sorry for hijacking this thread.
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?
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.
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?
sorry for hijacking this thread.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
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 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.jpegheh, 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?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.jpegheh, no need to worry, just do some self study and this stuff will be easy ^^
Good times will come, young padawan.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.jpegheh, no need to worry, just do some self study and this stuff will be easy ^^
ill send u the ebook? pm me ur email brah :Ddo you have ebook of heinemann spesh book? ;)
nope :(ill send u the ebook? pm me ur email brah :Ddo you have ebook of heinemann spesh book? ;)
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
Find x such that:Use the tan double angle formula then see where you can get to.
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
tan(2x)=(2tan(x))/(1-tan^2(x))
x also equals pi and 2pi from b^3's workingYeh I'm a bit rusty, I clicked post, then realised and my com froze.
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?
move everytihng to the one side then used null factor law.
4tan(x)-4tan^{3}(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan^{2}(x))=0
tan(x)=0 or -8-4tan^{2}(x)=0
x=0 or tan^{2}(x)=2
x=0, or or no solution
So x=0, ,
EDIT: forgot sols
YES! I did it in my head, and didn't type it, I've lost it already :(. Stupid holidays :|move everytihng to the one side then used null factor law.
4tan(x)-4tan^{3}(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan^{2}(x))=0
tan(x)=0 or -8-4tan^{2}(x)=0
x=0 or tan^{2}(x)=2
x=0, or or no solution
So x=0, ,
EDIT: forgot sols
legend! thanks bro!
just ah... wouldnt:
-8-4tan^{2}(x)=0
-4tan^{2}(x)=0 +8
tan^{2}(x)= 8 /-4
tan^{2}(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
TT, . Your attempt doesn't use the fact that it's a right angle triangle so that alerted me.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 ;)
What you should now do is compute the length:
Likewise we can show that
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?)
I don't get why the hypotenuse is a-b?
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
Which leads me to asking, the hypotenuse could be a-b or b-a yes?
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"?
Question must be wrong. What does the answer say?
This question has been puzzling me for a while :/.
Part of the graph with equation is shown below.
Find the area that is bounded by the curve and the x-axis. Give your answer in the form
, where a, b and c are the integers.
This question has been puzzling me for a while :/.
Hey, guys i really got stuck in this question :/, i tried yesterday night and this morning
this is where i got stuck in, please help
(http://imageshack.us/photo/my-images/528/spec2.jpg/)on what? lol
Could someone please help me on this one?!
If you click modify you can copy the image link and see it. Don't know why it didn't come up thought.(http://imageshack.us/photo/my-images/528/spec2.jpg/)on what? lol
Could someone please help me on this one?!
Chp7, MQ, pg 314,
Differential Equation question.
I got a bit stuck on it for a while
Q9. Verify that satisfies the differential equation.
(http://imageshack.us/photo/my-images/528/spec2.jpg/)here's a hint to help you start off
http://imageshack.us/photo/my-images/528/spec2.jpg
Could someone please help me on this one?!
no watch out, c = -(a+b)
so c.c = -(a+b).-(a+b)
No the algebra is correct, the only confusion arises from what is.if he writes c^2 and the expressions on a sac or exam, marks will definitely be taken off
Let's call the angle between between the vectors and by
Then (draw the diagram)
Thus hence the negative sign.
you cant square vectors, there is no such thing as c^2 if c is a vector.
you have to dot product the expressions
No the algebra is correct, the only confusion arises from what is.
Let's call the angle between between the vectors and by
Then (draw the diagram)
Thus hence the negative sign.
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.
Hope that helps.
Lol @ "acceleration due to air resistance" whose phrase is this? I'm sure the better way of saying it is the air resistance is upwards where 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 . (note downwards is taken as positive direction)Sorry its meant to be acceleration due to gravity, I'm sorry for not checking it :P twice.
b) just use the "fact" that and if I remember my calculus correctly (I havn't done that in a while, too busy with maths) you get something with log.
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
b) Find in terms of
c) Find the , that is, the maximum veloctiy
d) Find the distance fallen after 5 seconds (to the nearest metre)
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).
It's still basically the same way, you've worked out the inflow of x per second and the outflow of x per secondThis 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) =
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 :)
EDIT: Explaining the equationIts the same thing, just that you do and , 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 flow_{in}-flow_{out}.
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.
This is such an easy question, yet I keep getting the wrong answer :(
The radius of a spherical ball is and its volume is increasing at a rate of . At what rate is the radius increasing with respect to time?
My attempt was this:
But I got the answer wrong. I suspect that my flaw might have been in the part, but I'm not completely sure.
Please help me!
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.+1.
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 .
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 .
The equation of a curve C is where k is a constant.a few hints:
C does have a tangent parallel to the y axis.
Show that the y coordinate at the point of contact satisfies
EDIT: Beaten, dam latex is slower.
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 :PEDIT: 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
wow 2 months? I forget everything after 1 week after uni exams.
\underset{\sim}{j}
Do we need to know integration by parts?I heard integration by parts is useful for complicated derivatives via 'product rule' right? I think it would a good idea to learn it.
Should I learn it just incase?
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.
Do we need to know integration by parts?a quick tutorial if u ceebs wiki-ing it Re: Specialist Summer Holidays Question Thread!
Should I learn it just incase?
Do we need to know integration by parts?a quick tutorial if u ceebs wiki-ing it Re: Specialist Summer Holidays Question Thread!
Should I learn it just incase?
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?
pi, your way would have worked reasonably, except it's extremely long :P
Hey, that question involves partial fractions:
express the integral:
Now:
Using standard partial fractions techniques
now the integral is:
as required
pi's appraoch very valid as well but not on course
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)
Yeah, I see that, samad's way was much better, should've seen that (I'm really out of it lol)
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!
Btw, Year 12s, Tip: - If you see logs, think partial fractions
just equate the expression from part a) to 2 and solve for thetayar i did but i got stuck in the last part....
oh it's just general solutions, Re: Mathematical Methods Guides and Tips i used the exact same equation as an example loljust equate the expression from part a) to 2 and solve for thetayar 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
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 :'(oh it's just general solutions, Re: Mathematical Methods Guides and Tips i used the exact same equation as an example loljust equate the expression from part a) to 2 and solve for thetayar 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
Is there supposed to be a picture attached?
Is there supposed to be a picture attached?
was that at me? LOL?! i'm lost!
Express (2sqrt(3)+2i)/(1-3i) in polar form.
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.
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 , than it would result as , because we know that as general rule that , so if we expand out .
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.
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 , than it would result as , because we know that as general rule that , so if we expand out .
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
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
Thanks, but it gets harder:Do the same thing as above for finding the equation but for the flow out part, instead of using , the bottom part is your volume will change as time changes. So initally it was 400 and the volume us increasing at L/min.
"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 , the bottom part is your volume will change as time changes. So initally it was 400 and the volume us increasing at L/min.
So your volume will be 400+(1)t
i.e.
Then do the rest, i.e. dxin/dt-dxout/dt
For these questions where the flow in is not equal to the flow out, it is good to use this formula.So basically if you do chem you know that c=n/v (in M) or c=m/v
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 , think of it as c=mass/volume, mass is x and the volume is changing as t changes.
They're on the formula sheet, so you won't have to memorise them, but in all honesty, you might as well :P
Just wondering if we need to know Derivates of inverse circular functions?
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 :)
I'm not really sure why you drew that diagram that way...
Did you start with a triangle of forces?
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.
Did you start with a triangle of forces?
"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."
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."
The purpose of the lengths is to find the angle, which will ultimately help find the force.
(refer to the attached image).
Sorry guys, mustn't have attached it, it's attached in the original post now.(refer to the attached image).
We can't :P
If w= -(1/3) - ((square root 3)/3)i, express w^(-4) in polar form :D
If w= -(1/3) - ((square root 3)/3)i, express w^(-4) in polar form :D
Find the tangent to the curve x=2-3y^2, at point x=-1.
But it feels and looks sooo much better! :PFind 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........
Find the argument for 1-(i x square root of 3), in [0, 2 pi].
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?)
(Also: How do you get that Microsoft Word Equation Editor?)
anywho, i have another q. : Show that |z + w|^2 = |z|^2 + 2Re(z(conjugate of w)) + |w|^2
Isn't monkeywantsabananas way the same as wat truetears was talking about? :) and thanks paulsterio, made me look at it a different way :D
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. :Dahh 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)
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).Wow, so simple >.> Never thought of that 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.
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!
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!
z^{3} + i = 0
z^{3} = -i (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
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.
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!
z^{3} + i = 0
z^{3} = -i (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
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.
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!
z^{3} + i = 0
z^{3} = -i (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
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.
Ok, I don't know if I'm helping here, but for the example you showed:
, where
=>
Equating coefficients,
AND , where , and thus
So solving ,
and solving ,
now we pick out the common solutions, which are of course
and then sub back into
If this indeed what you were talking about, I'd still stick to the classic method because it doesn't involve trig (which I found deadly sometimes)
Have you guys done Vectors yet?Yeh they were quite hard last year.
Some of the ER questions in essentials make me sad.
Have you guys done Vectors yet?although bora in your dp makes me quite
Some of the ER questions in essentials make me sad.
I miss you, bebe. =(Have you guys done Vectors yet?although bora in your dp makes me quite
Some of the ER questions in essentials make me sad.happyhor..
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
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!
z^{3} + i = 0
z^{3} = -i (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
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.
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!
z^{3} + i = 0
z^{3} = -i (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
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 ?!
Thanks b^3, i know how to write pi in LaTeX now ;DToo bad you don't know the correct formula for the volume of a cylinder...
(quoted your post) :P
LOL!Thanks b^3, i know how to write pi in LaTeX now ;DToo bad you don't know the correct formula for the volume of a cylinder...
(quoted your post) :P
Kidding! (Don't take that the wrong way ;))
Too bad you don't know the correct formula for the volume of a cylinder...
Kidding! (Don't take that the wrong way ;))
Hmmm question:Converting to polar form we have:
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.)
Hmmm question:Converting to polar form we have:
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.)
Then, using De Moivre's theorem:
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... :sHmmm 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:Converting to polar form we have:
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.)
Then, using De Moivre's theorem:
,
Fourth quadrant, ,
, Sub it in.
Than you get, .
Using De Moivre's Theorem, .
Results as,
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.
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
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... :sHmmm 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:Converting to polar form we have:
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.)
Then, using De Moivre's theorem:
,
Fourth quadrant, ,
, Sub it in.
Than you get, .
Using De Moivre's Theorem, .
Results as,
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.
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
as required
as required
Complex number question: (From VCAA 2006 Exam 1)
Solve the quadratic equation , expressing your answers in exact cartesian form.
Must be a gap in my knowledge here, but fromWhen he wrote
and when you square rooted doesn't the argument change?
I'm confused because when you subbed in k=0, you got
but wouldn't it be because
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?
= =
Why is this invalid?Well firstly consider this statement
= =
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.
- cos^2(x)/sin(x) + sin(x) answer=cosec(x)
- Simplify tan((pi/2)-y) answer= cot(y)
Slightly shorter methodHaving 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.
- cos^2(x)/sin(x) + sin(x) answer=cosec(x)
First one:
Take sin(x) out as a factor
Could anyone prove this vector?a + b = AC = -c, .: a + b + c = 0
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)
It is easy to express and in terms of **, to express just write it as and again use your identities.
** http://mathworld.wolfram.com/Double-AngleFormulas.html second and third equation for
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..
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 :)
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. :(
Two vector questions that are doing my head in.90 degrees = perpendicular, so use dot product = 0 and solve for t. Or is that too simple.... lol
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). :(
Two vector questions that are doing my head in.90 degrees = perpendicular, so use dot product = 0 and solve for t. Or is that too simple.... lol
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). :(
Is question 1 asking for a vector resolute or something similar to that? lol i cant remember XD
Regarding the first question.
It does involve using vector resolutes.
(http://upload.wikimedia.org/wikipedia/commons/3/3e/Dot_Product.svg)
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
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
Unlatex'd. CA=OA-OC=a-[(a.b)/(b.b)]*b
Hope that helps, probably didn't explain it well.
Hi, could somebody help me with this question, describe the steps i should take or even show me the steps if possibleGiven 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!
Solve this equation over C: z^{2}=8+15i
Thanks!
the answer is D. what do you do to get rid of the 3?
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?
evaluate:
can somsone please guide me through that!
let u=
Ill just help you with the the intergal, then you can sub in the values yourself later.beautttty. thanks!
So we have
Now here there is something we can notice, we know that the derivatie of is , so if we let u=tan(x), then we can try and get the to cancel out.
So we have
Then sub in the values to find the definite integral.
Oh, and don't forget to change the terminals :)
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! :)
Why is it that = ?
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!
Why is it that = ?
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
In your case, the negative has been moved inside the integral :)
Why is it that = ?
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
In your case, the negative has been moved inside the integral :)
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 :)
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.
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
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
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
accordingly, depending on your choice, your integral becomes either
(1 - cosy)dy or (cosy - sqrt(3)/2)dy
The limits of integration stay the same, 0 and pi/6
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
accordingly, depending on your choice, your integral becomes either
(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?
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
accordingly, depending on your choice, your integral becomes either
(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
1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?Remember that and
(6a, Ex. 4D, Essential)
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)Let
(Question 3, Ex. 4D)
1) How do you show sin(theta) + cos(theta)i = cis(pi/2 - theta)?Remember that and
(6a, Ex. 4D, Essential)
So we have2) 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)Let
(Question 3, Ex. 4D)
So we know that
Now we know that the angles are restricted to and
So
Now you should be able to do the other half of the question, give it a go remembering that
Hope that helps :)
solve for
goddamit.
solve for
has no real solutions
edit: oops, forgot about the restriction:
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)
Solve and you'll see :)
Yeah, nice :) Why didn't you divide both sides by x then at the start?
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 + 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:
The dot product can be defined as follows:
where is the angle between the vectors.
Is it a calculator question?
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:
The dot product can be defined as follows:
where is the angle between the vectors.
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.
How do i draw a diagram for this:See attachment
For quadrilateral OABC , D is the point of trisection of OC nearer O and E is the point of trisection of AB nearer A.
I need help with question 8b! :D
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
Hey, spell check didn't pick it up :(
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
(Square root over the entire answer, I have no idea how to do that for LaTeX >.<)You just stick it around curly brackets
[tex]\sqrt{moooooooooooooooooo + roooooar}[/tex]
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?
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 itwell the formal definition is given on wiki http://en.wikipedia.org/wiki/Complex_number#Definition
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.
[tex]\int[/tex]
For the square root[tex]\sqrt{}[/tex]
Can someone show me how to do this:Hey, I assume you split the fraction into partial fractions with the denominators being 2-x and 2+x .
Find the solution to the differential eqn: =,given that y=2 when x=0
I keep getting it in the form y=log_{e}|(2-x)(2+x)|+c but answer says it should be in the form y=log_{e}||+c
Thanks
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;
That doesn't work with x^2 does it?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;
and you shouldn't go wrong.
That doesn't work with x^2 does it?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;
and you shouldn't go wrong.
I was referring to and his question of what to do with the negative in front of the x.That doesn't work with x^2 does it?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;
and you shouldn't go wrong.
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?
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??
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.
can someone help me with this Q? i don't really know where to really startThe 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.
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?
Can someone show me how to solve z^{3}-(2-i)z^{2}+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?
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 :)
How do i antidiff tan^{2}xsec^{2}x?
Could you show me why it's n=3?
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.
QuoteWhat 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).
Wikipedia's always good.
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
If anyone could help me with this question would be great!
√3z^{2} + √2z - i/2 = 0
Find the solutions to this equation
How do you antidiff xtan^{-1}x? Would something like this be in Exam 1 or 2?
Thanks
ok thanks, so i should just chuck it in my CAS?How do you antidiff xtan^{-1}x? 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?How do you antidiff xtan^{-1}x? 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.
What if it was diff xtan^{-1}x?
When asked to factorise say P(z)=z^{3}-5z^{2}+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?Not sure, but you might be thinking about how the three terms in brackets need to multiply together to make the last term.
Thanks
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≤LYep, is the deflection and is the gradient of the beam.
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
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≤LYep, is the deflection and is the gradient of the beam.
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
You have to integrate and then use the conditions given (when , , when , ) to find and or whatever you choose to label your constants after integration.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≤LYep, is the deflection and is the gradient of the beam.
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
Thanks, a little confused here... how do I solve for a specific solution with the DE and find k?
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?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.
And is there a geometric interpretation of that (do you count the sums backwards or something?)
Using the double angle formuals for coslegend, thanks mate!
Firstly I will take up as positive.
(positive as it is moving upwards initially)
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.Firstly I will take up as positive.
(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?
Could anyone please help me with this question:
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?
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.
I need help with this question:Original displacement;
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
I need help with this question: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
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
I need help with this question: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
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
Part b) Find the time the particle is travelling back towards its original position.I need help with this question: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
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
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!
Hmm, are you sure b^3, I personally think the question is asking when it changes direction?Thats the way I originally read it, but the OP did say the answer was 55/4 seconds.
i.e. 25/4 s
Its not a well worded question.I need help with this question: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
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
I think it should be worded: At what time does the particle REACH its initial position?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.
But OK :D
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.
I think it should be worded: At what time does the particle REACH its initial position?Not quite sure I follow this statement though?
But OK :D
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.
Could somebody explain to me why root of x^{2} = |x| ?
Thanks
Oh thanks, i kept thinking that |x| meant just +x and was wondering what happened to the -x. Yeh now i understand
Yeah, VCE physics is a joke.
It's funny how one part of one chapter of spec is basically all of physics unit 3 AOS 1 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 ;).
Hey could u help me with this question
implicity differentiate y^{2}=3xy+x^{2}
Thanks
its mainly the constructing the de and solving it, i can't do it the back of the book has weird answers
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)
the answer is (100-T)/(40)
I've got a gold one.
A cissoid (where and ) is rotated about it's asymptote to form a volume of revolution. Find this volume in terms of . 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. when it is rotated around the x-axis to form a solid volume of revolution, in terms of and .
Post your working/answers up here :)
I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be instead of ?
That would make sense to me but if not, how do you work out that asymptote?
I just sketched it for a=2 and the graph passed right through x=4 haha...
Is the function supposed to be instead of ?
That would make sense to me but if not, how do you work out that asymptote?
Oh shit, you're completely right. 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 :(
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.
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.Denote upwards to be positive and downwards to be negative.
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
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
i then FLIPPED
to get
when i integrated that w.r.t v,
i got
i dont understand how simplifying it to get two fractions, then flipping, gave me a different answer to not simplifying it and flipping initially ?
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!
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
Hello, just a quick question: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.
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 its from essentials, but i don't see the word "smooth" anywhere in the question.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.
Here's one thats bugging me: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 then dividing by x will only leave .
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?
if sin(x) = 1-a/1+a,0<x<pi/2 express cos(2x) in terms of a
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?
Vectors are not parallel if they're scalar multiples of each other, regardless of i,j,k
Also, are vectors parallel if they're scalar multiples of eachother, regardless of the signs the i, j or k's carry?
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 otherFor parallel vectors, is true. This is what it means for a vector to be a scalar multiple of another.
But you cant prove 2 vectors parallel just because their scalar multiples each other
They are parallel if vector a=k*vector b. From there, you can say their scalar multiples each otherFor parallel vectors, is true. This is what it means for a vector to be a scalar multiple of another.
But you cant prove 2 vectors parallel just because their scalar multiples each other
Your statement is a bit contradictory, you seem to be saying that they are parallel if , but you can't prove them parallel if they are That doesn't really make much sense to me.
Note that does imply that for when k is greater than 0. 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.
can i please have help with this definite integral
0 to pi/4 tan^3(x) dx
But you cant prove 2 vectors parallel just because their scalar multiples each other
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 otherFor parallel vectors, is true. This is what it means for a vector to be a scalar multiple of another.
But you cant prove 2 vectors parallel just because their scalar multiples each other
Your statement is a bit contradictory, you seem to be saying that they are parallel if , but you can't prove them parallel if they are That doesn't really make much sense to me.
Note that does imply that for when k is greater than 0. 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
If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallelI think this is just coming down to a matter of slightly different definitions and terminology really.
If generalkorn meant scalar multiplies of vectors, if k is positive => they are parallel. If k is negative => they are anti-parallelI 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 .
can i please have help with this definite integral
0 to pi/4 tan^3(x) dx
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 otherFor parallel vectors, is true. This is what it means for a vector to be a scalar multiple of another.
But you cant prove 2 vectors parallel just because their scalar multiples each other
Your statement is a bit contradictory, you seem to be saying that they are parallel if , but you can't prove them parallel if they are That doesn't really make much sense to me.
Note that does imply that for when k is greater than 0. 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
Dont make it so complicated. They are both parallel, anti-parallel just gives extra info that they are in opposite directionOh my bad, I misinterpreted what you meant then. I thought you were implying that 'anti-parallel' wasn't parallel at all.
can i please have help with this definite integral
0 to pi/4 tan^3(x) dx
Let and
Then and
I'm confused by this :/ Surely if k is negative the vectors are still parallel and the proof Paul is suggesting is valid.
^I didn't think of that either, haha. Seems quicker.
someone's enjoying the use of latex :Pcan i please have help with this definite integral
0 to pi/4 tan^3(x) dx
Let and
Then and
Wow, I didnt think of the method let y=cos (x) :)
What I do is
Edit: correct LaTex :P
Dont make it so complicated. They are both parallel, anti-parallel just gives extra info that they are in opposite directionOh my bad, I misinterpreted what you meant then. I thought you were implying that 'anti-parallel' wasn't parallel at all.
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.
someone's enjoying the use of latex :P
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.
The answer is T=16.8N.
The solutions to this question seem quite ... vague ... (Either that or I am quite stupid ;))Where abouts did you stop in your working? I'll just start at the last line you posted
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.
and
Hence,
'Expanded' by De Moivre's Theorem.
I really do not understand what the solutions did after 'expanding'.
Thanks for any help. :)
someone's enjoying the use of latex :Pcan i please have help with this definite integral
0 to pi/4 tan^3(x) dx
Let and
Then and
Wow, I didnt think of the method let y=cos (x) :)
What I do is
Edit: correct LaTex :P
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 -?
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
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 thisAlternatively, 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);
Methods 1: Put in CAS, find max is 2 :P
Method 2: Find derivative
There are 2 ways of doing thisAlternatively, 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);
Methods 1: Put in CAS, find max is 2 :P
Method 2: Find derivative
So the amplitude is given by;
Therefore max
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.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 xDAlright, considering it's not on the course I'll just show you the reasoning for a bit of fun.
We got taught that in spesh last year :OThat's because you went to MHS :P
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...
. Find in the form .
Thank you. :)
Would someone be able to show me how they would do this question?use double angle formula to find 1/1-2sin^2(pi/10) sub sin(pi/10) in and there you go. :)
I've done it but I feel as though my working is unconventional...
. Find in the form .
Thank you. :)
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.
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.
The forces acting on block 1 in the vertical direction.
The forces acting on block 1 in the horizontal direction.
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 , then the block won't move.
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...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
Thanks
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??
That would give the answer in terms of t though not x. wouldn't it?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??
Then you use formula
That would give the answer in terms of t though not x. wouldn't it?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??
Then you use formula
i wouldn't have realised too if i didn't try that earlier :PThat would give the answer in terms of t though not x. wouldn't it?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??
Then you use formula
lol stupid Jenny
I'm thinking we write 2e^0.3t in term of x and y then equate them
Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!I'm so sorry didn't mean to bother you. Thanks a lot though :D
This is the parametric equation of Bernoulli's (logarithmic) spiral
Its polar equation
(http://upload.wikimedia.org/wikipedia/commons/thumb/4/44/Logarithmic_spiral.png/213px-Logarithmic_spiral.png)
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!!!
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?
Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!I'm so sorry didn't mean to bother you.
This is the parametric equation of Bernoulli's (logarithmic) spiral
Its polar equation
(http://upload.wikimedia.org/wikipedia/commons/thumb/4/44/Logarithmic_spiral.png/213px-Logarithmic_spiral.png)
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!!!
:O wow that is amazing!! you are getting a 50 raw in spesh :O.Did you graph this parametric equation in CAS??? The graph looks like the spiral !!!I'm so sorry didn't mean to bother you.
This is the parametric equation of Bernoulli's (logarithmic) spiral
Its polar equation
(http://upload.wikimedia.org/wikipedia/commons/thumb/4/44/Logarithmic_spiral.png/213px-Logarithmic_spiral.png)
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!!!
Nah, you are not. I chose not to study at that day :(
Anyway, I figure it out
Sub (1) into
Therefore, the cartesian equation is
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.
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?
They had MB=AB-AM ? So are they wrong?
Ah ok, thanks jenny :)
Also, two more questions:
1. Consider the function defined by e^{x+y}=y+x^{2}+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
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?
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?
So you'd get , if you wanted to solved for y you'd need to integrate both sides with respect to x, i.e.: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?
Sorry, forgot about this!
e.g.
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
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
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.
^ 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
Bear in mind, that means buying a new calculator, then learning its ins and outs.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
Thanks b^3!Basically yeh.
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?
oo ok, that made sense, except when you said whats the smallest and biggest numbers we can get out for sin^4 (t),Since it's sin^4 the minimum is 0^4 (as -1^4=1)
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
wow.
haha thanks man!
your gonna kill spesh next year, sure you've already heard that ;p
The area under a velocity time graph will give displacement. So integrating from 0 to 50 will give the displacement of particle A from the starting position, at t=50s.How would you know which to take away from? Why couldn't it have been f(t)-g(t) ?
Now if we do the same for , 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
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) ?
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!
(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.
(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.
(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.
and
(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.
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 is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.
(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.
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 is measured to be the angle between u and v. I don't believe there's any domain restriction on that angle.
How do i find the time taken for the watch to reach the ground?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.
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+0.5at^{2}, but got t=3.54. The answer should be t=4.05, what have i done wrong?
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!
Q: why do we need to add the unit vectors of a and b?
THANKS
For the last step, the answer is
but why cannot I use the general solution i.e.
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 because the intervals between solutions in this equation are instead of
In the worked solution, they don't wanna separate solutions so they combine them
Length times cross-sectional area.
Wait, that would make it which isn't what I got. But I think their "show that" thing is wrong.
Well you're right, they have the same asymptotes. =/
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?
Umm, so it forms a square? I'm confused now, i really don't know :SIs 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
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.
Thanks a lot for your help :)u and v can be any complex numbers. So they won't necessarily only have one non-zero component.
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?
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.
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
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.
If it is the case, why simply just need to find the period of x component ?
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
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
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
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? :)
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 :)
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?
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.
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 V_{0}^{2} comes from in the 3rd step? I understand why the V_{0} 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.
Thanks in advance :)
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.
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
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
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.
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
Do we have to explicitly use u-substitution for anti differentiating f'(x)/f(x) functions?Or if the question is worth multiple marks. Often times the u substitution is involved in the marking criteria of integration questions.
i.e. would be sufficient working?
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).
Yes
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!
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
now you solve:
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! :)
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
So are you saying -i x (1+i) = -i+1 but.. :(
So are you saying -i x (1+i) = -i+1 but.. :(
but?
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...
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
2010 Exam 1 - Q2 b)
Why do you take the negative value of |(v-4)/4|?
2010 Exam 1 - Q2 b)"initially at rest", so at t = 0, v = 0.
Why do you take the negative value of |(v-4)/4|?
Hey could someone help me with MC question 11 from the VCAA 2011 exam please?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.
The magnitude of the sum of these three forces is equal to?
Don't really get how to do these force questions, thanks!
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 :)
2010 MC Q22: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.
(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 bit that's confusing me a bit.
The answer is D
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!
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!
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdfThey are just asking you to differentiate. They purposely gave that formula for that reason.
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 :)
Oh sorry i mean part c) ii) :)http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdfThey are just asking you to differentiate. They purposely gave that formula for that reason.
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 :)
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!
So we have something that looks like this right:
(http://i.imgur.com/dAwQR.png)
If the , we'll have something like this (pretend that the lines are of equal length):
(http://i.imgur.com/CcrWn.png)
But if we have , 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
Anyway to convert a parametric Equation to a cartesian equation on cas?
Yes but they replaced x(t) with (x/1+t) which is what is confusing me. Why did they do this?
substitute the from 5(c)(i) and stated in the question, simplify and eventually, it simplifies to and thus, satisfies the differential equation.
yeah, that's what you do to show it satisfies the condition
they substitute x(t) where the x was and that's why the denominator is instead ofoooohhh!! Thanks so much!!! Finally i understand it :D
You could do it on tiinspire before a recent update, just by typing in and pressing solve, but you can't do that anymore :(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.
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.
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
Vcaa 2011 2d.iiYou 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
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.
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?
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 :)
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)
Hint: try rearranging inside the sin and cos :)Spoilersin(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
Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)
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'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.
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.
I won't tell you how to do it exactly, but for hyperbolae with the general form , the asymptotes are given by . :)Remember that will be an ellipse, a hyperbola will have the form or
Hint: Rearrange like this:
Thats what I initially thought, and I got a answer ofThe answer should be , which I think is what you worked it out to be?
but the answer behind the book is
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?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.
Cheers
What do you guys think is the most challenging part of the spesh. course?
What do you guys think is the most challenging part of the spesh. course?
I won't tell you how to do it exactly, but for hyperbolae with the general form , the asymptotes are given by . :)
general circle formula stuff.What's that? I'm drawing a mind blank on what you mean
Dunno Pi, I always just remembered that asymptotes have the equation:
(y-k) = (+ or -)(b/a)(x-h)
:)
When a circle equation is in the form , 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
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 . To get this via another method, if you graph the original function, we can see it has asymptotes of and . 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 and always going to be less than (well more negative) than . 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 asymtptote). SO that is .
Anyways, thats just another way of looking at it, less mathematical though, but might help you understand it a little bit more.
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 ?
the answer is E, which is any real number in the interval , would that be the same as .
I'm just curious how you work out question b) part (i)
you need the series to converge, which occurs if
so, first you find r (common ratio) in terms of x,
then you apply the conditions for a series to be convergent and therefore possible for an infinite sum to exist
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 and the affect the y-coordinates, so it doesn't really move the vertical asymptote around.
It won't have an affect on the vertical asymptote though. The and the affect the y-coordinates, so it doesn't really move the vertical asymptote around.
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
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
EDIT (got confused at this point),
For the sine function we (conventionallyl) restrict the domain to [-pi/2, pi/2]
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.
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 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
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
if is a factor of , how would i find the remaining factors?
if is a factor of , how would i find the remaining factors?
Let
Expanding gives
So
and
Hence
Similarly, if you write out 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.
Assuming , then , so the result holds iff
thanks for your reply russ, but im still a little confused :/