac{cos^2(x)}{sin^2(x)}+1\right)$
$=sin(x)(cot^2(x)+1)$                          $cot^2(x)+1 = csc^2(x)$

$=sin(x)\cdot csc^2(x)$
$=csc(x)$

Slightly shorter method
$\frac{\cos^{2}(x)}{\sin(x)}+\sin(x)=\frac{\cos^{2}(x)+\sin^{2}(x)}{sin(x)}=\frac{1}{\sin(x)}=\csc(x)$

(as $\cos^{2}(x)+\sin^{2}(x)=1$)

For the others use
$\cos^{2}(x)+\sin^{2}(x)=1$ to find sin(x), then tan(x)=sin(x)/cos(x), then sec(x)=1/cos(x)

For sin(x+y), you've found cos(x) and sin(x) already, draw up a triangle with the opposite=5 and adjacent=12 with angle y so that you can find cos(y) an sin(y), then you can use the addition formula sin(x+y)=sin(x)cos(y)+cos(x)sin(y)

Then for tan(x+2y) we know that tan(a+b)=$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$, so we need tan(2y), which we can use the double angle formula for, $\tan(2y)=\frac{2\tan(y)}{1-\tan^{2}(y)}$
so then use $\tan(x+2y)=\frac{\tan(x)+\tan(2y)}{1-\tan(x)\tan(2y)}$ and sub in the values.

For tan(x)=2, draw up a traingle with opposite=2, adjacent=1, then find the hypotenus=root(4+1)=root(5). Now you can find sin(x)=2/root(5) and cos(x)=1/root(5).  Now you can use the double angle formulas $\sin(2x)=2\sin(x)\cos(x)$ and $\cos(2x)=1-2\sin^{2}(x)$ and take into account the quadrant.

For the last one $\frac{tan(\frac{x}{2})}{1-\tan^{2}(\frac{x}{2})}$, is just like you double angle formula for tan, so we can say that $\frac{tan(\frac{x}{2})}{1-\tan^{2}(\frac{x}{2})}=\tan(x)$

Hope that helps :) (and hope there aren't any mistakes there, eyes are tired so there might be)
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on March 06, 2012, 05:27:39 pm
Could anyone prove this vector?

It seems simple enough to do, AB + BC + CA = O and find another way to represent that and cancel down.

But for some reason, I can never seem to actually cancel down, (i always get -2c or something)
Title: Re: Specialist 3/4 Question Thread!
Post by: yawho on March 06, 2012, 08:17:50 pm
Could anyone prove this vector?

It seems simple enough to do, AB + BC + CA = O and find another way to represent that and cancel down.

But for some reason, I can never seem to actually cancel down, (i always get -2c or something)

a + b = AC = -c, .: a + b + c = 0
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on March 07, 2012, 05:08:39 pm
Thanks for that, yawho, seems to make more sense now!

And just one more question, when finding solutions of compex numbers, how would you know when you're meant to add the angle along with the 2πk rule?

Attached is what i mean:

I have a feeling you're meant to only use 2πk (and not add any angles) only for whole numbers such as 3, 6 and when it comes to cartesian/polar form (3-4i or rcis(2π)), you're meant to use the angle that it creates in addition to 2πk..
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on March 07, 2012, 05:21:32 pm
you always need to add the 2k(pi), think about it geometrically, you have angle theta, but any more full rotations around the circle will yield the same angle, hence the +2kpi
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on March 09, 2012, 07:24:56 pm
Would someone please be able to help me simplify the following:
It's part of an extended question and I have done previous working to get to this:

Find the following in terms of $cos(2\theta )$ and $cos(4\theta )$

$cos4\theta-cos^4\theta+6cos^2\theta sin^2\theta$
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on March 09, 2012, 09:07:11 pm
It is easy to express $\cos^2\theta$ and $\sin^2\theta$ in terms of $\cos(2\theta)$ **, to express $\cos^4\theta$ just write it as $\cos^2\theta\cos^2\theta$ and again use your identities.

** http://mathworld.wolfram.com/Double-AngleFormulas.html second and third equation for $\cos(2x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on March 09, 2012, 09:36:53 pm
It is easy to express $\cos^2\theta$ and $\sin^2\theta$ in terms of $\cos(2\theta)$ **, to express $\cos^4\theta$ just write it as $\cos^2\theta\cos^2\theta$ and again use your identities.

** http://mathworld.wolfram.com/Double-AngleFormulas.html second and third equation for $\cos(2x)$

Care to elaborate? The fact that it's cos(2x) = 2cos2(x)-1 is screwing me over a bit. Should I just add -1 + 1 to the expressions and simplify?
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on March 09, 2012, 10:10:56 pm
cos(4x) - cos^4(x) + 6cos^2(x)sin^2(x)
= cos(4x) - cos^2(x)cos^2(x) + 6 cos^2(x)sin^2(x)

now cos(2x) = 2cos^2(x) - 1, so cos^2(x) = [cos(2x) + 1]/2
also cos(2x) = 1-2sin^2(x), so sin^2(x) = [1-cos(2x)]/2

substitute and you're done.
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on March 09, 2012, 10:44:22 pm
mmm.. thanks, I seem to have overcomplicated it for myself.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 11, 2012, 01:34:35 pm
Two forces act on an object. They are described by the vectors: F1=$13i-5j$ and F2=$-4i+9j$. If a third force, F3 is applied to the body, and the resultant force is zero, Find F3.

The text book have an explanation that involved 3 forces, not the two.

I got the resultant force, for F1 and F2 to be $9i+4j$. If that helps.

Thanks. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: nubs on March 11, 2012, 02:41:11 pm
since we don't know what F3 is, let's assign the variable x and y to describe it

so let F3 = xi + yj

we know that the resultant force, F1 + F2 + F3 = 0
So:
13i -5j - 4i + 9j + xi +yj = 0
9i + 4j + xi + yj = 0
(9 + x)i + (4+y)j = 0

so 9 + x and 4 + y both have to equal 0
so x = -9 and y = -4
now we can sub these values back into F3 = xi + yj
F3 = -9i - 4j

You had already done most of the work when you found the resultant force of F1 and F2 to equal 9i + 4j, F3 just had to be a vector that cancelled them out to equal 0 when added together, so straight away you can tell from that that F3 needed to equal -9i -4j :)
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on March 11, 2012, 05:07:56 pm
In the storeroom of a fruit shop there were two boxes of apples, one of Golden Delicious and the other of Jonathons, which were to be sold at $2.80 and$3.50/kg, respectively. The apples were accidentally mixed together and, instead of sorting them out, the owner decides to sell them as they were. So as not to make  a loss, he sold the mixed apples at $3.10/kg. How many kilograms of each type of apple were there if together they weighed 35kg in total? thanks.. Title: Re: Specialist 3/4 Question Thread! Post by: Planck's constant on March 11, 2012, 05:23:47 pm In the storeroom of a fruit shop there were two boxes of apples, one of Golden Delicious and the other of Jonathons, which were to be sold at$2.80 and$3.50/kg, respectively. The apples were accidentally mixed together and, instead of sorting them out, the owner decides to sell them as they were. So as not to make a loss, he sold the mixed apples at$3.10/kg. How many kilograms of each type of apple were there if together they weighed 35kg in total?

thanks..

Let G be the original quantity of Golden Delicious apples [Kg]
Let J be the original quantity of Jonathon apples [Kg]

J + G = 35   (1)

2.80*G + 3.50*J = 3.10*(J + G)     (2)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 11, 2012, 08:13:50 pm
since we don't know what F3 is, let's assign the variable x and y to describe it

so let F3 = xi + yj

we know that the resultant force, F1 + F2 + F3 = 0
So:
13i -5j - 4i + 9j + xi +yj = 0
9i + 4j + xi + yj = 0
(9 + x)i + (4+y)j = 0

so 9 + x and 4 + y both have to equal 0
so x = -9 and y = -4
now we can sub these values back into F3 = xi + yj
F3 = -9i - 4j

You had already done most of the work when you found the resultant force of F1 and F2 to equal 9i + 4j, F3 just had to be a vector that cancelled them out to equal 0 when added together, so straight away you can tell from that that F3 needed to equal -9i -4j :)

Ah! That makes sense!

Thank you. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: powerrrr on March 13, 2012, 01:34:07 pm
I received my specialist sac marks last week and I just passed. 51%. This is terrble relative to other students. Have I already failed this course?

I've lost hope for a good SS. :(
Title: Re: Specialist 3/4 Question Thread!
Post by: abd123 on March 13, 2012, 04:03:25 pm
I received my specialist sac marks last week and I just passed. 51%. This is terrble relative to other students. Have I already failed this course?

I've lost hope for a good SS. :(

Naaah man, theres still a few more sacs that can make up your 51%.

It seems like an average score in spesh, after all spesh is one of the hardest subjects in school.

Anyways good luck, my brotheren.
Title: Re: Specialist 3/4 Question Thread!
Post by: domislong on March 22, 2012, 04:00:28 am
Hey I'm doing a spec-type unit at uni, mostly Methods revision at the moment, but I can't quite figure this question out. It asks to find values of x for which g(x) is positive, g(x) = (2x^2-1)/(x^2+6x-7).

I've already solved for the inequality g(x) > 0, but after looking at a graph that appears to be only a partial solution, I only found g(x) between the vertical asymptotes.

My question is, how can I find these x values analytically, i.e. without a graph? It is a multi-part question with asymptotes, limits and intercepts asked in later questions so I assume I can't use those.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on March 22, 2012, 01:57:37 pm
Solve 2 cases:

1. 2x^2-1 > 0 and x^2+6x-7>0

2. 2x^2-1 < 0 and x^2+6x-7<0
Title: Re: Specialist 3/4 Question Thread!
Post by: domislong on March 23, 2012, 12:48:10 am
Ah because the equation will be positive if numerator and denominator are both +ve or -ve., so simple yet I couldn't think of it!

Thanks TT
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on March 25, 2012, 10:00:36 am
OK, probably a very easy question but I forgot how to do it.

If tan(y)=5/12 y=[pi,3pi/2] how do you figure out tan(2y)?
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on March 25, 2012, 10:38:30 am
Never mind, I got it, double angle formulas.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 25, 2012, 11:14:03 am
Two vector questions that are doing my head in.

a=3i-6j+4k and b=2i+j-2k.
a) Find c, the vector component of a perpendicular to b

If I can understand how to get part a), I can do the rest of the question. :)

a=2I+3j-4k and b=2i-j+2k. c=2i+(1+3t)j+(-1+2t)k.
b) Find the values of t for which angle BCA=90 degrees

This is what I have so far:
CA=(2-3t)i+(-3-2t)k
CB=(-2-3t)j+(3-2t)k.

I'm guess we have to use the costheta=a.b/|a||b|. But I know the formula 'works' for two vectors, not the position vectors, and I guess we would use CA and CB. However, I did that and obtained a loooong quartic and I guess I did something wrong.

Could someone show me what I should be doing?

(Sorry for not putting it in LaTeX). :(
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on March 25, 2012, 11:28:33 am
Could anyone explain differential equations the 'k' value and the 'a' value?

Thanks.
Title: Re: Specialist 3/4 Question Thread!
Post by: Greatness on March 25, 2012, 12:26:53 pm
Two vector questions that are doing my head in.

a=3i-6j+4k and b=2i+j-2k.
a) Find c, the vector component of a perpendicular to b

If I can understand how to get part a), I can do the rest of the question. :)

a=2I+3j-4k and b=2i-j+2k. c=2i+(1+3t)j+(-1+2t)k.
b) Find the values of t for which angle BCA=90 degrees

This is what I have so far:
CA=(2-3t)i+(-3-2t)k
CB=(-2-3t)j+(3-2t)k.

I'm guess we have to use the costheta=a.b/|a||b|. But I know the formula 'works' for two vectors, not the position vectors, and I guess we would use CA and CB. However, I did that and obtained a loooong quartic and I guess I did something wrong.

Could someone show me what I should be doing?

(Sorry for not putting it in LaTeX). :(
90 degrees = perpendicular, so use dot product = 0 and solve for t. Or is that too simple.... lol
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on March 25, 2012, 01:18:51 pm
If sin(A)=1/sqrt(5) and A=[0,pi/2] then how do you workout sin(3A)?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on March 25, 2012, 01:20:54 pm
Use double angle formulas to get sin(2A) and then use sin(2A + A) to find sin(3A) :) N.B. you can find cos(A) by drawing a triangle and using Pythagoras' theorem.

(I would solve it but I don't remember the relevant formulas lol)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 25, 2012, 01:44:55 pm
Two vector questions that are doing my head in.

a=3i-6j+4k and b=2i+j-2k.
a) Find c, the vector component of a perpendicular to b

If I can understand how to get part a), I can do the rest of the question. :)

a=2I+3j-4k and b=2i-j+2k. c=2i+(1+3t)j+(-1+2t)k.
b) Find the values of t for which angle BCA=90 degrees

This is what I have so far:
CA=(2-3t)i+(-3-2t)k
CB=(-2-3t)j+(3-2t)k.

I'm guess we have to use the costheta=a.b/|a||b|. But I know the formula 'works' for two vectors, not the position vectors, and I guess we would use CA and CB. However, I did that and obtained a loooong quartic and I guess I did something wrong.

Could someone show me what I should be doing?

(Sorry for not putting it in LaTeX). :(
90 degrees = perpendicular, so use dot product = 0 and solve for t. Or is that too simple.... lol

I did that but doesn't a dot product give only a magnitude and no direction.

This is referring to Question 1, so If I did a.b is would be -8 but what would I do with the -8? :S
Title: Re: Specialist 3/4 Question Thread!
Post by: Greatness on March 25, 2012, 02:42:07 pm
Is question 1 asking for a vector resolute or something similar to that? lol i cant remember XD
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 25, 2012, 03:51:27 pm
Is question 1 asking for a vector resolute or something similar to that? lol i cant remember XD

:O, It's asking for a 'new' vector, vector C that is the vector component of vector a perpendicular to vector b.

:S, This is getting confusing. xD
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on March 25, 2012, 04:14:03 pm
Regarding the first question.

It does involve using vector resolutes.
So what you are being asked to do is find "the vector component of a perpendicular to b". I.e. find the perpendicular resolute. That is the dotted line in the diagram (ignore the |A|cos(theta), not needed).

Lets make where that right angle is point C.
I.e. we are looking for OC.

The vector resolute of a in the direction of b is $\overrightarrow{OC}=\frac{\underset{\sim}{a}.\underset{\sim}{b}}{\underset{\sim}{b}.\underset{\sim}{b}}\underset{\sim}{b}$

Unlatex'd. OC=[(a.b)/(b.b)]*b

That is representing the vector that is in the same direction as b, but has the lenght from the bottom right most point to where the right angle is, i.e. the projection.

Now we want the vector that is perpendicular, not parallel to b.

So that would be the dotted line, now the dotted line will be  $\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}
=\frac{\underset{\sim}{a}-\frac{\underset{\sim}{a}.\underset{\sim}{b}}{\underset{\sim}{b}.\underset{\sim}{b}}\underset{\sim}{b}$

Unlatex'd. CA=OA-OC=a-[(a.b)/(b.b)]*b

Hope that helps, probably didn't explain it well.
Title: Re: Specialist 3/4 Question Thread!
Post by: Greatness on March 25, 2012, 04:17:50 pm
yeah i think that is the vector resolute. read the theory on it in the textbook or look at some examples.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 25, 2012, 04:29:34 pm
Regarding the first question.

It does involve using vector resolutes.
So what you are being asked to do is find "the vector component of a perpendicular to b". I.e. find the perpendicular resolute. That is the dotted line in the diagram (ignore the |A|cos(theta), not needed).

Lets make where that right angle is point C.
I.e. we are looking for OC.

The vector resolute of a in the direction of b is $\overrightarrow{OC}=\frac{\underset{\sim}{a}.\underset{\sim}{b}}{\underset{\sim}{b}.\underset{\sim}{b}}\underset{\sim}{b}$

Unlatex'd. OC=[(a.b)/(b.b)]*b

That is representing the vector that is in the same direction as b, but has the lenght from the bottom right most point to where the right angle is, i.e. the projection.

Now we want the vector that is perpendicular, not parallel to b.

So that would be the dotted line, now the dotted line will be  $\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}
=\frac{\underset{\sim}{a}-\frac{\underset{\sim}{a}.\underset{\sim}{b}}{\underset{\sim}{b}.\underset{\sim}{b}}\underset{\sim}{b}$

Unlatex'd. CA=OA-OC=a-[(a.b)/(b.b)]*b

Hope that helps, probably didn't explain it well.

I see, Ah! The other text book I use ALWAYS says 'Find the vector resolute of a perpendicular to b' but this one 'jazzed' it up.

Thanks swarley and b^3. :)

EDIT: Just this really quick question

What is the quickest way to 'split' up this: 7pi/12? I can easily to the really simple stuff like pi/12 or 2pi/3. Hopefully that isn't too much of a basic question but thank you. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on March 27, 2012, 11:31:52 am
Hi, could somebody help me with this question, describe the steps i should take or even show me the steps if possible

Solve this equation over C: z2=8+15i

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 27, 2012, 12:24:44 pm
Move to one side and use the quadratic formula. ^

I cant do it right now as I'm at the state library and I have less than 1 minute remaining, sorry. :(

Sorry, thought there was a Z for 8. :S My bad.
Title: Re: Specialist 3/4 Question Thread!
Post by: John President on March 27, 2012, 01:45:19 pm
Hi, could somebody help me with this question, describe the steps i should take or even show me the steps if possible

Solve this equation over C: z2=8+15i

Thanks!
Given that 15^2 + 8^2 = 289 = 17^2, I suspect that they want you to convert 8+15i into polar form (modulus-argument form) before using De Moivre's theorem to find the roots. It looks like you'll need a calculator though - and don't forget that there will be two solutions!
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on March 27, 2012, 04:24:26 pm
its suppose to be calc free so i dono how to get the argument for it by hand ???
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on March 27, 2012, 04:57:16 pm
z^2 = 8 + 15i
(x+yi)^2 = 8 + 15i
x^2 - y^2 + 2xyi = 8 + 15 i
x^2 - y^2 = 8...(1)
2xy = 15 ...(2)
from (2), y = 15/(2x)
sub into (1)
x^2 - (15/(2x))^2 = 8
x^2 - 225/4x^2 = 8
4x^4 - 32x^2 - 225 =0
(2x^2 - 25) (2x^2 + 9)=0
2x^2 = 25/2 or 2x^2 + 9 = 0 (no sol for this one since x E R)
x^2 = 25/4
x = +-5/2
so y = 15/(+-5) = +-3
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on March 27, 2012, 07:43:16 pm
spec prac exam question attached

question is, how do you do it!?
i got B, but realised i actually didn't get any of the options listed! -        A/[3(x+c)]  +  B/(x+c)^2
:(

the answer is D. what do you do to get rid of the 3?
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on March 27, 2012, 07:56:53 pm
a/[3(x+c)]  +  B/(x+c)^2, let A=a/3 => A/(x+c) + B/(x+c)^2
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on March 27, 2012, 07:57:53 pm

the answer is D. what do you do to get rid of the 3?

You dont.
Its absorbed in A and B
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on March 27, 2012, 10:57:18 pm
For the complex numbers u = a+bi, v=c+di and w=p+qi prove that:

a) u+v=v+u
b) (u+v)+w = u(v+w)
c) uv=vu
d) (uv)w=u(vw)

Here's my attempt at a) and c)

1) (a+bi)(c+di)
3) ??

c)
1) (a+bi) + (c+di)
2) (a+c) + (b+d)i by definition of addition
3) ??

hehe
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on March 28, 2012, 06:28:29 pm
Just this vector q's at the bottom.

I got $\vec{OP}$=6m$i$-2m$j$.

$\vec{PS}$=$\vec{OS}$-$\vec{OP}$
=$8i-4j$-(6m$i$-2m($j$)
=$8i$-6m$i$)+($-4j$+2m$j$)
=8-6m$i$+(-4+2m)$j$

Dot Product of $\vec{OP}$.$\vec{PS}$
=(6m$i$-2m$j$).((8-6m)$i$+(-4+2m)$j$
=(6m(8-6m)+(-2m(-3+2m)))
=(48m-36m2+(6m-4m2)
=48m-36m2+6m-4m2
=54m-40m2
=2m(27-20m)

I am 100% sure my working out is wrong. Could someone look over it and explain? I usually get mixed up with Dot products when there is a variable. :S

Thanks. :)

(This questions from the VCAA 2002 Exam 2)
Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on April 01, 2012, 02:37:04 pm
Two questions:

If sin A = 1/sqrt5, how do i find sin 3A?

and

how to find exact values of sin (pi/8) using double angle formulas?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 01, 2012, 02:57:56 pm
Two questions:

If sin A = 1/sqrt5, how do i find sin 3A?

and

how to find exact values of sin (pi/8) using double angle formulas?

Hints only!

1) Find sin(2A) using relevant formula and then find sin(3A) using the sin(X+Y) formulas where X=A and Y=3A :)

2) Let 2A = pi/4, use sin(2A) identity to solve for sin(A) :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 03, 2012, 01:17:35 pm
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}$ $sec^2x tan^5x\, dx$

evaluate:

can somsone please guide me through that!
Title: Re: Specialist 3/4 Question Thread!
Post by: Ominous on April 03, 2012, 01:56:43 pm
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}$ $sec^2x tan^5x\, dx$

evaluate:

can somsone please guide me through that!

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}$ $sec^2x tan^5x\, dx$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{2}x \cdot \tan^{5}x \cdot dx=$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+tan^{2}(x)) \cdot \tan^{5}x \cdot dx$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+tan^{2}(x)) \cdot \tan^{5}x \cdot dx$

$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1+u^{2}) \cdot u^{5} \cdot dx$

let u= $1+ \tan^{2} x$, also change up the terminals and use u-subsitution $\int_{\frac{3+\sqrt{3}}{3}}^{1+\sqrt{3}} u^{5} \cdot u^{7} du$

and every would be straight forward.

EDIT: its just a matter of using the trig identites of $\sec^{2}x= 1+tan^2{x}$. and changing terminals through u subsitution.

Good luck
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 03, 2012, 06:10:06 pm
woah,
so ... i got as far as you did in the 3rd /4th line.

you've then let tan = u by the looks of it, but then said

let u= $1+ \tan^{2} x$

(http://t2.gstatic.com/images?q=tbn:ANd9GcSHUmUG6Q6ELAHY6aFsF6arknVmHW9Pxfbjbqq-aF6wfU9Pyy024t_9ioUy)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on April 03, 2012, 06:20:17 pm
Ill just help you with the the intergal, then you can sub in the values yourself later.
So we have $\int\sec^{2}(x)\tan^{5}(x)dx$
Now here there is something we can notice, we know that the derivative of $\tan(x)$ is $\sec^{2}(x)$, so if we let u=tan(x), then we can try and get the $\sec^{2}(x)$ to cancel out.

$\mathrm{Let\: u=\tan(x)}$
$\mathrm{Then\:}\frac{du}{dx}=\sec^{2}(x)$
So we have \begin{alignedat}{1}\int\sec^{2}(x)\tan^{5}(x)dx & =\intop u^{5}\sec^{2}(x)\times\frac{1}{\sec^{2}(x)}du
\\ & =\int u^{5}du
\\ & =\frac{1}{6}u^{6}+c
\\ & =\frac{1}{6}\tan^{6}(x)+c
\end{alignedat}

Then sub in the values to find the definite integral.
Oh, and don't forget to change the terminals :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 03, 2012, 06:24:58 pm
Ill just help you with the the intergal, then you can sub in the values yourself later.
So we have $\int\sec^{2}(x)\tan^{5}(x)dx$
Now here there is something we can notice, we know that the derivatie of $\tan(x)$ is $\sec^{2}(x)$, so if we let u=tan(x), then we can try and get the $\sec^{2}(x)$ to cancel out.

$\mathrm{Let\: u=\tan(x)}$
$\mathrm{Then\:}\frac{du}{dx}=\sec^{2}(x)$
So we have \begin{alignedat}{1}\int\sec^{2}(x)\tan^{5}(x)dx & =\intop u^{5}\sec^{2}(x)\times\frac{1}{\sec^{2}(x)}du
\\ & =\int u^{5}du
\\ & =\frac{1}{6}u^{6}+c
\\ & =\frac{1}{6}\tan^{6}(U)+c
\end{alignedat}

Then sub in the values to find the definite integral.
Oh, and don't forget to change the terminals :)
beautttty. thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on April 04, 2012, 10:16:13 am
Question: what does the second derivative tell you about the original graph? I read up on it but I'm very confused about concavity stuff? Thanks! :)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 04, 2012, 01:56:28 pm
Question: what does the second derivative tell you about the original graph? I read up on it but I'm very confused about concavity stuff? Thanks! :)

It 'usually' tells you (there are ambiguous exceptions, such as y = x^4) whether or not a particular point (based on the value of x) is a local max/min or a certain type of inflection :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 04, 2012, 03:40:15 pm
Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$  =  $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$   ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 04, 2012, 04:04:19 pm
Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$  =  $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$   ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

In general $\int^a_b f(x) \ dx = -\int^b_a f(x) \ dx = \int^b_a -f(x) \ dx$

In your case, the negative has been moved inside the integral :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on April 04, 2012, 04:06:14 pm
Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$  =  $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$   ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

In general $\int^a_b f(x) \ dx = -\int^b_a f(x) \ dx = \int^b_a -f(x) \ dx$

In your case, the negative has been moved inside the integral :)

10/10
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 04, 2012, 04:23:10 pm
Why is it that $\frac{1}{2} \int_{1}^{0} 1-u^2 \ du$  =  $\frac{1}{2} \int_{0}^{1} u^2 -1 \ du$   ?

is it because the latter equation is a reflection of the former, and thus the area will be the same ? cause i got the former as an answer, and since it was MC, i had trouble trying to justify why the latter is equivalent. so just confirming!

In general $\int^a_b f(x) \ dx = -\int^b_a f(x) \ dx = \int^b_a -f(x) \ dx$

In your case, the negative has been moved inside the integral :)

ahhh indeed! ok :) thanks!
handy lil tip right there!
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on April 04, 2012, 06:45:10 pm
Question: what does the second derivative tell you about the original graph? I read up on it but I'm very confused about concavity stuff? Thanks! :)

It 'usually' tells you (there are ambiguous exceptions, such as y = x^4) whether or not a particular point (based on the value of x) is a local max/min or a certain type of inflection :)

Could you explain this further? We got a worksheet where there were a few exercise to do on this and one of the question was 'if you draw a tangent on f(x) curve at the point f''(x)=0 what does it do and tell us?' Thank you for your help! :)
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on April 04, 2012, 07:04:46 pm
Resolve the following rational expressions into partial fractions.
(3x^2+2x+5) / (x^2+2)(x+1)
AND
(x^2+2x-13) / (2x^3+6x^2+2x+6)

Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on April 04, 2012, 09:16:56 pm
Use a double angle formula to show that the exact value of cos(pi/8)=sqrt(2+sqrt(2))/2
Explain why any values are rejected.
Title: Re: Specialist 3/4 Question Thread!
Post by: aznxD on April 04, 2012, 09:29:49 pm
Use a double angle formula to show that the exact value of cos(pi/8)=sqrt(2+sqrt(2))/2
Explain why any values are rejected.

Use cos(2x) = 2cos^2(x) - 1
Sub x=pi/8 and solve for cos(x)=pi/8
Reject the negative solution as pi/8 is in the first quadrant.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 05, 2012, 01:32:20 pm
Find the area between the curve the x-asus and given lines for each of the following functions:

$y=cos^-1 (x); x=\frac{\sqrt{3}}{2}, x = 1$

i know theres an easier way to do this, but i started off on the other foot, so i dont wanna start over the other way :p
( TJERE IS NO EASIER WAY TO DO THIS W.O CALC! GRRRR)
attached is a diagram of the situation described above, where the pink area shows what i want, and the green area shows the area which adds onto the pink area, to make a rectangle.

now, to find out the pink area, i thought i'd make a rectangle, so i found that:  DISREGARD AND READ NEXT POST PLZZZZ
length of rectangle = $\int_{\frac{\sqrt{3}}{2}}^{1} cos^-1 (x) dx$

width of rectangle =$\int_{0}^{\frac{\pi}{2}} cos(y) dy$                   - i found inverse function here

therefore total area of rectangle = $\int_{\frac{\sqrt{3}}{2}}^{1} cos^-1 (x) dx$  x $\int_{0}^{\frac{\pi}{2}} cos(y) dy$

this = 0.0465501..

Now to find green area, i wasnt really sure, but i thought that the area would be :
$\int_{0}^{\frac{\pi}{2}} cos(y) dy$

this = 1

therefore area of pink bit = area of rectangle - area of green bit
= 0.04655 - 1
= -0.04655
thus, area = 0.04655 units squared.

but, just wanted to confirm here that what i've done is right? or is it a fluke through misunderstanding of the concept?!

i know i couldve just done  $\int_{\frac{\sqrt{3}}{2}}^{1} cos^-1 (x) dx$ and gotten my answer, but yeah, :S  N.B. that'd involve using calc and 1 line working!

ta !
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 05, 2012, 06:01:40 pm
lulz, how dafuck did i get a length and width in terms of an integral? ladies and gentleman, i just shot myself in the foot..

anyways, New length = $1- \frac{\sqrt{3}}{2}$
new width = $\frac{\pi}{6}$

thus area of rectangle = (using calc)  0.07014893454
Area of the green bit was 0.5

thus area of pink bit = -0.429... bla bla

WHICH IS WRONG!

help please. im dying.. in vain, i cannot let integral calculus overcome my pride. BRUUUAHH
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on April 05, 2012, 08:00:58 pm
area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 05, 2012, 08:29:35 pm
i dont mean to be picky, but could you please  give the working in latex or upload a pic?!
i cant really workout some of the lines e.g (6 - pi*sqrt(3))/12)

thx

Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 05, 2012, 08:56:06 pm
area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

i think ive got it, dw then

So, 2 questions.
1) Why are the terminals $\frac{\pi}{6}$ to $0$ ? i thought it was the other way around
2) How did you get the area of the rectangle to be, $\frac{\sqrt{3}}{2} * \frac{\pi}{6}$ ?
i thought the length was $(1- \frac{\sqrt{3}}{2} )$

EDIT.

i think when you typed that out brightsky you werent thinking of latex coding ? cause i thought you meant the terminals were from pi/6 to 0.  :p

and i get why the length of the rectangle isnt what i said it is. !
so nvm!
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on April 06, 2012, 03:40:25 pm
The position of an object, r metres at time t, is given by: $r(t)$=2cos(3t)$i$+2sin(3t)$j$. Show that the object moves in a circular path of radius 2.

Here are my steps:
$(cos^-^1(x/2))/3=t$
$(sin^-^1(y/2))/3=t$
$x/2=(\sqrt(-(y-2)(y+2)))/2$
$-x^2=y^2-4$
$x^2+y^2=4$

Is there any way to do this question quicker? (I had to cut out a few lines)
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on April 06, 2012, 04:00:02 pm
well you have:
x = 2cos(3t)
y = 2sin(3t)
so x^2 + y^2 = 4(cos^2(3t) + sin^2(3t) = 4
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on April 07, 2012, 01:39:44 pm
area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 07, 2012, 01:47:15 pm
area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

i don't think that works for that specific question tho?
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on April 07, 2012, 02:12:23 pm
area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

i don't think that works for that specific question tho?

The second one does (gives you the pink)
The first one gives you the green
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on April 07, 2012, 02:30:25 pm
area = int^(pi/6)_(0) cos(y) dy - sqrt(3)/2 * pi/6 = 1/2 - sqrt(3)/2 * pi/6 = (6 - pi*sqrt(3))/12

I think you should ditch the 'rectangles', brightsky
It's an UNECESSARY method.

Best way to approach these problems is as an area between two curves.
One curve is x = cosy and for the other you have a choice of x = 1 or x = sqrt(3)/2

(1 - cosy)dy or (cosy - sqrt(3)/2)dy

The limits of integration stay the same, 0 and pi/6

i don't think that works for that specific question tho?

The second one does (gives you the pink)
The first one gives you the green

The key point is that you don't need the 'rectange' because the rectangle pops out naturally when you use area between two curves. It's a neater method and you don't have to give messy explanations like 'The area of the shaded region is the value of the integral blah blah blah minus the area of the rectangle ABCD (refer diagram) blah blah blah'
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on April 10, 2012, 06:18:02 pm
Having trouble with a few questions about partial fractions

6. Express the following in partial fractions.
g. (3x+1) / (x^3+x)
h. (3x^2+8) / x(x^2+4)
t. (x^3) / (x+1)(x-1)

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on April 10, 2012, 06:36:54 pm
Remember for partial fractions when you get a quadratic factor that cannot be expressed as linear factors you have you use $\frac{1}{x(x^{2}+b)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+b}$

So for the first one we start off with $\dfrac{3x+1}{x^{3}+x}=\dfrac{3x+1}{x(x^{2}+1)}$
Now that $x^{2}+1$ will result in the form I originally spoke about above, and the other x term will just give us the normal result.

So we are looking at $\dfrac{3x+1}{x^{3}+x}=\dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}$
Which then gives $3x+1=Ax^{2}+A+Bx^{2}+Cx$
$3x+1=x^{2}(A+B)+Cx+A$
Now if we equate the coefficients we get
C=3, A=1
A+B=0
B=-A
B=-1
So that gives $\dfrac{3x+1}{x^{3}+x}=\dfrac{1}{x}+\dfrac{-x+3}{x^2+1}$

The second one can be approached in the same method.

Now for the last one, because the greatest power of the numerator is equal to or greater than the greatest power of the denominator, you will have to long divide first, then apply the methods for partial fractions. So for the numerator the greatest power is $x^{3}$ and for the denominator the greatest power will be $x^{2}$, so we have to long divide first.

\begin{alignedat}{1} & \: x
\\ x^{2}-1 & |\overline{x^{3}\:\:\:}
\\ (-) & \underline{x^{3}-x}
\\ & x
\end{alignedat}

So that brings us to $\frac{x^{3}}{(x+1)(x-1)}=x+\frac{x}{x^{2}-1}$
Then from there just apply the normal rules again.
$\frac{x}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1}$
Then add the other x back in later.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on April 15, 2012, 11:42:39 am
Hey,

Find the cube roots of 1-$i \sqrt(3)$.

My teacher hasn't covered this question before. >.>.

Thanks. :)

NVM. I had to use another text book to find out how. :O
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on April 16, 2012, 08:15:57 am
1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?

(6a, Ex. 4D, Essential)

and

2) Show that if -pi/2 <Arg(z1) < pi/2 and -pi/2 < Arg(z2) < pi/2 then Arg(z1z2) = Arg(z1) + Arg(z2) and Arg(z1/z2) = Arg(z1) - Arg(z2)

(Question 3, Ex. 4D)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on April 16, 2012, 10:25:18 am
1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?

(6a, Ex. 4D, Essential)
Remember that $\sin(\theta)=\cos(\frac{\pi}{2}-\theta)$ and $\cos(\theta)=\sin(\frac{\pi}{2}-\theta)$
So we have \begin{alignedat}{1}\sin(\theta)+i\cos(\theta) & =\cos(\frac{\pi}{2}-\theta)+i\sin(\frac{\pi}{2}-\theta)
\\ & =\mathrm{cis}(\frac{\pi}{2}-\theta)
\end{alignedat}

2) Show that if -pi/2 <Arg(z1) < pi/2 and -pi/2 < Arg(z2) < pi/2 then Arg(z1z2) = Arg(z1) + Arg(z2) and Arg(z1/z2) = Arg(z1) - Arg(z2)

(Question 3, Ex. 4D)
Let
$z_{1}=r_{1}cis(\theta_{1})$
$z_{2}=r_{2}cis(\theta_{2})$
So we know that $Arg(z_{1}z_{2})=\theta_{1}+\theta_{2}$
Now we know that the angles are restricted to $-\frac{\pi}{2}<\theta_{1}<\frac{\pi}{2}$ and $-\frac{\pi}{2}<\theta_{2}<\frac{\pi}{2}$
So $-\pi<\theta_{1}+\theta_{2}<\pi$
$\therefore Arg(z_{1}z_{2})=Arg(z_{1})+Arg(z_{2})=\theta_{1}+\theta_{2}$

Now you should be able to do the other half of the question, give it a go remembering that $\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}cis(\theta_{1}-\theta_{2})$

Hope that helps :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 16, 2012, 03:56:28 pm
$sin^2 x cos^3 x = cos x$

solve for $-\pi

I can't get past a certain step, but i probs screwed it up before that:p

$sin^2 x cos^3 x = cos x$

$[\frac{1}{2} (1-cos(2x)](cosx)[\frac{1}{2} (1+cos(2x)] = cosx$

$(\frac{1-cos2x}{2})(\frac{1+cos2x}{2} = 1$

$let \frac{cos2x}{2} = A$

$(\frac{1}{2} - A)(\frac{1}{2} + A)$

$(\frac{1}{4} - A^2) = 1$

$-A^2 + \frac{1}{4} =1$

-$A^2 = -\frac{3}{4}$

yeah. wtf?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 16, 2012, 04:33:30 pm
I honestly don't remember enough of spesh to help you (ie. have no idea of identities anymore), but you definitely can't divide both sides by cos(x) as you are ridding the possible solutions of cos(x)=0

(I think)
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on April 16, 2012, 06:16:59 pm
1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?

(6a, Ex. 4D, Essential)
Remember that $\sin(\theta)=\cos(\frac{\pi}{2}-\theta)$ and $\cos(\theta)=\sin(\frac{\pi}{2}-\theta)$
So we have \begin{alignedat}{1}\sin(\theta)+i\cos(\theta) & =\cos(\frac{\pi}{2}-\theta)+i\sin(\frac{\pi}{2}-\theta)
\\ & =\mathrm{cis}(\frac{\pi}{2}-\theta)
\end{alignedat}

2) Show that if -pi/2 <Arg(z1) < pi/2 and -pi/2 < Arg(z2) < pi/2 then Arg(z1z2) = Arg(z1) + Arg(z2) and Arg(z1/z2) = Arg(z1) - Arg(z2)

(Question 3, Ex. 4D)
Let
$z_{1}=r_{1}cis(\theta_{1})$
$z_{2}=r_{2}cis(\theta_{2})$
So we know that $Arg(z_{1}z_{2})=\theta_{1}+\theta_{2}$
Now we know that the angles are restricted to $-\frac{\pi}{2}<\theta_{1}<\frac{\pi}{2}$ and $-\frac{\pi}{2}<\theta_{2}<\frac{\pi}{2}$
So $-\pi<\theta_{1}+\theta_{2}<\pi$
$\therefore Arg(z_{1}z_{2})=Arg(z_{1})+Arg(z_{2})=\theta_{1}+\theta_{2}$

Now you should be able to do the other half of the question, give it a go remembering that $\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}cis(\theta_{1}-\theta_{2})$

Hope that helps :)

Excellent, thanks! I have a little idea of what's going (like for the first one, think I tried to think of it as measuring the angle from the y-axis or something) but I have no idea how to show anything, guess part of the reason is I have a crappy foundation haha
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on April 16, 2012, 06:34:17 pm
$sin^2 x cos^3 x = cos x$

solve for $-\pi

$sin^2 x cos^3 x = cos x$
$(1-cos^2x)cos^3x = cosx$
$cos^3x-cos^5x-cosx = 0$
$-cosx[cos^4x-cos^2x+1] = 0$
$cosx=0$
$cos^4x-cos^2x+1=0$ has no real solutions
$\therefore x\in \{\frac{\pi}{2}+2k\pi, \frac{-\pi}{2}+2k\pi, k\in \mathbb{Z}\}$

edit: oops, forgot about the restriction:

$x\in \{\frac{-\pi}{2},\frac{\pi}{2},\frac{3\pi}{2} \}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 16, 2012, 06:45:06 pm
$sin^2 x cos^3 x = cos x$

solve for $-\pi

$sin^2 x cos^3 x = cos x$
$(1-cos^2x)cos^3x = cosx$
$cos^3x-cos^5x-cosx = 0$
$-cosx[cos^4x-cos^2x+1] = 0$
$cosx=0$
$cos^4x-cos^2x+1=0$ has no real solutions
$\therefore x\in \{\frac{\pi}{2}+2k\pi, \frac{-\pi}{2}+2k\pi, k\in \mathbb{Z}\}$

edit: oops, forgot about the restriction:

$x\in \{\frac{-\pi}{2},\frac{\pi}{2},\frac{3\pi}{2} \}$
goddamit.
How do people manage to pick such things up!

p.s.
I honestly don't remember enough of spesh to help you (ie. have no idea of identities anymore), but you definitely can't divide both sides by cos(x) as you are ridding the possible solutions of cos(x)=0

(I think)

how so ?  like, i dont get how thats algeabriclly incorrect
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 16, 2012, 06:50:47 pm
Solve $x^2 = x$ and you'll see :)

Spoiler
If you divide both sides by $x$, which is incorrect, you are assuming that $x \ne 0$ as you can't divide by 0 :) Similarly, in your question, if you divide by cos(x), then you are assuming that cos(x) can't equal 0 either, which is mathematically incorrect :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 16, 2012, 06:55:21 pm
Solve $x^2 = x$ and you'll see :)

x^2 -x =0
x(x-1) = 0

so x=0 and x=1

wait. (http://t.qkme.me/352qdk.jpg)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 16, 2012, 06:56:38 pm
Yeah, nice :) Why didn't you divide both sides by x then at the start?
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on April 16, 2012, 07:01:15 pm
Yeah, nice :) Why didn't you divide both sides by x then at the start?

Aw sheeeeet.

rookie mistake.

i get it ahaha.. :p
thanks vege! .  ;D
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on April 19, 2012, 07:42:28 pm
Herrow guaiz.

See attached!
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on April 19, 2012, 08:16:09 pm
Which question do you need answered?

I'll answer both - 'whydoihavenofriends', pretty obvious, you're too hot, the hotness that radiates from your dp scares away all the nubs.

As for the other question - tan(70*180/pi) is the gradient of the tangent.

now diff f(x), then find the x values for when the derivative of f(x) is equal to tan(70*180/pi) subject to the restriction of the domain. In other words, solve f'(x) = tan(70*180/pi)
Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on April 23, 2012, 07:41:03 pm
i + 2j - 3k and -9i +4j -k

^ find the Angle between the pair of vectors
(already found that the dot product is 2 from other part of the question)

And another question:

if (u-v) dot (u+v) = |v|^2, then:
A) u=v
B) u must be equal to the zero vector, O.
C) u perpendicular to v
D) u is multiple of v
E) None are true.

(nb: all us and vs etc have little squiggly things under them hehehe.
'dot' is that dot that just stands in space. If you get what i mean. The dot in dot product, the one that is off the floor, unlike this  . Why am i explaining all this, you prob figured anyway)
Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on April 23, 2012, 08:40:25 pm
i + 2j - 3k and -9i +4j -k

^ find the Angle between the pair of vectors
(already found that the dot product is 2 from other part of the question)

And another question:

if (u-v) dot (u+v) = |v|^2, then:
A) u=v
B) u must be equal to the zero vector, O.
C) u perpendicular to v
D) u is multiple of v
E) None are true.

(nb: all us and vs etc have little squiggly things under them hehehe.
'dot' is that dot that just stands in space. If you get what i mean. The dot in dot product, the one that is off the floor, unlike this  . Why am i explaining all this, you prob figured anyway)
Thanks :)

I am going to represent the vectors like this because it's easier to type:
$a_1i+a_2j+a_3k=(a_1,a_2,a_3)$

The dot product can be defined as follows:
$\tilde{a}\cdot \tilde{b}=|a||b|\cos{\theta}$ where $\theta$ is the angle between the vectors.

$\therefore \cos{\theta}=\frac{(1,2,-3)\cdot (-9,4,-1)}{|(1,2,-3)||(-9,4,-1)|}$

$\cos{\theta}=\frac{-9+8+3}{\sqrt{14}\cdot \sqrt{98}}$

$\cos{\theta}=\frac{2}{14\sqrt{7}}$

$\cos{\theta}=\frac{1}{7\sqrt{7}}$

$\therefore \theta=\cos^{-1}(\frac{1}{7\sqrt{7}})$     Is it a calculator question?

Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on April 23, 2012, 08:47:45 pm
i + 2j - 3k and -9i +4j -k

^ find the Angle between the pair of vectors
(already found that the dot product is 2 from other part of the question)

And another question:

if (u-v) dot (u+v) = |v|^2, then:
A) u=v
B) u must be equal to the zero vector, O.
C) u perpendicular to v
D) u is multiple of v
E) None are true.

(nb: all us and vs etc have little squiggly things under them hehehe.
'dot' is that dot that just stands in space. If you get what i mean. The dot in dot product, the one that is off the floor, unlike this  . Why am i explaining all this, you prob figured anyway)
Thanks :)

I am going to represent the vectors like this because it's easier to type:
$a_1i+a_2j+a_3k=(a_1,a_2,a_3)$

The dot product can be defined as follows:
$\tilde{a}\cdot \tilde{b}=|a||b|\cos{\theta}$ where $\theta$ is the angle between the vectors.

$\therefore \cos{\theta}=\frac{(1,2,-3)\cdot (-9,4,-1)}{|(1,2,-3)||(-9,4,-1)|}$

$\cos{\theta}=\frac{-9+8+3}{\sqrt{14}\cdot \sqrt{98}}$

$\cos{\theta}=\frac{2}{14\sqrt{7}}$

$\cos{\theta}=\frac{1}{7\sqrt{7}}$

$\therefore \theta=\cos^{-1}(\frac{1}{7\sqrt{7}})$     Is it a calculator question?

Thats what i got too! Yes it is a calc question, so i put that into my calc and got something like .. I dont know, it was really off frm what bob said.
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on April 23, 2012, 08:51:08 pm
i + 2j - 3k and -9i +4j -k

^ find the Angle between the pair of vectors
(already found that the dot product is 2 from other part of the question)

And another question:

if (u-v) dot (u+v) = |v|^2, then:
A) u=v
B) u must be equal to the zero vector, O.
C) u perpendicular to v
D) u is multiple of v
E) None are true.

(nb: all us and vs etc have little squiggly things under them hehehe.
'dot' is that dot that just stands in space. If you get what i mean. The dot in dot product, the one that is off the floor, unlike this  . Why am i explaining all this, you prob figured anyway)
Thanks :)

I am going to represent the vectors like this because it's easier to type:
$a_1i+a_2j+a_3k=(a_1,a_2,a_3)$

The dot product can be defined as follows:
$\tilde{a}\cdot \tilde{b}=|a||b|\cos{\theta}$ where $\theta$ is the angle between the vectors.

$\therefore \cos{\theta}=\frac{(1,2,-3)\cdot (-9,4,-1)}{|(1,2,-3)||(-9,4,-1)|}$

$\cos{\theta}=\frac{-9+8+3}{\sqrt{14}\cdot \sqrt{98}}$

$\cos{\theta}=\frac{2}{14\sqrt{7}}$

$\cos{\theta}=\frac{1}{7\sqrt{7}}$

$\therefore \theta=\cos^{-1}(\frac{1}{7\sqrt{7}})$     Is it a calculator question?

Thats what i got too! Yes it is a calc question, so i put that into my calc and got something like .. I dont know, it was really off frm what bob said.

I got 1.51678 radians or 86.9 degrees
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on April 25, 2012, 05:21:00 pm
How do i draw a diagram for this:
For quadrilateral OABC , D is the point of trisection of OC nearer O and E is the point of trisection of AB nearer A.

Also this question: ABC is a right angled triangle with the right angle at B. If AC= 2i+4j and AB is parallel to i+j, find AB.

Could someone help me asap, cause i have a test tmr. Cheers
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on April 25, 2012, 06:35:35 pm
How do i draw a diagram for this:
For quadrilateral OABC , D is the point of trisection of OC nearer O and E is the point of trisection of AB nearer A.
See attachment
$\vec{OD}=\frac{1}{3} \vec{OC}$  and  $\vec{AE}=\frac{1}{3} \vec{AB}$

I have to go now so hopefully someone else can help with your other question
Title: Re: Specialist 3/4 Question Thread!
Post by: Reckoner on April 25, 2012, 08:11:27 pm
f(x)=(Ax^2 +Bx)/(x+C), where A,B and C are constants. The line x=-1 and y=x+4 are asymptotes to the curve. Find A,B and C.

Could someone please show me how to do this question? Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on April 25, 2012, 08:40:52 pm
Going to go over a little bit of theory in the explanation.

So we start off with $f(x)=\frac{Ax^{2}+Bx}{x+C}$
To find the asymptotes we have to split it up into its fractions, in other words we can long divide as below.

\begin{alignedat}{1} & \underline{Ax+(B-AC)}
\\ x+C & |Ax^{2}+Bx
\\ (-) & \underline{Ax^{2}+ACx}
\\ & \:\:\:\:\:\:(B-AC)x
\\ (-) & \:\:\:\:\:\:\underline{(B-AC)x+(BC-AC^{2})}
\\ & \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\;\;\;\:\:\:\;\:\:(AC^{2}-BC)
\end{alignedat}

So that means that $\frac{Ax^{2}+Bx}{x+C}=Ax+(B-AC)+\frac{AC^{2}-BC}{x+C}$

Now we know there will be a vertical asymptote when the function results in something like $\frac{1}{0}$ which will be undefined. So that will be when $x+C=0$, so $x=-C$

So we have a vertical asymptote at $x=-C$. Now we know that $x=-1$ is an asymptote, so that means that here $C=1$.

Now for the oblique asymptote, what happens as the fraction part $\left(\frac{AC^{2}-BC}{x+C}\right)$ gets smaller and smaller and approaches 0? The function will get closer and closer to $Ax+(B-AC)$, in other words approach it. So now we know that our other asymptote will be $y=Ax+(B-AC)$, corresponding to $y=x+4$, so that means that $A=1$ and $(B-AC)=4$
Now solving for B
\begin{alignedat}{1}B-AC & =4
\\ B-(1)(1) & =4
\\ B & =5
\end{alignedat}

So that leaves us with the equation $f(x)=\frac{x^{2}+5x}{x+1}=x+4-\frac{4}{x+1}$

You don't need all that wordy stuff in the solution, but its just there to explain it (hopefully).

Hope that helps, and hope there are not any mistakes :)

EDIT: Fixed some of the incomprehendable sentences.

EDIT2: Just thought I'd add the graph so that you can visualise the asymptotic behaviour of the oblique asymptote :)
(https://s3.amazonaws.com/grapher/exports/bf6xghcama.png)
Title: Re: Specialist 3/4 Question Thread!
Post by: Reckoner on April 25, 2012, 09:14:25 pm
Great answer! Really helped, thanks a lot!
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on April 28, 2012, 06:42:22 pm
I need help with question 8b! :D
Title: Re: Specialist 3/4 Question Thread!
Post by: DisaFear on April 28, 2012, 07:03:42 pm
I need help with question 8b! :D

This isn't exactly spesh is it? D:

2P = 3a + 2(3b) + 2b + 3a = 6a + 8b (total perimeter)
A = 9ab + ab = 10ab

Answer must be in terms of a and P, so let's find b

8b + 6a = 2P
b= (2P - 6a)/8

Sub in b into the Area formula found before, A = 9ab + ab = 10ab

A = 10ab

A=(5aP - 15a^2)/2

Hope I didn't do anything wrong there. That'd be embarrassing
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 28, 2012, 07:09:29 pm
I need help with question 8b! :D

This isn't exactly spesh is it? D:

2P = 3a + 2(3b) + 2b + 3a = 6a + 8b (total perimeter)
A = 9ab + ab = 10ab

Answer must be in terms of a and P, so let's find b

8b + 6a = 2P
b= (2P - 6a)/8

Sub in b into the Area formula found before, A = 9ab + ab = 10ab

A = (10aP - 3a)/4

Hope I didn't do anything wrong there. That'd be embarassing

2 issues:
1) You spelt that word wrong, two 'r's
2) A = 10a(P-3a)/4 = (10aP-30a^2)/4 = (5aP-15a^2)/2

:D
Title: Re: Specialist 3/4 Question Thread!
Post by: DisaFear on April 28, 2012, 07:11:25 pm
Hey, spell check didn't pick it up :(
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on April 28, 2012, 07:13:20 pm
Hey, spell check didn't pick it up :(

That's what VCE English was for!

And of course if wouldn't pick up the error in the fraction ;)

Embarrassed? :P
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on April 28, 2012, 07:21:42 pm
HAHAHA! nice spot VegemitePi!
But thanks for your help DisaFear! :D
Title: Re: Specialist 3/4 Question Thread!
Post by: ashoni on April 28, 2012, 07:35:28 pm
Another question!

A man starts at 2 p.m. to walk to a place 26 km away. He walks at a constant speed until 4 p.m. when he increases his speed by 2 km/h. He reaches his destination at 5:30 p.m. At what speed did he walk for the first two hours?
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on April 28, 2012, 07:52:40 pm
$d=vt$ (or d = tv, that's how I'll set it up)

For the purposes of this question I'll be taking d in km, t in h and v in km/h

$d_1=v_1t_1=2v_1$

$d_2=v_2t_2=1.5(v_1+2) = 1.5v_1+3$

$d_t = d_1 + d_2 = 2v_1 + 1.5v_1 + 3 = 3.5v_1 +3 = 26$

$3.5v_1+3=26$

$3.5v_1=23$

$\frac{7}{2}v_1=23$

$v_1=\frac{23\times2}{7}=\frac{46}{7} \ km/h$

If you need me to explain each line, just ask :)

(should be right  :P)
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on April 30, 2012, 06:37:04 pm
show that:

$\int{\frac{dy}{y(1-y)}=\ln \left(\frac{y}{1-y}\right)+c$

for some constant $c$, where $0

edit: Haha, I got it, didn't realise it was a partial fractions question   -.-

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on April 30, 2012, 07:37:11 pm
Firstly split $\frac{1}{y(1-y)}$ into partial fractions.
\begin{alignedat}{1}\frac{1}{y(1-y)} & =\frac{A}{y}+\frac{B}{1-y}
\\ 1 & =A(1-y)+By
\\ 1 & =y(B-A)+A
\end{alignedat}

Equating coefficients
$A=1$
\begin{alignedat}{1}B-A & =0
\\ B-1 & =0
\\ B & =1
\end{alignedat}

So now we have
\begin{alignedat}{1}\intop\frac{dy}{y(1-y)} & =\intop\frac{1}{y}+\frac{1}{1-y}dy
\\ & =\ln(|y|)-\ln(|1-y|)+C
\\ & =\ln\left(\frac{|y|}{|1-y|}\right)+C
\\ & =\ln\left(\frac{y}{1-y}\right)+C\:(\mathrm{Remove\: the\: modulus\: as\: y>0\: and\:1-y>0\: for\:0\end{alignedat}

Hope that helps :)

EDIT: Just thought I'd add this, it might help.
For $\int\frac{1}{1-y}dy$
Let $u=1-y$
then $\frac{du}{dy}=-1$
So $dy=-du$
So we have
\begin{alignedat}{1}\int\frac{1}{1-y}dy & =\int\frac{1}{u}\times-du
\\ & =-\int\frac{1}{u}du
\\ & =-\ln(|1-y|)+C
\end{alignedat}

Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on May 07, 2012, 05:34:38 pm
Idk if people still post in this thread but with vector proof questions, generally i stumble around and find myself at the answer accidentally. Anyone know about mark allocation for these questions or is there a particular thought process i should follow for each question?  :s
Title: Re: Specialist 3/4 Question Thread!
Post by: nisha on May 07, 2012, 05:57:09 pm
Basically write down what you have to find, and what you are given. Then manipulate...somehow. In generally all the questions I have done so far, vector proofs can range from 2-4 mark questions.
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on May 07, 2012, 09:29:08 pm
Show that cos 3Ɵ = 4 cos^3 Ɵ - 3 cos Ɵ. Similar express sin3Ɵ  as a polynomial in sin Ɵ.

This would be an epic help!
I understand that you have to use the double angle formulas (and perhaps compound formulas), but I'm confused!

THANKS GUYS!
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on May 07, 2012, 09:50:06 pm
Firstly we know our Compound Angle Formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$
So..
\begin{alignedat}{1}\cos(3\theta) & =\cos(\theta+2\theta)
\\ & =\cos(\theta)\cos(2\theta)-\sin(\theta)\sin(2\theta)
\end{alignedat}

Now we also know our Double Angle Formula $cos(2\theta)=2\cos^{2}(\theta)-1=1-2\sin^{2}(\theta)=\cos^{2}(\theta)-\sin^{2}(\theta)$ and our one for sin $\sin(2\theta)=2\cos(\theta)\sin(\theta)$
So..
\begin{alignedat}{1}\cos(3\theta) & =\cos(\theta)\left(2\cos^{2}(\theta)-1\right)-\sin(\theta)\times2\cos(\theta)\sin(\theta)
\\ & =2\cos^{3}(\theta)-\cos(\theta)-2\cos(\theta)\sin^{2}(\theta)
\\ & =2\cos^{3}(\theta)-\cos(\theta)-2\cos(\theta)\left(1-\cos^{2}(\theta)\right)\:\left(\mathrm{as\:}\cos^{2}(\theta)+\sin^{2}(\theta)=1\right)
\\ & =2\cos^{3}(\theta)-\cos(\theta)-2\cos(\theta)+2\cos^{3}(\theta)
\\ & =4\cos^{3}(\theta)-3\cos(\theta)
\end{alignedat}

Now you can do a similar thing for $\sin(3\theta)$, using $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$, and some of the other compound angle and double angle formulas above.

EDIT: Just an extra note, the reason I subbed in the $2\cos^{2}(\theta)-1$ instead of any of the other compound angle formulas, was because the RHS is all in terms of cos(theta) to the power of something, so naturally subbing in the formula with the most cos terms and the least sine terms would be the better option.
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on May 07, 2012, 09:59:39 pm
Thanks so much man! Yeah, i know why you subbed in 2cos^2Ɵ -1, i did that as well, but I failed hardcore on the other steps :P

THANKS B^3, you're mah main man.
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on May 08, 2012, 12:34:31 pm
I have a SAC tomorrow and on the revision sheet given to us, the teacher talks about "sketching complicated circular functions". What would a complicated circ. function look like :/? Can someone give me some ideas as to what is considered "complicated" and worthy for an analysis task? Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Dominatorrr on May 08, 2012, 06:49:45 pm
I have a SAC tomorrow and on the revision sheet given to us, the teacher talks about "sketching complicated circular functions". What would a complicated circ. function look like :/? Can someone give me some ideas as to what is considered "complicated" and worthy for an analysis task? Thanks

I'm assuming graphs with a lot of transformations.

VECTORS) If AC = AD + DC and BD = BA + AD then how would we figure out the dot product of AC and BD? There are no values given, the letters are just points on a rhombus, it is part of a vector proof question and vectors AC and BD are obviously expressed as a sum of 2 vectors.
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on May 09, 2012, 04:55:32 pm
For Vector Calculus, when we have found the velocity vector and the question asks for the maximum and minimum speeds, do we equate the i or j components to  0?

I keep getting the i and j's mixed up when determining the minimum/maximum speeds and minimum/maximum heights..
Title: Re: Specialist 3/4 Question Thread!
Post by: Fishyiscool on May 09, 2012, 09:49:44 pm
A particle's position as a function of time is given by u = cos (2t)i + sin (2t)j.
a) Find the equation of the path
b) State the period of the motion of this particle

(P.S. Would this be a tech-assumed question?)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on May 09, 2012, 10:01:47 pm
Hmm.

If I assume is correct, For a), you want the cartesian form.

Let $x=cos(2t)$ and $y=sin(2t).$

$2t=cos^-^1(x)$
$t=cos^-^1(x)/2$

$2t=sin^-^1(y)$
$t=sin^-^1(y)/2$

$sin^-^1(y)/2=cos^-^1(x)/2$
$sin^-^1(y)=cos^-^1(x)$
$y=\sqrt1-x^2$

Someone check If I'm correct as I haven't done too many problems on vector calculus.

Hmm, I am not too sure about b). Sorry. :(

(Square root over the entire answer, I have no idea how to do that for LaTeX >.<)
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on May 09, 2012, 11:13:55 pm
(Square root over the entire answer, I have no idea how to do that for LaTeX >.<)
You just stick it around curly brackets

Code: [Select]
$$\sqrt{moooooooooooooooooo + roooooar}$$
$\sqrt{moooooooooooooooooo + roooooar}$

The curly bracket thing goes for most things in LaTeX when you want it to whatever you're doing to apply to the entire block, e.g. $X_{hello}$
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on May 25, 2012, 01:13:54 am
Express (1-i)^3 in cartesian form. isn't that already in cartesian form? if not, what is that form called?  the solution just expanded and simplified it
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on May 25, 2012, 01:42:19 pm
Also, more of an basic geometry question but I guess it came up in vectors - are two lines that are collinear also considered to be parallel?
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on May 28, 2012, 11:28:45 pm
Also, more of an basic geometry question but I guess it came up in vectors - are two lines that are collinear also considered to be parallel?

Two lines that are collinear would be the same line, so I would say yes, unless there is some technicality that states a line isn't parallel to itself...
Two vectors that are collinear would be parallel but not neccessarily the same vector, remember vectors are free to move in space.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on May 29, 2012, 10:48:42 pm
Express (1-i)^3 in cartesian form. isn't that already in cartesian form? if not, what is that form called?  the solution just expanded and simplified it
well the formal definition is given on wiki http://en.wikipedia.org/wiki/Complex_number#Definition

and so (1-i)^3 isn't in the correct form, you need to expand it
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on June 02, 2012, 04:50:21 pm
There are two ways to expand (1 - i)^3. Using a purely algebraic approach:
Using the formula (a - b)^3 = a^3 - 3(a^2)b + 3ab^2 - b^3
(1 - i)^3 = (1)^3 - 3(1^2)i + 3(1)(i^2) - (i)^3
= 1 - 3i - 3 + i
= -2 - 2i

Using the polar form approach:
(1 - i)^3
= (sqrt(2)*cis(-pi/4))^3
= 2*sqrt(2)*cis(-3pi/4)
= 2*sqrt(2)*(-1/sqrt(2) - 1/sqrt(2) i)
= -2 - 2i
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on June 07, 2012, 02:02:02 pm
^ Thanks guys
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on June 07, 2012, 06:05:18 pm
What's the difference between a parameter and a variable?
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on June 10, 2012, 01:03:54 pm
A parameter is like an added variable that is not part of the cartesian equation. For example:
x = cos(t) and y = sin(t)
The independent variable is x, the dependent variable is y and the parameter is t.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on June 10, 2012, 04:50:25 pm
I think the best way to define a "parameter" in general, is to just check the wiki entry on it, or simply use the definition provided in your text (which should apply thereon-forth), there is too many possible definitions of a parameter
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on June 13, 2012, 03:01:46 pm
^ I'm too dumb to fully understand the wiki article

This, I get
...a parameter is a quantity that serves to relate functions and variables using a common variable when such a relationship would be difficult to explicate with an equation.

This loses me
[/i]Mathematical functions have one or more arguments that are designated in the definition by variables, while their definition can also contain parameters. The variables are mentioned in the list of arguments that the function takes, but the parameters are not. When parameters are present, the definition actually defines a whole family of functions, one for every valid set of values of the parameters. [/i]
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on June 17, 2012, 01:31:47 pm
Simple question but i forgot the reason behind it: How do you determine the domain of something like y= sin-1(x/2)?
it's [-2,2] but why? why not [-0.5,0.5]? Can someone explain the logic behind it? Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on June 17, 2012, 01:36:57 pm
The way I would do it is that the x-cords of the endpoints would be:

x/2 = -1 and x/2 = 1   ie. whatever is inside the arcsin is either 1 or -1
Hence, x = -2 and x = 2

Hence domain = [-2, 2]

Alternatively, you could see from the equation that the simple y=arcsin(x) has been dilated and work the domain from there :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on June 17, 2012, 08:02:55 pm
I having trouble with this:
A is the point (-1,2) and B is (-5,1). Find P such that AP=2PB using a vector method.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on June 17, 2012, 08:23:44 pm
I having trouble with this:
A is the point (-1,2) and B is (-5,1). Find P such that AP=2PB using a vector method.

Let OP (x,y) (I dont know how to write vector on computer man :( just see whatsoever I wrote later as vectors)
AP = OP-OA = (x+1, y-2)
PB = OB-OP = (-5-x, 1-y)
Given: AP=2PB => x+1= 2(-5-x) and y-2=2(1-y) => DIY to solve x,y :P
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on June 20, 2012, 06:41:20 pm
Thanks :)

Can someone evaluate x root 1-x2 from 0 to 1.

I keep getting -1/3, but it should be 1/3.
Aslo can someone show me how to type the antidiff sign and other maths symbols using latex?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on June 20, 2012, 06:50:10 pm
\begin{alignedat}{1}\int_{0}^{1}x\sqrt{1-x^{2}}dx\end{alignedat}
Let $u=1-x^2$
$\frac{du}{dx}=-2x$
Changing the terminals $x=0$ then $\:u=1$ and if $x=1$ then $u=0$ (this may be why your answer is -ve, if you forgot this step)
The reason we change the terminals is because we are no longer integrating with respect to x, we are now integrating with respect to u.

\begin{alignedat}{1}\int_{0}^{1}x\sqrt{1-x^{2}}dx & =\int_{1}^{0}x\sqrt{u}\frac{1}{-2x}du
\\ & =\frac{-1}{2}\int_{1}^{0}u^{\frac{1}{2}}du
\\ & =-\frac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}
\\ & =-\frac{1}{2}\left(0-\frac{2}{3}\right)
\\ & =\frac{1}{3}
\end{alignedat}

As for the integral
Code: [Select]
$$\int$$For the square root
Code: [Select]
$$\sqrt{}$$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on June 20, 2012, 06:55:21 pm
Oh yes i forgot to change the terminals lol. Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on June 24, 2012, 10:08:29 am
Can someone show me how to do this:
Find the solution to the differential eqn: $\frac{dy}{dx}$=$\frac{1}{4-x^2}$,given that y=2 when x=0

I keep getting it in the form y=$\frac{1}{4}$loge|(2-x)(2+x)|+c but answer says it should be in the form  y=$\frac{1}{4}$loge|$\frac{2+x}{2-x}$|+c

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: HenryP on June 24, 2012, 03:00:14 pm
Can someone show me how to do this:
Find the solution to the differential eqn: $\frac{dy}{dx}$=$\frac{1}{4-x^2}$,given that y=2 when x=0

I keep getting it in the form y=$\frac{1}{4}$loge|(2-x)(2+x)|+c but answer says it should be in the form  y=$\frac{1}{4}$loge|$\frac{2+x}{2-x}$|+c

Thanks
Hey, I assume you split the fraction into partial fractions with the denominators being 2-x and 2+x .
I'm pretty sure the mistake you're making is that when integrating $\frac{1}{2-x}$ you should be getting $-\log_{e}|2-x|$
That should fix the problem you're having, hope this helps :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on June 24, 2012, 03:03:32 pm
Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?
Title: Re: Specialist 3/4 Question Thread!
Post by: HenryP on June 24, 2012, 03:06:45 pm
Probably best not to oversimplify it, but essentially yes. You can prove it to yourself by doing the substitution u= 2-x
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on June 24, 2012, 04:21:43 pm
Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?
Just follow;

$\int \frac{k}{ax+b} dx = \frac{k}{a}log_{e}|ax+b|$

and you shouldn't go wrong.
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on June 24, 2012, 04:41:51 pm
Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?
Just follow;

$\int \frac{k}{ax+b} dx = \frac{k}{a}log_{e}|ax+b|$

and you shouldn't go wrong.
That doesn't work with x^2 does it?
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on June 24, 2012, 04:52:17 pm
Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?
Just follow;

$\int \frac{k}{ax+b} dx = \frac{k}{a}log_{e}|ax+b|$

and you shouldn't go wrong.
That doesn't work with x^2 does it?

The integral of 1/x^2 is -1/x, so nope.

Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on June 24, 2012, 04:55:04 pm
Oh ok thanks, so whenever there's a negative in front of the x, we have to put a negative in front of the log?
Just follow;

$\int \frac{k}{ax+b} dx = \frac{k}{a}log_{e}|ax+b|$

and you shouldn't go wrong.
That doesn't work with x^2 does it?
I was referring to $\frac{1}{2-x}$ and his question of what to do with the negative in front of the x.

i.e. $k = 1, a = -1, b = 2$
Title: Re: Specialist 3/4 Question Thread!
Post by: HenryP on June 24, 2012, 05:19:19 pm
I believe there is nothing wrong with that derivation. As to why $x^2$ doesn't work is because the formula should only work such the case the power of x is 1
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on June 24, 2012, 09:57:13 pm
\begin{alignedat}{1}2-x^{2} & =x^{2}
\\ 2 & =2x^{2}
\\ x & =\pm1
\end{alignedat}

$x=1\:\mathrm{then}\: y=1$
The region is the overlapping region that is in the first quadrant.
(https://s3.amazonaws.com/grapher/exports/dohhkzgtrd.png)
So we need to look at rotating around the y-axis, that is we need to look at it in two parts. That is the region b/w y=x^2 and the y-axis for 0 to 1 and the region between y=2-x^2 and the y-axis for y=1 to 2.
Now the formula we have for rotating around the y-axis is \begin{alignedat}{1}V & =\pi\int_{a}^{b}x^{2}dy\end{alignedat}
So we need x^2.
So
$x^{2}=y$ and $x^{2}=2-y$
Now to rotate that
\begin{alignedat}{1}V & =\pi\int_{0}^{1}ydy+\pi\int_{1}^{2}2-ydy
\\ & =\pi\left[\frac{1}{2}y^{2}\right]_{0}^{1}+\pi\left[2y-\frac{1}{2}y^{2}\right]_{1}^{2}
\\ & =\pi\left(\frac{1}{2}-0+4-\frac{4}{2}-\left(2-\frac{1}{2}\right)\right)
\\ & =\pi\left(\frac{1}{2}+2-\frac{3}{2}\right)
\\ & =\pi\left(2-1\right)
\\ & =\pi\:\mathrm{cubic\: units}
\end{alignedat}

Now I did that in a rush, so there probably is a mistake in there somewhere, so if there is someone pick me up on it. Now to watch the Grand Prix....

Anyways hope that helps.

Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on June 24, 2012, 10:20:01 pm
Hey, I think we can calculate one area and then double it because the upper and below parts are both equal

Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on June 24, 2012, 10:38:08 pm
That's an interesting question. In order to avoid confusing myself, I'm going to start with a much simplified version of that problem and then work my way upwards: Suppose they only wanted you to calculate the volume of the solid formed when the area between the curve y = 2 - x^2 and the lines x = 0 and x = 1 was rotated around the y-axis.
How would I calculate that volume?
Title: Re: Specialist 3/4 Question Thread!
Post by: Mao on June 24, 2012, 10:45:07 pm
That's an interesting question. In order to avoid confusing myself, I'm going to start with a much simplified version of that problem and then work my way upwards: Suppose they only wanted you to calculate the volume of the solid formed when the area between the curve y = 2 - x^2 and the lines x = 0 and x = 1 was rotated around the y-axis.
How would I calculate that volume?

Break it down into two parts:

The bottom half 0<x<1, 0<y<1, is just a simple cylinder. Volume is thus $V_{bot} = \pi r^2 h = \pi$

The top half can be constructed from a series of thin horizontal disks. Volume of each of these disks are $dV = \pi x^2 dy$, thus we must integrate over y.

We know that:
- the region we're integrating is from y=1 to y=2 (this is the curved part)
- $x = \sqrt{2-y}$

Thus, $V_{top} = \int_{1}^{2} \pi x^2\, dy = \int_{1}^{2} \pi (2-y) \, dy = \frac{\pi}{2}$

Total volume is thus $V = \frac{3\pi}{2}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on June 26, 2012, 08:08:36 pm
what is the general approach to finding the complex roots of a polynomial if you are given a real root eg. z=a?
if you sub in that factor in the form (z-a) into the original polynomial, is that going to help in any way??
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on June 26, 2012, 08:34:20 pm
what is the general approach to finding the complex roots of a polynomial if you are given a real root eg. z=a?
if you sub in that factor in the form (z-a) into the original polynomial, is that going to help in any way??

It will usually be a cubic, divide your cubic by z - a in order to get a quadratic, then just complete the square, or use the quadratic formula or factorise, whichever is appropriate.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on June 26, 2012, 08:50:05 pm
what is the general approach to finding the complex roots of a polynomial if you are given a real root eg. z=a?
if you sub in that factor in the form (z-a) into the original polynomial, is that going to help in any way??

It will usually be a cubic, divide your cubic by z - a in order to get a quadratic, then just complete the square, or use the quadratic formula or factorise, whichever is appropriate.

haha thanks a lot. forgot how to do it :S
Title: Re: Specialist 3/4 Question Thread!
Post by: mxglasses on June 28, 2012, 11:53:56 pm
can someone help me with this Q? i don't really know where to really start

A vessel, filled with liquid, is being emptied. The rate at which the liquid is flowing out at any instant is proportional to the volume remaining at that instant. If one quarter of the vessel is emptied in 5mins, what fraction remains after 10mins?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on June 29, 2012, 12:26:05 am
can someone help me with this Q? i don't really know where to really start

A vessel, filled with liquid, is being emptied. The rate at which the liquid is flowing out at any instant is proportional to the volume remaining at that instant. If one quarter of the vessel is emptied in 5mins, what fraction remains after 10mins?

The key here is to know that if the rate is proportional to the volume, then it is equal to constant K mutipled by the volume. Then to integrate it you need to flip it and you'll get t in terms of k, then rearrange.
\begin{alignedat}{1}\frac{dV}{dt} & \propto V
\\ \frac{dV}{dt} & =kV
\\ \frac{dt}{dV} & =\frac{1}{kV}
\\ t & =\int\frac{1}{kV}dV
\\ t & =\frac{1}{k}\ln|V|+C
\\ k\left(t-C\right) & =\ln|V|
\\ V & =\pm e^{k\left(t-C\right)}
\\ V>0,\:\mathrm{take\: positive\: sol}
\\ \mathrm{Let\:}A=e^{-kC}
\\ V & =Ae^{kt}
\end{alignedat}

Then from that you have two points $(0,V_{i})$ and $(5,\frac{3}{4}V_{i})$, from that you should be able to find $k$ and $A$, then a fraction of the initial and sub in $t=10$ mins. Hope that helps :)
Title: Re: Specialist 3/4 Question Thread!
Post by: mxglasses on June 29, 2012, 12:45:35 am
that's so smart! Man.. i should really get use to the wording of these problems.
anyways that's plenty of help! thankyou  :D
Title: Re: Specialist 3/4 Question Thread!
Post by: mxglasses on July 02, 2012, 09:45:50 pm
2 more Qs

2) If the water in a tank escapes through a hole in the bottom, the rate of flow is proportional to the square root of the depth of the water in the tank. Show that if the tank is cylindrical with its axis vertical, the time required for three-quarters of the water to flow out is equal to the time required for the remaining one-quarter to flow out.

3) Water is leaking from a cylindrical vessel at a rate that is proportional to the square root of the depth of water remaining. If the depth drops from 36cm to 25cm in one minute, how much longer will it take the vessel to empty?

is the starting point for both of them dv/dt= -k$\sqrt{h}$ ? , i don't get what goes next after this
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 02, 2012, 11:02:03 pm
Basically yes it is the starting point, then you need to use the chain rule.
\begin{alignedat}{1}\frac{dV}{dt} & =-k\sqrt{h},\: k>0
\\ \frac{dh}{dt} & =\frac{dV}{dt}\frac{dh}{dV}
\end{alignedat}

As it is a cylinder.
$V=\pi r^{2}h$
Now we don't know $r$, but we know that as the water drains out that it isn't changing, i.e. it is a constant, so we can diff that equation like so \begin{alignedat}{1}\frac{dV}{dh} & =\pi r^{2}\end{alignedat}
That will give $\frac{dh}{dt}=\frac{-k\sqrt{h}}{\pi r^{2}}$
From there flip it and integrate it (remebering that $r$ is constant because its a cylinder (it won't be for things like a cone e.t.c). Then you will have $t$ in terms of $h$ and you have your two points. Then see if you can get it from there.

Just want to stress that $r$ will only be constant for cylinders standing upwards, for cones you would need to use ratios.

Anyway, hope that helps :)

EDIT: To make it a bit simpler, as $\pi r^{2}$ is constant you could just represent it with $A$ (as it is the cross-sectional area after all). So you would have $A=\pi r^{2}$ so $\frac{dh}{dt}=\frac{-k\sqrt{h}}{A}$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 04, 2012, 02:38:49 pm
How do you construct a differential eqn for this:
The gradient of a normal to a curve at any point (x,y) is three times the gradient of the line joining the same point to the origin.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on July 04, 2012, 02:49:35 pm
-dx/dy=3y/x
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 04, 2012, 02:50:50 pm
The gradient will be given by the derivative, then remembering that tangents and normals are perpendicular, so $m_{1}m_{2}=-1$
So the gradient of the normal is
\begin{alignedat}{1}\frac{dy}{dx}m_{normal} & =-1
\\ m_{normal} & =\frac{-1}{\frac{dy}{dx}}
\end{alignedat}

The gradient of the line joining the point to the origin will just be rise over run, that is
$m=\frac{y-0}{x-0}=\frac{y}{x}$
So then the gradient of the normal is 3 times this
So we have
\begin{alignedat}{1}\frac{-1}{\frac{dy}{dx}} & =3\frac{y}{x}
\\ -\frac{x}{y} & =3\frac{dy}{dx}
\\ \frac{dy}{dx} & =\frac{-x}{3y}
\end{alignedat}

EDIT: Like brightsky has done, you probably wouldn't have to rearrange for dy/dx as it would still be a differential equation.
Title: Re: Specialist 3/4 Question Thread!
Post by: tonyduong on July 04, 2012, 03:11:11 pm
I have a problem finding the cartesian equation of |z+2i|=3^1/2*|z-3|. I tried to substitute z=x+yi except the equations were looking ugly. If you could help that would be great. Thanks. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on July 04, 2012, 04:07:19 pm
I assume you mean:

$|z+2i|^2=3|z-3|^2$

$x^2+(y+2)^2=3(x-3)^2+3y^2$

$x^2+y^2+4y+4=3x^2-18x+27+3y^2$

$0=2x^2-18x+2y^2-4y+23$

Now complete the square to $2x^2-18x$ and $2y^2-4y$ and at the end of the day you will actually get some circle.
Title: Re: Specialist 3/4 Question Thread!
Post by: mxglasses on July 05, 2012, 12:54:26 pm
Ty b^3! i followed the steps and integrated it, but i don't know if the values i'm subbing in are the right ones. The ones i subbed in were t=1 when h= 25, and t=0 when h=36. i keep getting 5 mins as my answer but the solution says it's 6mins.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 05, 2012, 03:29:15 pm
Integrating you should end up with $t=-\frac{2A}{k}\sqrt{h}+C$
Did you add the $+C$? :P
You have the two points, which give the correct answer of 6 mins.
Solving for $C$ you should get $C=\frac{12A}{k}$.
Then you should be able to solve for $k$, getting $k=2A$.
Then substituting that those into the equation, you should get $t=6-\sqrt{h}$, then you want when the vessel is empty, that is when $h=0$, putting that in you should get $t=6\:mins$.

You might of forgot the C or swapped the h values around for the two coordinates.

EDIT: Just had another look, you probably did swap the h values when you worked it out, it should be as you said $t=1$ when $h=25$ and $t=0$ when $h=36$. Using those you get the correct answer of 6 mins, swapping them around will give you the 5 mins that you got.

Hope that helps :)
Title: Re: Specialist 3/4 Question Thread!
Post by: mxglasses on July 05, 2012, 04:33:23 pm
hahaha nice observation  :D i did switch the h values LOL .. anyways ty again
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 06, 2012, 10:05:36 pm
The function f(x)=A+Bsin-1(x), where A and B are integers, passes through the points with coordinates (1/2 , (18-pi)/6 ) and ( -1, (6+pi)/2 .
Find the values of A and B.

When i subbed -1 in to the eqn, the sin-1(-1), does that become 3pi/2 or -pi/2 ? Cause they give different values for A and B, and the solutions used pi/2 but why?
Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 08, 2012, 01:10:01 pm
Can someone show me how to solve z3-(2-i)z2+z-2+i=0 for z E C ,without using calculator. I forgot how to start, do we sub in random numbers to find a factor?
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on July 08, 2012, 01:18:33 pm
Can someone show me how to solve z3-(2-i)z2+z-2+i=0 for z E C ,without using calculator. I forgot how to start, do we sub in random numbers to find a factor?

You need to use factorization by recognition, or something like that. It's called factorisation by grouping(?). First:

z^3 - (2-i)z^2 + z - 2 + i = 0
z^2(z-2+i)+1(z-2+i)=0
(z^2+1)(z-2+i) = 0
(z-i)(z+i)(z-2+i)=0

z = i, -i, 2-i
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on July 08, 2012, 01:19:24 pm
The function f(x)=A+Bsin-1(x), where A and B are integers, passes through the points with coordinates (1/2 , (18-pi)/6 ) and ( -1, (6+pi)/2 .
Find the values of A and B.

When i subbed -1 in to the eqn, the sin-1(-1), does that become 3pi/2 or -pi/2 ? Cause they give different values for A and B, and the solutions used pi/2 but why?
Thanks :)

You use -pi/2 because the arcsin (inverse sine function) is restricted to the domain of -pi/2 to pi/2
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on July 08, 2012, 02:08:20 pm
All inverse trig functions (arcsin, arccos, arctan, arcsec, arccsc and arccot) are within the domain [-pi/2, pi] and as close to 0 as possible.
So with arcsin(-1), it is -pi/2, since that is the only value within that domain to make sin(-pi/2) = -1.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 08, 2012, 04:06:17 pm
Oh ok, thanks a lot guys :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 08, 2012, 08:33:41 pm
How do i antidiff tan2xsec2x?
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on July 08, 2012, 08:39:41 pm
How do i antidiff tan2xsec2x?

$\int tan^2(x) \cdot sec^2(x) \ dx$

$let \ u = tan(x)$

$\frac{du}{dx} = sec^2(x)$

$\int u^2 \frac{du}{dx} \ dx = \int u^2 \ du = \frac{1}{3} u^3 + c$

$= \frac{1}{3} tan^3(x) + c$

edit: fixed mistake, forgot to power of 3 :/
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 08, 2012, 08:41:22 pm
Thank you
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 09, 2012, 06:33:13 pm
How do i antidiff secx?
Why cant i do this: root(sec2x), so antidiff is root(tanx)?

Also, why is secx=1/cox not equal to cos-1x?
Thanks  :)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 09, 2012, 06:45:18 pm
Remember that $\cos^{-1}(x)$ is the inverse of $\cos(x)$ that is if $y=\cos(x)$ then $x=\cos^{-1}(y)$. $\frac{1}{\cos(x)}$ is the reciprocal of $\cos(x)$ that is 1 divided by $\cos(x)$. Draw the two graphs out and you will see (they aren't the same).

As for the first one, this is quite a hard question (doubt you would ever get it on a VCAA exam) but the hint is
\begin{alignedat}{1}\int\sec(x)dx & =\int\sec(x)\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx\end{alignedat} then see what $u$ subsitution you can do.

EDIT: pit's way for int sec(x) dx is easier, follow that instead :)

Moderator action: removed real name, sorry for the inconvenience
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on July 09, 2012, 06:47:37 pm
As explained by b^3, ignore the "-1" as a power/index, it's not, it shows inverse and inverse only (you will NEVER see negative powers being written there for trig functions). It's just notation you'll have to get used to. Such is maths I suppose :P

I've got a simpler way for integral of sec(x), write it as 1/cos(x) and then multiply both sides by cos(x)/cos(x). Then let u=sin(x) after substituting 1-sin^2(x)=cos^2(x). And the rest is elementary.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 10, 2012, 04:02:06 pm
How do you find the volume of a shape rotated about the y axis?

e.g The region S is in the first quadrant of the x-y plane is bounded by the axes, the line x=3 and the curve $y= \sqrt{1+x^2}$. Find the volume of the solid when S is rotated around about the y axis.

Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on July 10, 2012, 04:22:53 pm
Sorry, me being a noob  ::)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 10, 2012, 04:36:56 pm
Also be careful of which region you are rotating, with what Hancook has you will get the area rotated that is left of the curve (and with y=root(10)) in the graph below, but we want the volume of the solid of the area on the right that is between the curve, the axis and x=3 to be rotated.
(https://s3.amazonaws.com/grapher/exports/tvhstfwtn2.png)
So we can split it into two volumes, one from the top part that os rotated on the right side of the graph and one for the ccylinder below it.

So you would have to do \begin{alignedat}{1}V & =\pi\int\end{alignedat}
x_{2}^{2}-x_{1}^{2}dy

So that would be \begin{alignedat}{1}V_{1} & =\pi\int\end{alignedat}
_{1}^{\sqrt{10}}3^{2}-\left(y^{2}-1\right)dy

Then you need to add the cylinder thats still left for y=0 to y=1.
So that would be $V_{2}=\pi*r^{2}h=9\pi$
Total voulme would then be \begin{alignedat}{1}V & =\pi\int\end{alignedat}
_{1}^{\sqrt{10}}3^{2}-\left(y^{2}-1\right)dy+9\pi

Which gives $V=\frac{20\sqrt{10}\pi}{3}-\frac{2\pi}{3}$ cubic units

EDIT: Should be fixed now.

EDIT2: You could of also have rotated the line x= aroud the y-axis. then taken away the voulme that Hancook found.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 10, 2012, 04:46:58 pm
If it helps, then answer is [((20root10)pi)/3]-(2pi)/3

Thanks heaps :D
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 10, 2012, 04:52:15 pm
Should be fixed now.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 10, 2012, 08:03:31 pm
z=2cis(pi/3)
Find the smallest possible integer n for which zn is a real number

I let sin[(n X pi)/3]=0 and solved to get n=0 , but the answer should be n=3.
What have i done wrong?
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on July 10, 2012, 09:21:17 pm
The actual correct answer is that no such smallest integer exists. I'm pretty sure the question should've been "smallest POSITIVE integer", then n=3 is the correct answer, if it was "smallest non-negative integer" then you are right
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 11, 2012, 01:27:15 pm
My bad, its suppose to be "smallest positive integer" not "smallest possible integer." Could you show me why it's n=3?
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on July 11, 2012, 03:02:40 pm
Could you show me why it's n=3?

You had the right idea with $\sin\left(\frac{n\pi}{3}\right)=0$
from that, we get

$\frac{n\pi}{3}=k\pi$   for $k\in \mathbb{Z}$

$\therefore n=3k$

to get the answer of $n=0$  we just used $k=0$

to find the smallest positive integer solution, we just use $k=1$  which gives $n=3$
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on July 12, 2012, 04:09:53 pm
For differentiation, we have a definition from first principles, that is

$\frac{d}{dx}(f(x))=\lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

Is there some sort of definition/expression like this, but for integration?
I have come across the definition regarding the definite integral as the limit of an infinite sum, but unlike the first principle definition for differentiation, this didn't seem to actually help in directly calculating an integral... Like, it makes intuitive sense and I understand it, I just don't really see how you can plug values in and arrive at the indefinite integral in the way that you can with the above expression.

What is the connection between $\int^b_a{f(x)}dx=\lim_{n \to \infty}\sum_{i=1}^{n} f(x_{i}^{*})\Delta x$
and the actual method of computation of an integral - that is, doing the opposite of what we do for differentiation.

I hope that's clear...
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on July 12, 2012, 05:04:08 pm
The fundamental theorem of calculus is what links antidifferentiation to area calculation. Basically, the theorem has two parts, the first of which shows that A'(x) = f(x), where A(x) is the area between a and x, where a<x<b. The second part of the theorem proves that A(b) = F(b) - F(a), where F(x) is the antiderivative of f(x). This is the 'formula' we use nowadays to calculate areas under curves.

It is important to understand that the Riemann sum definition does not relate area calculation to calculus at all. It is due to the fundamental theorem of calculus that the formula Methods students often take for granted holds.
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on July 12, 2012, 05:38:39 pm
Quote
What is the connection between \int^b_a{f(x)}dx=\lim_{n \to \infty}\sum_{i=1}^{n} f(x_{i}^{*})\Delta x
and the actual method of computation of an integral - that is, doing the opposite of what we do for differentiation.

You're right, the definition of a (Riemann) Integral is (sort of) that what you wrote down (i.e a limit, but there are in fact some more technical conditions). Computationally it's easier to use Fundamental Theorem of Calculus in many occasions. It's just like with differentiation, you don't always differentiate from first principles but mostly with product rule, chain rule etc. (which you do prove from the limit definition, just as you can prove FTC from the Riemann Sum limit definition).
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on July 12, 2012, 06:17:39 pm
Quote
What is the connection between \int^b_a{f(x)}dx=\lim_{n \to \infty}\sum_{i=1}^{n} f(x_{i}^{*})\Delta x
and the actual method of computation of an integral - that is, doing the opposite of what we do for differentiation.

You're right, the definition of a (Riemann) Integral is (sort of) that what you wrote down (i.e a limit, but there are in fact some more technical conditions). Computationally it's easier to use Fundamental Theorem of Calculus in many occasions. It's just like with differentiation, you don't always differentiate from first principles but mostly with product rule, chain rule etc. (which you do prove from the limit definition, just as you can prove FTC from the Riemann Sum limit definition).

Thanks for answering guys, I didn't realise the distinction between the Riemann sum and the FTC.
Do you know where I could find a proof of what I highlighted? I think actually seeing that could definitely clear things up.

Also, do you guys have a preferred texts or online resources that go into more depth about this stuff? I think I have a copy of Stewart's Calculus on my external HDD for what it's worth.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on July 12, 2012, 06:22:12 pm
Wikipedia's always good.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on July 12, 2012, 06:29:16 pm
Wikipedia's always good.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

haha thanks.

And so the endless night of Wikipedia link-clicking begins.....

Last time this happened I started on Taylor series and ended up reading about the Soviet Space program with about 140 tabs open.. -.-

Also could you or kamil perhaps have a look at my other few questions here?
Title: Re: Specialist 3/4 Question Thread!
Post by: Shiney94 on July 12, 2012, 07:12:12 pm
If anyone could help me with this question would be great!

√3z2 + √2z - i/2 = 0
Find the solutions to this equation
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on July 12, 2012, 07:33:54 pm
If anyone could help me with this question would be great!

√3z2 + √2z - i/2 = 0
Find the solutions to this equation

It could get a bit messy because the LaTex server seems to be down, so I grabbed this from WolframAlpha:
(http://i.imgur.com/W6Vt4.gif)

Feel free to ask any further questions.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on July 12, 2012, 07:37:27 pm
Or just chuck it all into the quadratic formula :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 14, 2012, 11:58:30 am
How do you antidiff xtan-1x? Would something like this be in Exam 1 or 2?
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on July 14, 2012, 12:04:20 pm
How do you antidiff xtan-1x? Would something like this be in Exam 1 or 2?
Thanks

I can do it with integration by parts, but it's not on the course. So you won't have to do something like this for VCE.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 14, 2012, 12:05:57 pm
How do you antidiff xtan-1x? Would something like this be in Exam 1 or 2?
Thanks

I can do it with integration by parts, but it's not on the course. So you won't have to do something like this for VCE.
ok thanks, so i should just chuck it in my CAS?

What if it was diff xtan-1x?
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on July 14, 2012, 12:15:59 pm
How do you antidiff xtan-1x? Would something like this be in Exam 1 or 2?
Thanks

I can do it with integration by parts, but it's not on the course. So you won't have to do something like this for VCE.
ok thanks, so i should just chuck it in my CAS?

What if it was diff xtan-1x?

You can do that by using product rule
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 14, 2012, 12:19:04 pm
oh yeh! thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 14, 2012, 01:27:45 pm
How do i show that g'(x)>0 for >0 if g'(x)=(2tan-1x)/(1+x2)? Can someone describe what to do or show the steps? Thanks:)
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on July 14, 2012, 02:06:48 pm
look at the numerator and denominator separately. recall the graph of arctan(x). for x>0, arctan(x) > 0 from the graph. it is obvious that 1+x^2>0 for x>0 (in fact for all real x). since the numerator and denominator are both positive for positive x, then the whoe function is positive for x>0.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 15, 2012, 09:06:51 pm
A bowl is modeled by rotating a curve y=x2 for 0<x<1 around the y axis.

part b) If liquid is poured into a bowl at rate of R units of volume per second, find the rate of increases of the depth of liquid in the bowl when the depth is 1/4.

I dono how to start it.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on July 15, 2012, 09:19:38 pm
dy/dt = dy/dV * dV/dt
that should get you started.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 16, 2012, 10:11:22 pm
When asked to factorise say P(z)=z3-5z2+8z-6 and i know z-3 is a factor, how do you find the quadratic bit of it without doing long division. I remember there was a way by just inspection, something like looking at the last term of the original p(z) but i forgot it. Can someone explain it to me?

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 16, 2012, 10:16:55 pm
When asked to factorise say P(z)=z3-5z2+8z-6 and i know z-3 is a factor, how do you find the quadratic bit of it without doing long division. I remember there was a way by just inspection, something like looking at the last term of the original p(z) but i forgot it. Can someone explain it to me?

Thanks
Not sure, but you might be thinking about how the three terms in brackets need to multiply together to make the last term.

(say you had (x-3)(x-2)(x+1) you know the last term of the expanded polynomial will be 6 as -3 x -2 x 1 = 6)

Meaning the remaining two must multiple to make 2 (as we already have -3 and 2 x -3 = -6).

So something like $1+i$ and $1-i$ (as $(1+i)(1-i)=2$) could work, and does.

edit: Oh I misinterpreted your question, my apologies.

You have;

$z^3-5z^2+8z-6$

You know $z-3$ is a factor

$(z-3)(az^2+bz+c)$

We know that the coefficient of the $z^3$ term is $1$, which is calculated by $z \times z^2$ therefore $a = 1$

$(z-3)(z^2+bz+c)$

We also know that the last term in the expanded polynomial is $-6$, which will be - $3 \times c$, thus $c = 2$

$(z-3)(z^2+bz+2)$

Lastly the coefficient of the $z^2$ is $-5$, which is calculated by the addition of $-3 \times z^2$ and $bz \times z$, therefore $b = -2$ and hence;

$z^3-5z^2+8z-6=(z-3)(z^2-2z+2)$

Giving our desired roots of $z=3, 1+i , 1-i$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 16, 2012, 10:39:45 pm
I'm having trouble getting the middle term though, any ideas on how to do it?
Title: Re: Specialist 3/4 Question Thread!
Post by: anthony1n on July 16, 2012, 11:09:40 pm
so P(Z) = z3-5z2+8z-6
So you know (z-3) is a factor, the first term is z3 which means you have to mutiply z by z2 to get the first term.
Similarly you can do that to get that last term of 2
which leaves you with
(z-3)(z2 + ____ +2)
Ok so to get the middle term pick up either of the middle terms lets say (8z) in this example. So only two of the possible terms can be multiplied to give 8z, the first being z x 2 = 2z
The other term being the (-3) so: 2z - 3 = 8z => -3 = 6z , z=-2 So the middle term is -2z.
Alternatively since 2z + x = 8z , x=6z
What can (-3) be multiplied to, to give an answer of 6z, -3 * -2z
Sorry its kinda long and was also hard to explain online but this is what I do to solve the middle term

Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 17, 2012, 12:02:25 am
A flexible beam is supported at its ends, which are the same horizontal level and at a distance L apart. The deflection y of the beam, measured downwards from the horizontal, satisfies the differential equation: 16 d^2y/dx^2 = L-3x, 0≤x≤L
The deflection of the beam and its inclination to the horizontal are both zero at x=0
Use calculus to find an expression for y in terms if x.

Is deflection d^2y/dx^2 and inclination dy/dx?

Thanks
Yep, $\frac{d^2y}{dx^2}$ is the deflection and $\frac{dy}{dx}$ is the gradient of the beam.
Title: Re: Specialist 3/4 Question Thread!
Post by: |ll|lll| on July 17, 2012, 04:47:14 pm
A flexible beam is supported at its ends, which are the same horizontal level and at a distance L apart. The deflection y of the beam, measured downwards from the horizontal, satisfies the differential equation: 16 d^2y/dx^2 = L-3x, 0≤x≤L
The deflection of the beam and its inclination to the horizontal are both zero at x=0
Use calculus to find an expression for y in terms if x.

Is deflection d^2y/dx^2 and inclination dy/dx?

Thanks
Yep, $\frac{d^2y}{dx^2}$ is the deflection and $\frac{dy}{dx}$ is the gradient of the beam.

Thanks, a little confused here... how do I solve for a specific solution with the DE and find k?
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 17, 2012, 04:56:21 pm
A flexible beam is supported at its ends, which are the same horizontal level and at a distance L apart. The deflection y of the beam, measured downwards from the horizontal, satisfies the differential equation: 16 d^2y/dx^2 = L-3x, 0≤x≤L
The deflection of the beam and its inclination to the horizontal are both zero at x=0
Use calculus to find an expression for y in terms if x.

Is deflection d^2y/dx^2 and inclination dy/dx?

Thanks
Yep, $\frac{d^2y}{dx^2}$ is the deflection and $\frac{dy}{dx}$ is the gradient of the beam.

Thanks, a little confused here... how do I solve for a specific solution with the DE and find k?
You have to integrate and then use the conditions given (when $x = 0$, $\frac{dy}{dx} = 0$, when $x = 0$, $\frac{d^2y}{dx^2}=0$) to find $C_1$ and $C_2$ or whatever you choose to label your constants after integration.

When I get home, I'll type up the solution if you still need help :)

Also, not sure what you mean by "k"?
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on July 19, 2012, 07:02:33 pm
How do you prove the a the tangent to a point on any circle is always perpendicular to the radius? I can do it fine if it's not a translated circle, multiplying the derivative and the slope of the radius just gives me -1 but when I tried differentiating a translated circle, I get $\frac{dy}{dx}=-(\frac{x-a}{y-b})$. I'm not sure how to deal with that....probably made an error when I differentiated but if I did, I can't spot it
Title: Re: Specialist 3/4 Question Thread!
Post by: kamil9876 on July 19, 2012, 10:44:02 pm
Well for me at least, I think it's enough to prove it for a circle with centre at the origin, and then just notice that translations preserve angles.

But oh well if you want to do it directly: You've already done the hard part (I assume you used implicit differentiation or something of the sort). Then just notice that the radial segment between $(a,b)$ and $(x,y)$ has gradient $\frac{y-b}{x-a}$.

I think another neat way of doing it is to parametrize the circle: $p(t)=(r\sin(t),r\cos(t))$ and notice that $p'(t)=(r\cos(t),-r\sin(t))$ then just take the dot product and notice it is zero. Calculating $p''(t)$ then also shows you that the acceleration is towards the centre - something that is often mentioned but not proved in VCE physics :)
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on July 24, 2012, 10:20:54 pm
When using definite integrals to find areas below the x-axis, could you change the terminals around to make it positive rather than use the modulus?

And is there a geometric interpretation of that (do you count the sums backwards or something?)

EDIT: Oh and thanks kamil, forgot I asked that! :p
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 24, 2012, 10:26:25 pm
When using definite integrals to find areas below the x-axis, could you change the terminals around to make it positive rather than use the modulus?

And is there a geometric interpretation of that (do you count the sums backwards or something?)
Yes, if you are asked to represent the area between the curve and the x-axis by an integral and the graph is below the axis, you either alter the terminals or place a negative in front.

e.g. Use a definite integral to represent the area between the curve $x^2-4$ and the x-axis.

$A=-\int^{2}_{-2}x^2-4 \ dx$

or

$A=\int^{-2}_{2}x^2-4 \ dx$

I'm not sure I can answer your second question with extreme certainty, so I'll leave that for someone else.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on July 25, 2012, 08:43:34 pm
$r(t) = (sin \pi t)\tilde{i}+ (4cos2\pi t)\tilde{j}$

so $x=(sin \pi t)$
and $y=(4cos2\pi t)$

how would i go about finding cartesian equation ?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 25, 2012, 09:17:22 pm
Using the double angle formuals for cos
\begin{alignedat}{1}y & =4\cos(2\pi t)
\\ & =4\left(1-2\sin^{2}(\pi t)\right)
\\ & =4-8x^{2}
\end{alignedat}
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on July 25, 2012, 09:19:04 pm
Using the double angle formuals for cos
\begin{alignedat}{1}y & =4\cos(2\pi t)
\\ & =4\left(1-2\sin^{2}(\pi t)\right)
\\ & =4-8x^{2}
\end{alignedat}

legend, thanks mate!
Title: Re: Specialist 3/4 Question Thread!
Post by: cassettekid on July 28, 2012, 02:54:06 pm
(sorry I don't know how to make all the math symbols come up!)

Many chemical reactions follow the law which states that the velocity of the reaction is proportional to the different between the initial concentration of the reagent and the amount transformed at any time.

dx/dt= k(a-x)   where, x is an element [0,a]

Where a is the original concentration and x is the amount transformed at time t. If a = 10 and x = 4 after 2 minutes, find the amount left after 5 minutes.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 28, 2012, 05:11:00 pm
I'm stuck with this:

From a balloon ascending with a velocity of 10m/s a stone was dropped and reached the ground in 12 seconds. Given that the gravitational acceleration is 9.8m/s2. Find the height of the balloon when the stone was dropped.

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 28, 2012, 05:21:46 pm
Firstly I will take up as positive.
$u=10\:m/s$ (positive as it is moving upwards initially)
$t=12\:s$
$a=-9.8\:m/s^{2}$ (negative as acceleration due to gravity is downwards)
$s=?$

\begin{alignedat}{1}s & =ut+\frac{1}{2}at^{2}
\\ s & =\left(10\right)\left(12\right)+\frac{1}{2}\left(-9.8\right)\left(12\right)^{2}
\\ s & =120-\left(72\times9.8\right)
\\ s & =-585.6\: m
\end{alignedat}

i.e. it dropped (downwards) by $585.6\:m$
So the height of the balloon when the stone was dropped was $585.6\:m$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 28, 2012, 05:46:47 pm
Firstly I will take up as positive.
$u=10\:m/s$ (positive as it is moving upwards initially)

I dont understand why the initial velocity is 10? It ascends at a velocity of 10m/s but then when the stone drops, wouldn't the velocity be instantaneously at rest for it to switch directions and fall downwards?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 28, 2012, 06:02:08 pm
Firstly I will take up as positive.
$u=10\:m/s$ (positive as it is moving upwards initially)

I dont understand why the initial velocity is 10? It ascends at a velocity of 10m/s but then when the stone drops, wouldn't the velocity be instantaneously at rest for it to switch directions and fall downwards?
When it is initially released, the balloon and the stone are moving upwards with a velocity of 10 m/s, the stone starts to accelerate towards the ground, but for a short period of time, it is still moving upwards, as it will take time for the stones velocity to change from 10 m/s upwards to 0 m/s. Here it is instantaneously at rest, but it's position will be above that of where it started. Then its velocity will increase in the negative direction (i.e. downwards) until it hits the ground.

We know that the total time for it to reach the ground is 12 seconds, that is when it travels upwards initially, then travels downwards and hits the ground.

It will be instantaneously at rest at some point, but as this point will be above the starting position (the position we want) we can't use it, and the time taken from that point at which it is instantaneously at rest to hit the ground, won't be the 12 seconds.

Draw a diagram of the situation, it should help.

Hopefully the above helps too :)
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on July 28, 2012, 06:07:37 pm
Just think of a ball being thrown into the air, by your logic, soccerboi, it would stop the moment it falls out of your hand, but that is not true, it continues going upwards, decreasing in speed until it reaches zero, then it continues back down. Velocity and acceleration can be in two different directions, in this case, up and down, this will cause the velocity to decrease.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 28, 2012, 06:11:59 pm
Thanks a lot guys, i finally got my head around it :D
Title: Re: Specialist 3/4 Question Thread!
Post by: Shiney94 on July 30, 2012, 06:31:48 pm
A car is accelerating at a constant rate from rest, it then reaches a speed of 60km/h after travelling 50 m.
At what speed will the car be travelling at the end of the 11th second?
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 30, 2012, 06:44:20 pm
A car is accelerating at a constant rate from rest, it then reaches a speed of 60km/h after travelling 50 m.
At what speed will the car be travelling at the end of the 11th second?
$v^2 = u ^2 + 2as$

$a = \frac{v^2 - u^2}{2s}$

$u = 0 \ m/s$

$v = 60 \ km/h = \frac{50}{3} \ m/s$

$s = 50 \ m$

$a = ?$

$a = \frac{\left(\frac{50}{3}\right)^2 - 0^2}{2(50)}$

$a = \frac{25}{9} \ m/s^2$

$v = u + at$

$u = 0 \ m/s$

$a = \frac{25}{9} \ m/s^2$

$t = 11s$

$v = ?$

$v = 0 + \frac{25}{9} \times 11 = \frac{275}{9} \approx 30.6 \ m/s$
Title: Re: Specialist 3/4 Question Thread!
Post by: Shiney94 on July 30, 2012, 08:07:27 pm
Thanks for that. Just another question,
A car travels at a constant velocity of 75km/h, passing a stationary police car which immediately sets off in pursuit at a constant rate of acceleration. It takes the police car 80 seconds to catch the car, find:
a) The acceleration of the police car
b) The speed of the police car when the other car has been reached.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on July 30, 2012, 08:20:33 pm
Thanks for that. Just another question,
A car travels at a constant velocity of 75km/h, passing a stationary police car which immediately sets off in pursuit at a constant rate of acceleration. It takes the police car 80 seconds to catch the car, find:
a) The acceleration of the police car
b) The speed of the police car when the other car has been reached.

75km/h= 125/6 m/s
When the police car catches the car, the distances are both the same
At t=80s, distance of the car= 80. 125/6=5000/3 (m)
Let v= velocity of the police car when catching the car
5000/3=1/2.80.v => v=125/3 (m/s)
a. Acceleration of the police car = 125/3:80=25/48 (m/s2)
b. Speed of the police car=125/3 (m/s)

Correct me if I'm wrong
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 31, 2012, 02:59:40 pm
I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 31, 2012, 03:08:37 pm
I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle
Original displacement;

$s = vt = 4 \times 12 = 48m$

To return its original position;

$s = ut + \frac{1}{2}at^2$

$-48 = (4)(20) + \frac{1}{2}a(20)^2$

$a = -0.64 m/s^2$

i.e. 0.64 m/s2 in the opposite direction to the original motion, given the original motion is defined as positive (which I did)

edit: Though I'd just mention you can also define certain variables as positive or negative, when told which direction they are in and it will just give you the magnitude.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 31, 2012, 03:10:56 pm
^ The answer is somehow -0.64m/s2

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 31, 2012, 04:59:18 pm
I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle
How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on July 31, 2012, 05:12:14 pm
I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle
How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s

can you please rephrase that -  doesnt make any sense to me haha, or give the question ?
from what you say, it sounds like the question was the time the particle takes to get back to its initial position, which would just be 20 seconds cause thats given..
but i dont think thats the question!
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on July 31, 2012, 06:24:47 pm
I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle
How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s

can you please rephrase that -  doesnt make any sense to me haha, or give the question ?
from what you say, it sounds like the question was the time the particle takes to get back to its initial position, which would just be 20 seconds cause thats given..
but i dont think thats the question!
Part b) Find the time the particle is travelling back towards its original position.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 31, 2012, 08:25:07 pm
Firstly you need to find when it changes direction, i.e. when it comes to rest.
So taking when the acceleration is first applied as $t=0\:seconds$
$a=-0.64\: m\s^{2}$
$u=4\:m/s$
$v=0\:m/s$
$t=?$

\begin{alignedat}{1}v & =u+at
\\ t & =\frac{v-u}{a}
\\ t & =\frac{0-4}{-0.64}
\\ t & =6.25\: seconds
\\ t & =\frac{25}{4}\: seconds
\end{alignedat}

So the total time it was decelerating was 20 seconds, and we know that after 6.25 seconds it changed direction, so the time left is the time that it was travelling back to its starting position.
So that is
\begin{alignedat}{1}t & =20-\frac{25}{4}
\\ t & =\frac{80-25}{4}
\\ t & =\frac{55}{4}\: seconds
\end{alignedat}
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on July 31, 2012, 08:37:39 pm
Hmm, are you sure b^3, I personally think the question is asking when it changes direction?

i.e. 25/4 s
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 31, 2012, 08:46:21 pm
Hmm, are you sure b^3, I personally think the question is asking when it changes direction?

i.e. 25/4 s
Thats the way I originally read it, but the OP did say the answer was 55/4 seconds.
I need help with this question:
A particle travels in a straight line with a constant velocity of 4m/s for 12 seconds.It is then subjected to a constant acceleration in the opposite direction for 20 seconds which returns the particle to its original position.Find:
a)the acceleration of the particle
How would i find the time the particle is traveling back towards its initial position? I thought it was just 12+20=32 s, but the answer is 55/4 s
Its not a well worded question.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on July 31, 2012, 09:00:14 pm
I think it should be worded: At what time does the particle REACH its initial position?

But OK :D
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on July 31, 2012, 09:13:36 pm
I think it should be worded: At what time does the particle REACH its initial position?

But OK :D
For that the answer would be 12+20=32 seconds wouldn't it?, as you need to include the first leg before the acceleration as well, and then still get back to the starting position, not where it turns around. It should be something along the lines of "After coming to rest, how long does it take for the particle to return to its original position?" As that specifies that only the time taken for the return journey is needed.

Probably looking into this too much but anyway...
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on July 31, 2012, 09:23:32 pm
Hmm, are you sure b^3, I personally think the question is asking when it changes direction?

i.e. 25/4 s
Part b) Find the time the particle is travelling back towards its original position.
I would interpret this as it asking for the time period in which the object is travelling in the negative direction i.e. back to its original position from being stationary.

To calculate this you would do as b^3 has done by subtracting the time it decelerated to a momentarily stationary position from the total time of acceleration in the negative direction.

Although, I do agree it is worded rather ambiguously and if it was on a VCAA exam it would be open to skepticism, I believe this is the correct interpretation of the question.

I think it should be worded: At what time does the particle REACH its initial position?

But OK :D
Not quite sure I follow this statement though?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on July 31, 2012, 09:42:20 pm
Part b) Find the time the particle is travelling back towards its original position.
I would interpret this as it asking for the time period in which the object is travelling in the negative direction i.e. back to its original position from being stationary.

I agree, because questions given: 20 seconds which returns particle back to its original position => assumed 20s from the opposite direction to the starting point
If the answer is 12+20=32, the question should be the total time takes the particle to travel back the original position
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 06, 2012, 07:08:27 pm
Could somebody explain to me why root of x2 = |x| ?
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on August 06, 2012, 07:25:10 pm
Could somebody explain to me why root of x2 = |x| ?
Thanks

|x| = + or - x.

If x is 2 for example...

2*2 = 4

If x is -2..

-2*-2 = 4

Therefore (+-2)^2 = x^2, or root(4)=x or +-2=|x| :).
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 06, 2012, 07:44:10 pm
Oh thanks, i kept thinking that |x| meant just +x and was wondering what happened to the -x. Yeh now i understand
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on August 06, 2012, 07:49:46 pm
Oh thanks, i kept thinking that |x| meant just +x and was wondering what happened to the -x. Yeh now i understand

x can be positive or negative but |x| is always positive
Title: Re: Specialist 3/4 Question Thread!
Post by: WhoTookMyUsername on August 06, 2012, 08:05:22 pm
It's funny how one part of one chapter of spec is basically all of physics unit 3 AOS 1 LOL
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 06, 2012, 08:08:24 pm
Yeah, VCE physics is a joke.
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on August 06, 2012, 08:11:29 pm
Yeah, VCE physics is a joke.

Really? Please tell me more about your feelings about Physics, as it is a previously unexplored topic which you have never really gone into specific detail about ;).
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on August 06, 2012, 11:57:18 pm
It's funny how one part of one chapter of spec is basically all of physics unit 3 AOS 1 LOL

Last year I was so scared of this chapter but now I really like kinematics :P I wish I studied Physics unit 3 LOL

Yeah, VCE physics is a joke.

Really? Please tell me more about your feelings about Physics, as it is a previously unexplored topic which you have never really gone into specific detail about ;).

Vegemite is a physics hater so you wont get any interests of this subject from him btw

Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on August 07, 2012, 06:17:40 pm
Could someone not do the question but tell me how I should attempt it?

I know we are supposed to transpose the equation and let x be the subject and use the intervals on the y-axis but transposing the equation to let x be the subject seems to be pretty tedious. On my CAS calculator, I get 4 different equations...

Note: The region described in part c is the area enclosed by the y-axis and the line y=3.

I should say this is a 5 mark question ...
Title: Re: Specialist 3/4 Question Thread!
Post by: xZero on August 07, 2012, 06:31:42 pm
I'm a but rusty with the solid of revolution, you have to make x^2 the subject right? Try square rooting the whole equation then expand x out, I got a pretty neat equation for x^2 after transposing.
Title: Re: Specialist 3/4 Question Thread!
Post by: chris19021 on August 07, 2012, 07:34:24 pm
Hey could u help me with this question
implicity differentiate y2=3xy+x2

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on August 07, 2012, 07:56:06 pm
Hey could u help me with this question
implicity differentiate y2=3xy+x2

Thanks

2y dy/dx=2x+3x.dy/dx+3y
dy/dx=(3y+2x)/(2y-3x)
Title: Re: Specialist 3/4 Question Thread!
Post by: 1i1ii1i on August 12, 2012, 09:42:28 pm
I have 4 questions could anyone help me please i have a SAC tommorow :/
1. On a cattle station there were p head of cattle at time t years after January 1 2005. The populations naturally increases at a rate proportional to p. every year 1000 head of cattle are withdrawn from the herd.
a) show that dp/dt =kp-1000 where k is a constant
b) If the herd initially had 5000 head of cattle, find an expression for t in terms of k and p
c) the population increased to 6000 head of cattle afte 5 years
i) show that 5k=loge((6k-1)/(5k-1))

2. In the main lake of a trout farm the trout population is N at t days afater 1 January 2005, the number of fish harvested on a particular day is proportional to the number of fish in the lake at that time. Every day 100 trout are added to the lake
a) construct a differentialy equation with dN/dt in terms of N and k where k is a constant
b) originally the trout population was 1000. find an expression for t in terms of K and N
c) If the procedure at the farm remains unchanged find the eventual trout population in the lake

3. The water in a hot water tank cools at a rate proportional to (T-t0) where Tdegrees celcius is the temperature of the water
at time t minutes and To the temperature of the surrounding air. When T is 60 the water is cooling at 1degrees celcius per minute
When switched on the heater supplies sufficient heat to raise the temperature by 2degrees celciuseach minute(neglecting heat loss by cooling) If T =20 when the heater is switched on and To=20
a) construct a differential equation for dT/dt as afunction of T (heating and cooling are both taking place)
b) solve the differential equation
c) find the temperature of the water 30minutes after turning it on

4. The rate of growth of a population of iguanas on an islan is given by dW/dt =0.04W where W is the number of iguana alive after
t years. Initially there were 350 iguanas
i) solve the differential equation
ii)give the value of W when t=50
b) if dW/dT =kW and there are 350 iguanas initially find the value of k if the population is to remain constant
c)a more realistic rate of growth for the iguanas is determined by the differential euqation dW/dT=(0.04-0.00005W)W initially there were 350 iguanas
i) solve the differential equation
II) find the population after 50 years

Any help would be appreciated :)
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on August 12, 2012, 09:44:19 pm
To be honest, if you're going to ask a question, you should be specific.

What have you been able to do?

What are you stuck on?

What do you need help on?

It's not that I don't want to do those questions for you, it won't help you to just read solutions!
Title: Re: Specialist 3/4 Question Thread!
Post by: 1i1ii1i on August 12, 2012, 09:47:51 pm
i can't get my answers in the form that the back of the book has it
Title: Re: Specialist 3/4 Question Thread!
Post by: 1i1ii1i on August 12, 2012, 09:49:02 pm
its mainly the constructing the de and solving it, i can't do it the back of the book has weird answers
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on August 12, 2012, 09:55:18 pm
What is your DE that you have written, and what is the form at the back of the book - is this for all the questions?
Title: Re: Specialist 3/4 Question Thread!
Post by: 1i1ii1i on August 12, 2012, 09:57:15 pm
Yes for all of them okay i will show you wwhat i did
Title: Re: Specialist 3/4 Question Thread!
Post by: 1i1ii1i on August 12, 2012, 10:03:20 pm
well thats really awkward....i figured the first 1 out by redoing it =.=
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on August 12, 2012, 10:05:55 pm
its mainly the constructing the de and solving it, i can't do it the back of the book has weird answers

If you cant do it, read the examples in the textbook again and apply the formula.
If you have SAC tomorrow, you should do all of those earlier, dont just leave until the day before the SAC.
Title: Re: Specialist 3/4 Question Thread!
Post by: 1i1ii1i on August 12, 2012, 10:13:20 pm
Wow that's really messed up i figured them all out, sometimes when i come back to a question i get it so weird >.<
um i couldn't figure out how to construct a de for the water tank question i used newtons law of cooling
and did dT/dt=k(T-20)
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on August 13, 2012, 06:21:18 pm
Wow that's really messed up i figured them all out, sometimes when i come back to a question i get it so weird >.<
um i couldn't figure out how to construct a de for the water tank question i used newtons law of cooling
and did dT/dt=k(T-20)

dT/dt=k(T-20)
At T=60, dT/dt=1
Sub: 1=k(60-20) => k=1/40

Remember the question asks: cooling and heating are both taking place.
dT/dt=k (100-T)= (100-T)/40 because water is boiled at 100 degrees and its cooled down over time
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on August 14, 2012, 08:11:05 pm
Hey does anyone have some tricky 'Volumes of solids of revolution' questions? I have a SAC tomorrow that is supposed to feature the topic.
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 14, 2012, 08:16:43 pm
I've got a gold one.

A cissoid $y^2(2a-x)=x^3$ (where $x \ge 0$ and $a>0$) is rotated about it's asymptote to form a volume of revolution. Find this volume in terms of $a$. You don't really need a CAS until the very last part to get the answer. TYPO FIXED, SORRY! D:

:)

edit: another good exercise is finding the volume of a torus, ie. $(y-a)^2 + x^2 = r^2$ when it is rotated around the x-axis to form a solid volume of revolution, in terms of $a$ and $r$.

Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on August 14, 2012, 09:27:59 pm
I've got a gold one.

A cissoid $y^2(2a-3)=x^3$ (where $x \ge 0$ and $a>0$) is rotated about it's asymptote to form a volume of revolution. Find this volume in terms of $a$. You don't really need a CAS until the very last part to get the answer.

:)

edit: another good exercise is finding the volume of a torus, ie. $(y-a)^2 + x^2 = r^2$ when it is rotated around the x-axis to form a solid volume of revolution, in terms of $a$ and $r$.

What's the asymptote???

Yeah I did that earlier today and got a solution for the volume in terms of the inner and outer radii of the torus.
It's a good question nonetheless, maybe someone else can try put up a solution, I already spent some time on it and verified my results so there isn't really any point in me doing it.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 14, 2012, 09:30:35 pm
haha, you're supposed to find that out! The asymptote is $x=2a$ though in case you couldn't find a way (graph it to help though) :)

The solution to the other one is $V=2\pi^2ar^2$
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on August 14, 2012, 09:35:57 pm
for second question:
volume = pi* int^(r)_(-r) (sqrt(r^2 + x^2) + a)^2 dx - pi*int^(r)_(-r) (-sqrt(r^2 + x^2) + a)^2 dx
= pi*int^(r)_(-r) (4a sqrt(r^2+x^2)) dx
= 4*pi*a*int*(r)_(-r) sqrt(r^2 +x^2) dx
the integral we know equals to the area of a semicircle pi/2 r^2
so volume = 4*pi*a*pi/2 * r^2 = 2a*pi^2*r^2

hopefully that's right..
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 14, 2012, 09:36:58 pm
Yep, all good :)
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on August 14, 2012, 09:41:35 pm
I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be $y^2(2a-x)=x^3$ instead of $y^2(2a-3)=x^3$ ?
That would make sense to me but if not, how do you work out that asymptote?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 14, 2012, 09:43:33 pm
I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be $y^2(2a-x)=x^3$ instead of $y^2(2a-3)=x^3$ ?
That would make sense to me but if not, how do you work out that asymptote?

Oh shit, you're completely right. $y^2(2a-x)=x^3$ is the right one, no idea how i copied it wrong from my notes book.

Sorry if I've just wasted your last couple of hours :(
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on August 14, 2012, 10:18:48 pm
I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be $y^2(2a-x)=x^3$ instead of $y^2(2a-3)=x^3$ ?
That would make sense to me but if not, how do you work out that asymptote?

Oh shit, you're completely right. $y^2(2a-x)=x^3$ is the right one, no idea how i copied it wrong from my notes book.

Sorry if I've just wasted your last couple of hours :(

Haha, it's alright, I now know what a cissoid is so I wouldn't call it a waste...

As for the question...

The volume of the cissoid rotated about x=2a is equivalent to the volume of the cissoid, translated by -2a units in the x direction rotated about the y axis

the new equation is thus
$y^2(2a-(x+2a))=(x+2a)^3$
so
$y^2=\frac{-(x+2a)^3}{x}$

I'm a bit stuck now and I'm about to go to bed (2 SACs tomorrow)... do you reckon you could either give me a big hint that will give it away or actually show exactly how to get the answer otherwise I swear I won't be able to sleep haha...
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 14, 2012, 10:36:37 pm
It's a very tough question and I think it's beyond the yr12 course, so don't stress on it.

As for a hint, think of it in halves (top and bottom half). Now think of the top half being made of infinite number of cylinders, such that their "diameter" is $2a-x$ and hence, their individual "height" is $dy$. Using the formula for a cylinder, sub those in and see what happens when you take the sum of those cylinders when $dy$ approaches 0. ie. $\lim_{dy \rightarrow \infty} \sum{(formula \ for \ volume)}$ :)
Title: Re: Specialist 3/4 Question Thread!
Post by: luke_rulz94 on August 15, 2012, 05:34:24 pm
hey guys,

can someone show me how to do this :

A tank has a volume v cm3 at a depth of h m given by v =  pix4h^3/3 + h. The tank is initially empty but water is pumped in at the rate of 0.03 cubic metres per second and flows out from the bottom of the tank at 0.04x root of h cubic metres per second.

1   Find the rate, in metres per second, at which the depth is increasing when the depth is 0.5 metres, giving your answer correct to two significant figures.

2   Express the time taken for the depth to rise to 0.5 metres as a definite integral and evaluate it correct to the nearest second.

3   At what depth does the water ultimately stabilise in the tank?
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on August 15, 2012, 05:52:42 pm
When you are asked to implicitly differentiate something like (2/x)+(1/y)=4 why is it incorrect to get rid o the denominators first ie 2y+x=4xy i know you loset some information regardin thw domaim and range bu t shouldnt it still be similar ouldnt you write down the dom and ran?
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 15, 2012, 06:38:03 pm
hey guys,

can someone show me how to do this :

A tank has a volume v cm3 at a depth of h m given by v =  pix4h^3/3 + h. The tank is initially empty but water is pumped in at the rate of 0.03 cubic metres per second and flows out from the bottom of the tank at 0.04x root of h cubic metres per second.

1   Find the rate, in metres per second, at which the depth is increasing when the depth is 0.5 metres, giving your answer correct to two significant figures.

Is the answer to part a) = 0.0072 m/s ?

Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 15, 2012, 09:23:16 pm
A cricket ball is thrown upwards from the top of a 10m high building at a speed of 2m/s. Find the exact value of the speed at which the ball strikes the ground.

Solution used u=2 , a= -9.8 , s= -10 and got v=10root2

I dont understand why s=-10? I though that whatever you choose as the upwards direction will be opposite sign to the acceleration. so if a is negative should s be positive?
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on August 15, 2012, 09:28:09 pm
assuming rectilinear motion, we can define upwards direction as positive and downwards direction as negative (from the point at which you throw the ball). so the top of the building is 0. using these definitions then the bottom of the building is -10. displacement = final position - initial position = -10 - 0 = -10.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on August 15, 2012, 09:28:40 pm
A cricket ball is thrown upwards from the top of a 10m high building at a speed of 2m/s. Find the exact value of the speed at which the ball strikes the ground.

Solution used u=2 , a= -9.8 , s= -10 and got v=10root2

I dont understand why s=-10? I though that whatever you choose as the upwards direction will be opposite sign to the acceleration. so if a is negative should s be positive?
Thanks
Denote upwards to be positive and downwards to be negative.

u = 2, a = -9.8 and x = -10.

the displacement is negative because the ground is "lower" than the building (assuming we defined the top of the building as our origin), it is clearly a negative displacement.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 15, 2012, 09:30:39 pm
Thanks guys :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 12:50:04 pm
Hey
got some questions i couldnt get out  -- ONLY 6a and 7a
because i can do the rest once i can do part a)
i thought i was doing it right but yeah, nah.... :p
for q6, a)  i got 12(x-5)^2 m/s^2
which is wrong

then for q7a) i used $\frac{d(\frac{v^2}{2})}{dx}$
and got something like $v=\sqrt{18x-\frac{8x^3}{3} - \frac{41}{3}$

Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on August 18, 2012, 12:52:00 pm
Am I missing something but what are the actual questions Bhootnike?
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 12:54:32 pm
oops
forgot to attach
/attached in op!
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on August 18, 2012, 01:00:35 pm
The derivative of velocity is acceleration

EDIT: b^3 just pointed out the rookie mistake I did. Ignore this post.

(we can't use a = dv/dt since we don't have an expression in terms of time).

6a.
$v = 4(x-5)^2$ and when $x = 0, v = 0$

$\frac{dv}{dx} = 12(x-5)^2$is right? Is it possible that the answer was just an expanded out form of it?

7a.

$a = 9 - 4x$, when $x = 2, v=1$ (letting x > 0 being to the right of the origin)

$v = \int (9-4x) \ dx = 9x - 2x^2 + c$

$1 = 9(2) - 2(2)^2 + c = 18 - 8 + c = 10 + c$

$c = 1 - 10 = -9$

$v = 9x - 2x^2 - 9$

?

Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on August 18, 2012, 01:13:47 pm
6a)

a = v * (dv/dx) = (dx/dt)(dv/dx) = dv/dt = a

dv/dx = 12(x-5)^2

a = (4(x-5)^3)*(12(x-5)^2)) = 48(x-5)^5 m/s

7a)

a = 9 - 4x

0.5v^2 = int (9-4x) dx
0.5v^2 = 9x - 2x^2 + c
v^2 = 18x - 4x^2 + c

v = 1, x = 2

1 = 36 - 4(4) + c
1 = 36 - 16 + c
1 = 20 + c
c = -19

v^2 = 18x - 4x^2 - 19
v = (18x - 4x^2 - 19)^0.5

Taken positive root because when x = 2, v = +1

and is defined where 18x - 4x^2 - 19 > 0

-4x^2 + 18x - 19 = 0
4x^2 - 18x + 19 = 0

x = (18 +/- (18^2 - 4(4)(19))^0.5) / 8
x = (1/4)(9 +/- 5^0.5)

Therefore, velocity is defined between (1/4)(9-root5) =< x =< (1/4)(9+root5)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 01:14:45 pm
i got the same ans. as you for q6a,
but the answer is $48(5-x)^5 m/s^2$

LOL.

and for 7, nope
its $\sqrt{-4x^2 + 18x - 19}$

EDIT. What Hancock got :)
thanks guys.
ill try and digest it now!
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 01:19:23 pm
for q7
i just realised...
i accidentally wrote down 9-4x^2 rather than just 9-4x
ffs... no wonder i was getting it wrong

for q6
that's awesome! i didnt even think about that...
your told v, and you can find out dv/dx
so you multiply them together to find a.
genius!

Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 01:45:55 pm
alright
im so bad at this i couldnt do the next one either...

this time i employed the use of v. dv/dx and did it
and i got x = 2log|v| + 3v^2/2  + c
which is not right cos once you find c, and solve for v, it doesnt give the right answer..

Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on August 18, 2012, 01:53:20 pm
a = (1/6)(3v^2 + 2)
v(dv/dx) = (1/6)(3v^2 + 2)
dv/dx = (1/6)(3v + 2/v)

dv/dx = (1/6v)(3v^2 + 2)
FLIP
You didn't flip here and integrated with respect to v, which screwed it up :)

dx/dv = 6v / (3v^2 + 2)

x = int 6v/(3v^2 + 2) .dv
x = ln |3v^2 + 2| + c
x + c = ln |3v^2 + 2|
e^(x+c) = |3v^2 + 2|
e^(x)*e^(c) = |3v^2 + 2|

Let e^c = +/- A
Ae^x = 3v^2 + 2

3v^2 = Ae^(x) - 2
v = 3 when x = 0

3(9) = Ae^0 - 2
27 = A - 2
A = 29

3v^2 = 29e^(x) - 2
v = ([29e^(x) - 2]/3)^0.5
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 03:07:03 pm
this is what i did:

a= (1/6)(3v^2 + 2)
v(dv/dx) = (1/6)(3v^2 + 2)
dv/dx = (1/6)(3v + 2/v)

dv/dx = (1/6v)(3v^2 + 2)

so from here,
i simplified the fraction to get $\frac{v}{2} + \frac{1}{3v}$
i then FLIPPED
to get $\frac{2}{v} + \frac{3v}{1}$
when i integrated that w.r.t v,
i got $2log_ex + \frac{3v^2}{2} +c$

i dont understand how simplifying it to get two fractions, then flipping, gave me a different answer to not simplifying it and flipping initially ?
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on August 18, 2012, 03:10:49 pm
this is what i did:

a= (1/6)(3v^2 + 2)
v(dv/dx) = (1/6)(3v^2 + 2)
dv/dx = (1/6)(3v + 2/v)

dv/dx = (1/6v)(3v^2 + 2)

so from here,
i simplified the fraction to get $\frac{v}{2} + \frac{1}{3v}$
i then FLIPPED
to get $\frac{2}{v} + \frac{3v}{1}$
when i integrated that w.r.t v,
i got $log_ex + \frac{3v^2}{2} +c$

i dont understand how simplifying it to get two fractions, then flipping, gave me a different answer to not simplifying it and flipping initially ?

They problem is that you have to flip after you simplify into one fraction, not flip both.

Example:

2 = 1/2 + 3/2

Does flipping both sides now work?

1/2 =/= 2 + 2/3

However,

2 = 4/2 (simplified RHS fractions into one term)
1/2 = 2/4
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on August 18, 2012, 03:20:28 pm
ahhh dammit'
you cant divide additive expressions ,.....
*facepalm*

thanks man!
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on August 20, 2012, 03:53:38 pm
Not too sure if this is a noob question.

The solutions state that $m_2=tan(\alpha+2\pi/3)$. Why exactly can they say this?

Thanks. :)

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on August 20, 2012, 05:00:44 pm

Now we know that $\tan(\theta)=m$ where $\theta$ is the angle the line makes with the positive direction of the x-axis. Here that angle is $\alpha+\frac{2\pi}{3}$
So $m_{2}=\tan\left(\alpha+\frac{2\pi}{3}\right)$

(originally I looked at this wrongly as if the m's were representing lengths, and tried to overcomplicate it..., might have been what you were thinking too :P)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on August 20, 2012, 06:52:31 pm
Haha, yeah ... I was thinking of lengths.  :-[

Thanks $b^3$. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 21, 2012, 09:30:41 pm
In a lift that is accelerating downwards at 1m/s2, a spring balance shows the apparent weight of the body to be 2.5 kg wt. What would be the reading if the lift was:
a) at rest?
b) accelerating upwards at 2 m/s2

I can do part a) and got 2.784 kg wt, but i don't know how they got 3.35 kg wt for part b)
Can someone show me part b) ?
Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on August 21, 2012, 11:33:47 pm
accelerating upwards at 2 ms^(-2) means there is a force of magnitude 2m acting downwards in addition to the weight force of the obect mg. so anwer is 2m + mg N (plug in the value of m you obtained for part 1).
Title: Re: Specialist 3/4 Question Thread!
Post by: stephanieteddy on August 26, 2012, 06:22:09 pm
Hello, just a quick question:

What exactly are we doing when we resolve vector a in the direction of vector b?

I know how to do it, just not sure why!
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on August 26, 2012, 06:35:39 pm
^ the same way i know how to use a^ but not know what i'm actually doing. I've got so many whys in spesh.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 26, 2012, 06:40:00 pm
An engine of mass 40 tonnes is pulling a truck of mas 8000kg up a plane inclined at theta to the horizontal where sin(theta)=1/8. If the tractive force exerted by the engine is 60000N, calculate blahblahblah

Is there suppose to be frictional force behind the engine and behind the truck? The solutions didn't consider friction but i'm confused as to why?

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on August 26, 2012, 06:49:20 pm
Hello, just a quick question:

What exactly are we doing when we resolve vector a in the direction of vector b?

I know how to do it, just not sure why!

YEAH, if someone can explain this, i'd be very happy =)
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on August 26, 2012, 07:07:50 pm
An engine of mass 40 tonnes is pulling a truck of mas 8000kg up a plane inclined at theta to the horizontal where sin(theta)=1/8. If the tractive force exerted by the engine is 60000N, calculate blahblahblah

Is there suppose to be frictional force behind the engine and behind the truck? The solutions didn't consider friction but i'm confused as to why?

Thanks

Assuming that this is from Essential, it says smooth plane.
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on August 26, 2012, 07:26:26 pm
Hello, just a quick question:

What exactly are we doing when we resolve vector a in the direction of vector b?

I know how to do it, just not sure why!
If you want a kind of intuitive explanation, technically we call it a vector "projection", this name makes more sense as it is essentially just "projecting" the vector's tail onto the other vector and finding an expression for that vector, see diagram: http://en.wikipedia.org/wiki/File:Projection_and_rejection.png, note that a_1 is the "projection" formed.

It is very useful in many applications (especially in linear algebra) because we are often interested in the shapes that are formed when things in 3D space are "projected" onto a 2D plane (ie, shadows from buildings etc). The more technical (and general) definition of vector projections are found here: http://en.wikipedia.org/wiki/Vector_projection#Generalizations
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on August 26, 2012, 09:06:13 pm
An engine of mass 40 tonnes is pulling a truck of mas 8000kg up a plane inclined at theta to the horizontal where sin(theta)=1/8. If the tractive force exerted by the engine is 60000N, calculate blahblahblah

Is there suppose to be frictional force behind the engine and behind the truck? The solutions didn't consider friction but i'm confused as to why?

Thanks

Assuming that this is from Essential, it says smooth plane.
Yeah its from essentials, but i don't see the word "smooth" anywhere in the question.
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on August 27, 2012, 03:34:49 pm
Here's one thats bugging me:

a = sin(t)i+cos(t)j
b = cos(t)i + j

Determine when a and b are at right-angles.

Therefore, I used the dot product and came to this step:

sin(t)cos(t)+cos(t)=0,

and I thought we'd divide both sides by cos and solve for sin, yet, when I look at the answers it does this:
i) cos(t)(sin(t)+1))=0, solving cos(t) =0 and sin(t)+1=0

Is there a reason they've factorised and solved seperately?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on August 27, 2012, 04:05:38 pm
If you divide by $\cos(t)$ then you'll be losing solutions (i.e. the solutions for $\cos(t)=0$, hence why they factored the $\cos(t)$ out and then used the null factor law, then solved seperately, as it keeps these solutions.
Title: Re: Specialist 3/4 Question Thread!
Post by: tony3272 on August 27, 2012, 04:07:19 pm
Here's one thats bugging me:

a = sin(t)i+cos(t)j
b = cos(t)i + j

Determine when a and b are at right-angles.

Therefore, I used the dot product and came to this step:

sin(t)cos(t)+cos(t)=0,

and I thought we'd divide both sides by cos and solve for sin, yet, when I look at the answers it does this:
i) cos(t)(sin(t)+1))=0, solving cos(t) =0 and sin(t)+1=0

Is there a reason they've factorised and solved seperately?
The reason you need to factorise it out is so that you don't get rid of any possible solutions. For example, if you want to solve $x^2 +x=0$ then dividing by x will only leave $x+1=0$.
Hence, you only end up with $x=-1$ and no longer have the solution $x=0$. Doing this with cos(t) is the same idea.
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on August 27, 2012, 06:54:38 pm
how can an object be in two places at once? /Y/ to quantum mechanics!
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on August 27, 2012, 07:55:11 pm
This is almost the exact question to Ringwood Secondary's Specialist SAC last year (and I presume, this year too). But yeah, you just combine the two position vectors for each spinning wheel, and make sure you take into account the rotation direction.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 27, 2012, 07:59:31 pm
What was the question? ??? :O
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on August 27, 2012, 08:03:56 pm
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on August 27, 2012, 08:04:41 pm
Hmm, I think it was deleted actually, nothing wrong with you haha
Title: Re: Specialist 3/4 Question Thread!
Post by: shaiga95 on August 27, 2012, 08:25:24 pm
if sin(x) = 1-a/1+a,0<x<pi/2 express cos(2x) in terms of a
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on August 27, 2012, 08:59:13 pm
if sin(x) = 1-a/1+a,0<x<pi/2 express cos(2x) in terms of a

use double angle formula
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on September 01, 2012, 11:30:36 am
For a question such as, |v|=(13-12sin(t))1/2, why is that when |v|=-1, there is a maximum? From the sin graph i can only see +1 as when the graph is a maximum.

Also, are vectors parallel if they're scalar multiples of eachother, regardless of the signs the i, j or k's carry?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 03:55:52 pm
For a question such as, |v|=(13-12sin(t))1/2, why is that when |v|=-1, there is a maximum? From the sin graph i can only see +1 as when the graph is a maximum.

Also, are vectors parallel if they're scalar multiples of eachother, regardless of the signs the i, j or k's carry?

|v| cant be equal a negative number, maybe they mean v=-1?

Vectors are not parallel if they're scalar multiples of each other, regardless of i,j,k
For example:
a= 2i+j+k => |a|=sqr(6)
b=2sqr(5)i+sqr(2)j+sqr(2)k => |b|=2sqr(6)

|b|=2|a| but vector a is not parallel to vector b

Btw, does anyone know how to type vectors and signs of i,j,k in LaTex? Thanks!

Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on September 01, 2012, 04:12:33 pm
Vectors are not parallel if they're scalar multiples of each other, regardless of i,j,k

Yes they are, the whole definition of parallel is that a is // to b if a = kb where k is a real constant.

Also, are vectors parallel if they're scalar multiples of eachother, regardless of the signs the i, j or k's carry?

only if the signs are the same or opposite. For example, 2i + 2j is parallel to 10i + 10j. We agree on that. The value of k is 5. If the value of k was -5, then 2i + 2j would be parallel to -10i - 10j, in this case, they are all opposite. However, 2i + 2j would not be parallel to 10i - 10j or -10i + 10j.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 04:26:14 pm
Vectors are not parallel if they're scalar multiples of each other, regardless of i,j,k

Yes they are, the whole definition of parallel is that a is // to b if a = kb where k is a real constant.

They are parallel if vector a=k*vector b. From there, you can say their scalar multiples each other

But you cant prove 2 vectors parallel just because their scalar multiples each other
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on September 01, 2012, 05:30:46 pm
They are parallel if vector a=k*vector b. From there, you can say their scalar multiples each other

But you cant prove 2 vectors parallel just because their scalar multiples each other
For parallel vectors, $\mathbf{a} = k \cdot \mathbf{b}$ is true. This is what it means for a vector to be a scalar multiple of another.

Your statement is a bit contradictory, you seem to be saying that they are parallel if $\mathbf{a} = k \cdot \mathbf{b}$, but you can't prove them parallel if they are $\mathbf{a} = k \cdot \mathbf{b}$ That doesn't really make much sense to me.

Note that $\mathbf{a} = k \cdot \mathbf{b}$ does imply that $|\mathbf{a}| = k|\mathbf{b}|$ for when k is greater than 0. $|\mathbf{a}| = k|\mathbf{b}|$ being true doesn't imply that they're parallel vectors. The example you gave in your previous post is an example of this. I think that's what you're getting at, and I think the confusion is whether we're talking about the vectors a and b, or if we're talking about the magnitudes of vectors a and b.

For parallel vectors, we are talking about the vectors.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 05:59:02 pm
They are parallel if vector a=k*vector b. From there, you can say their scalar multiples each other

But you cant prove 2 vectors parallel just because their scalar multiples each other
For parallel vectors, $\mathbf{a} = k \cdot \mathbf{b}$ is true. This is what it means for a vector to be a scalar multiple of another.

Your statement is a bit contradictory, you seem to be saying that they are parallel if $\mathbf{a} = k \cdot \mathbf{b}$, but you can't prove them parallel if they are $\mathbf{a} = k \cdot \mathbf{b}$ That doesn't really make much sense to me.

Note that $\mathbf{a} = k \cdot \mathbf{b}$ does imply that $|\mathbf{a}| = k|\mathbf{b}|$ for when k is greater than 0. $|\mathbf{a}| = k|\mathbf{b}|$ being true doesn't imply that they're parallel vectors. The example you gave in your previous post is an example of this. I think that's what you're getting at, and I think the confusion is whether we're talking about the vectors a and b, or if we're talking about the magnitudes of vectors a and b.

For parallel vectors, we are talking about the vectors.

Yeah, I was talking about multiplies of magnitudes

If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallel
Title: Re: Specialist 3/4 Question Thread!
Post by: studynotes on September 01, 2012, 06:20:33 pm
can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 06:38:35 pm
can i please have help with this definite integral

0 to pi/4 tan^3(x) dx

$tan^3(x)=tan(x) \times tan^2(x) =tan(x) \times (sec^2(x)-1)$

Let tan(x)=u, find $\frac{du}{dx}$

Afterthat you integrate function of u

You can do it by yourself then.

Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on September 01, 2012, 06:57:07 pm
But you cant prove 2 vectors parallel just because their scalar multiples each other

That's the only way to prove that 2 vectors are parallel, that they are scalar multiples of eachother. If you're saying that you CAN'T prove 2 vectors are parallel to eachother by this method then how do you show that 2 vectors are parallel?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 07:41:20 pm
But you cant prove 2 vectors parallel just because their scalar multiples each other

That's the only way to prove that 2 vectors are parallel, that they are scalar multiples of eachother. If you're saying that you CAN'T prove 2 vectors are parallel to eachother by this method then how do you show that 2 vectors are parallel?

They are parallel if vector a=k*vector b. From there, you can say their scalar multiples each other

But you cant prove 2 vectors parallel just because their scalar multiples each other
For parallel vectors, $\mathbf{a} = k \cdot \mathbf{b}$ is true. This is what it means for a vector to be a scalar multiple of another.

Your statement is a bit contradictory, you seem to be saying that they are parallel if $\mathbf{a} = k \cdot \mathbf{b}$, but you can't prove them parallel if they are $\mathbf{a} = k \cdot \mathbf{b}$ That doesn't really make much sense to me.

Note that $\mathbf{a} = k \cdot \mathbf{b}$ does imply that $|\mathbf{a}| = k|\mathbf{b}|$ for when k is greater than 0. $|\mathbf{a}| = k|\mathbf{b}|$ being true doesn't imply that they're parallel vectors. The example you gave in your previous post is an example of this. I think that's what you're getting at, and I think the confusion is whether we're talking about the vectors a and b, or if we're talking about the magnitudes of vectors a and b.

For parallel vectors, we are talking about the vectors.

Yeah, I was talking about multiplies of magnitudes

If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallel
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on September 01, 2012, 07:53:46 pm
If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallel
I think this is just coming down to a matter of slightly different definitions and terminology really.

I would define parallel as "The non-zero vectors u and v are said to be parallel if there exists k ∈ R\{0} such that u = kv" (quoted from essentials) or in other words "Two vectors are parallel if they have the same direction or are in exactly opposite directions." (quoted from http://tutorial.math.lamar.edu/Classes/CalcII/VectorArithmetic.aspx)

Going by that definition, "anti-parallel" vectors are just parallel vectors that are in the opposite direction (http://en.wikipedia.org/wiki/Antiparallel_(mathematics)#Antiparallel_vectors). You also see the terminology "like parallel vectors" (facing the same direction) and "unlike parallel vectors" (facing the opposite direction) being used to refer to the same thing.

I did a bit of searching but I can't really find a source that doesn't consider those vectors that face in the opposite direction to not be parallel.

That might not make sense if you're defining parallel vectors as only being scalar multiples that face that same direction - but going by most sources it seems the definition of a parallel vector is that it would be scalar multiples that have the same or opposite direction (in other words just $\mathbf{a} = k\mathbf{b}$.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 08:06:21 pm
If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallel
I think this is just coming down to a matter of slightly different definitions and terminology really.

I would define parallel as "The non-zero vectors u and v are said to be parallel if there exists k ∈ R\{0} such that u = kv" (quoted from essentials) or in other words "Two vectors are parallel if they have the same direction or are in exactly opposite directions." (quoted from http://tutorial.math.lamar.edu/Classes/CalcII/VectorArithmetic.aspx)

Going by that definition, "anti-parallel" vectors are just parallel vectors that are in the opposite direction (http://en.wikipedia.org/wiki/Antiparallel_(mathematics)#Antiparallel_vectors). You also see the terminology "like parallel vectors" (facing the same direction) and "unlike parallel vectors" (facing the opposite direction) being used to refer to the same thing.

I did a bit of searching but I can't really find a source that doesn't consider those vectors that face in the opposite direction to not be parallel.

That might not make sense if you're defining parallel vectors as only being scalar multiples that face that same direction - but going by most sources it seems the definition of a parallel vector is that it would be scalar multiples that have the same or opposite direction (in other words just $\mathbf{a} = k\mathbf{b}$.

Dont make it so complicated. They are both parallel, anti-parallel just gives extra info that they are in opposite direction
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on September 01, 2012, 08:26:36 pm
To me 'parallel' just means linearly dependent, i.e. one is a scalar multiple of the other, whether negative or positive.

can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on September 01, 2012, 08:32:46 pm
But you cant prove 2 vectors parallel just because their scalar multiples each other

That's the only way to prove that 2 vectors are parallel, that they are scalar multiples of eachother. If you're saying that you CAN'T prove 2 vectors are parallel to eachother by this method then how do you show that 2 vectors are parallel?

They are parallel if vector a=k*vector b. From there, you can say their scalar multiples each other

But you cant prove 2 vectors parallel just because their scalar multiples each other
For parallel vectors, $\mathbf{a} = k \cdot \mathbf{b}$ is true. This is what it means for a vector to be a scalar multiple of another.

Your statement is a bit contradictory, you seem to be saying that they are parallel if $\mathbf{a} = k \cdot \mathbf{b}$, but you can't prove them parallel if they are $\mathbf{a} = k \cdot \mathbf{b}$ That doesn't really make much sense to me.

Note that $\mathbf{a} = k \cdot \mathbf{b}$ does imply that $|\mathbf{a}| = k|\mathbf{b}|$ for when k is greater than 0. $|\mathbf{a}| = k|\mathbf{b}|$ being true doesn't imply that they're parallel vectors. The example you gave in your previous post is an example of this. I think that's what you're getting at, and I think the confusion is whether we're talking about the vectors a and b, or if we're talking about the magnitudes of vectors a and b.

For parallel vectors, we are talking about the vectors.

Yeah, I was talking about multiplies of magnitudes

If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallel

I'm confused by this :/ Surely if k is negative the vectors are still parallel and the proof Paul is suggesting is valid.
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on September 01, 2012, 08:38:02 pm
Dont make it so complicated. They are both parallel, anti-parallel just gives extra info that they are in opposite direction
Oh my bad, I misinterpreted what you meant then. I thought you were implying that 'anti-parallel' wasn't parallel at all.

Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 01, 2012, 09:05:53 pm

can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$

Wow, I didnt think of the method let y=cos (x) :)

What I do is

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$let u=tan(x) => \frac{du}{dx}=sec^2(x) => dx=\frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1) \frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(1- \frac{1}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{\frac{\pi }{4}}\tan (x) - \frac{tan(x)}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{1} u - \frac{u}{1+u^2} \mathrm{d}u$

$= [\frac{u^2}{2} - \frac{1}{2} ln(u^2+1)]_ 0^1$

$= \frac{1}{2}-\frac{1}{2}ln(2)+\frac{1}{2}ln(1)$

$= \frac{1}{2}(1-ln(2))$

Edit: correct LaTex :P
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on September 01, 2012, 09:55:07 pm
^I didn't think of that either, haha. Seems quicker.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on September 01, 2012, 10:22:37 pm
I'm confused by this :/ Surely if k is negative the vectors are still parallel and the proof Paul is suggesting is valid.

Yeah, I'm confused as well. Some things I want to point out:

1) Jenny, when you say scalar multiple, it is automatically assumed that you mean scalar multiple of a vector - like why would you want the scalar multiple of the magnitude - that doesn't make sense.

2) Don't worry about parallel or anti-parallel, you're getting too caught up in words - just stick to the basic concept, parallel is parallel, it means the same gradient, the same slope, the same angle made with a particular axis.

3) Why do you keep making things more complicated, just stick to the basic rules and you'll be fine, just note that when a vector is scaled by a constant amount (i.e. a scalar) it will be parallel to the original vector, i.e. a = kb, then a//b - don't worry about parallel, anti-parallel and all that other bs, just stick to the basic definition that is used in VCE.
Title: Re: Specialist 3/4 Question Thread!
Post by: BlueSky_3 on September 01, 2012, 11:20:58 pm
^I didn't think of that either, haha. Seems quicker.

Want to see something even quicker?

Well: S tan(x) x ( sec^2(x) - 1)dx, and let sec^2(x) -1 = u
Then dx= 2tan(x)sec^2(x) du
Next, I = .5S u/(u+1) du
= .5S (u+1-1)/(u+1) du
= .5S 1 -1/(u+1) du
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on September 02, 2012, 01:09:47 am

can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$

Wow, I didnt think of the method let y=cos (x) :)

What I do is

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$let u=tan(x) => \frac{du}{dx}=sec^2(x) => dx=\frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1) \frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(1- \frac{1}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{\frac{\pi }{4}}\tan (x) - \frac{tan(x)}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{1} u - \frac{u}{1+u^2} \mathrm{d}u$

$= [\frac{u^2}{2} - \frac{1}{2} ln(u^2+1)]_ 0^1$

$= \frac{1}{2}-\frac{1}{2}ln(2)+\frac{1}{2}ln(1)$

$= \frac{1}{2}(1-ln(2))$

Edit: correct LaTex :P

someone's enjoying the use of latex :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 02, 2012, 08:50:32 pm
Dont make it so complicated. They are both parallel, anti-parallel just gives extra info that they are in opposite direction
Oh my bad, I misinterpreted what you meant then. I thought you were implying that 'anti-parallel' wasn't parallel at all.

Nah, it was my fault actually, I didnt expain clearly and actually, anti-parallel means they are in opposite sense, not direction

Some things I want to point out:

1) Jenny, when you say scalar multiple, it is automatically assumed that you mean scalar multiple of a vector - like why would you want the scalar multiple of the magnitude - that doesn't make sense.

In my understanding, scalar multiplies just means "times with a number" (of a quantity, having only magnitude, no direction. In this situation, this number is k. Thus, why cant I say |a|=k|b|?

Anyway, its just due to my misunderstanding the question. I read "regardless i,j,k" in the question and I thought without i,j,k => implies no direction, no sense at all => regardless vectors

2) Don't worry about parallel or anti-parallel, you're getting too caught up in words - just stick to the basic concept, parallel is parallel, it means the same gradient, the same slope, the same angle made with a particular axis.

I just wanna be specific a bit, vectors have direction. Without direction, its just a line
"Parallel means same gradient, same slope, same angle made with a particular axis" is applied for gradient of lines or parallel vectors in the same direction, same sense

Gradients of parallel vectors=$(f_x, f_y,f_z)$
The tangent plane to the surface given by f(x,y,z) while with an antiparallel vector given by f(-x,-y,-z)
If they are anti-parallel, they will have opposite slope and gradient, NOT same

3) Why do you keep making things more complicated, just stick to the basic rules and you'll be fine, just note that when a vector is scaled by a constant amount (i.e. a scalar) it will be parallel to the original vector, i.e. a = kb, then a//b - don't worry about parallel, anti-parallel and all that other bs, just stick to the basic definition that is used in VCE.

I dont wanna make things complicated but in VCE, parallel vectors are not defined accurately

Parallel vectors should be considered when they have: same direction + same sense
Antiparallel vectors have: same direction +opposite sense

Direction just means a horizontal line
Sense: can be to the right or left

What I said before about k, I meant: +if k is positive, they are parallel (same direction, same sense)
+if k is negative, they are antiparallel (same direction, opposite sense)

Anyway, as you said, just stick with VCE level, so just ignore what I mentioned earlier about antiparallel.
Its too much and unecessary for the exams

someone's enjoying the use of latex :P

Haha, actually I like LaTex now  :D
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 03, 2012, 08:51:31 pm
Can somebody help me with this:
A train is moving with uniform acceleration is observed to take 20s and 30s to travel successive half kilometres. How much farther will it travel before coming to rest if the acceleration remains constant?

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: InsaneMcFries on September 03, 2012, 09:10:59 pm
At the moment Dynamics is doing my head in. My first problem is a simple pulley problem with two particles.

The mass of one particle is 1.5kg and the other is 2kg. They hang vertically around a pulley. There are two Tension forces, one on each side of the pulley rope.

The first question is to find the value of the tension.

My main issue is I don't understand how to set out the vector equations for this setup, as I set them up as T-1.5g=1.5a and T-2g=2a.

Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 03, 2012, 09:19:51 pm
Can somebody help me with this:
A train is moving with uniform acceleration is observed to take 20s and 30s to travel successive half kilometres. How much farther will it travel before coming to rest if the acceleration remains constant?

Thanks

Let's say after 20 s, the train has travelled 0.5 km = 500 m, then after 20 + 30 = 50 s, it has travelled 1000 m. Consider the initial time and distance to both be 0. Use the equation s = ut + 1/2at^2. Your unknowns are u and a, which you can find since you have two sets of values for s and t (i.e. two equations). Once you have u and a, use v^2 = u^2 + 2as to find s, then subtract 1000 m.

At the moment Dynamics is doing my head in. My first problem is a simple pulley problem with two particles.

The mass of one particle is 1.5kg and the other is 2kg. They hang vertically around a pulley. There are two Tension forces, one on each side of the pulley rope.

The first question is to find the value of the tension.

My main issue is I don't understand how to set out the vector equations for this setup, as I set them up as T-1.5g=1.5a and T-2g=2a.

You have to decide which direction is positive.
Then the equations should be: T-1.5g=1.5a and 2g-T=2a
Or: 1.5g-T=1.5a and T-2g=2a

Solve it, you get T=16.8N
Title: Re: Specialist 3/4 Question Thread!
Post by: InsaneMcFries on September 03, 2012, 10:21:59 pm
Oh right! I didn't know I had to decide a direction. Thanks! :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on September 05, 2012, 08:54:31 am
The solutions to this question seem quite ... vague ... (Either that or I am quite stupid  ;))

I got two marks, out of 4, for this question. (Kilbaha 2009 Specialist Exam 1)

I converted each .... form (Can't think of correct terminology), into polar form.

$z=2cis(\pi/6)$ and $z=2cis(-\pi/6)$

Hence, $(2cis(\pi/6))^m-(2cis(-\pi/6))^m=0$

'Expanded' by De Moivre's Theorem.

$2^mcis(m\pi/6)-2^m(cis(-m\pi/6)=0$

I really do not understand what the solutions did after 'expanding'.

Thanks for any help. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: tony3272 on September 05, 2012, 10:05:14 am
The solutions to this question seem quite ... vague ... (Either that or I am quite stupid  ;))

I got two marks, out of 4, for this question. (Kilbaha 2009 Specialist Exam 1)

I converted each .... form (Can't think of correct terminology), into polar form.

$z=2cis(\pi/6)$ and $z=2cis(-\pi/6)$

Hence, $(2cis(\pi/6))^m-(2cis(-\pi/6))^m=0$

'Expanded' by De Moivre's Theorem.

$2^mcis(m\pi/6)-2^m(cis(-m\pi/6)=0$

I really do not understand what the solutions did after 'expanding'.

Thanks for any help. :)
Where abouts did you stop in your working? I'll just start at the last line you posted

You have $2^m(cis(\frac{m\pi}{6})-cis(-\frac{m\pi}{6}))=0$
And since we know $2^m\neq 0$ we can say
$cis(\frac{m\pi}{6})-cis(-\frac{m\pi}{6})=0$

$\therefore cos(\frac{m\pi}{6})+isin(\frac{m\pi}{6})-cos(-\frac{m\pi}{6})-isin(-\frac{m\pi}{6})=0$

$cos(\frac{m\pi}{6})+isin(\frac{m\pi}{6})-(cos(\frac{m\pi}{6}))-(-isin(\frac{m\pi}{6}))=0$

$\therefore 2isin(\frac{m\pi}{6})=0$

$\therefore sin(\frac{m\pi}{6})=0$   solve to find m

Is this what you were looking for?
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on September 05, 2012, 03:51:05 pm
Yep, that's exactly it!

Thanks tony. :)

Title: Re: Specialist 3/4 Question Thread!
Post by: Sweetpotato on September 07, 2012, 12:48:40 pm

can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$

Wow, I didnt think of the method let y=cos (x) :)

What I do is

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$let u=tan(x) => \frac{du}{dx}=sec^2(x) => dx=\frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1) \frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(1- \frac{1}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{\frac{\pi }{4}}\tan (x) - \frac{tan(x)}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{1} u - \frac{u}{1+u^2} \mathrm{d}u$

$= [\frac{u^2}{2} - \frac{1}{2} ln(u^2+1)]_ 0^1$

$= \frac{1}{2}-\frac{1}{2}ln(2)+\frac{1}{2}ln(1)$

$= \frac{1}{2}(1-ln(2))$

Edit: correct LaTex :P

someone's enjoying the use of latex :P

Hey help with this qs please!!!!!!!!!

A particle P of unit mass moves on the positive x axis. At time t, the velocity of the
particle is v and the force F acting on the particle is given by
F= 50/(25+v) for 0<t< and including 50
F=  −v^2/1000 for t>50
Initially the particle is at rest at the origin O.
a Show that v=50 when t=50.
b Find the distance of P from O when v = 50.
c Find the distance of P from O when v=25 and t>50.
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on September 07, 2012, 08:59:37 pm
When you have these types of questions:

A particle moves in a straight line. When its displacement from a fixed origin is x m, its velocity is v m/s and its acceleration is a m/s^2.

Given that a=16x and that v = -5 when x = 0, the relation between v and x is?

I got two answers for v but I used the positive square root instead of the negative square root. How does one work out if it's + or -?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 07, 2012, 10:02:28 pm
When you have these types of questions:

A particle moves in a straight line. When its displacement from a fixed origin is x m, its velocity is v m/s and its acceleration is a m/s^2.

Given that a=16x and that v = -5 when x = 0, the relation between v and x is?

I got two answers for v but I used the positive square root instead of the negative square root. How does one work out if it's + or -?

The original velocity is -5 so the motion of particle is negative
Title: Re: Specialist 3/4 Question Thread!
Post by: monkeywantsabanana on September 07, 2012, 10:08:52 pm
When you have these types of questions:

A particle moves in a straight line. When its displacement from a fixed origin is x m, its velocity is v m/s and its acceleration is a m/s^2.

Given that a=16x and that v = -5 when x = 0, the relation between v and x is?

I got two answers for v but I used the positive square root instead of the negative square root. How does one work out if it's + or -?

The original velocity is -5 so the motion of particle is negative

danke!
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on September 07, 2012, 10:41:30 pm
Trig question here,
Find the maximum value of 4sin(3x)+3cos(3x)-3

I have no idea how to do this haha, thanks for the help!
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 07, 2012, 11:09:13 pm
Trig question here,
Find the maximum value of 4sin(3x)+3cos(3x)-3

I have no idea how to do this haha, thanks for the help!

There are 2 ways of doing this

Methods 1: Put in CAS, find max is 2 :P

Method 2: Find derivative

$\frac{dy}{dx}=12 cos(3x) - 9sin(3x) = 0$ so that its the maximum

$12 cos(3x)=9 sin(3x)$

$tan(3x)= \frac{12}{9}= \frac{4}{3}$

So you need to draw a triangle to see that it has 3 sides with lengths 3,4,5

$sin(3x)= \frac{4}{5}$

$cos(3x)= \frac{3}{5}$

Maximum of $4 sin(3x) + 3 cos(3x) - 3 = 4 \times \frac{4}{5} + 3 \times \frac{3}{5} -3 = 2$

Thus, maximum value of function is 2
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on September 07, 2012, 11:19:54 pm
There are 2 ways of doing this

Methods 1: Put in CAS, find max is 2 :P

Method 2: Find derivative
Alternatively, you could find it the same way you would convert the function into a single sine or cosine (but only do the amplitude related part);

So the amplitude is given by;

$A^2 = 4^2 + 3^2$

$A = 5$

Therefore max $= 5 - 3 = 2$
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 07, 2012, 11:26:04 pm
There are 2 ways of doing this

Methods 1: Put in CAS, find max is 2 :P

Method 2: Find derivative
Alternatively, you could find it the same way you would convert the function into a single sine or cosine (but only do the amplitude related part);

So the amplitude is given by;

$A^2 = 4^2 + 3^2$

$A = 5$

Therefore max $= 5 - 3 = 2$

yeah, thats true we can square the amplitude instead
Its so much easier
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on September 08, 2012, 02:24:44 pm
Didn't know the amplitude was given by $\sqrt{A^2+B^2}$
Thanks guys :)
Title: Re: Specialist 3/4 Question Thread!
Post by: WhoTookMyUsername on September 08, 2012, 05:02:03 pm
D: what's the full reasoning behind that?  I don't get it xD
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on September 08, 2012, 05:09:50 pm
D: what's the full reasoning behind that?  I don't get it xD

It's not on the spesh course, but if two sinusoidal curves are in phase, they will sum to give another sinusoidal curve.
Title: Re: Specialist 3/4 Question Thread!
Post by: WhoTookMyUsername on September 08, 2012, 05:42:05 pm
Ahk thanks, so this would function solely as a checking method?
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on September 08, 2012, 06:05:16 pm
It's not on the spesh course, but if two sinusoidal curves are in phase, they will sum to give another sinusoidal curve.
Oh really? Oops. I just assumed it was because I had to learn it for my foundation maths (specialist equivalent) bridging unit for engineering.

D: what's the full reasoning behind that?  I don't get it xD
Alright, considering it's not on the course I'll just show you the reasoning for a bit of fun.

Let's say you have the function $4 \sin(3x)+3 \cos(3x)-3$ and you want it in the form $A \sin(3x+\alpha)-3$.

$A \sin(3x+\alpha)-3 = \sin(3x) \cos(\alpha)+\cos(3x) \sin(\alpha) - 3$ (using sine addition formula)

So we have;

$A \sin(3x) \cos(\alpha)+A \cos(3x) \sin(\alpha) - 3 = 4 \sin(3x)+3 \cos(3x) - 3$

or $A \cos(\alpha) \sin(3x)+A \sin(\alpha) \cos(3x) = 4 \sin(3x)+3 \cos(3x)$

For the above to be true;

$A \cos(\alpha) = 4$

$A \sin(\alpha) = 3$

Once you have these you can just use generic formula to work out $A$ and $\alpha$, but I'll show the reasoning below.

To obtain the amplitude you can eliminate $\alpha$ by squaring both sides of each of the equations and then adding them together;

$(A \cos(\alpha))^2 + (A \sin(\alpha))^2 = 4^2 + 3^2$

$A^2(\cos^2(\alpha) + \sin^2(\alpha)) = 4^2 + 3^2$

$A^2(1) = 4^2+3^2$

$A = 5$ (as amplitude is always positive)

To find the phase angle you divide one equation by the other;

$\tan(\alpha) = \frac{3}{4}$

$\alpha = \tan^{-1} \left(\frac{3}{4}\right)$

Both sine and cosine are positive, so the phase angle must be located in the first quadrant, which it is and hence no addition or subtraction of $\pi$ is needed, as this is the principal phase angle.

So yeah, you end up with;

$4 \sin(3x)+3 \cos(3x)-3 = 5 \sin\left(3x + \tan^{-1} \left(\frac{3}{4}\right)\right) - 3$

I'm actually pretty surprised it's not on the specialist course to be honest  :-X

edit: removed the italics on sin/cos/tan for b^3 :P
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on September 08, 2012, 06:13:33 pm
We got taught that in spesh last year :O
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on September 08, 2012, 06:18:21 pm
We got taught that in spesh last year :O
That's because you went to MHS :P
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on September 08, 2012, 06:20:14 pm
So does Bazza haha :P

I think my teacher was just an awesome crazy Russian who liked going beyond the course :D
Title: Re: Specialist 3/4 Question Thread!
Post by: BlueSky_3 on September 08, 2012, 09:56:08 pm
Phhh...learnt that in GMA  ;)
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on September 10, 2012, 04:28:15 pm
Would someone be able to show me how they would do this question?

I've done it but I feel as though my working is unconventional...

$sin(\pi/10)=(\sqrt(5)-1)/4$. Find $sec(\pi/5)$ in the form $a\sqrt(5)+b$.

Thank you. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on September 10, 2012, 05:13:51 pm
Would someone be able to show me how they would do this question?

I've done it but I feel as though my working is unconventional...

$sin(\pi/10)=(\sqrt(5)-1)/4$. Find $sec(\pi/5)$ in the form $a\sqrt(5)+b$.

Thank you. :)

sec(pi/5) = 1/cos(pi/5) = 1/cos(2*pi/10)
= 1/1-2sin^2(pi/10), and then you just sub in your value for sin(pi/10) and it's just algebra. Eventually you get sec(pi/5) = root(5) - 1, where a=1 and b=-1.

The only other way I can think to do it is to sketch a triangle and use one of the cos double angle formulas with cos in them, but that would be dumb :P.
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on September 10, 2012, 05:17:20 pm
Would someone be able to show me how they would do this question?

I've done it but I feel as though my working is unconventional...

$sin(\pi/10)=(\sqrt(5)-1)/4$. Find $sec(\pi/5)$ in the form $a\sqrt(5)+b$.

Thank you. :)
use double angle formula to find 1/1-2sin^2(pi/10) sub sin(pi/10) in and there you go. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: luke_rulz94 on September 10, 2012, 07:01:43 pm
Can someone please show me how to do these two questions :

A body of mass 10 kg is attached to a second body of mass 14 kg by a light string passing over a smooth pulley attached to the edge of a horizontal table. The lighter body is on the table and the heavier body hangs vertically. The coefficient of friction between the lighter body and the table is 0.25. The system is held at rest by a horizontal force F applied to the lighter body. What is the least value of F for the body to be in limiting equilibrium?

4   A box of mass 4 kg lies on a level floor. The coefficient of friction between the floor and the box is 0.45. A horizontal force of 15 N is applied to the box. Show that the box does not move.

Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on September 10, 2012, 07:47:45 pm
Can someone please show me how to do these two questions :

A body of mass 10 kg is attached to a second body of mass 14 kg by a light string passing over a smooth pulley attached to the edge of a horizontal table. The lighter body is on the table and the heavier body hangs vertically. The coefficient of friction between the lighter body and the table is 0.25. The system is held at rest by a horizontal force F applied to the lighter body. What is the least value of F for the body to be in limiting equilibrium?

4   A box of mass 4 kg lies on a level floor. The coefficient of friction between the floor and the box is 0.45. A horizontal force of 15 N is applied to the box. Show that the box does not move.

Clue: For the system to be in limiting equilibrium / your box not moving, your resultant force has to = 0.
That means that if you resolve your forces (for example) vertically and horizontally, Upwards Force = Downwards Force and Right Force = Frictional Force. (alternatively, Upwards Force - Downwards Force = 0, Right Force - Frictional Force = 0).

See if that helps you at all.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on September 10, 2012, 08:00:33 pm
EDIT: Teewreck's given you the hints, but since I've made the diagram and typed all this out..., I'm going to post it anyway, although try with teewreck's hints before you have a look below.

First of all, diagrams are your best friend in this situation :)
(http://i1082.photobucket.com/albums/j373/mclaren200800/dynamics-1.png)
Labelling all the forces, assuming that if block 2 were moving it would be moving downwards. We have a normal and weight force acting on block 1,as well as the force F that is to hold it in equlibirum and a frictional force. This frictional force will be in the opposite direction to which the block will travel. So block 1 will move to the right if its not held, so the fricitonal force is to the left. There is a weigth force acting on block 2, now there is also tension in the rope over the pulley.

So now we want it to be in limiting equilibirum, that is it is on the point of moving. So our net force on each block is 0 N.

The forces acting on block 2.
\begin{alignedat}{1}m_{2}g-T & =m_{2}a
\\ 14g-T & =0
\\ T & =14g....[1]
\end{alignedat}

The forces acting on block 1 in the  vertical direction.
\begin{alignedat}{1}N & =m_{1}g
\\ N & =10g
\end{alignedat}

The forces acting on block 1 in the horizontal direction.
\begin{alignedat}{1}T-F-Fr & =m_{1}a
\\ T-F-\mu m_{1}g & =m_{1}a
\\ 14g-F-(0.25\times10\times g) & =0
\\ F & =14g-2.5g
\\ F & =22.5g
\\ F & =112.7\: N
\end{alignedat}

For the second question, again draw a diagram out (including your forces). The key for this one is that if the force applied to the block is not greater than $\mu N$, then the block won't move.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on September 10, 2012, 08:28:40 pm
^Why isn't this guy writing a book? The diagram alone is more than impressive! :O
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 10, 2012, 09:33:59 pm
EDIT: Teewreck's given you the hints, but since I've made the diagram and typed all this out..., I'm going to post it anyway, although try with teewreck's hints before you have a look below.

First of all, diagrams are your best friend in this situation :)
(http://i1082.photobucket.com/albums/j373/mclaren200800/dynamics-1.png)
Labelling all the forces, assuming that if block 2 were moving it would be moving downwards. We have a normal and weight force acting on block 1,as well as the force F that is to hold it in equlibirum and a frictional force. This frictional force will be in the opposite direction to which the block will travel. So block 1 will move to the right if its not held, so the fricitonal force is to the left. There is a weigth force acting on block 2, now there is also tension in the rope over the pulley.

So now we want it to be in limiting equilibirum, that is it is on the point of moving. So our net force on each block is 0 N.

The forces acting on block 2.
\begin{alignedat}{1}m_{2}g-T & =m_{2}a
\\ 14g-T & =0
\\ T & =14g....[1]
\end{alignedat}

The forces acting on block 1 in the  vertical direction.
\begin{alignedat}{1}N & =m_{1}g
\\ N & =10g
\end{alignedat}

The forces acting on block 1 in the horizontal direction.
\begin{alignedat}{1}T-F-Fr & =m_{1}a
\\ T-F-\mu m_{1}g & =m_{1}a
\\ 14g-F-(0.25\times10\times g) & =0
\\ F & =14g-2.5g
\\ F & =22.5g
\\ F & =112.7\: N
\end{alignedat}

For the second question, again draw a diagram out (including your forces). The key for this one is that if the force applied to the block is not greater than $\mu N$, then the block won't move.

You should write Spesh and Methods books for 2013 students, b^3!!!!!!!!!
The diagram is so awesome.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 12, 2012, 07:30:42 pm
I'm doing an extended response question for differential eqns and it asks for a prediction of a flock of emus after 5 yrs. I got 1296.42 but do i round up to 1297 or round down to 1296? The answers rounded down but i disagree as there is 0.42 of an emu left which can't just disappear...

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on September 12, 2012, 07:52:50 pm
I'm doing an extended response question for differential eqns and it asks for a prediction of a flock of emus after 5 yrs. I got 1296.42 but do i round up to 1297 or round down to 1296? The answers rounded down but i disagree as there is 0.42 of an emu left which can't just disappear...

Thanks
But then again you can't pull out 0.58 of an emu from no where :P With these kinds of questions you need to round down even if the decimal is larger than 0.5
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on September 12, 2012, 07:53:22 pm
They would be specific about an answer on an exam though, i.e. "rounded down" or "to the nearest ten X." So don't worry too much.
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on September 12, 2012, 09:56:00 pm
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 12, 2012, 10:02:48 pm
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??

$x=2e^{0.3t} cos(2t) \implies cos(2t) = \frac{x}{2e^{0.3t}}$

$y=2e^{0.3t} sin(2t) \implies sin(2t) = \frac{y}{2e^{0.3t}}$

Then you use formula

$(sin(2x))^2 +(cos(2x))^2 = 1$
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on September 12, 2012, 10:06:29 pm
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??

$x=2e^{0.3t} cos(2t) \implies cos(2t) = \frac{x}{2e^{0.3t}}$

$y=2e^{0.3t} sin(2t) \implies sin(2t) = \frac{y}{2e^{0.3t}}$

Then you use formula

$(sin(2x))^2 +(cos(2x))^2 = 1$
That would give the answer in terms of t though not x. wouldn't it?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 12, 2012, 10:25:40 pm
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??

$x=2e^{0.3t} cos(2t) \implies cos(2t) = \frac{x}{2e^{0.3t}}$

$y=2e^{0.3t} sin(2t) \implies sin(2t) = \frac{y}{2e^{0.3t}}$

Then you use formula

$(sin(2x))^2 +(cos(2x))^2 = 1$
That would give the answer in terms of t though not x. wouldn't it?

lol  stupid Jenny

I'm thinking we write 2e^0.3t in term of x and y then equate them
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on September 12, 2012, 10:28:01 pm
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??

$x=2e^{0.3t} cos(2t) \implies cos(2t) = \frac{x}{2e^{0.3t}}$

$y=2e^{0.3t} sin(2t) \implies sin(2t) = \frac{y}{2e^{0.3t}}$

Then you use formula

$(sin(2x))^2 +(cos(2x))^2 = 1$
That would give the answer in terms of t though not x. wouldn't it?

lol  stupid Jenny

I'm thinking we write 2e^0.3t in term of x and y then equate them
i wouldn't have realised too if i didn't try that earlier :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 12, 2012, 11:58:48 pm
Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!

$x=2e^{0.3t} cos(2t)$

$y=2e^{0.3t} sin(2t)$

This is the parametric equation of Bernoulli's (logarithmic) spiral

Its polar equation  $r=2e^{0.3t}$

Anyway, I bet it won't be in Spesh exam so dw about that
Just realise I spend like 2 hrs to figure out wth its and now haven't reviewed anything for Spesh SAC tomorrow LOL. Damn it!!!
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on September 13, 2012, 12:08:54 am
Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!

$x=2e^{0.3t} cos(2t)$

$y=2e^{0.3t} sin(2t)$

This is the parametric equation of Bernoulli's (logarithmic) spiral

Its polar equation  $r=2e^{0.3t}$

Anyway, I bet it won't be in Spesh exam so dw about that
Just realise I spend like 2 hrs to figure out wth its and now haven't reviewed anything for Spesh SAC tomorrow LOL. Damn it!!!
I'm so sorry didn't mean to bother you. Thanks a lot though  :D
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on September 13, 2012, 06:52:19 pm
When you integrate a second order DE twice and you have to make a substitution, why is it wrong to keep the initial substitution and use it for the second integration? So if you subbed in U, how come it's wrong to integrate that U again? And how would you get around this (aside from changing back to x's)

Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on September 14, 2012, 08:45:56 pm
When you integrate a second order DE twice and you have to make a substitution, why is it wrong to keep the initial substitution and use it for the second integration? So if you subbed in U, how come it's wrong to integrate that U again? And how would you get around this (aside from changing back to x's)

Example?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 15, 2012, 02:09:11 am
Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!

$x=2e^{0.3t} cos(2t)$

$y=2e^{0.3t} sin(2t)$

This is the parametric equation of Bernoulli's (logarithmic) spiral

Its polar equation  $r=2e^{0.3t}$

Anyway, I bet it won't be in Spesh exam so dw about that
Just realise I spend like 2 hrs to figure out wth its and now haven't reviewed anything for Spesh SAC tomorrow LOL. Damn it!!!
I'm so sorry didn't mean to bother you.

Nah, you are not. I chose not to study at that day :(

Anyway, I figure it out

$(sin(2t))^2 +(cos(2t))^2 = 1$

$x^2 + y^2 = (2e^{0.3t})^2$

$\sqrt {x^2 + y^2}= 2e^{0.3t}$

$\frac {\sqrt {x^2 + y^2}}{2} = e^{0.3t}$

$t= \frac {10}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}) (1)$

Sub (1) into $y=2e^{0.3t} sin(2t)$

$y=2e^{\frac{3}{10} \times \frac {10}{3} ln (\frac {\sqrt {x^2 + y^2}}{2})} sin(2 \times \frac {10}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}))$

$y= \sqrt{x^2+y^2} sin(\frac {20}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}))$

Therefore, the cartesian equation is   $y= \sqrt{x^2+y^2} sin(\frac {20}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}))$

Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on September 15, 2012, 04:04:23 pm
Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!

$x=2e^{0.3t} cos(2t)$

$y=2e^{0.3t} sin(2t)$

This is the parametric equation of Bernoulli's (logarithmic) spiral

Its polar equation  $r=2e^{0.3t}$

Anyway, I bet it won't be in Spesh exam so dw about that
Just realise I spend like 2 hrs to figure out wth its and now haven't reviewed anything for Spesh SAC tomorrow LOL. Damn it!!!
I'm so sorry didn't mean to bother you.

Nah, you are not. I chose not to study at that day :(

Anyway, I figure it out

$(sin(2t))^2 +(cos(2t))^2 = 1$

$x^2 + y^2 = (2e^{0.3t})^2$

$\sqrt {x^2 + y^2}= 2e^{0.3t}$

$\frac {\sqrt {x^2 + y^2}}{2} = e^{0.3t}$

$t= \frac {10}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}) (1)$

Sub (1) into $y=2e^{0.3t} sin(2t)$

$y=2e^{\frac{3}{10} \times \frac {10}{3} ln (\frac {\sqrt {x^2 + y^2}}{2})} sin(2 \times \frac {10}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}))$

$y= \sqrt{x^2+y^2} sin(\frac {20}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}))$

Therefore, the cartesian equation is   $y= \sqrt{x^2+y^2} sin(\frac {20}{3} ln (\frac {\sqrt {x^2 + y^2}}{2}))$

:O wow that is amazing!! you are getting a 50 raw in spesh :O.
Thanks a lot!!
Title: Re: Specialist 3/4 Question Thread!
Post by: nisha on September 16, 2012, 01:47:54 pm
I forgot how to do a question like this.
1. Two objects of equal mass, 1.0kg are connected by a light string slung over a frictionless pulley. They lie on two identical surfaces and are at the point of slipping. Both are situated on a triangle, with Mass 2 slipping, and mass 1 on the surface, directly under a 30 degree angle. Mass 2 is adjacent to a 60 degree angle.
(a) calculate the coefficient of friction for the two surfaces
(b)Mass 2 is now replaced with a mass of 2.0kg. Find the acceleration of the system
(c)Find the magnitude of the tension in the string connecting the two masses
-----------
q2. Two ferries, A and B are travelling at constant velocities, have position and velocity vectors at 10am given by:
r(a)=6i-3j
r(b)=-2i+j
v(a)=-2i+3j
v(b)=2i+j
The distance is measured in km and time in hours.

Show that the ferries collide if they maintain their current velocities and determine the time when this happens.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 16, 2012, 08:25:42 pm
I forgot how to do a question like this.
1. Two objects of equal mass, 1.0kg are connected by a light string slung over a frictionless pulley. They lie on two identical surfaces and are at the point of slipping. Both are situated on a triangle, with Mass 2 slipping, and mass 1 on the surface, directly under a 30 degree angle. Mass 2 is adjacent to a 60 degree angle.
(a) calculate the coefficient of friction for the two surfaces
(b)Mass 2 is now replaced with a mass of 2.0kg. Find the acceleration of the system
(c)Find the magnitude of the tension in the string connecting the two masses
-----------
q2. Two ferries, A and B are travelling at constant velocities, have position and velocity vectors at 10am given by:
r(a)=6i-3j
r(b)=-2i+j
v(a)=-2i+3j
v(b)=2i+j
The distance is measured in km and time in hours.

Show that the ferries collide if they maintain their current velocities and determine the time when this happens.

Q1. You need to draw the diagram first. I don't know how to do it here though so its hard to explain
a) You find R1= ai = (T - g x sin30 - uN1)i + (N1 - g x cos30)j
a = T - g sin30 - u x g cos30 (1)
R2= ai = (T-g x sin60 - uN2)i + (N2 - g cos60)j
a = T - g sin60 - u x g cos60 (2)
From (1) and (2), you solve to find u (coefficient friction)

b+c)  Similar as (a), you write again the resultant force of R2
Then substitute u (you found in (a))and find a and T

Q2) Given velocity => you find r(a) and r(b) by antidifferentiation, then substitite t=10, and r(a),r(b) given to find value of c
They collide when they have the same position at the same time.
After you find r(a) and r(b), you equate them to find t. The same value of t for both is the time they collide
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 16, 2012, 08:49:39 pm
Three points, A, B and C, have coordinates A(1, 1, 1), B(2, 3, –6) and C(5, –3, –3),
respectively. If M is the midpoint of AC , use a vector method to show that MB is
perpendicular to AC.

Worked solution
They calculated vectors AB,AC,AM.
Then calculated MB by subtracting AM from AB
Then multiplied AC and MB and it equaled zero.

Im confused about how they got vector MB, can someone explain please?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 16, 2012, 08:56:03 pm
Three points, A, B and C, have coordinates A(1, 1, 1), B(2, 3, –6) and C(5, –3, –3),
respectively. If M is the midpoint of AC , use a vector method to show that MB is
perpendicular to AC.

Worked solution
They calculated vectors AB,AC,AM.
Then calculated MB by subtracting AM from AB
Then multiplied AC and MB and it equaled zero.

Im confused about how they got vector MB, can someone explain please?

So you understand they calculate vectors AB, AC, AM yeh?
Then, what they mean is: $\vec{MB} = \vec{MA} + \vec{AB} = \vec{AB} - \vec{AM}$

Edit: I typed wrong
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 16, 2012, 08:59:40 pm
They had MB=AB-AM ? So are they wrong?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 16, 2012, 09:04:06 pm
They had MB=AB-AM ? So are they wrong?

They are right. Check my post above again, I typed wrong.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 16, 2012, 09:10:39 pm
Ah ok, thanks jenny :)

Also, two more questions:

1. Consider the function defined by ex+y=y+x2+e-1.
Show that y=1 when x=0.
The solutions just subbed (0,1) in and showed that LHS and RHS both equal e.

Does their way answer the question properly? By showing that the LHS=RHS, are they actually showing that y=1 when x=0?

2. Basic question: y=sin-1(2x).
Whats the domain? i forget how u know when to multiply [-1,1] by 2 and when to divide it by 2
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 16, 2012, 09:24:49 pm
Ah ok, thanks jenny :)

Also, two more questions:

1. Consider the function defined by ex+y=y+x2+e-1.
Show that y=1 when x=0.
The solutions just subbed (0,1) in and showed that LHS and RHS both equal e.

Does their way answer the question properly? By showing that the LHS=RHS, are they actually showing that y=1 when x=0?

2. Basic question: y=sin-1(2x).
Whats the domain? i forget how u know when to multiply [-1,1] by 2 and when to divide it by 2

1. I think it should be fine or maybe someone else can clarify for you

2. the domain is always within [-1,1] so 2x has to be between -1 and 1
To find domain of x, you divide by 2
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on September 16, 2012, 09:34:19 pm
I like solving the domain like this.
-1<_x<_1

-1<_2x<_1 => (make it x again by dividing it all by 2) -1/2<_x<_1/2

Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 16, 2012, 09:54:27 pm
To sketch the graph y= sec(2x-(pi/2)) , we change it to y=1/ cos(2x-(pi/2)).
Then to find the stationary points, the solutions let cos(2x-(pi/2))=plus minus 1, why?
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on September 16, 2012, 10:23:52 pm
To sketch the graph y= sec(2x-(pi/2)) , we change it to y=1/ cos(2x-(pi/2)).
Then to find the stationary points, the solutions let cos(2x-(pi/2))=plus minus 1, why?

Have a think about it. Let's just say y = 1/A

and A is a number between -1 and 1 inclusive (A = cos(...))

so when will y be smallest? when A = -1

when will y be greatest? when A = 1
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on September 26, 2012, 01:16:03 pm
When you integrate a second order DE twice and you have to make a substitution, why is it wrong to keep the initial substitution and use it for the second integration? So if you subbed in U, how come it's wrong to integrate that U again? And how would you get around this (aside from changing back to x's)

Example?

e.g. $\frac{d^2y}{dx^2}=\frac{x}{(4-x^2)^\frac{3}{2}}$

-----

Another question, how would you find an expression for T1 and T2 (tension), in terms of the angle (alpha?) for this (see attached image, it's supposed to be a hanging mass). m=5

---

@b^3: How the heck did you do that crazy diagram on page 47?!
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on September 27, 2012, 11:39:31 am
When you integrate a second order DE twice and you have to make a substitution, why is it wrong to keep the initial substitution and use it for the second integration? So if you subbed in U, how come it's wrong to integrate that U again? And how would you get around this (aside from changing back to x's)

Example?

e.g. $\frac{d^2y}{dx^2}=\frac{x}{(4-x^2)^\frac{3}{2}}$

So you'd get $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{\sqrt{u}}+c$, if you wanted to solved for y you'd need to integrate both sides with respect to x, i.e.:

$y=\int (\frac{1}{\sqrt{u}}+c)\mathrm{d}x$

So you can see that would be problematic without changing it back to x first. If you integrated it with respect to u you'd get a different function:

$\int \frac{\mathrm{d} y}{\mathrm{d} x}\mathrm{d}x\neq \int \frac{\mathrm{d} y}{\mathrm{d} x}\mathrm{d}u$
Title: Re: Specialist 3/4 Question Thread!
Post by: VCEstudentguy on September 27, 2012, 07:27:12 pm
How come if I try to solve the definite integral in the image below using Casio Classpad it comes up as undefined? It's a VCAA question on the 2006 Exam 2 Q3iii.
Title: Re: Specialist 3/4 Question Thread!
Post by: meemz on September 27, 2012, 08:28:10 pm
How come if I try to solve the definite integral in the image below using Casio Classpad it comes up as undefined? It's a VCAA question on the 2006 Exam 2 Q3iii.
not sure, but i tried it on the ti-nspire and it worked
maybe you're putting dx instead of dv or you're not multiplying the 9600 by v, but just typing 9600v
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on September 27, 2012, 08:49:55 pm
How come if I try to solve the definite integral in the image below using Casio Classpad it comes up as undefined? It's a VCAA question on the 2006 Exam 2 Q3iii.
not sure, but i tried it on the ti-nspire and it worked
maybe you're putting dx instead of dv or you're not multiplying the 9600 by v, but just typing 9600v

Wtf X________________X. My classpad doesn't solve it either.. Classpads are so shit lol. Hope this stuff doesn't happen on the end of year exams :/
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 27, 2012, 10:15:33 pm
^ You should buy ti-nspire!!! Somtimes Classpad can't solve for combination in methods as well. It'll be disadvantages for you in the maths exams
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on September 27, 2012, 11:04:49 pm
Easier said than done!
It takes quite a long time to get used to a new calculator. You can't just pick it up and become a super beast.

Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on September 27, 2012, 11:38:04 pm
Easier said than done!
It takes quite a long time to get used to a new calculator. You can't just pick it up and become a super beast.

From now until the exams, you have 6 weeks.
If you use the new CAS when doing trials during hols, I'm sure you will get used to
It took me 3 days in year 11 to know how to use CAS but you already use Classpad so it will be quicker for you anyway
Title: Re: Specialist 3/4 Question Thread!
Post by: VCEstudentguy on September 28, 2012, 12:17:35 am
^ You should buy ti-nspire!!! Somtimes Classpad can't solve for combination in methods as well. It'll be disadvantages for you in the maths exams

Yeah I think we discovered yesterday with the Methods question and today with the Spesh question that the TI seems to be superior haha. I think I'll stick with the Classpad though, too late to back away now after years of use.

Title: Re: Specialist 3/4 Question Thread!
Post by: nisha on September 28, 2012, 12:22:25 am
Easier said than done!
It takes quite a long time to get used to a new calculator. You can't just pick it up and become a super beast.

From now until the exams, you have 6 weeks.
If you use the new CAS when doing trials during hols, I'm sure you will get used to
It took me 3 days in year 11 to know how to use CAS but you already use Classpad so it will be quicker for you anyway
Bear in mind, that means buying a new calculator, then learning its ins and outs.

^^Go to your teacher and ask them why it isn't working. I'm sure there is a way.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on September 28, 2012, 12:43:26 am
It might be worth shooting a PM to Paul Sterio about it, he used the Casio in VCE :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Conic on September 28, 2012, 01:34:07 pm
It might be worth watching some online comparisons between the Calculators and reading through the TI nSpire guides before you decide whether or not to buy the Calculator.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 28, 2012, 05:23:27 pm

A box of mass 5kg rests on a rough horizontal floor. The coefficient of friction between the box and the floor is 0.1 . A boy applies a horizontal dragging force of D newtons to the box in an attempt to move it.

a.   Find the values of D if the box is not at the point of moving across the floor.

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on September 28, 2012, 05:48:52 pm
For the box to not be at the point of moving across the floor (I'm assuming its going to stay stationary), then we need the force $D$ to be less than the frictional force at which the box would be at the point of moving. As when the force applied to the box is less than this, it won't move and the friction force will be equal to the force we are applying.

So we need $D<\mu N$

(http://i1082.photobucket.com/albums/j373/mclaren200800/dynamics3-1.png)

Forces in the vertical direction.
$N-m_{1}g=0$
i.e. $N=5g....[1]$

So that means $0 (assuming D is always in that direction).
So $0 (I think you missed the value of mu)

NOTE: If I've interpretted the question wrong and it just wants the value of which its not at the point of moving, but includes when  it actually moves, then it will be not equal to the value found, but I'm pretty sure it asked what I did above (but just in case anyway).
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on September 28, 2012, 06:32:54 pm
Thanks b^3!
And yeh i missed the mu value, should be 0.1.

So if D > mu N, it moves to the right, and if D< mu N, it stays where it is? What if D=mu N, does it just stay?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on September 28, 2012, 06:37:54 pm
Thanks b^3!
And yeh i missed the mu value, should be 0.1.

So if D > mu N, it moves to the right, and if D< mu N, it stays where it is? What if D=mu N, does it just stay?
Basically yeh.

if $D>\mu N$ then the box will move to the right (for the diagram we drew, well it will move in the direction that D is in, its just that I drew D to the right)
if $D=\mu N$ then the box will be 'on the point of moving' but won't move.
if $D<\mu N$ then it will stay where it is as you aren't applying enough force to overcome the friction force opposing it.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 01, 2012, 03:29:56 pm
A ball of mass 1/5 kg is projected vertically upwards at a speed 10 ms−1 from the top of a 20 m high cliff. Air
resistance is negligible. Correct to two decimal places, the ball will hit the foot of the cliff with a momentum, in kg.ms−1, of
A. 3.42
B. 3.96
C. 4.44
D. 17.10
E. 22.12

Also, the solutions said that by using calculus you can get v=root 492, but what did they mean? How do you find v by using calculus?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 01, 2012, 03:44:52 pm
Remember that u is the initial velocity where as v is the final velocity.
So the initial velocity will be 10 m/s up as this is where we are taking time 0 as when it is being projected vertically upwards.

If we were interested in when the ball got to it's maximum heigth, then v would be 0 m/s. But we want when it hits the foot of the cliff.

Now with $v^{2}=u^{2}+2as$ (sometimes instead of s they will use x), s here isn't distance, its displacement. I.e. Where you end up in reference to the original point. So when the ball hits the foot of the cliff it will be 20 m below our starting reference point, i.e. s=-20 m.

If you were asked to find the distance it falls from the maximum heigth then it would be the 20 m plus the extra distance.

So remember, u is initial velocity, v is final velocity and 's' (I'd rather use x but anyways) is displacement.

Hope that helps :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 01, 2012, 03:51:23 pm
a particle moves such that its pos. vector is given by: r = sin(2t)i +2cos(t)j
show that the distance of the particle, $|r(t)|$ is given by $2\sqrt{1-sin^4(t)}$

so i started by rooting and squaring r.

$|r| = \sqrt{(sin2t)^2 + (2cost)^2}
$

$|r| = \sqrt{(2sintcost)^2 + (2cost)^2}
$

is it right to take out cost as a factor. if not.. whats the next step!>?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 01, 2012, 03:56:51 pm
Yeh its fine, we're only taking it out as a factor, so it's all good.
\begin{alignedat}{1}|\underset{\sim}{r}(t)| & =\sqrt{\sin^{2}(2t)+2^{2}\cos^{2}(t)}
\\ & =\sqrt{4\sin^{2}(t)\cos^{2}(t)+4\cos^{2}(t)}
\\ & =\sqrt{4\cos^{2}(t)\left(\sin^{2}(t)+1\right)}
\\ & =2\sqrt{\left(1-\sin^{2}(t)\right)\left(\sin^{2}(t)+1\right)}
\\ & =2\sqrt{1-\sin^{4}(t)}
\end{alignedat}
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 01, 2012, 04:05:44 pm
ok im also stuck on the next bit :\
it asks for, 'HENCE, find the maximum distance of the particle from the origin and state where this occurs'

these questions always stump me..
i could integrate it, might take a while though.
theres a way to do using sin graphs ... right? or something like that?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 01, 2012, 04:15:38 pm
Think about it this way, if you make $1-\sin^{4}(t)$ as big as you can then you are making $2\sqrt{1-\sin^{4}(t)}$ as big as you can right?

So now to look at $1-\sin^{4}(t)$.
Think about what values $\sin(t)$ can be.
The smallest we can get out is $-1$, the largest is $1$
Now if we look at $\sin^{4}(t)$ whats the smallest and biggest numbers we can get out?
They will be $0$ and $1$ right?

So now to make $1-\sin^{4}(t)$ as large as possible, we need to make $\sin^{4}(t)$ as small a possible.
So the smallest value of $\sin^{4}(t)$ is $0$ so we have
Max distance=$2\sqrt{1-0}=2$ (then the units).

If you get a $sin^{2}(t)$ its a similar deal.

Anyway hope that makes sense, probably over explained it a little.

EDIT: fixed it up, was missing bits of sentences, guess thats what happens when you put latex in the middle of sentences...
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 01, 2012, 04:42:00 pm
oo ok, that made sense, except when you said whats the smallest and biggest numbers we can get out for sin^4 (t),
i dont get how its 0 and 1 ,  - i wouldve said -1 and 1.
im not in the maths mood i think at the moment, so please excuse my crappy maths atm :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Conic on October 01, 2012, 05:25:21 pm
oo ok, that made sense, except when you said whats the smallest and biggest numbers we can get out for sin^4 (t),
i dont get how its 0 and 1 ,  - i wouldve said -1 and 1.
im not in the maths mood i think at the moment, so please excuse my crappy maths atm :P
Since it's sin^4 the minimum is 0^4 (as -1^4=1)
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 01, 2012, 05:46:41 pm
i see!
thanks guys.

another question:

given that $z^8 - 2cos(x)z^4 + 1 = (z^4 - cis(x))(z^4-cis(-x))$,
solve the equation$z^8 - z^4 + 1 =0,$ giving your answers in the form cis(A), where $a E (-\pi,\pi]$

I Don't even know where to start.
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 01, 2012, 06:38:01 pm
Let cos(x) = 1/2
Title: Re: Specialist 3/4 Question Thread!
Post by: oneoneoneone on October 03, 2012, 03:01:05 pm
There are 4 intersections, But at when the first particle is in the second quadrant, the second particle is in the fourth quadrant and vice versa. So their paths cross at 4 points, but the actualy particles only meet when theyre at the same spot at the same time, which is only in the first and 3rd quadrants.
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 03, 2012, 10:12:02 pm
consider $arctan(\frac{1}{m}) + arctan(\frac{1}{n}) = \frac{\pi}{4}$ , where m and n are non-zero positive integers

show that  $(m-1)(n-1) =2$

dafuk?
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on October 03, 2012, 10:28:47 pm
let x = arctan(1/m) and y = arctan(1/n)
m = 1/tanx and n = 1/tany
tan(x+y) = (tanx +tany)/(1-tanx tany) = 1
tan x + tany = 1 - tanx tany

(m-1)(n-1)
=mn - m - n + 1
=1/(tanx tany) - 1/tanx - 1/tany + 1
= 1/(tanx tany) - (tanx + tany)/(tanx tany)+1
= (1 - tanx -tany + tanx tany)/(tanx tany)
= (1 - (1-tanx tany) + tanx tany)/(tanx tany)
= 2tanx tany/(tanx tany)
= 2
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 03, 2012, 10:30:31 pm
wow.
haha thanks man!

your gonna kill spesh next year, sure you've already heard that ;p
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on October 03, 2012, 10:48:21 pm
wow.
haha thanks man!

your gonna kill spesh next year, sure you've already heard that ;p

He was killing spesh in year 9 X_x.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on October 04, 2012, 03:28:24 pm
Advanced Maths Question: A mass of 3 kg is suspended from a beam by two strings, 5 cm and 12 cm long. If the
strings are perpendicular to each other, find the tension in each string.
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 04, 2012, 04:20:21 pm
Here, hopefully that's clear enough. I would have typed it here but I'd rather have the diagram there and I'm not b^3 so yeah. :P

It says "13 cm" at the top, it was cut off slightly.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on October 04, 2012, 06:18:33 pm
Thanks for your help! but we haven't learnt about "arctan" yet in our advanced class.  Is there possibly another way of solving such problems? Btw the answers given are 11.31 N; 27.14 N.
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 04, 2012, 06:24:40 pm
arctan is just the inverse tan function. I think you would know about it by now? And you could possibly do it without using arctan, by being careful with your geometry and finding values for the relevant sines of angles, but you'll have to learn about inverse trigonometric functions eventually so I guess you should start now. :P
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 05, 2012, 10:01:34 pm
I need help with this multiple choice please
Cheers
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 05, 2012, 10:13:26 pm
The area under a velocity time graph will give displacement. So integrating $V_{A}(t)$ from 0 to 50 will give the displacement of particle A from the starting position, at t=50s.

Now if we do the same for $V_{B}(t)$, we will get the displacement of particle B. (Initially B is moving in the opposite direction until it turns around. When the integral is negative then it is to the left of O, i.e. it starts moving in the opposite direction at t=20 seconds but is still to the left of the starting position until the area under the graph and the area above the grap cancels out). <--- Probably didn't need that last bit but the explanation may help a little.

So basically if we know the displacement of each particle from the original position at t=50, take one away from the other and take the absolute value of that, we will get the distance between the particles. (since they are moving in a striaght line, drawing a number line may help visualise their movement and the distance between them). Hence E.

Hope that makes sense. I guess the key thing to remember here is $\int v(t)dt=\Delta x$

EDIT: Sorry if that didn't make sense, my explanations are all over the place tonight... not on enough sleep atm.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 06, 2012, 02:21:18 pm
The area under a velocity time graph will give displacement. So integrating $V_{A}(t)$ from 0 to 50 will give the displacement of particle A from the starting position, at t=50s.

Now if we do the same for $V_{B}(t)$, we will get the displacement of particle B. (Initially B is moving in the opposite direction until it turns around. When the integral is negative then it is to the left of O, i.e. it starts moving in the opposite direction at t=20 seconds but is still to the left of the starting position until the area under the graph and the area above the grap cancels out). <--- Probably didn't need that last bit but the explanation may help a little.

So basically if we know the displacement of each particle from the original position at t=50, take one away from the other and take the absolute value of that, we will get the distance between the particles. (since they are moving in a striaght line, drawing a number line may help visualise their movement and the distance between them). Hence E.

Hope that makes sense. I guess the key thing to remember here is $\int v(t)dt=\Delta x$

EDIT: Sorry if that didn't make sense, my explanations are all over the place tonight... not on enough sleep atm.
How would you know which to take away from? Why couldn't it have been f(t)-g(t) ?
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on October 06, 2012, 02:28:04 pm
How would you know which to take away from? Why couldn't it have been f(t)-g(t) ?

Given the modulus, it actually can be f(t)-g(t)
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on October 08, 2012, 06:54:58 pm
I've just got two questions that are bugging me:

a) 'On a rough inclined plane, a mass of mg is being pushed off the top of the plane with initial velocity 1ms', sorry for butchering the question, but I was assuming that you would use the formula R=ma, and integrate that to get velocity, et cetera, but the question seems to assume we can use R=Ma, find the acceleration, and then use the four formulas, as it the solutions say because it has constant acceleration, is that even allowed on rough inclines?

b) For position vectors such as, r(t) = (4t-20)i + 2t j -+ (80 - 2t) k, if it asks for the angle when it hits the ground, wouldn't there be NO angle, because the vertical height would be zero (as it hits the ground?)...

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 08, 2012, 07:11:29 pm
I've just got two questions that are bugging me:

a) 'On a rough inclined plane, a mass of mg is being pushed off the top of the plane with initial velocity 1ms', sorry for butchering the question, but I was assuming that you would use the formula R=ma, and integrate that to get velocity, et cetera, but the question seems to assume we can use R=Ma, find the acceleration, and then use the four formulas, as it the solutions say because it has constant acceleration, is that even allowed on rough inclines?

b) For position vectors such as, r(t) = (4t-20)i + 2t j -+ (80 - 2t) k, if it asks for the angle when it hits the ground, wouldn't there be NO angle, because the vertical height would be zero (as it hits the ground?)...

Thanks!

a) yep
b) because it hits the ground so height = 0. You find angle based on i, j directions
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 08, 2012, 08:01:48 pm
I've got two questions

(1)  Is is true that a complex number is rotated by "θ" (anticlockwise) when it is multiplied by cis(θ)?

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,

e.g OA.OB=OA.OB.cosθ (1)

OA.BO=OA.BO.cosθ (2)

The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on October 08, 2012, 08:18:49 pm
(1)  Is is true that a complex number is rotated by "θ" (anticlockwise) when it is multiplied by cis(θ)?
Yep. You can see and draw it out it for yourself.

$z_1 = rcis(0)$ and $z_2 = rcis(\theta)$

$z_1z_2 = r^2cis(0 + \theta)$

(you could also do it in rectangular form if that's more intuitive to you)

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,

e.g OA.OB=OA.OB.cosθ (1)

OA.BO=OA.BO.cosθ (2)

The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.
$\mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|cos(\theta)$
Just emphasising that it's the magnitudes of the vectors in that equation. Other than that, I don't see anything wrong with the second one. The $\theta$ is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 08, 2012, 08:34:06 pm
(1)  Is is true that a complex number is rotated by "θ" (anticlockwise) when it is multiplied by cis(θ)?
Yep. You can see and draw it out it for yourself.

$z_1 = rcis(0)$ and $z_2 = rcis(\theta)$

$z_1z_2 = r^2cis(0 + \theta)$

(you could also do it in rectangular form if that's more intuitive to you)

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,

e.g OA.OB=OA.OB.cosθ (1)

OA.BO=OA.BO.cosθ (2)

The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.
$\mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|cos(\theta)$
Just emphasising that it's the magnitudes of the vectors in that equation. Other than that, I don't see anything wrong with the second one. The $\theta$ is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.

for example, OA=i+2j OB=i-j  BO=-i+j ; In this case the dot product of OA and OB vs.OA and BO is different, hence the angle cannot be the same

so the starting point of two vectors must be the same i.e. OA and OB

I am not sure if i my thinking is correct
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 08, 2012, 08:45:07 pm
Of course their angles are different. It must be tail-to-tail as well
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on October 08, 2012, 08:48:53 pm
Oh you were asking if dot products (1) and (2) were equivalent. Sorry, didn't pick up on that.

Yeah what Ennjy said, OB and BO are different vectors, I think forgetting that is where you're getting confused.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 08, 2012, 09:17:02 pm
A hot air balloon is ascending with uniform velocity of 5m/s. When the balloon is 60 m above the ground pone of the occupants drops his watch. Assuming that the watch experiences no air resistance, find the maximum height reached by the watch.

I used v2=u2+2as, but had a=0 as it says uniform velocity but ended up not being able to solve for s. Why is the a=-9.8? I thought that uniform velocity meant velocity stays the same so acceleration is 0.

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 08, 2012, 09:19:25 pm
When it is initialy rising before we release it, then it is ascending with uniform velocity. But as soon as we release it (at $t=0\:s$), it will start falling under tha influence of gravity alone, and so the acceleration will be the acceleartion due to gravity, $a=-9.8\:m/s^{2}$.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 08, 2012, 09:36:45 pm
How do i find the time taken for the watch to reach the ground?

I considered the point from when it was at its max height to when it hit the ground. So i had s=61.28,t=?,a=9.8, u=0
And used s=ut+0.5at2, but got t=3.54. The answer should be t=4.05, what have i done wrong?
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on October 08, 2012, 09:45:53 pm
(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,

e.g OA.OB=OA.OB.cosθ (1)

OA.BO=OA.BO.cosθ (2)

The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.
$\mathbf{a}.\mathbf{b} = |\mathbf{a}||\mathbf{b}|cos(\theta)$
Just emphasising that it's the magnitudes of the vectors in that equation. Other than that, I don't see anything wrong with the second one. The $\theta$ is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.

Hmm, there is a bit of an issue though, the angle between u and v and the angle between v and u will be different (they are supplementary), so yeah, watch what they're asking you I guess.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 08, 2012, 09:52:49 pm
How do i find the time taken for the watch to reach the ground?

I considered the point from when it was at its max height to when it hit the ground. So i had s=61.28,t=?,a=9.8, u=0
And used s=ut+0.5at2, but got t=3.54. The answer should be t=4.05, what have i done wrong?

You will need to add the time it takes to get to max height in the first place aswell. So you found the time for the downwards part of it's motion.
To find the time for the upwards part you will have.
u=5 m/s
v=0 m/s (stationary at max height)
a=-9.8 m/s
t=?
v=u+at
t=(v-u)/a
t=(0-5)/9.8=0.51 seconds

Total time T=0.51+3.54=4.05 seconds

-OR-
If you didn't want to work it out in parts like that, and just do it in one shot you could do this.
We know that it is initially 60 m above the ground. So when it hits the ground it will be 60 m below its starting point.
So we would have
s=-60 m
u=+5 m/s
a=-9.8 m/s^2
t=?
s=ut+0.5at2
\begin{alignedat}{1}-60 & =5t+\frac{1}{2}\times-9.8\times t^{2}
\\ -4.9t^{2}+5t+60 & =0
\\ t & =\frac{-5\pm\sqrt{5^{2}-4(-4.9)(60)}}{2(-4.9)}
\\ t & =\frac{-5\pm34.655}{-9.8}
\\ t & =-3.03,\:4.05
\end{alignedat}

Ignore the negative solution, so $\therefore t=4.05\:s$

EDIT2: Beaten to get the edit in for the easier way.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on October 08, 2012, 09:57:53 pm
s = 60m
a = 9.8ms-2
u = -5ms-1
t = ?

s = ut + 1/2at^2

60 = -5t + 4.9t^2

CAS gives me: t = -3.03s or 4.05s

Obviously we know it's 4.05s because of the domain of the time variable - might be easier doing it this way as opposed to b^3's, although both methods are right.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 08, 2012, 10:09:17 pm
Ok thanks guys but why is a negative? If its falling downwards, isn't it going in the direction of gravity?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 08, 2012, 10:11:50 pm
It depends whether you define positive as up or down.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 08, 2012, 10:13:33 pm
It just depends which way you denote as positive. If you take up as positive, then be consistent with it. I.e. Paul took down as positive, where as I took up as positive (as I was sticking with things from the previous question). It won't matter in the end as long as you are consistent with everything in that part of the question. E.g. if down is positive, then a and s will be positive but the initial velocity u will be negative.

I got used to just always taking up as positive, as it meant I wouldn't make mistakes, but that was just me, what ever works best for you (as long as you have everything relative to each other in the right direction) :)

EDIT: beaten.
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on October 09, 2012, 08:24:32 am
So, for the position vector, r(t) = (4t-20)i + 2t j -+ (80 - 2t) k, the worked solution still confuses me. The question is, 'At what angle does the kite hit the ground?', they seem to be using k still, but Im assuming that k=0....

Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 09, 2012, 04:32:23 pm
The angle at which an object hits the ground will be dependent on the velocity vector, not the position vector.
That is
\begin{alignedat}{1}\underset{\sim}{r}(t) & =(4t-20)\underset{\sim}{i}+2t\underset{\sim}{j}+(80-2t)\underset{\sim}{k}
\\ \underset{\sim}{\dot{r}}(t) & =4\underset{\sim}{i}+2\underset{\sim}{j}-2\underset{\sim}{k}
\end{alignedat}

So the angle that it hits the ground at will be given by (draw the triangle out, the hypotenus will be the magnitude of the vector and the Opposite side will be equal to the magnitude of the k component)
(http://i1082.photobucket.com/albums/j373/mclaren200800/vectors.png)
\begin{alignedat}{1}\sin(\theta) & =\frac{O}{H}
\\ & =\frac{2}{\sqrt{4^{2}+2^{2}+(-2)^{2}}}
\\ \theta & =\sin^{-1}(\frac{2}{\sqrt{24}})
\\ \theta & =\sin^{-1}(\frac{1}{\sqrt{6}})
\\ \theta & =24.09^{o}
\end{alignedat}

EDIT: Added the image, didn't turn out to be what I wanted it to, but it shows it anyway.

EDIT2: Forgot to mention something.
Once you get the velocity vector, you will need to get the velocity vector at time t where it hits the ground, that would be when the k component is 0, then you sub in that t into the velocity vector. But in this case the velocity vector is not dependent on t, as it is moving with constant velocity, so it didn't matter in this situation, but it's just something I thought I would point out :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on October 09, 2012, 06:37:56 pm
I have a quick question:
Question 10: VCAA 2007. I really suck with these types of questions:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

I did something funky D= It's wrong, I gave myself 0/3 for it, but I was hoping someone could help me understand what I did wrong =)
{SEE ATTACHED IMAGE}

@b^3  I'd like to publicly thank you for all of your efforts in regards to helping the VN community. You're an absolute legend. I MEAN, LOOK AT THOSE DIAGRAMS!
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 09, 2012, 07:26:55 pm
I have a quick question:
Question 10: VCAA 2007. I really suck with these types of questions:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

I did something funky D= It's wrong, I gave myself 0/3 for it, but I was hoping someone could help me understand what I did wrong =)
{SEE ATTACHED IMAGE}

@b^3  I'd like to publicly thank you for all of your efforts in regards to helping the VN community. You're an absolute legend. I MEAN, LOOK AT THOSE DIAGRAMS!

You did wrong at line 3. It should be

$2sin(x)=\frac{4\sqrt{2}}{7} \times \frac{1-2sin^2(x)}{cos(x)}$
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 09, 2012, 07:36:21 pm
$\cos(2x)=1-2\sin^{2}(x)=\frac{7}{\sqrt{(4\sqrt{2})^{2}+7^{2}}}=\frac{7}{9}$ (use triangle)

$\Rightarrow 1-\frac{7}{9}=\frac{2}{9}=2\sin^{2}(x)$

$\Rightarrow \sin^{2}(x)=\frac{1}{9}$

$\Rightarrow \sin(x)=\frac{1}{3}$ as $x\in[0,\frac{\pi}{4})$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 09, 2012, 08:33:42 pm
A particle moves along a curved path in such a way that its position vector r(t) at any time t is given by r(t)=acos(bt) i + asin(bt) j, where a,b E R+

How would i "verify that the acceleration vector is always directed towards the origin" ?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 09, 2012, 08:56:15 pm
If a particle is moving in a circle, then it's velocity is tangential to it's path.
Now if we find the acceleration vector.
\begin{alignedat}{1}\underset{\sim}{r}(t) & =a\cos(bt)\underset{\sim}{i}+a\sin(bt)\underset{\sim}{j}
\\ \underset{\sim}{\dot{r}}(t) & =-ab\sin(bt)\underset{\sim}{i}+ab\cos(bt)\underset{\sim}{j}
\\ \underset{\sim}{\ddot{r}}(t) & =-ab^{2}\cos(bt)\underset{\sim}{i}-ab^{2}\sin(bt)\underset{\sim}{j}
\\ & =-b^{2}\left(a\cos(bt)\underset{\sim}{i}+a\sin(bt)\underset{\sim}{j}\right)
\\ & =-b^{2}\underset{\sim}{r}(t)
\end{alignedat}

So this means that the acceleration vector is the displacement vector mu,tipled by the constant -b2. I.e. The acceleration vector will be in the opposite direction to the displacement vector, and as the displacement vector starts at the origin, this means that the acceleration vector is always directed towards the origin.

NOTE: I feel there is something I've missed above, maybe even missed the point of the question completely, if there is someone pick me up on it (I went on the path of finding that the velocity vectors and acceleration vectors are perpendicular then realised that wasn't needed either).... I might just be going crazy a bit..

Anyway, hope that helps :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 09, 2012, 09:00:44 pm
^ OMG, you type so quickly b^3
I was typing and now delete LOL Hate LaTex now
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 10, 2012, 05:00:04 pm
I sketched the graph of y=2sin-1(x/3) and noticed that the domain is [-3,3]. However, i always thought that the domain of a sin-1 graph is always between [-1,1]. Can someone please explain where i'm lost at?
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 10, 2012, 05:05:44 pm
The factor in front of the x in the sine is $\frac{1}{3}$
\begin{alignedat}{1}x & =\frac{1}{3}x'
\\ x' & =3x
\\ (x,y)\rightarrow & (3x,y)
\end{alignedat}

That is we are dilating the graph of $sin^{-1}(x)$ by a factor of 3 from the y-axis, this changes the domain of the function we have. (there is also a dilation from the x-axis involved in this one too which I haven't considered in the above working, this dilation doesn't change the domain, but changes the range).

We could also look at it this way.
$\sin^{-1}(x)$ can 'take' values between -1 and 1 as input values.
So if we have $\frac{x}{3}$, then this can be between -1 and 1.
$-1\leq\frac{x}{3}\leq1$
So solving for x, our domain will become
$-3\leq x\leq3$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 10, 2012, 06:19:26 pm
Thanks b^3!

Could someone explain to me why Friction is pointing the way it is in the diagram below?

EDIT: Nevermind, i think i got it. It's because it says "Albert is on the verge of moving down the slide."
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 10, 2012, 06:50:59 pm
The info in the above digram relates to this question that i'm stuck on.

Take the bottom end of the slide as the origin,i  as the unit vector horizontally to the right and jas the unit vector vertically up.
d. Find the velocity vector v(t) which represents the velocity of Albert t seconds after he leaves the slide.

Note: in part c) i found that the speed of Albert as he reaches the end of the slide is 7m/s.
The answer to part d) is v (t) = (3.5 root3) i − (3.5+ gt) j
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 10, 2012, 07:09:19 pm
Assuming that Albert was travelling with a constant velocity once he reached the end of the slide, then once he leaves the slide the only force that will be acting on him is gravity.
Now since the slide is at an angle, we can make the velocity vector at the end of the slide into it's x (well i) and y (well j) components.
(http://i1082.photobucket.com/albums/j373/mclaren200800/vectocomps.png)
So our initial velocity vector will be $\underset{\sim}{u}(t)&=V_{x}\underset{\sim}{i}+V_{y}\underset{\sim}{j}$. (you will need to find these :P)
Now the as we said before the only force acting on the object is gravity at this point, so that means $a=-g\underset{\sim}{j}$
So
\begin{alignedat}{1}\frac{d\underset{\sim}{v}(t)}{dt} & =-g\underset{\sim}{j}
\\ \underset{\sim}{v}(t) & =\int-g\underset{\sim}{j}dt+\underset{\sim}{u}(t)
\end{alignedat}

So from that I think you should be able to get it :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 10, 2012, 07:51:54 pm
What i initially did was just subbed in t=0, v(t)=7 into v(t)=-gt i+c to solve for c, but why is this wrong? Why does the initial velocity have to be changed into i,j from?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 10, 2012, 08:01:40 pm
Remember $\underset{\sim}{v}(t)$ is a vector, not a scalar. 7 is just the magnitude of the vector, so we need to make it into it's components. Also be careful with your coordinate system, the gt should be on j not i as gravity acts downwards not sidewards :P
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 10, 2012, 10:11:46 pm
Q: why do we need to add the unit vectors of a and b?

THANKS

Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 10, 2012, 11:21:25 pm
Q: why do we need to add the unit vectors of a and b?

THANKS

Its the formula of unit vector bisects angle of 2 vectors. It takes long time to type LaTex and I'm doing homework but if you want I can post the proof tomorrow

I post now :P

They add $\underset {\sim}{\hat a} and \underset {\sim}{\hat b}$ by using parallelogram rule

Before that, they find unit vectors so that their magnitudes are equal because the diagonals of a rhombus bisects the angle at vertex.

(http://03.edu-cdn.com/files/static/mcgrawhill-images/9780071416504/f0048-01.jpg)

$\underset {\sim}{c}= \underset{\sim}{\hat a} + \underset {\sim}{\hat b}$ is vector bisects the angle of vector $\underset {\sim}{a}$ and $\underset {\sim}{b}$

$\Rightarrow \underset {\sim}{\hat c}= \frac{\underset{\sim}{\hat a} + \underset {\sim}{\hat b}}{ |\underset{\sim}{\hat a} + \underset {\sim}{\hat b}|}$
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 11, 2012, 05:11:58 pm
thank u :)
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 12, 2012, 10:29:41 pm
For the last step, the answer is $\frac{m\pi}{4}=\frac{(2k+1)\pi}{2}$

but why cannot I use the general solution i.e. $\frac{m\pi}{4}=\frac{\pi}{2}+2n\pi or \frac{-\pi}{2}+2n\pi$

Is that related to the restricted domain of the argument of the complex number ??

thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 12, 2012, 11:14:40 pm
For the last step, the answer is $\frac{m\pi}{4}=\frac{(2k+1)\pi}{2}$

but why cannot I use the general solution i.e. $\frac{m\pi}{4}=\frac{\pi}{2}+2n\pi or \frac{-\pi}{2}+2n\pi$

Is that related to the restricted domain of the argument of the complex number ??

thanks

You can use the general solutions, but it will be $\frac{m\pi}{4}=\frac{\pi}{2}+2n\pi and \frac{-\pi}{2}+2n\pi$ because the intervals between solutions in this equation are $\pi$ instead of $2\pi$

In the worked solution, they don't wanna separate solutions so they combine them

$\frac{\pi}{2} + k\pi = \frac{(2k+1)\pi}{2}$
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 13, 2012, 12:06:05 am
You can use the general solutions, but it will be $\frac{m\pi}{4}=\frac{\pi}{2}+2n\pi and \frac{-\pi}{2}+2n\pi$ because the intervals between solutions in this equation are $\pi$ instead of $2\pi$

In the worked solution, they don't wanna separate solutions so they combine them

$\frac{\pi}{2} + k\pi = \frac{(2k+1)\pi}{2}$

My worked general solution is the same as urs $\frac{m\pi}{4}=\frac{\pi}{2}+2n\pi and \frac{-\pi}{2}+2n\pi$ :)

so in the real exam can i write like that or I have to combine the two answers together? ennjy
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 13, 2012, 12:15:20 am
^ I think it should be fine for both

Like for solving general solutions of sin(x), there is an alternate formula as well
The formula look different but when you substitute values of k or n, the solutions will be the same
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 13, 2012, 12:18:27 am
Thanks ennjy
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on October 16, 2012, 07:49:04 am
Would anyone be able to explain the following question from NEAP:

I've always just considered using a maxima/minima approach, using the boundary of S as being (x,y) coordinates and just using distance formula, but what they've done is us |u-w|, which I thought would give the distance between the circles centre and w...
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on October 16, 2012, 01:04:43 pm
Hey,

Question in the diagram.

For where the arrow is pointing, I don't really understand how they can make that statement.

Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 16, 2012, 01:10:38 pm
Basically equate $z^{8}-z^{4}+1$ with $z^{8}-2\cos(\theta)z^{4}+1$, then you get $-2\cos(\theta)=-1$ by equating coefficients.
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 16, 2012, 06:08:57 pm
For part b) I don't understand how they use the area formula to gain the volume?

Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: 99.00 on October 16, 2012, 06:31:51 pm
^Use sector of the circle minus triangle.
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 16, 2012, 07:36:46 pm
Area of sector = $\frac{\alpha }{2\pi}\cdot \pi\cdot 1^{2}=\frac{1}{2}\alpha$   (since $\frac{\alpha }{2\pi}$ is the proportion of the circle that the sector takes up)

Area of triangle = $\frac{1}{2}\cdot1\cdot1\cdot\sin(\alpha)=\frac{1}{2}\sin(\alpha)$   (rule for area of a triangle)

Therefore cross-sectional area = $\frac{1}{2}\alpha-\frac{1}{2}\sin(\alpha)=0.5(\alpha-\sin(\alpha))$
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 16, 2012, 08:40:10 pm
umm.. but how does the cross sectional area relate to the volume in this case??
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 16, 2012, 10:00:28 pm
Length times cross-sectional area.

Wait, that would make it $\alpha-\sin(\alpha)$ which isn't what I got. But I think their "show that" thing is wrong.
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 16, 2012, 10:25:00 pm
Length times cross-sectional area.

Wait, that would make it $\alpha-\sin(\alpha)$ which isn't what I got. But I think their "show that" thing is wrong.

I think they made an error..not a good question.. Also, could u help me with this MC question?? I work out the answer B and C are both correct. the solution also states that B and C have the same answer but weird....they suggest C is the only answer

Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 16, 2012, 10:30:36 pm
Well you're right, they have the same asymptotes. =/
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 17, 2012, 06:23:50 pm
Well you're right, they have the same asymptotes. =/

thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on October 19, 2012, 04:49:50 pm
For  dependence, how do we know which of these formulas to use?
i) ma+nb+oc=0
ii) a=mb+nc

I used (i) for the 2008 VCAA exam,  and got into a messy situation, and the solutions suggested using (ii), but i'm having trouble working out the difference on when switch between both .......
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on October 19, 2012, 05:03:57 pm
With ma+nb+oc=0, if m = n = o = 0 is the only solution, then the vectors are linearly independent.

If a=mb+nc is true, then the vectors are linearly dependent. This should make sense, because you have one of the vectors being expressed in terms of the other vectors (i.e. it "depends" on the other vectors). You can derive this expression from the first equation too.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 20, 2012, 10:15:57 am
VCAA 2011 MC 7 EX2
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf

Does anyone have any fast ways of doing this question?
I graphed each graph to see which intersects twice but it takes a long time to do it this way.
Also, i tried using the solve command, but when i solved the eqn with option c, how come it only gave one coordinate instead of two?

Thanks for any help :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 20, 2012, 12:37:33 pm
I kind of just knew what the graphs looked like from experience, so they didn't take long to do quick sketches of. Do you know the perpendicular bisector thing? For lack of a better word =P
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 20, 2012, 12:45:30 pm
I think i came across it once, but i don't rmb much about it, could you perhaps explain to to me please?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 20, 2012, 01:36:34 pm
|z - u| = |z - v| (where u and v are complex numbers) represents the perpendicular bisector of the line joining u and v.

So for option B in that question they had |z + 3| = |z - 3i|, which is equivalent to |z - (-3)| = |z - 3i|. So it's the perpendicular bisector of the line joining -3 and 3i, which is the line Im(z) = -Re(z) (or alternatively, y = -x).

Option C is the perpendicular bisector of the line joining 3 and 3i, which is the line Im(z) = Re(z) (y = x). I think this was the correct answer.

Try and come up with a worded explanation of why that general expression gives the perpendicular bisector. If you can't get it then I'll tell you, but have a go.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 20, 2012, 07:25:19 pm
Thanks a lot for your help :)

Okay, I'll try: Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 20, 2012, 07:50:53 pm
Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?

what do you mean "the point (u,v)"?
Its the distance of z to the point u equal to the point v. Have a go again :P
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 20, 2012, 08:07:30 pm
Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?

what do you mean "the point (u,v)"?
Its the distance of z to the point u equal to the point v. Have a go again :P
Umm, so it forms a square? I'm confused now, i really don't know :S
Title: Re: Specialist 3/4 Question Thread!
Post by: Conic on October 20, 2012, 09:26:17 pm
This might seem a little vague, but what does ' the acceleration is always directed towards the origin' exactly mean?
What context exactly? I guess it could refer to vertical motion (i.e., objects being thrown directly upwards) always accelerating downwards due to gravity.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 20, 2012, 09:34:39 pm
I'm guessing this is referring to the motion of an object moving in a circular path (about the origin). In which the displacement of the particle is the circular path, the velocity vector is tangential to the path, and the acceleration vector is always towards the origin. This is because this acceleration is needed to keep moving the object in the circular path, if no force were to act on it, it would just keep travelling in a straight line, a force is needed to keep 'pulling it around' to make it move in the circular path.
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 21, 2012, 12:07:41 pm
Thanks a lot for your help :)

Okay, I'll try: Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?
u and v can be any complex numbers. So they won't necessarily only have one non-zero component.

Try this sort of thing:

Let $\underset{\sim}u=u_{1}\underset{\sim}i+u_{2}\underset{\sim}j$, $\underset{\sim}v=v_{1}\underset{\sim}i+v_{2}\underset{\sim}j$ and let $\underset{\sim}z=x\underset{\sim}i+y\underset{\sim}j$ be the position vector of any point on the perpendicular bisector of the line joining the points $(u_{1},u_{2})$ and $(v_{1},v_{2})$. What do $\underset{\sim}z-\underset{\sim}u$ and $\underset{\sim}z-\underset{\sim}v$ represent? What do their magnitudes represent?
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 21, 2012, 05:21:32 pm
I am confused with the part c). The answer gives that frictional force acts in the same direction as the tension. The friction must be opposed to the tension, isn't it?

my answer is $T-f-\sin20\times8g=0$
$T-0.3\times\cos20\times8g-\sin20\times8g=0$
$\therefore T=48.9 N$
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 21, 2012, 05:31:58 pm
We are trying to prevent the box from moving down the slope (as gravity is trying to pull it down), that means friction has to be in the opposite direction to which the block is wanting to move (this is the static friction type of friction). So that would be up the slope, in this case we are also appying a force up the slope, which is the tension.

In some cases the friction will be opposed to tension (e.g. a block on a flat slope being pulled by a rope), others it won't be (as in this case), it just depends on what the situation is.
Title: Re: Specialist 3/4 Question Thread!
Post by: |ll|lll| on October 23, 2012, 01:27:29 pm
a) Given z + 1/z = k, where k us a real constant, show that z either lies on the Re(z) axis, or the unit circle centred at the origin.
b) if z lies on the Re(z) axis, show that |k| ≥ 2.

What will everyone's approach be on this? I proved it using polar form, but not in a very elegant way so could someone shed some light? Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 23, 2012, 03:18:34 pm
We are trying to prevent the box from moving down the slope (as gravity is trying to pull it down), that means friction has to be in the opposite direction to which the block is wanting to move (this is the static friction type of friction). So that would be up the slope, in this case we are also appying a force up the slope, which is the tension.

In some cases the friction will be opposed to tension (e.g. a block on a flat slope being pulled by a rope), others it won't be (as in this case), it just depends on what the situation is.

thank you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on October 23, 2012, 04:14:26 pm
a) Given z + 1/z = k, where k us a real constant, show that z either lies on the Re(z) axis, or the unit circle centred at the origin.
b) if z lies on the Re(z) axis, show that |k| ≥ 2.

What will everyone's approach be on this? I proved it using polar form, but not in a very elegant way so could someone shed some light? Thanks
$z+\frac{1}{z} = k, k\in \mathbb{R}$
let $z=x+yi$
so $x+yi+\frac{1}{x+yi}=k$

$=x+yi + \frac{x-yi}{x^2+y^2}$

but this must equal to a real constant, so the imaginary part must equal zero

$\therefore y-\frac{y}{x^2+y^2}=0$

$\frac{y(x^2+y^2)-y}{x^2+y^2}=0$

$\therefore y(x^2+y^2-1)=0$

so then we get y=0 which implies that z lies on the Re(z) axis

or we get $x^2+y^2=1$ which shows that z lies on the unit circle centred at the origin

just give me a sec to look at the second part
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on October 23, 2012, 04:40:36 pm
If you consider the real component,

$\text{Re}(z) = x + \frac{x}{x^2+y^2}$
For z to lie on the Re(z) axis, y=0

$x + \frac{x}{x^2} = k$

$x + \frac{1}{x} =k$

$\frac{x^2+1}{x} =k$

$x^2 - kx +1 =0$

x must be a real number otherwise it won't lie on the Re(z) axis, therefore, there must be at least 1 real solution

$k^2 - 4 \ge 0$

$(k-2)(k+2) \ge 0$

$k \le -2 \; \text{or} \; k \ge 2$

$|k| \ge 2$
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on October 23, 2012, 04:48:17 pm
For part b) Polar's answer is a nice way to do it (although it should be $\ge$ instead of $>$), but I'm bored, so for something more visual:

z lies on the Re(z) axis implies that y=0

$\therefore x+yi+\frac{1}{x+yi}=k$ reduces to

$x+\frac{1}{x}=k$

now if we look at $x+\frac{1}{x}=k$ then we realise that the range of $x+\frac{1}{x}$ is all values that k can take. So now we need to find the range of $x+\frac{1}{x}$

The graph of $y=x+\frac{1}{x}$ looks like this:

(http://i.imgur.com/LGcSQ.png)

We can verify the values of those turning points by letting  $\frac{d}{dx}\left(x+\frac{1}{x}\right)=0$
From that we get $x=\pm 1$ which gives us turning points at $(x,y)\in \{(1,2),(-1,-2) \}$

So it follows that the range of $x+\frac{1}{x}$ is $k \in (-\infty,-2]\cup [2,\infty)$

which is equivalent to the statement: $|k|\ge 2$

For a proper exam response you might want a bit more explanation but whatever..
Title: Re: Specialist 3/4 Question Thread!
Post by: Bhootnike on October 24, 2012, 08:41:44 pm
(http://imgur.com/mLyVd)

http://imgur.com/mLyVd
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on October 24, 2012, 08:57:18 pm
I'm not sure if I've intepretted what you're asking right but anyways, hope it helps.

When we had a repeated root, we need to split the partial fractions to have a $\frac{A}{x+a}$ and a $\frac{B}{(x+a)^{2}}$ term.
Then you multiply by the factors on the bottom, expand out, collecting like terms and equate the coefficents so that you can find A, B and C.
For the example above
\begin{alignedat}{1}\frac{3x^{2}+10x-5}{(x+1)^{2}(x-2)} & =\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}
\\ 3x^{2}+10x-5 & =A(x+1)^{2}+B(x-2)(x+1)+C(x-2)
\\ 3x^{2}+10x-5 & =A(x^{2}+2x+1)+B(x^{2}-x-2)+C(x-2)
\\ 3x^{2}+10x-5 & =x^{2}\left(A+B\right)+x\left(2A-B+C\right)+\left(A-2B-2C\right)
\\ A+B & =3
\\ B & =3-A...[1]
\\ 2A-B+C & =10...[2]
\\ A-2B-2C & =-5...[3]
\\ {}
[1]\: into\:[2]
\\ 2A-3+A+C & =10
\\ 3A+C & =13...[4]
\\ {} [1]\: into\:[3]
\\ A-6+2A-2C & =-5
\\ 3A-2C & =1...[5]
\\ {} [4]-[5]
\\ 3C & =12
\\ \therefore C & =4
\\ {} into \: [4]
\\ 3A+4 & =13
\\ 3A & =9
\\ \therefore A & =3
\\ {} into \: [1]
\\ B & =3-3
\\ \therefore B & =0
\\ \therefore\frac{3x^{2}+10x-5}{(x+1)^{2}(x-2)} & =\frac{3}{x-2}+\frac{4}{(x+1)^{2}}
\end{alignedat}
Title: Re: Specialist 3/4 Question Thread!
Post by: StumbleBum on October 24, 2012, 09:03:09 pm
I'm not sure if I've intepretted what you're asking right but anyways, hope it helps.
That's how i interpreted what he was asking.
Also im impressed, you did all the latex faster than i did it by hand and scanned it onto my computer. Was just about to upload it and then you posted.
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on October 24, 2012, 09:19:54 pm
This is the method I use for partial fractions, I find it a lot quicker and easier: http://en.wikipedia.org/wiki/Heaviside_cover-up_method

$\frac{3x^{2}+10x-5}{(x+1)^{2}(x-2)} =\frac{A}{(x-2)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^{2}}$

Multiply all sides by (x-2) --> this will get A by itself, and let x = 2, this will get rid of the B and C terms (stuff multiplied by zero is zero).

This means we end up with $A = \frac{3x^2+10x-5}{(x+1)^2} = \frac{3(2)^2+10(2)-5}{(2+1)^2} = \frac{12+20-5}{9} = \frac{27}{9} = 3$

Multiply all sides by (x+1)^2 and let x = -1
$C = \frac{3x^{2}+10x-5}{(x-2)} = \frac{3(-1)^2+10(-1)-5}{(-1-2)} = \frac{3-10-5}{-3} = \frac{-12}{-3} = 4$

Now for B:
Multiply all sides by (x+1)
We get: $\frac{3x^{2}+10x-5}{(x+1)(x-2)} = B + \frac{C}{(x+1)}$
If let x = -1, we'll have division by zero. This is a problem.

Let's go back to $\frac{3x^{2}+10x-5}{(x+1)^{2}(x-2)} =\frac{A}{(x-2)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^{2}}$

We know the values for A and C now though. Let's say we let x = 0, and sub that into the original thingo here (as well as A = 3 and C = 4):

(I'm not showing all my working here, it's just a bit of subbing in and simplifying things down)
$\frac{5}{2} =-\frac{3}{2}+B+4 = B + \frac{5}{2}$

Easy enough to find B now. From there we can find $B = 0$

So $\frac{3x^{2}+10x-5}{(x+1)^{2}(x-2)} =\frac{3}{(x-2)}+\frac{4}{(x+1)^{2}}$
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 25, 2012, 04:14:25 pm
Question 3c) from vce 2008 exam2, why it is unnecessary to consider the y-component ?

my solving process: $\sin(\frac{t}{3})=0$  $gives t=0, 3\pi , 6\pi ...$

and  $\frac{1}{2}\sin(\frac{2t}{3})=0$  $gives t=0, \frac{3\pi}{2} , 3\pi ...$

Therefore it takes $3\pi$secs for the train to complete one circuit of the track as its start point is (0,0)

_____________________________________________________________________________________________

Umm... I probably see my problem...
because the train passes (0,0) twice, so $3\pi$ is the time taken for completing a half of the journey?

If it is the case, why simply just need to find the period of x component ?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 25, 2012, 08:13:51 pm
If it is the case, why simply just need to find the period of x component ?

Its about how you understand the question and show your working, not about how you get the answer $6\pi$

If you find the period of x component, then you can get the same answer (you should include the explanation as well, just to be clear in this case, the train starts from O) but in other cases, I don't think you can use this method
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 25, 2012, 09:20:43 pm
If it is the case, why simply just need to find the period of x component ?

Its about how you understand the question and show your working, not about how you get the answer $6\pi$

If you find the period of x component, then you can get the same answer (you should include the explanation as well, just to be clear in this case, the train starts from O) but in other cases, I don't think you can use this method

what's the correct method to solve this question? :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 26, 2012, 08:38:40 am
If it is the case, why simply just need to find the period of x component ?

Its about how you understand the question and show your working, not about how you get the answer $6\pi$

If you find the period of x component, then you can get the same answer (you should include the explanation as well, just to be clear in this case, the train starts from O) but in other cases, I don't think you can use this method

what's the correct method to solve this question? :)

I suggest you do the 1st method (x, y component) to be safe  :P
Title: Re: Specialist 3/4 Question Thread!
Post by: martin1106 on October 26, 2012, 10:54:57 pm
Thanks Ennjy :)
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on October 27, 2012, 10:55:11 am
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009specmaths2-w.pdf

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/spec2_assessrep_09.pdf

Q 3 f) Why does the assessment report square the magnitude of a?
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on October 27, 2012, 02:12:41 pm
^ When you calculate magnitude of acceleration, you take square root of i,j, k components right?
Eg: a= xi+yj+zk
|a| = sqrt(x^2+y^2+z^2)
They don't take square root of i,j,k components so they have to square both sides
|a|^2= x^2+y^2+z^2

sr, I'm using mobile so its hard to type LaTex :(

Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 27, 2012, 03:56:17 pm
If the square of |a| is constant then so is |a|, so you don't have have to square root it, but of course it's not wrong to square root it.
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on October 29, 2012, 07:36:19 pm
$\frac{dy}{dx}=sin(x^3)$ and y=2 when x=0, then the value of y correct to three decimal places when x=3.

Ok so I'm having trouble integrating this, CAS just gives me the exact same thing. So how would I go about diong this? thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on October 29, 2012, 07:41:42 pm
$y(3)-y(0)=\int_{0}^{3}\sin(x^{3})\mathrm{d}x$

$\Rightarrow y(3)=\int_{0}^{3}\sin(x^{3})\mathrm{d}x+y(0)$

$\Rightarrow y(3)=\int_{0}^{3}\sin(x^{3})\mathrm{d}x+2$
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on October 29, 2012, 08:09:39 pm
$\frac{dy}{dx}=sin(x^3)$ and y=2 when x=0, then the value of y correct to three decimal places when x=3.

Ok so I'm having trouble integrating this, CAS just gives me the exact same thing. So how would I go about diong this? thanks :)

Note that there in fact isn't actually an analytical solution to $\int{\sin{x^3}}dx$ so your calculator has to solve the original question using some kind of numerical method, like taking the sum of areas of very small rectangles or trapeziods, which will end up being a decent, but importantly not exact, approximation. It's like the difference between finding an exact solution to a differential equation or finding an approximation using Euler's rule, which we did as a part of DEs.
Title: Re: Specialist 3/4 Question Thread!
Post by: generalkorn12 on October 31, 2012, 09:22:11 am
For Vector Proofing (such as, parallelogram and rhombus), whats the difference between a=b and |a|=|b|, I seem to get these confused...

An example would be from the 2008 VCAA Exam 1:

A (1,0,5), B(-1,2,4), C(3,5,2), D(x,y,z). Determine the Points of D such that ABCD is a Parallelogram.

The worked solutions use DC=AB, but I figured, since its a parallelogram wouldn't the lengths be equal?
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on October 31, 2012, 11:25:55 am
For Vector Proofing (such as, parallelogram and rhombus), whats the difference between a=b and |a|=|b|, I seem to get these confused...

An example would be from the 2008 VCAA Exam 1:

A (1,0,5), B(-1,2,4), C(3,5,2), D(x,y,z). Determine the Points of D such that ABCD is a Parallelogram.

The worked solutions use DC=AB, but I figured, since its a parallelogram wouldn't the lengths be equal?

Remember that a vector has both direction and magnitude. If you say a = b, that means that both the magnitude and the direction of vectors a and b are the same - it means that a and b are the same vector. An example of this would be if both a and b were 2i + 3j (for example).

If you say that |a| = |b|, it means that the magnitudes of a and b are the same, but their directions might not be, for example, if a = i + 3j and b = 3i + j, they have the same magnitudes, but different directions and thus are different vectors.

If you want to prove that something is a parallelogram, you would have to prove DC = AB because the opposite sides have to be both EQUAL IN LENGTH and PARALLEL (same direction).
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 03, 2012, 12:15:33 pm
When plotting roots of a complex number on an argand plane, why is it incorrect to connect the point with the origin?
e.g VCAA 2006 ex 2, last question.

Also, can someone please explain to me how to do the attached question from VCAA 2006 ex2.
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 03, 2012, 12:36:23 pm
the complex number $z = a +bi$ is just a point at $(a,b)$ because it is a number and not a line. if you drew a line from $(0,0)$ to $(a,b)$ that would be the line $\text{Im}(z) = \frac{b}{a}\text{Re}(z)$ which is not just the complex number being represented anymore

the line joining $z_1$ and $-z_1$ is the line $\text{Re}(z) = \text{Im}(z)$ but it needs to be written in terms of $\bar{z_1}$

\begin{aligned} \text{Since} \; \; z_1 + \bar{z_1} &= 2\text{Re}(z)
\\ |z_1 + \bar{z_1}| &= 2\text{Re}(z)
\\ z_1 - \bar{z_1} &= 2\text{Im}(z)
\\ |z_1 - \bar{z_1}| &= 2\text{Im}(z)
\\ |z_1 + \bar{z_1}| & = |z_1 - \bar{z_1}|
\\ 2\text{Re}(z) &= 2\text{Im}(z)
\\ \text{Re}(z)&= \text{Im}(z) \end{aligned}

note that in this case, it was $z_1 - \bar{z_1}$ and $z_1 + \bar{z_1}$ which meant that the real and imaginary components were positive, thus, didn't require the modulus signs
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 05, 2012, 12:07:40 pm
A set of vectors are linearly dependent if one of them can be expressed as a sum of non zero multiples of another two vectors.

In VCAA 2011 exam 1, Q 9 c)ii), They have it as 3c-a=nb. This subtraction is not a"sum," so why is it linearly dependent?
I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 05, 2012, 12:28:26 pm
If one can be expressed as a linear combination of the other two, then they're linearly dependent. Doesn't matter if the coefficients are negative.

The geometric interpretation for dependence of three vectors in three dimensions is that they lie in the same plane.
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on November 05, 2012, 12:31:22 pm
I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 05, 2012, 12:33:28 pm
ok thanks :)

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath1-w.pdf

For Q 7b),
I did mg+T2+T1=0
mg+98+(98/root3)=0
and solved for m, getting a negative answer...

Whats wrong with me setting in out like this?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 05, 2012, 12:38:05 pm
You need to take into account the directions
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 05, 2012, 12:47:13 pm
Here's the diagram i worked from
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 05, 2012, 01:04:11 pm
If you use a vector method you need to use a vector sum. Define i and j components and go from there.

Though, it's easiest to use Lami's theorem. $\frac{mg}{\sin({90}^{\circ})}=\frac{T_{2}}{\sin({150}^{\circ})}=\frac{T_{1}}{\sin({120}^{\circ})}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Niskii on November 05, 2012, 04:45:42 pm
Hey guys, quick question regarding VCAA 2010 Exam 2 M/C Q.22

Just confused with how they got D. I understand the iTute worked solutions for the question for the most part, just not sure where the V02 comes from in the 3rd step? I understand why the V0 is there, just not why it's squared! I managed to find a previous thread on the same question but I'm still a bit confused.

Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 05, 2012, 06:27:48 pm
Hey guys, quick question regarding VCAA 2010 Exam 2 M/C Q.22

Just confused with how they got D. I understand the iTute worked solutions for the question for the most part, just not sure where the V02 comes from in the 3rd step? I understand why the V0 is there, just not why it's squared! I managed to find a previous thread on the same question but I'm still a bit confused.

I have no idea how they did what they did, but I'll show you how I solved it:

F = ma where F = F(x)
F(x) = ma
a = F(x) / m
d(1/2 v^2) / dx = F(x) / m
1/2 v^2 = 1/m * Integral of F(x) dx
Let Integral of F(x) dx = G(x) + C
1/2 v^2 = 1/m * (G(x) + C)
When v = v0, x = x0
1/2 (v0)^2 = 1/m * (G(x0) + C)
m/2 (v0)^2 = G(x0) + C
C = m/2 (v0)^2 – G(x0)
Therefore 1/2 v^2 = 1/m * (G(x) + m/2 (v0)^2 – G(x0))
v^2 = 2/m * (G(x) + m/2 (v0)^2 – G(x0))
When x = x1, v = v1
(v1)^2 = 2/m * (G(x1) + m/2 (v0)^2 – G(x0))
(v1)^2 = 2/m * (G(x1) – G(x0)) + (v0)^2
Since G(x1) – G(x0) = definite integral from x0 to x1 of F(x) dx
Then (v1)^2 = 2/m * definite integral from x0 to x1 of F(x) dx + (v0)^2
v1 = square root of all of that
OPTION D
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 05, 2012, 07:46:48 pm
Given z=(root2 - 1) + i (root2)

Find both values of root(1+z) in polar form.

I dont understand how they got (root2)cis(-7pi/8) as the second value. Anyone know how?
Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 05, 2012, 08:06:08 pm
I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

I have a question regarding this. For this general formular how do you know which of the three equations to make substitute in to find m and n when it just asks if this are linearly dependent. Does it matter which to equations to would add? Thank you! Any help is really appreciated
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on November 05, 2012, 08:07:34 pm
Given z=(root2 - 1) + i (root2)

Find both values of root(1+z) in polar form.

I dont understand how they got (root2)cis(-7pi/8) as the second value. Anyone know how?
Thanks

$Z = \sqrt{2} - 1 + \sqrt{2}i$

$\therefore Z + 1 = \sqrt{2} + \sqrt{2}i = 2cis\left(\frac{\pi}{4}\right) = 2cis\left(\frac{\pi}{4} + 2\pi n\right)$

$Z^{\frac{1}{2}} = \sqrt{2cis\left(\frac{\pi}{4} + 2\pi n\right)} = \sqrt{2}cis\left(\frac{\pi}{8}+\pi n \right)$

Look for which have an angle between $\pi$ and $- \pi$:

$\sqrt{2}cis\left(\frac{\pi}{8}\right) \ and \ \sqrt{2}cis\left(-\frac{7\pi}{8}\right)$

Note: For the above $n$ is an element of $Z$ (the set of real integers)
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 05, 2012, 11:32:18 pm
I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

I have a question regarding this. For this general formular how do you know which of the three equations to make substitute in to find m and n when it just asks if this are linearly dependent. Does it matter which to equations to would add? Thank you! Any help is really appreciated

It makes no difference. If they ask whether a, b and c are linearly dependant, you can make any of the following tests:
ma + nb = c
a + mb = nc
ma + b = nc
You only need to test out one of them and it really doesn't make a difference which one you choose. If the equation is solvable, then they are linearly dependent, otherwise they are linearly independent.
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 05, 2012, 11:56:39 pm
Not sure if I should bring this up but the determinant of $\begin{bmatrix}a_{x}&a_{y}&a_{z}\\b_{x}& b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{bmatrix}$ is zero if and only if $\underset{\sim}a$, $\underset{\sim}b$ and $\underset{\sim}c$ are linearly dependent. It's a fairly quick calculator test for a multiple choice question but it's probably a bad idea to show it in your working since it's outside the course.
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 06, 2012, 08:06:46 am
I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

I have a question regarding this. For this general formular how do you know which of the three equations to make substitute in to find m and n when it just asks if this are linearly dependent. Does it matter which to equations to would add? Thank you! Any help is really appreciated

It makes no difference. If they ask whether a, b and c are linearly dependant, you can make any of the following tests:
ma + nb = c
a + mb = nc
ma + b = nc
You only need to test out one of them and it really doesn't make a difference which one you choose. If the equation is solvable, then they are linearly dependent, otherwise they are linearly independent.

Thank you so much! Just another thing to clear up, if two of the three equations are parallel or collinear does that affect where the equations go?

also Is it different when it's just two equations? Do you only look for a scalar multiple I think I've read? Which means they are linearly dependent? Thanks, really appreciated! :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 06, 2012, 09:12:35 am
Yeah for two vectors, it's whether they're parallel/antiparallel
Title: Re: Specialist 3/4 Question Thread!
Post by: Niskii on November 06, 2012, 10:26:09 am
I have no idea how they did what they did, but I'll show you how I solved it:

F = ma where F = F(x)
F(x) = ma
a = F(x) / m
d(1/2 v^2) / dx = F(x) / m
1/2 v^2 = 1/m * Integral of F(x) dx
Let Integral of F(x) dx = G(x) + C
1/2 v^2 = 1/m * (G(x) + C)
When v = v0, x = x0
1/2 (v0)^2 = 1/m * (G(x0) + C)
m/2 (v0)^2 = G(x0) + C
C = m/2 (v0)^2 – G(x0)
Therefore 1/2 v^2 = 1/m * (G(x) + m/2 (v0)^2 – G(x0))
v^2 = 2/m * (G(x) + m/2 (v0)^2 – G(x0))
When x = x1, v = v1
(v1)^2 = 2/m * (G(x1) + m/2 (v0)^2 – G(x0))
(v1)^2 = 2/m * (G(x1) – G(x0)) + (v0)^2
Since G(x1) – G(x0) = definite integral from x0 to x1 of F(x) dx
Then (v1)^2 = 2/m * definite integral from x0 to x1 of F(x) dx + (v0)^2
v1 = square root of all of that
OPTION D

Perfect! Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 06, 2012, 11:08:21 am
I find that questions involving straight line motion with initial conditions are often made simpler by applying the fundamental theorem of calculus and then the substitution rule

$a=v\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{F(x)}{m}$

$\Rightarrow \int_{x_{0}}^{x_{1}}v\frac{\mathrm{d} v}{\mathrm{d} x}\mathrm{d}x=\frac{1}{m}\int_{x_{0}}^{x_{1}}F(x)\mathrm{d}x$ (*)

The key part here is understanding why I integrated with respect to x in particular. On the left hand side we had $\frac{\mathrm{d}v}{\mathrm{d}x}$, and we know from the substitution rule that the dxs will 'cancel';

$\int_{a}^{b}f(u)\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d}x=\int_{g(a)}^{g(b)}f(u)\mathrm{d}u$, where $u=g(x)$

So applying this to the left hand side of (*):

$\int_{v_{0}}^{v_{1}}v\mathrm{d}v=\frac{1}{m}\int_{x_{0}}^{x_{1}}F(x)\mathrm{d}x$ (make sure the terminals match up)

$\Rightarrow \left [ \frac{1}{2}v^{2} \right ]_{v_{0}}^{v_{1}}=\frac{1}{m}\int_{x_{0}}^{x_{1}}F(x)\mathrm{d}x$

$\Rightarrow \frac{1}{2}v_{1}^{2}-\frac{1}{2}v_{0}^{2}=\frac{1}{m}\int_{x_{0}}^{x_{1}}F(x)\mathrm{d}x$

$\Rightarrow v_{1}^{2}=\frac{2}{m}\int_{x_{0}}^{x_{1}}F(x)\mathrm{d}x+v_{0}^{2}$

$\Rightarrow v_{1}=\sqrt{\frac{2}{m}\int_{x_{0}}^{x_{1}}F(x)\mathrm{d}x+v_{0}^{2}}$, assuming $v_{1}$ is positive
Title: Re: Specialist 3/4 Question Thread!
Post by: d3stiny on November 06, 2012, 02:16:18 pm
Do we have to explicitly use u-substitution for anti differentiating f'(x)/f(x) functions?
i.e. would $\int \frac{x}{1-x^2}dx = -\frac{1}{2}\log_{e}|1-x^2| + c$ be sufficient working?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 06, 2012, 02:42:06 pm
Yeah, if you're capable of going straight there it's fine, unless they specifically ask you to use substitution.
Title: Re: Specialist 3/4 Question Thread!
Post by: Niskii on November 06, 2012, 06:20:25 pm
@ClimbTooHigh: Ah yes that works too! Originally I was attempting to do it that way but it looks like I made an error with the left hand side. Thank you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 06, 2012, 06:39:34 pm
Yea, using FTC and substitution usually takes a lot of care
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 07, 2012, 08:01:15 am
Do we have to explicitly use u-substitution for anti differentiating f'(x)/f(x) functions?
i.e. would $\int \frac{x}{1-x^2}dx = -\frac{1}{2}\log_{e}|1-x^2| + c$ be sufficient working?
Or if the question is worth multiple marks. Often times the u substitution is involved in the marking criteria of integration questions.
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on November 07, 2012, 04:12:27 pm
Hey guys,

Does anyone know if we are allowed to use Half-Angle formulas for trig?

I ... can't seem to recall how I know them ... -_-
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 07, 2012, 05:22:39 pm
Don't see why not
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on November 07, 2012, 05:43:26 pm
Hey guys which VCAA paper from the last few years do you think has been the most difficult? For both exam 1 and exam 2.
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 07, 2012, 06:41:08 pm
2008 both
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 07, 2012, 06:57:59 pm
^agreed
Title: Re: Specialist 3/4 Question Thread!
Post by: Hutchoo on November 07, 2012, 07:43:51 pm
I don't really understand what's wrong with my brain.. but I did a lot better on the VCAA 2008 than I did on last years D= Last years was "the easiest spesh exam" as well.
Title: Re: Specialist 3/4 Question Thread!
Post by: StumbleBum on November 07, 2012, 09:12:21 pm
What is this 'k' constant from integration and where did they get it from...?

$t+c=-\frac{1}{2}log|g-2v|$
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 07, 2012, 09:18:34 pm
Normally "+c" is referred to as the constant of integration, but in this context it can be put inside the log since log(a) + log(b) = log(ab).
Title: Re: Specialist 3/4 Question Thread!
Post by: StumbleBum on November 07, 2012, 09:20:28 pm
Normally "+c" is referred to as the constant of integration, but in this context it can be put inside the log since log(a) + log(b) = log(ab).

So they just treated the constant as $log_e(k)$ then simplified that way?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 07, 2012, 09:22:39 pm
Yes
Title: Re: Specialist 3/4 Question Thread!
Post by: StumbleBum on November 07, 2012, 09:28:14 pm
Yes

Would my way yield the same result?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 07, 2012, 09:47:00 pm
It would, yes. You'd just have c equal to ln(k).
Title: Re: Specialist 3/4 Question Thread!
Post by: StumbleBum on November 07, 2012, 09:57:03 pm
I men't like this:

\begin{align} \int \frac{1}{g-2v} dv &= \int dt \\ -\frac{1}{2}\log_e|g-2v| &= t+c \\ -\frac{1}{2}\log_e|g-4g| &= c , \text{initial condition} t=0, v=2g \\ -\frac{1}{2}\log_e|-3g| &= c \\ -\frac{1}{2}\log_e|g-2v| &= t-\frac{1}{2}\log_e|-3g| \\ \frac{1}{2}\log_e|-3g| -\frac{1}{2}\log_e|g-2v| &= t \\ \frac{1}{2}\log_e|\frac{-3g}{g-2v}| &= t \end{align}

It doesn't give the same answer? or am I doing something wrong...

EDIT: OH GAWDD, realised that they are equal... didn't see the answers negative at the front
Title: Re: Specialist 3/4 Question Thread!
Post by: WhoTookMyUsername on November 08, 2012, 05:02:51 pm
yep effectively
a constant can be anything that is constant
k^1000000
ln(k)
log_100000_(K)

(k-1)^e^2
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 08, 2012, 05:16:19 pm
Can someone help me to solve:

z^3 - (2+2i)z^2 + 3iz + 1 - i = 0 given (z-i) is factor?

I tried division (short method) but doesn't seem to be working for me...

Any help would be reaally appreciated. Thank you!
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on November 08, 2012, 05:35:46 pm
Can someone help me to solve:

z^3 - (2+2i)z^2 + 3iz + 1 - i = 0 given (z-i) is factor?

I tried division (short method) but doesn't seem to be working for me...

Any help would be reaally appreciated. Thank you!

You can do long division but normally I prefer other methods

Method1: You can try with other solutions like z=-i,1,-1 etc then you can get one more solution
In this case, z-1 is another solution
so you have (z-i)(z-1)(z-a) => from there you can easily find a

Method2: I like this one :P because it works in all cases without trying to substitute many values of z or if solutions are not easily found like 1,-1,i,-i
$(z-i)(z^2+az+1+i)$

$a-i=-2-2i \implies a=-2-i$

now you solve: $z^2- (2+i)z+1+i =0 \implies z=1 or 1+i$
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 08, 2012, 05:44:33 pm
Can someone help me to solve:

z^3 - (2+2i)z^2 + 3iz + 1 - i = 0 given (z-i) is factor?

I tried division (short method) but doesn't seem to be working for me...

Any help would be reaally appreciated. Thank you!

You can do long division but normally I prefer other methods

Method1: You can try with other solutions like z=-i,1,-1 etc then you can get one more solution
In this case, z-1 is another solution
so you have (z-i)(z-1)(z-a) => from there you can easily find a

Method2: I like this one :P because it works in all cases without trying to substitute many values of z or if solutions are not easily found like 1,-1,i,-i
$(z-i)(z^2+az+1+i)$

$a-i=-2-2i \implies a=-2-i$

now you solve: $z^2- (2+i)z+1+i =0 \implies z=1 or 1+i$

Thank you so much! Just a question about the method 2 though, how did you know (z^2 + az + 1 + i) is the other factor? As in where did the 3z in the equation go and how do you know it was 1+i ? Thank you! :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on November 08, 2012, 05:50:04 pm
Just a question about the method 2 though, how did you know (z^2 + az + 1 + i) is the other factor? As in where did the 3z in the equation go and how do you know it was 1+i ? Thank you! :)

The constant part is (1-i) so (-i) x *a number*= 1-i => this number = 1+i (my explanation sucks  :()
Hope you can get what I mean :S
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 08, 2012, 05:55:11 pm
Just a question about the method 2 though, how did you know (z^2 + az + 1 + i) is the other factor? As in where did the 3z in the equation go and how do you know it was 1+i ? Thank you! :)

The constant part is (1-i) so (-i) x *a number*= 1-i => this number = 1+i (my explanation sucks  :()
Hope you can get what I mean :S

Sorry for being annoying... I'm still confused. So do you multiply the constant by the solution you know?

So are you saying -i x (1+i) = -i+1 but.. :(
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on November 08, 2012, 05:59:40 pm
So are you saying -i x (1+i) = -i+1 but.. :(

but?
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 08, 2012, 06:02:35 pm
So are you saying -i x (1+i) = -i+1 but.. :(

but?

Sorry! as in it doesn't equal to 1+i like you said...
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on November 08, 2012, 06:13:45 pm
So are you saying -i x (1+i) = -i+1 but.. :(

but?

Sorry! as in it doesn't equal to 1+i like you said...

-i x (1+i)= -i-i^2= -i+1
Title: Re: Specialist 3/4 Question Thread!
Post by: nina_rox on November 08, 2012, 06:22:57 pm
So are you saying -i x (1+i) = -i+1 but.. :(

but?

Sorry! as in it doesn't equal to 1+i like you said...

-i x (1+i)= -i-i^2= -i+1

Nevermind, thanks a load anyway. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: aznxD on November 08, 2012, 06:24:39 pm
2010 Exam 1 - Q2 b)

Why do you take the negative value of |(v-4)/4|?

When t=0, v=0, sub in these values and you'll see that only $v=-4e^{\frac{t}{2}} + 4$ satisfies the initial conditions.
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on November 08, 2012, 06:26:50 pm
2010 Exam 1 - Q2 b)

Why do you take the negative value of |(v-4)/4|?
"initially at rest", so at t = 0, v = 0.

We end up getting to a step with $v = 4 \pm 4e^{\frac{t}{2}}$

Sub in t = 0, $v = 4 \pm 4$. We know that v must equal zero, it's the negative case that'll get us this solution.

edit: oh didn't see that there was another page with aznxD's post.
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 08, 2012, 09:19:01 pm

The path of a particle is given by r = sin(t)*i - cos(2t)*j, 0 ≤ t ≤ 5pi/4
Find a vector in the direction of motion of the particle at t = pi/4
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on November 08, 2012, 09:28:25 pm
The vector in the direction of motion, will be in the direction of the velocity vector.
So finding the velocity vector
\begin{alignedat}{1}\underset{\sim}{r}(t) & =\sin(t)\underset{\sim}{i}-\cos(2t)\underset{\sim}{j}
\\ \dot{\underset{\sim}{r}}(t) & =\cos(t)\underset{\sim}{i}+2\sin(2t)\underset{\sim}{j}
\end{alignedat}

At $t=\frac{\pi}{4}$
\begin{alignedat}{1}\dot{\underset{\sim}{r}}(\frac{\pi}{4}) & =\cos(\frac{\pi}{4})\underset{\sim}{i}+2\sin(2\times\frac{\pi}{4})\underset{\sim}{j}
\\ & =\frac{\sqrt{2}}{2}\underset{\sim}{i}+2\underset{\sim}{j}
\end{alignedat}

We might have to find the unit vector, depending on the question, well the vector above gives the direction anyway, but I'll do it just in case.
\begin{alignedat}{1}\dot{|\underset{\sim}{r}}(\frac{\pi}{4})| & =\sqrt{\left(\frac{\sqrt{2}}{2}\right)^{2}+2^{2}}
\\ & =\sqrt{\frac{2}{4}+4}
\\ & =\sqrt{\frac{9}{2}}
\\ & =\frac{3\sqrt{2}}{2}
\\ \mathrm{\underset{\sim}{r}} & =\frac{2}{3\sqrt{2}}\left(\frac{\sqrt{2}}{2}\underset{\sim}{i}+2\underset{\sim}{j}\right)
\\ & =\frac{1}{3}\underset{\sim}{i}+\frac{2\sqrt{2}}{3}\underset{\sim}{j}
\end{alignedat}
Title: Re: Specialist 3/4 Question Thread!
Post by: Special At Specialist on November 08, 2012, 09:47:55 pm
^^ Thanks. I get it now.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 09, 2012, 01:32:21 pm
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf

With MC Q21, If i assign acceleration as positive in the downward direction, i get :

Mg-T=M(7/5)
T-5g=5(-7/5)

Why does this not get me the value of M?

Instead when i solve using simultaneous eqns with :
Mg-T=M(7/5)
T-5g=5(7/5)
I get the correct answer. So why does the second eqn have positive acceleration when i have assigned the downwards direction as positive. Shouldn't it be negative acceleration since the 5kg mass is moving up?

Also, with MCQ22, can someone explain to me how to start it off?
I don't understand the first line of Itute solutions for it. a=F(x)/m
Thanks :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 09, 2012, 01:44:34 pm
If the downward direction is positive then 5g should be positive and T should be negative
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 09, 2012, 01:47:52 pm
F = ma is newton's second law, F(x) is a function for a variable force

\begin{aligned} F(x) &= ma
\\ a &= \frac{F(x)}{m}\end{aligned}

they equate two expressions for acceleration in the first line
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 09, 2012, 03:33:27 pm
Dot products are only applied to vectors in i,j,k form right? so if vectors are written in a,b,c notation, we can't use dot product? Just need clarification, thanks.
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on November 09, 2012, 03:37:48 pm
You can't apply the dot product to $\underset{\sim}{a}$ as you would with components, if you don't have the components, but you can still 'dot' them, you just end up with the term $\underset{\sim}{a}.\underset{\sim}{b}$ and may have to work with that, or if you dot a vector with itself you get the magnitude which you may need to manipulate to show something.
e.g.
\begin{alignedat}{1}\underset{\sim}{a}.\underset{\sim}{a} & =|\underset{\sim}{a}|^{2}
\\ \left(x_{1}\underset{\sim}{i}+x_{2}\underset{\sim}{j}+x_{3}\underset{\sim}{k}\right).\left(y_{1}\underset{\sim}{i}+y_{2}\underset{\sim}{j}+y_{3}\underset{\sim}{k}\right) & =x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}
\end{alignedat}

Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on November 09, 2012, 03:39:27 pm
That's correct, unless you know the dot products of a, b and c.

Because the dot product distributed over brackets: a.(b+c) = a.b + a.c,

and we know the dot products of the unit vectors: i.j = j.k = k.i = 0 and i.i = j.j = k.k = 1,

that's why we can use the dot product in that way, but if we don't know the dot products of the vectors then we can't use the dot product in that way.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 09, 2012, 04:36:25 pm
With connected mass problems, at the instant the string breaks, does tension exist?

Also, with questions involving a person jumping out of a plane, at the instant the parachute opens, does the magnitude of the velocity immediately change? e.g velocity of man falling is 80km/hr, then parachute opens, his velocity is immediately 50km/hr or will it decrease from 80 to 50 after a certain amount of time?

One more: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf
With Q 2e) Where in the question did it indicate that acceleration is constant, so that the constant acceleration formula can be used?

Many thanks for all the help :)
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on November 09, 2012, 06:27:43 pm
Hey could someone help me with MC question 11 from the VCAA 2011 exam please?

$\underset{1}F=-\sqrt{3}j$      $\underset{2}F=i+\sqrt{3}j$      $\underset{3}F=-\frac{1}{2}i+\frac{\sqrt{3}}{2}j\$

The magnitude of the sum of these three forces is equal to?

Don't really get how to do these force questions, thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on November 09, 2012, 06:30:52 pm
Hey could someone help me with MC question 11 from the VCAA 2011 exam please?

$\underset{1}F=-\sqrt{3}j$      $\underset{2}F=i+\sqrt{3}j$      $\underset{3}F=-\frac{1}{2}i+\frac{\sqrt{3}}{2}j\$

The magnitude of the sum of these three forces is equal to?

Don't really get how to do these force questions, thanks!
Just think of them as vectors really, the forces is really just giving the maths some context I guess. Add them up and find the magnitude. You'll be doing vector addition, so you'd be taking into account the direction of the forces that way.

$R = (1-\frac{1}{2})i + (\sqrt{3} - \sqrt{3} + \frac{\sqrt{3}}{2})j$

$|R| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$

edit: Then just go through each option and look for the one that has a magnitude of 1, that happens to be E.
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on November 09, 2012, 06:50:57 pm
With connected mass problems, at the instant the string breaks, does tension exist?

Also, with questions involving a person jumping out of a plane, at the instant the parachute opens, does the magnitude of the velocity immediately change? e.g velocity of man falling is 80km/hr, then parachute opens, his velocity is immediately 50km/hr or will it decrease from 80 to 50 after a certain amount of time?

One more: http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010specmath2-w.pdf
With Q 2e) Where in the question did it indicate that acceleration is constant, so that the constant acceleration formula can be used?

Many thanks for all the help :)

1. Yes. Tension doesn't exist after that
2. I think velocity changes because as parachute opens, the air resistance increases, thus body deccelerates => affects velocity
3. Use a=dv/dt then integrate v from t=0 to t=3
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on November 10, 2012, 03:21:02 pm
2010 MC Q22:
(http://i.imgur.com/tfg51.png)

Can someone help me with this question? I had a vague idea what I was doing and got it correct, but what is a methodical way of approaching this question? It's mostly the $v_0^2$ bit that's confusing me a bit.

EDIT: Don't worry, I got it!
Good old F=ma...

$F=m\cdot \frac{d}{dx}\left(\frac{v^2}{2}\right)$

$\therefore F\cdot \frac{2}{m}=\frac{d}{dx}(v^2)$

$\frac{2}{m}\int^{x_1}_{x_0}F(x) dx=\int^{v_1}_{v_0}\frac{d}{dx}(v^2)dx$

$\therefore {v_1}^2-{v_0}^2=\frac{2}{m}\int^{x_1}_{x_0}F(x) dx$

$\therefore v_1=\sqrt{\frac{2}{m}\int^{x_1}_{x_0}F(x) dx+{v_0}^2}$

Edit II:
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 10, 2012, 03:43:59 pm
2010 MC Q22:
(http://i.imgur.com/tfg51.png)

Can someone help me with this question? I had a vague idea what I was doing and got it correct, but what is a methodical way of approaching this question? It's mostly the $v_0^2$ bit that's confusing me a bit.

Ok you use definite integrals on both sides were F(x)=ma a=vdv/dx you make v0 and x0 both lower limits for the integrals and v1 and x1 are upper limits.
Title: Re: Specialist 3/4 Question Thread!
Post by: kensan on November 10, 2012, 04:25:04 pm
Could someone help with with question 5b please? I don't really get how to show that in this scenario, thank you.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 10, 2012, 04:33:12 pm
\begin{aligned} \frac{dx}{dt} &= R_{in} - R_{out}
\\ &= 20e^{-0.2t} - \frac{10x}{10 + (20-10)t}
\\ &= 20e^{-0.2t} - \frac{x}{t+1}
\\ \frac{dx}{dt} + \frac{x}{1+t} &= 20e^{-0.2t}\end{aligned}
Title: Re: Specialist 3/4 Question Thread!
Post by: Mr. Study on November 11, 2012, 12:22:18 pm

Are we allowed to use the third derivative to confirm a point of inflection?
Title: Re: Specialist 3/4 Question Thread!
Post by: sodapop on November 11, 2012, 12:36:05 pm
Hey guys, could someone explain this answer (the correct answer is in red)? I understand that the region defined by |z|= 2 is a circle with r=2. By why is |z| >or= |z-(-sqrt(3) + i ) | defined by that line? And why is the required region above the line?
Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 11, 2012, 01:08:38 pm
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2004specmaths2.pdf

final 2 parts of the complex number question.
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on November 11, 2012, 01:13:58 pm
Hey guys, could someone explain this answer (the correct answer is in red)? I understand that the region defined by |z|= 2 is a circle with r=2. By why is |z| >or= |z-(-sqrt(3) + i ) | defined by that line? And why is the required region above the line?
Thanks!

$\left\{ z: |z| \le 2 \right\} \cap \left\{ z: |z| \ge |z - z_1| \right\}$

So we have $\left\{ z: |z| \ge |z - z_1| \right\}$ bit. If we had $|z| = |z - z_1|$ That's just the perpendicular bisector of the line joining $(-\sqrt{3},1)$ and the origin.

With perpendicular bisectors, each point on it is the same distance from the end points of the line that it's cutting. I think I've linked this before, but I like the picture here http://mathworld.wolfram.com/PerpendicularBisectorTheorem.html. That's what $|z| = |z-z_1|$ is saying isn't it? We have the line from the origin to $(-\sqrt{3},1)$ - so those are our end points. So z being a set of complex numbers, |z| is the distance of some complex number from the origin, and it happens to be equal to the distance between some complex number and the point $(-\sqrt{3},1)$. Hence that's why we know it's that line. (I hope that reasoning is okay, hopefully someone else will be able to confirm).

That's what the parts of the question before this one was pushing you towards. If you couldn't recognise that straight away, the question before had you state that $|z| = |z - z_1|$ is equivalent to $y = \sqrt{3}x + 2$ (which you could have shown by subbing in z=x+yi, but that'd just be going the long way around).

Now note the greater than, that's why we're looking at the region that's above that line.

What about the $\left\{ z: |z| \le 2 \right\}$ bit? Well that's just a circle with radius 2.

Now the $\cap$ --> intersection. The region we want is the region when $\left\{ z: |z| \le 2 \right\}$  and $\left\{ z: |z| \ge |z - z_1| \right\}$  overlap.
Title: Re: Specialist 3/4 Question Thread!
Post by: sodapop on November 11, 2012, 01:44:25 pm
Ah, thanks laseredd, that's a brilliant explanation, I completely get why the equation defines that particular line. But I'm still confused why the "greater than" part would specify that the region is above the line though.. (The equation is y >or= sqrt(3)x + 1) Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on November 11, 2012, 01:54:34 pm
Ah, thanks laseredd, that's a brilliant explanation, I completely get why the equation defines that particular line. But I'm still confused why the "greater than" part would specify that the region is above the line though.. (The equation is y >or= sqrt(3)x + 1) Thanks!
$\left\{ z: |z| \ge |z - z_1| \right\}$

So we have something that looks like this right:
(http://i.imgur.com/dAwQR.png)

If the $|z| = |z-z_1|$, we'll have something like this (pretend that the lines are of equal length):
(http://i.imgur.com/CcrWn.png)

But if we have $|z| \ge |z-z_1|$, then we can have the |z| being longer than the |z-z_1| line:

(http://i.imgur.com/iGLsz.png)

In other words, the point z could be any point above that line.

edit: fixed error on diagram 3
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 11, 2012, 02:02:33 pm
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths2_assessrep_11.pdf

VCAA 2011 EXAM 2 Q5c)i)

To satisfy the differential eqn, on the LHS, why was (x/1+t) replaced with the expression for x(t) ? Are the two the same? How? Why?

Thank you for any assistance :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 11, 2012, 02:11:40 pm
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths2_assessrep_11.pdf

VCAA 2011 EXAM 2 Q5c)i)

To satisfy the differential eqn, on the LHS, why was (x/1+t) replaced with the expression for x(t) ? Are the two the same? How? Why?

Thank you for any assistance :)
They are just asking you to differentiate. They purposely gave that formula for that reason.
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 11, 2012, 02:12:53 pm
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths2_assessrep_11.pdf

VCAA 2011 EXAM 2 Q5c)i)

To satisfy the differential eqn, on the LHS, why was (x/1+t) replaced with the expression for x(t) ? Are the two the same? How? Why?

Thank you for any assistance :)
They are just asking you to differentiate. They purposely gave that formula for that reason.
Oh sorry i mean part c) ii) :)
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 11, 2012, 02:19:59 pm
nah, they're not the same. when you solve the differential equation $\frac{dx}{dt} + \frac{x}{t+1} = 20e^{-0.2t}$, x(t) is the solution
Title: Re: Specialist 3/4 Question Thread!
Post by: sodapop on November 11, 2012, 02:35:28 pm
Ah, thanks laseredd, that's a brilliant explanation, I completely get why the equation defines that particular line. But I'm still confused why the "greater than" part would specify that the region is above the line though.. (The equation is y >or= sqrt(3)x + 1) Thanks!
$\left\{ z: |z| \ge |z - z_1| \right\}$

So we have something that looks like this right:
(http://i.imgur.com/dAwQR.png)

If the $|z| = |z-z_1|$, we'll have something like this (pretend that the lines are of equal length):
(http://i.imgur.com/CcrWn.png)

But if we have $|z| \ge |z-z_1|$, then we can have the |z| being longer than the |z-z_1| line:

(http://i.imgur.com/iGLsz.png)

In other words, the point z could be any point above that line.

edit: fixed error on diagram 3

Ahh, okay! Get it now. Thanks laseredd, you're a genius. :D
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 11, 2012, 02:36:20 pm
I'm still confused about it. I don't understand how to verify that x(t) satisfies the differential eqn. I don't see what they did on the report. Could someone perhaps provide a description of what they did for me?
Also, with satisfying the initial condition, we just sub t=0 and make it equal 0, and this would satisfy it because at t=0 the amount in the tank was 0?
Title: Re: Specialist 3/4 Question Thread!
Post by: WhoTookMyUsername on November 11, 2012, 02:37:57 pm
Anyway to convert a parametric Equation to a cartesian equation on cas?
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on November 11, 2012, 02:40:00 pm
Anyway to convert a parametric Equation to a cartesian equation on cas?

I know of a way with the Classpad, but I don't think there's a way with the nSpire.
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 11, 2012, 02:44:30 pm
$\frac{dx}{dt} + \frac{x}{1+t} = 20e^{-0.2t}$

substitute the $\frac{dx}{dt}$ from 5(c)(i) and $x(t)$ stated in the question, simplify and eventually, it simplifies to $20e^{-0.2t}$ and thus, satisfies the differential equation.

yeah, that's what you do to show it satisfies the condition
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 11, 2012, 02:47:32 pm
$\frac{dx}{dt} + \frac{x}{1+t} = 20e^{-0.2t}$

substitute the $\frac{dx}{dt}$ from 5(c)(i) and $x(t)$ stated in the question, simplify and eventually, it simplifies to $20e^{-0.2t}$ and thus, satisfies the differential equation.

yeah, that's what you do to show it satisfies the condition
Yes but they replaced x(t) with (x/1+t) which is what is confusing me. Why did they do this?
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 11, 2012, 02:49:25 pm
they substitute x(t) where the x was and that's why the denominator is $(1+t)^2$ instead of $1+t$
Title: Re: Specialist 3/4 Question Thread!
Post by: soccerboi on November 11, 2012, 02:51:27 pm
they substitute x(t) where the x was and that's why the denominator is $(1+t)^2$ instead of $1+t$
oooohhh!! Thanks so much!!! Finally i understand it :D
Title: Re: Specialist 3/4 Question Thread!
Post by: WhoTookMyUsername on November 11, 2012, 03:04:08 pm
Anyway to convert a parametric Equation to a cartesian equation on cas?

I know of a way with the Classpad, but I don't think there's a way with the nSpire.
You could do it on tiinspire before a recent update, just by typing in and pressing solve, but you can't do that anymore :(
Also is piSx^2dy
An acceptable line of notation? As your not integrating with respect to x, but with respect to y?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on November 11, 2012, 03:06:08 pm
It's probably not the best notation for VCAA exams (even though I used it in my own notes etc.) unless you define x=... first. Play it safe imo.
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 11, 2012, 05:20:06 pm
i do parametric equations by hand and if that isn't possible then i just use the parametric equations as they are. You can sketch them on a ti now btw.
Title: Re: Specialist 3/4 Question Thread!
Post by: rife168 on November 11, 2012, 06:33:03 pm
Anyway to convert a parametric Equation to a cartesian equation on cas?

I know of a way with the Classpad, but I don't think there's a way with the nSpire.

How do you do it with the Classpad?
Title: Re: Specialist 3/4 Question Thread!
Post by: supermanflyaway on November 11, 2012, 08:50:17 pm
Please explain VCAA 2007 4.c) and 2011 MC 18 (why find x2).
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 11, 2012, 09:07:20 pm
for 4) c) get the velocity vector then you use tan inverse to come up with the angle.
MC 18) it says scalar resolute in the direction of u so you find u unit vector and then get the dot product with v and then you make it equal to 1 and so a=3/2
Title: Re: Specialist 3/4 Question Thread!
Post by: supermanflyaway on November 11, 2012, 09:09:51 pm
for 4) c) get the velocity vector then you use tan inverse to come up with the angle.
MC 18) it says scalar resolute in the direction of u so you find u unit vector and then get the dot product with v and then you make it equal to 1 and so a=3/2

What values do you use tan inverse on?

Sorry, it was 2011 MC
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 11, 2012, 09:14:50 pm
k value over the value of the line between i and j (use pythagoras theorem).
Title: Re: Specialist 3/4 Question Thread!
Post by: Biceps on November 11, 2012, 09:16:45 pm
as for MC 2011 question 18 all you have to do is eulers formula but you use x as your y and t as your usual x.
Title: Re: Specialist 3/4 Question Thread!
Post by: Wingtips on November 12, 2012, 09:39:58 am
Vcaa 2011 2d.ii

How is z=x+ix?

I worked it out with something like
Re(z)=Im(z)
And subbed in values to get the required line.
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on November 12, 2012, 10:50:24 am
Vcaa 2011 2d.ii

How is z=x+ix?

I worked it out with something like
Re(z)=Im(z)
And subbed in values to get the required line.
You did pretty much the same thing, instead you used the values as Re(z) = Im(z). Since z = x + yi, Re(z) = x, Im(z) = y

So we want to express the line y = x in the form $z = a \bar{z}$

$x + yi = a(x - yi)$

sub in y = x
$x + xi = a(x - xi)$

and so on
Title: Re: Specialist 3/4 Question Thread!
Post by: zvezda on November 18, 2012, 06:12:25 pm
Use the double angle formula for tan2x, and the fact that tan(pi/4)=1, to find the exact value of tan(pi/8).
Having a bit of trouble with this one; how would I go about doing this?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on November 18, 2012, 06:21:22 pm
Use the double angle formula for tan2x, and the fact that tan(pi/4)=1, to find the exact value of tan(pi/8).
Having a bit of trouble with this one; how would I go about doing this?

Hint: let 2x = pi/4 and then rearrange and algebraically have a go at it :)
Title: Re: Specialist 3/4 Question Thread!
Post by: zvezda on November 18, 2012, 06:50:47 pm
Use the double angle formula for tan2x, and the fact that tan(pi/4)=1, to find the exact value of tan(pi/8).
Having a bit of trouble with this one; how would I go about doing this?

Hint: let 2x = pi/4 and then rearrange and algebraically have a go at it :)

that's sneaky. Thanks for that
Title: Re: Specialist 3/4 Question Thread!
Post by: zvezda on November 18, 2012, 08:03:07 pm
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on November 18, 2012, 08:26:06 pm
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

Hint: try rearranging inside the sin and cos :)

Spoiler
sin(theta) + i cos(theta)
= cos(pi/2 - theta) + i sin(pi/2 - theta)    <-- standard trig rearranging*
= cis(pi/2 - theta)
QED

*Proof:
sin(a-b) = sin(a)*cos(b) - sin(b)*cos(a)
Let a =  pi/2
sin(pi/2 - b) = sin(pi/2)*cos(b) - sin(b)* cos(pi/2)
sin(pi/2 - b) = 1*cos(b) - sin(b)*0
sin(pi/2 - b) = cos(b)
Title: Re: Specialist 3/4 Question Thread!
Post by: zvezda on November 18, 2012, 09:11:48 pm
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

Hint: try rearranging inside the sin and cos :)

Spoiler
sin(theta) + i cos(theta)
= cos(pi/2 - theta) + i sin(pi/2 - theta)    <-- standard trig rearranging*
= cis(pi/2 - theta)
QED

*Proof:
sin(a-b) = sin(a)*cos(b) - sin(b)*cos(a)
Let a =  pi/2
sin(pi/2 - b) = sin(pi/2)*cos(b) - sin(b)* cos(pi/2)
sin(pi/2 - b) = 1*cos(b) - sin(b)*0
sin(pi/2 - b) = cos(b)

had to look at the spoiler lol. You had (pi/2 - theta) instead of (theta - pi/2) in your working though? Am I missing something? Or did you just misread? :p
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on November 18, 2012, 09:43:39 pm
had to look at the spoiler lol. You had (pi/2 - theta) instead of (theta - pi/2) in your working though? Am I missing something? Or did you just misread? :p

You're right, I did + instead of -, if you do - you just "switch" what's inside cis due to cos(-x)=cos(x) and sin(-x)=-sin(x) :) Sorry about that haha
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on November 18, 2012, 09:54:48 pm
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

$sin(\theta)-icos(\theta)= cos (\frac{\pi}{2}-\theta) - isin(\frac{\pi}{2}-\theta) = cos(\theta- \frac{\pi}{2}) +isin(\theta-\frac{\pi}{2}) = cis(\theta - \frac{\pi}{2}) (QED)$
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on November 19, 2012, 01:33:43 pm
just another way of looking at the question:

\begin{aligned} \text{Using the fact that if \;} z &= \text{cis}(\theta), \; \text{then,} \; \bar{z} = \text{cis}(-\theta),
\\ \text{Let} \; z &= \sin(\theta) - i\cos(\theta)
\\ \; \bar{z} &= \sin(\theta) + i\cos(\theta)
\\ &= \cos(\frac{\pi}{2} - \theta) + i\sin(\frac{\pi}{2}-\theta)
\\ &= \text{cis}(\frac{\pi}{2} - \theta)
\\ z &= \text{cis}(-(\frac{\pi}{2}-\theta))
\\ z &= \text{cis}(\theta - \frac{\pi}{2}}) \end{aligned}
Title: Re: Specialist 3/4 Question Thread!
Post by: zvezda on November 19, 2012, 04:55:52 pm
thanks for the help guys
Title: Re: Specialist 3/4 Question Thread!
Post by: #1procrastinator on November 26, 2012, 07:25:48 am
Show that there is no intersection point of the line y=x+5 and the ellipse x^2+y^2/4=1

Is it a valid method to square the linear equation, equate the two and use the discrimnant to show that the resulting quadratic has no solutions? The book subs the linear equation into the equation of the ellipse and shows that the squared term cannot be equal to the negative term.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on November 26, 2012, 11:16:54 am
You can't just square one equation afaik.
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on November 26, 2012, 01:28:11 pm
Show that there is no intersection point of the line y=x+5 and the ellipse x^2+y^2/4=1

Is it a valid method to square the linear equation, equate the two and use the discrimnant to show that the resulting quadratic has no solutions? The book subs the linear equation into the equation of the ellipse and shows that the squared term cannot be equal to the negative term.

If you square each side you'd have to multiply it so that it's y^2=(x+5)^2, not y^2=x^2 + 5^2

In future these questions will get harder - you'll have 2 simultaneous equations in terms of 3 variables, i.e. (easy start) y=kx + 2 and y=3x+k, and the question will be "find the values of k for there to be no solutions, infinite solutions or 1 unique solution".
Title: Re: Specialist 3/4 Question Thread!
Post by: paulsterio on November 26, 2012, 02:26:49 pm
If you square each side you're not actually multiplying each side by the same number - i.e. you're multiplying the LHS by "y" and the R.H.S. by (x+5), so although y=x+5, y^2=/=(x+5)^2.

What? y and x+5 are the same number, however.
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on November 26, 2012, 02:48:56 pm
If you square each side you're not actually multiplying each side by the same number - i.e. you're multiplying the LHS by "y" and the R.H.S. by (x+5), so although y=x+5, y^2=/=(x+5)^2.

What? y and x+5 are the same number, however.

Derp, I think I/he meant to say y^2=x^2+5^2 (so you times each individual item by their values).

Ignore what I said before haha, y^2=(x+5)^2, but y^2=/=x^2 + 5^2. 2 weeks out of school and my brain is already done lol
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 07, 2012, 05:42:11 pm
How would i find the asymptote for the hyperbola with equation $\frac{2x^{2}}{3}-\frac{y^{2}}{2}=1$
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 07, 2012, 05:47:13 pm
Hint: Rearrange like this:

$\frac{x^2}{\left(\sqrt{\frac{3}{2}}\right)^2} - \frac{y^2}{\left(\sqrt{2}\right)^2} = 1$
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 07, 2012, 05:48:16 pm
I won't tell you how to do it exactly, but for hyperbolae with the general form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the asymptotes are given by $y=\pm\frac{bx}{a}$. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 07, 2012, 05:51:19 pm
I won't tell you how to do it exactly, but for hyperbolae with the general form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the asymptotes are given by $y=\pm\frac{bx}{a}$. :)
Remember that will be an ellipse, a hyperbola will have the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 07, 2012, 05:51:55 pm
Whoops, that was a typo. :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 07, 2012, 06:01:55 pm
Hint: Rearrange like this:

$\frac{x^2}{\left(\sqrt{\frac{3}{2}}\right)^2} - \frac{y^2}{\left(\sqrt{2}\right)^2} = 1$

Thats what I initially thought, and I got a answer of $\frac{2(3)^{1/2}}{3}$

but the answer behind the book is $\frac{4}{3^{1/2}}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Phy124 on December 07, 2012, 06:19:58 pm
Thats what I initially thought, and I got a answer of $\frac{2(3)^{1/2}}{3}$

but the answer behind the book is $\frac{4}{3^{1/2}}$

The answer should be $y = \pm \frac{2\sqrt{3}x}{3}$, which I think is what you worked it out to be?

They're supposed to have $y = \pm \frac{2x}{\sqrt{3}}$ if they have it in that form rather than $y = \pm \frac{4x}{\sqrt{3}}$
Title: Re: Specialist 3/4 Question Thread!
Post by: sin0001 on December 07, 2012, 06:21:04 pm
Are we expected to sketch Inverse circular & reciprocal circular functions by hand (like including transformed functions)? If so, then how frequently do these types of  questions appear on exams?
Cheers
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on December 07, 2012, 06:28:35 pm
Are we expected to sketch Inverse circular & reciprocal circular functions by hand (like including transformed functions)? If so, then how frequently do these types of  questions appear on exams?
Cheers
Yes. Quite often there'll probably at least be an inverse one, and you probably can't rule out the reciprocal ones either. e.g. it's something you can't ignore, there's no reason that it couldn't come up on an exam and you probably do need to know how to do it.

With the reciprocal ones, if you get stuck, just treat it like sketching any other reciprocal function to figure it out. You probably would want to remember what they look like in your head anyway, it can make certain questions easier if you have a visual picture of what's going on in your head.

For the inverse ones, you can treat it like sketching any other function - know the basic form and then apply the transformations.
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 07, 2012, 08:14:51 pm
Reciprocal circular functions can get a bit tricky so spend a bit of extra time and draw the regular graph first. You probably won't need to do this for inverse graphs though. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 07, 2012, 11:52:03 pm
I don't think you need to draw the regular graph for reciprocal circular functions. There are enough exercises in books to practice sketching them thinking of them as y=sec(x) or whatever.

For normal reciprocal graphs (y=1/f(x)) I'd dot the outline of f(x) first of course :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 08, 2012, 09:49:15 am
I don't know, I tend to draw the curves incorrectly for sec and cosec (ie everything is opposite what it should be). I find taking a few extra seconds to draw the regular graph helpful, but then again, I haven't practiced it too much and I will probably get better with time. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: sin0001 on December 08, 2012, 11:55:54 am
What do you guys think is the most challenging part of the spesh. course?
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on December 08, 2012, 01:20:41 pm
What do you guys think is the most challenging part of the spesh. course?

Personally: nothing in particular... I find that in every area of the course, the more advanced questions can be really challenging. In terms of the mundane theory work, nothing is too hard imo.
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 08, 2012, 01:23:42 pm
What do you guys think is the most challenging part of the spesh. course?

Found some areas of vectors to be the most challenging. Vector proofs didn't do down so well with me.
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 08, 2012, 01:32:10 pm
Probably differential equation slope fields and some of the more advanced areas of complex numbers

(also agree with pi re vector proofs)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 08, 2012, 01:38:25 pm
I won't tell you how to do it exactly, but for hyperbolae with the general form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the asymptotes are given by $y=\pm\frac{bx}{a}$. :)

Easier way is if you have $\pm \frac{(x-h)^2}{a^2} \mp \frac{(y-k)^2}{b^2} = 1$, just solve for $y$ using the equation $\pm \frac{(x-h)^2}{a^2} \mp \frac{(y-k)^2}{b^2}=0$ :)

Saves you remembering asymptote formulas that you might muddle up during exam pressures.
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 08, 2012, 08:32:16 pm
Dunno Pi, I always just remembered that asymptotes have the equation:

(y-k) = (+ or -)(b/a)(x-h)

:)
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 08, 2012, 08:55:56 pm
It's not that hard to remember, unlike the general circle formula stuff.
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on December 08, 2012, 08:57:16 pm
general circle formula stuff.
What's that? I'm drawing a mind blank on what you mean
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 08, 2012, 09:00:07 pm
When a circle equation is in the form $ax^2+by^2+Dx+Ey+F=0$, there are some generic formulae which give you to the centre and radius of the circle without having to complete the square.
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 08, 2012, 09:11:00 pm
Dunno Pi, I always just remembered that asymptotes have the equation:

(y-k) = (+ or -)(b/a)(x-h)

:)

Good try, but you don't look anything like the 'remembering' type :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 08, 2012, 09:15:05 pm
When a circle equation is in the form $ax^2+by^2+Dx+Ey+F=0$, there are some generic formulae which give you to the centre and radius of the circle without having to complete the square.

This will never be a circle equation unless a=b
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 08, 2012, 09:32:21 pm
There danger with trying to remember too many formulas in maths is that you don't develop the critical thinking that comes with deriving things.
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 08, 2012, 10:33:19 pm

This will never be a circle equation unless a=b

Well of course. :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 10, 2012, 10:35:13 pm
The equation $x^{2}+ax+y^{2}+1=0,$, where a is a real constant, will represent a circle if? I'm not sure how to approach this question lol :/
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 10, 2012, 10:40:28 pm
Firstly, complete the square so that we can get it into the form for the equation of a circle, $x^{2}+y^{2}=r^{2}$
Spoiler
\begin{alignedat}{1}x^{2}+ax+y^{2}+1 & =0
\\ x^{2}+ax+\left(\frac{a}{2}\right)^{2}-\left(\frac{a}{2}\right)^{2}+y^{2} & =-1
\\ \left(x+\frac{a}{2}\right)^{2}+y^{2} & =\frac{a^{2}-4}{4}
\end{alignedat}
Now for this to be a circle, the RHS of the equation cannot be $0$ and has to be positive, otherwise we would have a radius of $0$ and well, we would have no circle.
Spoiler
\begin{alignedat}{1}\frac{a^{2}-4}{4} & >0
\\ a^{2} & >4
\\ a<-2 & \: or\: a>2
\\ a & \in R\backslash\left[-2,2\right]
\end{alignedat}

EDIT: Stuffed, up fixed it now.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 10, 2012, 10:41:27 pm
Also, $P$ is any point on the hyperbola with equation $x^{2}-\frac{y^{2}}{4}=1$.

If $m$ is the gradient of the hyperbola at $P$, then $m$ could be:
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on December 10, 2012, 11:04:23 pm
using implicit differentiation
\begin{aligned} x^2 - \frac{y^2}{4} &= 1
\\ \frac{d}{dx}(x^2-\frac{y^2}{4}) &= \frac{d}{dx}(1)
\\ \frac{d}{dx}(x^2) - \frac{d}{dx}(\frac{y^2}{4}) &= 0
\\ 2x - \frac{d}{dx}(\frac{y^2}{4}) &= 0 \end{aligned}

but we need to differentiate with respect to x in the end and not y, so we introduce a new variable and apply the chain rule

\begin{aligned}\\ \text{Let\;} u=\frac{y^2}{4}, \; \frac{du}{dy} &= \frac{y}{2}
\\ \frac{du}{dx} &= \frac{du}{dy}\frac{dy}{dx}
\\ \frac{du}{dy}\frac{dy}{dx} &= \frac{y}{2}\frac{dy}{dx}
\\ 2x - \frac{du}{dx} &= 0
\\ 2x - \frac{y}{2}\frac{dy}{dx} &= 0
\\ \frac{y}{2}\frac{dy}{dx} &= 2x
\\ \frac{dy}{dx} &= \frac{2x}{\frac{y}{2}}
\\ \frac{dy}{dx} &= \frac{4x}{y}
\\ m &= \frac{4x}{y}\end{aligned}

another method
\begin{aligned} x^2 - \frac{y^2}{4} &= 1
\\ \frac{y^2}{4} &= x^2 -1
\\ y &= \pm\sqrt{4(x^2-1)}
\\ \frac{dy}{dx} &= \pm\frac{2x}{\sqrt{x^2-1}}\end{aligned}
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 10, 2012, 11:20:51 pm
The options I am provided with are

a) any real number
b) any real number in the interval (-2,2)
c) any real number in the interval [-2,2]
d) any real number in the number R/(-2,2)
e) any real number in the interval R/[-2,2]
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 10, 2012, 11:28:32 pm
I might have a slightly different interpretation of this, initally I went to use implicit differentiation too, then realised that they probably haven't come across it yet. But the way its worded, it may be asking for the values that m can be, which you can get from the second method, graph it and you can see that the gradient can be $R\backslash\left[-2,2\right]$. To get this via another method, if you graph the original function, we can see it has asymptotes of $y=2x$ and $y=-2x$. And when we look at the graph, looking at the right arm, the gradient is going to always be greater than 2 when it approaches $y=2x$ and always going to be less than (well more negative) than $y=-2x$. For the left arm the gradient on the top part of the curve is under the asymptote and moving away from it, so that the gradient is always less than (more negative) than -2 and for the bottom half, always going to be greater than 2 (for the $y=2x$ asymtptote). SO that is $m\in R\backslash\left[-2,2\right]$.

Anyways, thats just another way of looking at it, less mathematical though, but might help you understand it a little bit more.

Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 11, 2012, 10:12:22 am
I was looking at the implicit differentiation there and was all like O_O. :P But the second method and b^3's explanation has really covered it well. Well done, guys. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: barydos on December 11, 2012, 11:34:07 am
Simple question here!
(http://i.imgur.com/mrtP6.png)
I'm just curious how you work out question b) part (i)
I can only get x = R\{2} but that's not right haha
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 11, 2012, 12:10:12 pm
I might have a slightly different interpretation of this, initally I went to use implicit differentiation too, then realised that they probably haven't come across it yet. But the way its worded, it may be asking for the values that m can be, which you can get from the second method, graph it and you can see that the gradient can be $R\backslash\left[-2,2\right]$. To get this via another method, if you graph the original function, we can see it has asymptotes of $y=2x$ and $y=-2x$. And when we look at the graph, looking at the right arm, the gradient is going to always be greater than 2 when it approaches $y=2x$ and always going to be less than (well more negative) than $y=-2x$. For the left arm the gradient on the top part of the curve is under the asymptote and moving away from it, so that the gradient is always less than (more negative) than -2 and for the bottom half, always going to be greater than 2 (for the $y=2x$ asymtptote). SO that is $m\in R\backslash\left[-2,2\right]$.

Anyways, thats just another way of looking at it, less mathematical though, but might help you understand it a little bit more.

the answer is E, which is any real number in the interval $\frac{R}{[-2,2]}$, would that be the same as $m\in R\backslash\left[-2,2\right]$.
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 11, 2012, 12:21:48 pm
I'm just curious how you work out question b) part (i)
I can only get x = R\{2} but that's not right haha

Is the answer [-2,2)  by any chance ?
Title: Re: Specialist 3/4 Question Thread!
Post by: barydos on December 11, 2012, 01:10:45 pm

Is the answer [-2,2)  by any chance ?

-2 is not inclusive, but yes otherwise it is correct! TEACH ME YOUR WAYS
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 11, 2012, 01:25:57 pm

the answer is E, which is any real number in the interval $\frac{R}{[-2,2]}$, would that be the same as $m\in R\backslash\left[-2,2\right]$.

Note that "R/[-2,2]" isn't a fraction :P It means all real numbers ("R") excluding ("\") the interval of -2 to 2 inclusive ("[-2,2]") :)
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on December 11, 2012, 01:39:53 pm
I'm just curious how you work out question b) part (i)

you need the series to converge, which occurs if $-1 < r < 1$

so, first you find r (common ratio) in terms of x,
\begin{aligned} \frac{\frac{x}{2}}{1} = \frac{\frac{x^2}{4}}{\frac{x}{2}} = ... &= \frac{x}{2}
\\ r &= \frac{x}{2} \end{aligned}

then you apply the conditions for a series to be convergent and therefore possible for an infinite sum to exist
\begin{aligned} -1 < &r < 1
\\ -1 < &\frac{x}{2} < 1
\\ -2 < &x < 2\end{aligned}
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 11, 2012, 02:05:02 pm
you need the series to converge, which occurs if $-1 < r < 1$

so, first you find r (common ratio) in terms of x,
\begin{aligned} \frac{\frac{x}{2}}{1} = \frac{\frac{x^2}{4}}{\frac{x}{2}} = ... &= \frac{x}{2}
\\ r &= \frac{x}{2} \end{aligned}

then you apply the conditions for a series to be convergent and therefore possible for an infinite sum to exist
\begin{aligned} -1 < &r < 1
\\ -1 < &\frac{x}{2} < 1
\\ -2 < &x < 2\end{aligned}

Yes.

Also, using the formula for the sum of a geometric series with a=1, r=x/2

S = [1-(x/2)^n]/[1-(x/2)]

Therefore, if you dont want S to bugger off to infinity as n-> infinity,

|x/2| < 1

giving the result above.
I initially thought that x=-2 would be OK, but it can be seen that in this case S will oscillate between the valus of 0 and 1 as n-> infinity, which does not make the series convergent.
Title: Re: Specialist 3/4 Question Thread!
Post by: barydos on December 11, 2012, 02:17:45 pm
AWESOME  thanks both of you :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 10:39:06 am
$f(x)=\sqrt{3} \sec (2x+\frac{\Pi }{2})-2$

The function $f(x)$ has vertical asymptote where $x$ is equal to:
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 12, 2012, 10:43:21 am
If you think about it, the asymptotes of sec(x) occurs when cos(x) = 0.

Therefore, $cos(x) = 0$
implies
$x = \pi/2, 3\pi/2, 5\pi/2$

But the function f has been transformed. So we need to know the new value of x.

Let the inner function (the bracketed function) be equal to the value of x which gave the asymptote locations of the untransformed function (sec(x))
Let $2x + \pi/2 = \pi/2, 3\pi/2, 5\pi/2$

Solving this will tell us which x value will give the location of the new asymptote.

$x = \pi/2, \pi, 3\pi/2 ....$
Title: Re: Specialist 3/4 Question Thread!
Post by: aimhigh on December 12, 2012, 10:59:23 am
What would happen to the $-2 , \sqrt{3}$?
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 12, 2012, 11:06:49 am
Yes, they would have a slight effect on it as well. If you need to, draw the regular cosine graph and find your x-intercepts. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 11:19:12 am
what would be the cosine equation? :/

cos(2x+pi/2)    or          root(3)cos(2x+pi/2)-2
Title: Re: Specialist 3/4 Question Thread!
Post by: polar on December 12, 2012, 11:21:37 am
if you're trying to find the asymptotes, then use the first one

$\sqrt{3}\sec(2x + \frac{\pi}{2}) - 2 = \frac{\sqrt{3}}{\cos(2x + \frac{\pi}{2})} - 2$

thus, asymptotes occur at the x-intercepts of $y= \cos(2x + \frac{\pi}{2}})$
Title: Re: Specialist 3/4 Question Thread!
Post by: Lasercookie on December 12, 2012, 11:21:56 am
Yes, they would have a slight effect on it as well. If you need to, draw the regular cosine graph and find your x-intercepts. :)
It won't have an affect on the vertical asymptote though. The $-2$ and the $\sqrt{3}$ affect the y-coordinates, so it doesn't really move the vertical asymptote around.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 11:47:27 am
It won't have an affect on the vertical asymptote though. The $-2$ and the $\sqrt{3}$ affect the y-coordinates, so it doesn't really move the vertical asymptote around.

Oh yes offcourse haha, thanks laseredd
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 12:00:41 pm
Also how would i find the turning point of the function. I know it'd be the same as the reciprocal function (cosine function), but idk what to use lol :/
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 12, 2012, 12:09:56 pm
Half way between your asymptotes. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 03:04:12 pm
Thanks for your help guys :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 10:03:19 pm
how would i simplify this $sin(x)^{4}-cos(x)^{4}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 12, 2012, 10:07:47 pm
= $sin^4(x) - cos^4(x)$

DOPS
= $(sin^2(x) - cos^2(x))(sin^2(x) + cos^2(x))$

Knowing that $sin^2(x) + cos^2(x) = 1$
= $(sin^2(x) - cos^2(x))$

Knowing that $cos^2(x) - sin^2(x) = cos(2x)$
= $-cos(2x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 10:28:12 pm
I don't understand the last step, cos2(x)-sin2(x)=cos(2x)
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 12, 2012, 10:32:30 pm
It's one of the double angle formulae. Have you learnt this yet? :)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 12, 2012, 10:33:01 pm
See the formula sheet :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 12, 2012, 10:52:45 pm
Nah i havent, plus the answers is sin2(x)-cos2(x), but will look through it :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 12, 2012, 10:57:04 pm
That's the problem with simplifying questions - you can come to all sorts of conclusions. That's why I prefer proving questions. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 12, 2012, 11:00:37 pm
If that's the answer Jai, you just stop before my last step. Otherwise, the final step is done by seeing the formula sheet like others have said. Hope it helped.
Title: Re: Specialist 3/4 Question Thread!
Post by: Truck on December 13, 2012, 12:23:33 am
If it was an exam, -cos(2x) would probably be the one to get it to :) .
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 14, 2012, 09:41:45 am
If $sin(A)=\frac{1}{\sqrt{5}}$ and $x\epsilon [o,\frac{\pi }{2}]$ find the exact values of $sin (3A)$

Title: Re: Specialist 3/4 Question Thread!
Post by: polar on December 14, 2012, 10:18:23 am
\begin{aligned} \sin(3A) &= \sin(A+2A)
\\ &= \sin(A)\cos(2A) + \sin(2A)\cos(A)
\\ &= \sin(A)(1-2\sin^2(A)) + (2\sin(A)\cos(A))\cos(A)
\\ &= \sin(A)(1-2\sin^2(A)) + 2\sin(A)\cos^2(A)
\\ &= \sin(A)(1-2\sin^2(A)) + 2\sin(A)(1-\sin^2(A))
\\ \text{Since} \; \sin(A) = \frac{1}{\sqrt{5}}, \; \sin(3A) &= \frac{1}{\sqrt{5}}\times\left(1-2\left(\frac{1}{\sqrt{5}}\right)^2\right) + 2\times\frac{1}{\sqrt{5}}\left(1-\left(\frac{1}{\sqrt{5}}\right)^2\right)
\\ &= \frac{1}{\sqrt{5}}\left(\left(1-\frac{2}{5}\right) + 2\left(1-\frac{1}{5}\right)\right)
\\ &= \frac{1}{\sqrt{5}}\times\frac{11}{5}
\\ &= \frac{11\sqrt{5}}{25} \end{aligned}
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 14, 2012, 04:35:02 pm
Thanks polar, however I am stuck on another one of these questions :/

if $cos(B)=0.7$ and $x\epsilon [o,\frac{\pi }{2}]$ then $cos(\frac{B}{2})$ equals?

Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 14, 2012, 04:36:04 pm
Here's a hint - use the double angle formula using cos(B) rather than cos(2B).
Title: Re: Specialist 3/4 Question Thread!
Post by: TrueTears on December 14, 2012, 04:36:34 pm
cos(2B) = 2cos^2(B)-1

cos(B) = 2cos^2(B/2)-1
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 14, 2012, 04:39:31 pm
smaartt :) thanks truetears and stick
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 15, 2012, 12:04:52 pm
how would i find the implied domain of $cos^{-1}(sin2x)$
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 15, 2012, 12:22:10 pm
-1 =< sin2x =< 1
which is true for all x
so implied domain is R
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 15, 2012, 12:50:27 pm
the answer is $[\frac{-\pi }{4},\frac{\pi }{4}]$ :/
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 15, 2012, 12:52:02 pm
however, if I solve for x from -1 =< sin2x =< 1 i would get the answer. Thanks Brightsky :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 15, 2012, 01:02:24 pm
For the inverse of a function to be defined, the function has to be one-to-one (else the inverse will fail the vertical line test and it will not be a function)
For this reason we restrict the domain of functions so that the inverse may be defined.

EDIT (got confused at this point),
For the sine function we (conventionally) restrict the domain to [-pi/2 , pi/2]

With this in mind, if we apply brightsky's method, we obtain jai's result
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 15, 2012, 01:49:38 pm
however, if I solve for x from -1 =< sin2x =< 1 i would get the answer. Thanks Brightsky :)
For the inverse of a function to be defined, the function has to be one-to-one (else the inverse will fail the vertical line test and it will not be a function)
For this reason we restrict the domain of functions so that the inverse may be defined.

EDIT (got confused at this point),
For the sine function we (conventionally) restrict the domain to [-pi/2 , pi/2]

With this in mind, if we apply brightsky's method, we obtain jai's result

hmm...calculator seems to agree with my answer. i don't see a need to restrict the sine function....
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 15, 2012, 01:54:28 pm

EDIT (got confused at this point),
For the sine function we (conventionally) restrict the domain to [-pi/2 , pi/2]

With this in mind, if we apply brightsky's method, we obtain jai's result

I think theoretically  [-pi/4 , pi/4] is right, however when you sketch it on cas its R :/
Title: Re: Specialist 3/4 Question Thread!
Post by: Jenny_2108 on December 15, 2012, 06:19:26 pm
EDIT (got confused at this point),
For the sine function we (conventionallyl) restrict the domain to [-pi/2, pi/2]

Why do we have to restrict domain of sine function?

Its a composite function so it'll be defined when range of sin(2x) E domain of inverse cos (which is what brightsky did above)

-1 =< sin(2x) =< 1
-pi/2 +2npi =< 2x =< pi/2 + 2npi (n is integer)
-pi/4 + npi =< x =< pi/4 + npi

I think the ans should be R
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 15, 2012, 07:01:59 pm
Yes, I am now also convinced that the implied domain is R
Sorry for the confusion.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 17, 2012, 06:01:23 pm
solve 4cos^3(t)-3cos(t) = cos(t) x belongs to [0,2pi]
I keep on getting 0,pi,2pi, but the answer has more to it :/
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 17, 2012, 06:08:37 pm
$4cos^3(t) - 3cos(t) = cos(t)$
$4cos^3(t) - 4cos(t) = 0$
$cos^3(t) - cos(t) = 0$
$cos(t)(cos^2(t) - 1) = 0$

So Null-Factor law states that either $cos(t) = 0$ or $cos^2(t) = 1$

$cos(t) = 0$
$t = \pi/2, 3\pi/2$

$cos(t) = 1$
$t = 0, 2\pi$

$cos(t) = -1$
$t = \pi$

Therefore the solutions for the system is:
$0,\pi/2, \pi, 3\pi/2, 2\pi$

I think I know what you did. You can't divide by $cos(t)$ on both sides of the equation because it just so happens that cos(t) = 0 in our solution set. Since you can't divide by zero, you lost some solutions and it screwed up from there. It's always better to do factorization when you are dealing with functions to avoid this problem. It's also noted in the study design (or recap) for 2011.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 17, 2012, 06:38:34 pm
Yes you're right Hancock, haha i did divide by cos(t) and only obtained cos(t)=+or-1. Thanks again :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 17, 2012, 06:39:42 pm
No problems bud, keep the questions coming haha.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 19, 2012, 08:52:18 pm
if z=1+i find z^4

I know I could work it out by multiplying (1+i)(1+i)(1+i)(1+i). Is there any other method I could use?

Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 19, 2012, 08:54:21 pm
Have you learnt De Moivre's theorem?
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 19, 2012, 08:55:36 pm
so would i need to convert it into polar form?
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 19, 2012, 08:58:03 pm
Yes, that would be it. 1+i is fairly straightforward (I just know it by heart lol).
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 19, 2012, 09:00:20 pm
yeee ive got the answer using de moivre's theorem, but this question was like 2 exercise before the one where we learn the theorem, so I thought there would be a different method :/
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 19, 2012, 09:00:35 pm
Convert into polar form, apply de'movires theroem, solve it back out.
Spoiler
\begin{alignedat}{1}|z| & =\sqrt{1^{2}+1^{2}}
\\ & =\sqrt{2}
\\ \sin(\theta) & =\frac{1}{\sqrt{2}}
\\ \cos(\theta) & =\frac{1}{\sqrt{2}}
\\ \theta & =\frac{\pi}{4}
\\ z & =\sqrt{2}\mathrm{cis}\left(\frac{\pi}{4}\right)
\\ z^{4} & =\left(\sqrt{2}\right)^{4}\mathrm{cis}\left(4\frac{\pi}{4}\right)
\\ & =4\mathrm{cis\left(\pi\right)}
\\ & =4\left(\cos(\pi)+i\sin(\pi)\right)
\\ & =-4
\end{alignedat}

EDIT:
yeee ive got the answer using de moivre's theorem, but this question was like 2 exercise before the one where we learn the theorem, so I thought there would be a different method :/
In that case, if you didn't know it then yeh, expand it out, then it should come to the same result, although once you know de'movires, its better to stick with it.
Spoiler
\begin{alignedat}{1}\left(1+i\right)^{4} & =\left(1+i\right)^{2}\left(1+i\right)^{2}
\\ & =\left(1+2i+i^{2}\right)\left(1+2i+i^{2}\right)
\\ & =\left(2i+1-1\right)\left(2i+1-1\right)
\\ & =\left(2i\right)\left(2i\right)
\\ & =4i^{2}
\\ & =-4
\end{alignedat}
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 19, 2012, 09:01:09 pm
yeee ive got the answer using de moivre's theorem, but this question was like 2 exercise before the one where we learn the theorem, so I thought there would be a different method :/

Haha, just use the method you would use in an exam :P Which is to use De Moivre's theorem as you did :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 19, 2012, 09:01:25 pm
If that's the case, you'd probably just have to expand it normally. Using Pascal's Triangle might speed things up though. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 19, 2012, 09:06:39 pm
haha yeah coz THEN the question asks us to find out z^12. This would take like forever considering de moivre's theorem is 2 exercise away and we aren't supposed to know it yet. Maybe it wants us to work things out on the calculator? :/

but thanks anyways guys
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 19, 2012, 09:07:48 pm
Yeah, it might be a good idea to practice your calculator skills then. :)
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 19, 2012, 09:11:30 pm
haha yeah coz THEN the question asks us to find out z^12. This would take like forever considering de moivre's theorem is 2 exercise away and we aren't supposed to know it yet. Maybe it wants us to work things out on the calculator? :/

but thanks anyways guys

Well, not really :P

If (1+i)^12 = ((1+i)^4)^3 = (-4)^3 = -64 :)

I plucked (1+i)^4 out because it was in the previous part of the question :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 19, 2012, 09:12:05 pm
smarty pants
Title: Re: Specialist 3/4 Question Thread!
Post by: Nobby on December 21, 2012, 01:10:57 am
Hey guys, some help for this thing would be great:

z2 = a + bi
Let z = x + yi
Thus (x+yi)2 = a + bi
x2-y2 + 2xyi = a + bi
Therefore x2-y2 = a
and 2xy = b

Find x2 in terms of a and b?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 21, 2012, 01:25:51 am
x^2-y^2 = a   and    2xy = b
Hence, y = b/2x
=> y^2 = b^2/4x^2
Hence, x^2 - b^2/4x^2 = a
=> x^4 - b^2/4 = ax^2
=> x^4 - ax^2 - b^2/4 = 0

That's a quadratic for x^2, but I think the solution is going to be a bit ugly haha :P
Title: Re: Specialist 3/4 Question Thread!
Post by: Nobby on December 21, 2012, 01:29:26 am
x^2-y^2 = a   and    2xy = b
Hence, y = b/2x
=> y^2 = b^2/4x^2
Hence, x^2 - b^2/4x^2 = a
=> x^4 - b^2/4 = ax^2
=> x^4 - ax^2 - b^2/4 = 0

That's a quadratic for x^2, but I think the solution is going to be a bit ugly haha :P

Yeah I got there but would there any neater ways to do it?
My working is going to look disgusting :S

EDIT: actually it looks okay
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 22, 2012, 05:31:59 pm
if $z-3i$ is a factor of $2z^{4}-4z^{3}+21z^{2}-36z+27$, how would i find the remaining factors?
Title: Re: Specialist 3/4 Question Thread!
Post by: pi on December 22, 2012, 05:36:05 pm
if $z-3i$ is a factor of $2z^{4}-4z^{3}+21z^{2}-36z+27$, how would i find the remaining factors?

Steps:
- Well, as (z-3i) is a factoryou then know that (z+3i) is also a factor.
- Hence, you would know that (z-3i)(z+3i)=(z^2 + 9) is a factor.
- From here, you can long divide or factorise in another way (your choice) to find the other quadratic factor
- Then factorise that (or jump to the quadratic formula) to find the final two linear factors :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Stick on December 22, 2012, 05:36:51 pm
I'm not exactly sure, but I think you could apply long division, the factor theorem and the remainder theorem (much like cubics in Methods). Give it a go and let me know if it works. :)

EDIT: pi confirmed what I was thinking anyway. ^_^
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 22, 2012, 05:39:40 pm
Since the coefficients of the expression are real, then each complex factor will have a conjugate, that is also a factor, i.e. $z+3i$ is also a factor. From that if you divide through by the two factors we have you should get a quadratic, that you can factorise to find the last 2 factors.

EDIT: Beaten, shouldn't have tried to spend time LaTeXing the long division before getting lazy...
Title: Re: Specialist 3/4 Question Thread!
Post by: Planck's constant on December 22, 2012, 05:42:36 pm
if $z-3i$ is a factor of $2z^{4}-4z^{3}+21z^{2}-36z+27$, how would i find the remaining factors?

The important piece of theory is that if a polynomial with REAL coefficient has a complex root, the conjugate of this root is also a root of the polynomial.

Here 3i is a root, therefore so is -3i
x-3i is a factor, x+3i is a factor
(x-3i)(x+3i) is a factor, therefore x^2 + 9 is a factor.

One way to proceed from here is long division of x^2 + 9 into the original polynomial.
The remainder is a quadratic which you can easily factorise

Alternatively, you can ignore my answer and go the one above :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 22, 2012, 07:53:42 pm
thanks heaps guys. I worked out the z^2+9 but i wasn't comfortable doing the long division. could someone please help me from there :)
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 22, 2012, 08:49:00 pm
Let $2z^{4}-4z^{3}+21z^{2}-36z+27=(z^{2}+9)(2z^{2}+az+b)$

Expanding gives $(z^{2}+9)(2z^{2}+az+b)=2x^{4}+az^{3}+bz^{2}+18z^{2}+9az+9b$
$=2x^{4}+az^{3}+(b+18)z^{2}+9az+9b$

So $a=-4$
and $9b=27\Rightarrow b=3$

Hence $2z^{4}-4z^{3}+21z^{2}-36z+27=(z^{2}+9)(2z^{2}-4z+3)$
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 23, 2012, 10:24:30 am
I get it. thanks guys :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Jaswinder on December 23, 2012, 11:25:07 am
Also by using the quadratic formula or by using the method of factorizing of polynomials in C,

(2z^2-4z+3) would break down into $(z-1\pm \frac{\sqrt{2}}{2}i)$

so in total there are 4 solutions $(z+3i)$$(z-3i)$$(z-1\ + \frac{\sqrt{2}}{2}i)(z-1\ - \frac{\sqrt{2}}{2}i)$

Title: Re: Specialist 3/4 Question Thread!
Post by: jadams on December 27, 2012, 01:51:50 am
Let $2z^{4}-4z^{3}+21z^{2}-36z+27=(z^{2}+9)(2z^{2}+az+b)$

Expanding gives $(z^{2}+9)(2z^{2}+az+b)=2x^{4}+az^{3}+bz^{2}+18z^{2}+9az+9b$
$=2x^{4}+az^{3}+(b+18)z^{2}+9az+9b$

So $a=-4$
and $9b=27\Rightarrow b=3$

Hence $2z^{4}-4z^{3}+21z^{2}-36z+27=(z^{2}+9)(2z^{2}-4z+3)$

Similarly, if you write out $(z^{2}+9) (...)$ you can "by inspection" work out the other factor, by imagining what the expanded product of the two factors would look like. (Exact same method as that of Climbtoohigh, but could be used if you can manage the numbers in your head/scrap piece of paper, could possibly be quicker.)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 27, 2012, 09:55:21 am
If $w=1+ai$

show that $\left | w^2 \right |=(1+a^{2})^{\frac{3}{2}}$
Title: Re: Specialist 3/4 Question Thread!
Post by: Russ on December 27, 2012, 11:28:56 am
Assuming $a \in \mathbb{R}$, then $|w^2|=|w|^2 = 1 + a^2$, so the result holds iff $a = 0$
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 27, 2012, 02:14:13 pm
Similarly, if you write out $(z^{2}+9) (...)$ you can "by inspection" work out the other factor, by imagining what the expanded product of the two factors would look like. (Exact same method as that of Climbtoohigh, but could be used if you can manage the numbers in your head/scrap piece of paper, could possibly be quicker.)
It's somewhat risky to do that on an exam 1, though in retrospect I guess it was somewhat easy to see that b = 3.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 27, 2012, 02:48:45 pm
Assuming $a \in \mathbb{R}$, then $|w^2|=|w|^2 = 1 + a^2$, so the result holds iff $a = 0$

thanks for your reply russ, but im still a little confused :/
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 27, 2012, 03:28:09 pm
Was it meant to be $w^3$ rather than $w^2$? It's a question from a VCAA exam.
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on December 27, 2012, 03:33:25 pm
ClimbTooHigh is right. I remember the question.
Title: Re: Specialist 3/4 Question Thread!
Post by: jadams on December 27, 2012, 07:26:11 pm
thanks for your reply russ, but im still a little confused :/

Russ is saying that the statement is only true/proven if $a=0. ie. (1+a^2)=(1+a^2)^{\frac{3}{2}}$ only holds true when a=0, which would answer the question you posted.

Most likely though, the question is a little different than what you posted.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 28, 2012, 01:29:30 pm
I think there is a printing mistake :) as climbtoohigh and hancock said its ment to be w^3 instead of w^2, but thanks for your help guys
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 29, 2012, 01:50:43 pm
just a theoretical question related to relations and regions of the complex plane. When drawing rays suppose {z:arg(z)=theta} why is there an open circle at the origin, showing 0+0i as not a member of the set?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 29, 2012, 02:00:13 pm
You can't define the argument of zero, it's like trying to define the direction of the zero vector.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 29, 2012, 02:49:41 pm
if the ray was translated suppose {z:arg(z+a+bi)=theta,a,b belong to R} would the point (-a-bi) be a member of the set?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 29, 2012, 02:56:36 pm
If you translate an undefined point it's still undefined
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 29, 2012, 03:04:45 pm
yeh, coz i got confused since the book didnt have an open circle :/
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 29, 2012, 03:09:42 pm
Is wrong
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 30, 2012, 10:19:54 pm
when converting {z: |z+3|+ |z+1|=4}  into Cartesian form, i keep getting $3x^2+4x+4y^2=0$, but I can never transform it to  $\frac{(x+2)^2}{4}+\frac{y^2}{3}=1$ as shown in the answer :/
Title: Re: Specialist 3/4 Question Thread!
Post by: brightsky on December 30, 2012, 10:25:18 pm
complete the square
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 30, 2012, 10:36:42 pm
$|z+3|+|z+1|=4$

$\implies \sqrt{(x+3)^2+y^2}+\sqrt{(x+1)^2+y^2}=4$

$\implies \sqrt{x^2+6x+9+y^2}=4-\sqrt{x^2+2x+1+y^2}$

$\implies x^2+6x+9+y^2=16-8\sqrt{x^2+2x+1+y^2}+x^2+2x+1+y^2$ (squared both sides)

$\implies 8\sqrt{x^2+2x+1+y^2}=8-4x$

$\implies 2\sqrt{x^2+2x+1+y^2}=2-x$

$\implies 4(x^2+2x+1+y^2)=4-4x+x^2$

$\implies 3x^2+12x+4y^2=0$

$\implies 3x^2+12x+12+4y^2=12$

$\implies 3(x^2+4x+4)+4y^2=12$

$\implies 3(x+2)^2+4y^2=12$

$\implies \frac{(x+2)^2}{4}+\frac{y^2}{3}=1$
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 30, 2012, 10:45:01 pm
$\implies 8\sqrt{x^2+2x+1+y^2}=8-4x$

wouldn't it be -8-4x?
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on December 30, 2012, 10:49:42 pm
Are you sure thats the right equation? because the diagram shows something slightly different (x axis semi major length). Also I don't think the expressions you've got to is quite right, it nearly is, just one term is off, as long as I haven't made a mistake. Anyway, heres what I have
Spoiler
\begin{alignedat}{1}|z+3|+|z+1| & =4
\\ \mathrm{Let}\; z=x+iy
\\ |\left(x+3\right)+iy|+|\left(x+1\right)+iy| & =4
\\ \sqrt{\left(x+3\right)^{2}+y^{2}}+\sqrt{\left(x+1\right)^{2}+y^{2}} & =4
\\ \sqrt{\left(x+3\right)^{2}+y^{2}} & =4-\sqrt{\left(x+1\right)^{2}+y^{2}}
\\ x^{2}+6x+9+y^{2} & =16-2\left(4\right)\sqrt{\left(x+1\right)^{2}+y^{2}}+x^{2}+2x+1+y^{2}
\\ 4x-8 & =-8\sqrt{\left(x+1\right)^{2}+y^{2}}
\\ 2-x & =2\sqrt{\left(x+1\right)^{2}+y^{2}}
\\ 4-4x+x^{2} & =4\left(x^{2}+2x+1+y^{2}\right)
\\ 0 & =3x^{2}+12x+4y^{2}
\\ 3\left(x^{2}+4x+2^{2}-2^{2}\right)+4y^{2} & =0
\\ 3\left(\left(x+2\right)^{2}-4\right)+4y^{2} & =0
\\ 3\left(x+2\right)^{2}+4y^{2} & =12
\\ \frac{\left(x+2\right)^{2}}{4}+\frac{y^{2}}{3} & =1
\end{alignedat}

EDIT: Beaten. but I typed all this out... soo..
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on December 30, 2012, 10:57:54 pm
what I did was

$4x-8=-8\sqrt{(x+1)^2+y^2}$
$(4x-8)^2=64[{(x+1)^2+y^2}]$
$16x^2-64x+64=64x^2+128x+64+64y^2$
$48x^2-64x-64y^2=0$
$16(3x^2-4x-4y^2)=0$
$3x^2-4x-4y^2=0$
:/
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on December 30, 2012, 11:24:53 pm
Should be $-48x^2$, and you didn't subtract $128x$ properly
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on January 03, 2013, 10:08:23 am
If i write {z: Im(z)Re(z-2)=-1+2Re(z-2)} instead of {z:Im(z-2)Re(z-2)=-1} would it mean the same if i was to convert it into Cartesian form?
Title: Re: Specialist 3/4 Question Thread!
Post by: BubbleWrapMan on January 03, 2013, 11:28:48 am
Assuming you meant $\left\{z:\mathrm{Im}(z-2i)\mathrm{Re}(z-2)=-1\right\}$, yes
Title: Re: Specialist 3/4 Question Thread!
Post by: sin0001 on January 03, 2013, 11:46:48 pm
Came across this exam question from heinemann:
Find an anti derivative of  (2+6x)/sqrt(4-x^2)
Thanks!
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 03, 2013, 11:49:00 pm
Hint, split it up into $\frac{2}{\sqrt{4-x^{2}}}+\frac{6x}{\sqrt{4-x^{2}}}$ then the first part will become a sin inverse, and try a u substitution for the second part.
Title: Re: Specialist 3/4 Question Thread!
Post by: sin0001 on January 04, 2013, 12:04:11 am
Omg never thought of that. Ty!
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on January 06, 2013, 04:32:15 pm
not sure how to do dis   :-\
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on January 06, 2013, 04:38:58 pm
$\frac{A}{x-c} + \frac{B}{2x+c}$

$=\frac{A(2x+c)+B(x-c)}{(x-c)(2x+c)}$

$=\frac{2Ax+Ac+Bx-Bc}{(x-c)(2x+c)}$

$=\frac{x(2A+B)+c(A-B)}{(x-c)(2x+c)}$

Let $a = 2A+B$ and $b = c(A-B)$

$=\frac{ax+b}{(x-c)(2x+c)}$

So, answer b if I'm not mistaken.
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on January 06, 2013, 04:48:10 pm
thanks man, i got upto here $=\frac{x(2A+B)+c(A-B)}{(x-c)(2x+c)}$ and then didn't know what to do :/ btw, what if you cant identify that a = 2A+B and b = c(A-B)
Title: Re: Specialist 3/4 Question Thread!
Post by: Hancock on January 06, 2013, 04:49:18 pm
You should be able to identify that a = 2A + B and so on because 2A+B is just a scalar number. It makes sense to introduce a new variable 'a' to simplify it :)
Title: Re: Specialist 3/4 Question Thread!
Post by: Homer on January 07, 2013, 02:01:34 pm
We have to find the second derivative of the question attached. The answer i get is slightly different to the answer provided. I have upload my workings (http://sphotos-c.ak.fbcdn.net/hphotos-ak-snc6/254695_390068854410942_1240762991_n.jpg) it'd be great if someone could pin point my mistakes. Thanks
Title: Re: Specialist 3/4 Question Thread!
Post by: b^3 on January 07, 2013, 02:51:41 pm
The answer is equivalent to what the calculator gives, its just in a different form, so you haven't made a mistake, just not got it in the same form. So it's correct, I'll post up in a couple of minutes to show that they are the same.