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#### TrueTears

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##### VCE Specialist 3/4 Question Thread!
« on: November 26, 2011, 10:07:42 pm »
+13

If you have general questions about the VCE Specialist Maths course or how to improve in certain areas, this is the place to ask!

Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

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There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you, including TuteSmart tutors! So you may even get multiple answers from different people offering their insights - very cool.

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similar to the methods one Methods [3/4] Summer Holidays Question Thread! post away your questions from your summer holidays self-studying, everyone can discuss and benefit! I'll try answer as much questions as possible too ^^
« Last Edit: February 26, 2020, 03:22:03 pm by PhoenixxFire »
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#### Special At Specialist

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##### Re: Specialist 3/4 Question Thread!
« Reply #1 on: November 27, 2011, 12:00:02 pm »
0
Let w = 2cis(θ) and z = w + 1/w
Show that z lies on the ellipse with equation (x^2)/25 + (y^2)/9 = 1/4
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#### brightsky

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##### Re: Specialist 3/4 Question Thread!
« Reply #2 on: November 27, 2011, 12:28:30 pm »
+4
z = w + 1/w = 2cist + 1/(2cist) = 2cist + 1/(2(cost + isint)) = 2cist + (cost - i sint)/(2) = 5/2 cost + 3/2 i sint
parameters:
x = 5/2cost
y =3/2 sint
convert into cartesian form and you get the equation of the ellipse.
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#### brightsky

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##### Re: Specialist 3/4 Question Thread!
« Reply #3 on: November 27, 2011, 08:47:43 pm »
+1
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.
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#### Special At Specialist

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##### Re: Specialist 3/4 Question Thread!
« Reply #4 on: November 27, 2011, 08:48:49 pm »
+2

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

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

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

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

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

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

Ye know too much for a year 10 student!
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#### Special At Specialist

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##### Re: Specialist 3/4 Question Thread!
« Reply #5 on: November 27, 2011, 09:11:07 pm »
+1
Someone please tell me where I went wrong in this question!

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

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

Where did I go wrong? I was supposed to get the answer 5 but instead I got 4cos(θ)
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#### brightsky

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##### Re: Specialist 3/4 Question Thread!
« Reply #6 on: November 27, 2011, 09:42:39 pm »
+4
|z - 2i| = sqrt((2cos(θ) - 5/2)^2)
|z - 2i| = 2cos(θ) - 5/2
here's the error. cosθ =< 1, which means  2cosθ - 5/2 =< -1/2
so sqrt((2cos(θ) - 5/2)^2) = abs(2cos(θ) - 5/2) = 5/2 - 2cosθ
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#### Special At Specialist

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##### Re: Specialist 3/4 Question Thread!
« Reply #7 on: November 27, 2011, 09:51:46 pm »
0
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?
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#### TrueTears

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##### Re: Specialist 3/4 Question Thread!
« Reply #8 on: November 27, 2011, 09:55:27 pm »
0
oh absolute values!

if it were ur alternative case, then itd be cos(t) <=1 hence cos(t) -1/2 <= -1/2
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#### brightsky

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##### Re: Specialist 3/4 Question Thread!
« Reply #9 on: November 27, 2011, 09:56:32 pm »
0
I never knew you had to do that...
What if it was sqrt((cosθ - 1/2)^2) ?
How would you know whether to take the positive or negative solution?

depends on the restrictions of θ. if none are specified, then you'll have two solutions.
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#### Special At Specialist

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##### Re: Specialist 3/4 Question Thread!
« Reply #10 on: November 27, 2011, 10:00:12 pm »
0
That makes sense... actually I think I remember a similar problem when taking an integral which ended up with an absolute value and I was unsure whether to write ln(5 - x) or ln(x - 5) so I had to use the initial conditions to see which solution it was.
This is basically the same concept.
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#### TrueTears

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##### Re: Specialist 3/4 Question Thread!
« Reply #11 on: November 27, 2011, 10:00:23 pm »
+4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

and so the rest is trivial

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

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

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

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

« Last Edit: November 27, 2011, 10:05:50 pm by TrueTears »
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#### kensan

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##### Re: Specialist 3/4 Question Thread!
« Reply #12 on: November 27, 2011, 10:12:51 pm »
+5
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
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#### TrueTears

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##### Re: Specialist 3/4 Question Thread!
« Reply #13 on: November 27, 2011, 10:19:26 pm »
0
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
heh, no need to worry, just do some self study and this stuff will be easy ^^
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