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September 21, 2020, 01:29:06 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 1321559 times)  Share 

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coldairballoon

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9690 on: August 02, 2020, 12:01:29 am »
0
Y'all I'm sorry this must be a really stupid question but for some reason my brain's just not working with this question and my tutor isn't replying :(
I tried doing the normal method where you convert the cos^2 to 1-sin^2 and then substitute but it didn't work? I've attached it below (hopefully it works), help please!!
[2019] Biology - 50, Methods - 44
[2020] Chemistry - ?, English Language - ?, Latin - ?, Specialist - ?

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9691 on: August 02, 2020, 12:37:06 am »
+2
Y'all I'm sorry this must be a really stupid question but for some reason my brain's just not working with this question and my tutor isn't replying :(
I tried doing the normal method where you convert the cos^2 to 1-sin^2 and then substitute but it didn't work? I've attached it below (hopefully it works), help please!!


Not are if there's another way but I saw symbolab's solution and its pretty neat

https://www.symbolab.com/solver/integral-calculator/%5Cint%20%5Cfrac%7Bsin%5Cleft(2x%5Cright)%5E%7B3%7D%7D%7Bcos%5Cleft(2x%5Cright)%5E%7B6%7D%7Ddx

schoolstudent115

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9692 on: August 02, 2020, 11:42:40 am »
+1
Y'all I'm sorry this must be a really stupid question but for some reason my brain's just not working with this question and my tutor isn't replying :(
I tried doing the normal method where you convert the cos^2 to 1-sin^2 and then substitute but it didn't work? I've attached it below (hopefully it works), help please!!
Convert sin^2(2x) to 1-cos^2(2x). Then let u = 2x > dx = 1/2 * du
You will then have 1/2 the integral of (1-cos^2(u))/cos^2(u)) = 1/2 integral of (sec^2(u)-1).
You might know that the derivative of tan(x) is sec^2(x) so the rest should be simple.
(2020) Year 11: Physics 3/4

dskel

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9693 on: August 03, 2020, 10:38:42 am »
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Hey could anyone explain how to do Q9 from 2007 exam 1? specifically i don't understand how they manipulated dv/dt and dy/dx ... to find the acceleration. Thanks.

1729

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9694 on: August 03, 2020, 11:12:23 am »
+2
Hey could anyone explain how to do Q9 from 2007 exam 1? specifically i don't understand how they manipulated dv/dt and dy/dx ... to find the acceleration. Thanks.

Is this the question you were talking about?

Derivative of position vector = velocity vector, if x'=-y and y'=x try using that to find the acceleration vector.


Was this the solution you were talking about?
Use a = dv/dt = -yi + xj = -xi - yj = -(xi + yj) = -r. Instead, no dy/dx involved.

And theres no dy/dx in that solution, for one thing, its pretty much the same thing but just written out with more detail (and more leibniz notation instead of lagrange notation)
« Last Edit: August 03, 2020, 11:15:54 am by 1729 »

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9695 on: August 03, 2020, 06:24:23 pm »
+1
Another approach is to notice that the differential equations x' = -y and y' = x are solved by x = cos(t) and y = sin(t). Hence x'' = -cos(t) = -x and y'' = -sin(t) = -y, so r'' = -r.

schoolstudent115

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9696 on: August 04, 2020, 03:34:32 pm »
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Another approach is to notice that the differential equations x' = -y and y' = x are solved by x = cos(t) and y = sin(t). Hence x'' = -cos(t) = -x and y'' = -sin(t) = -y, so r'' = -r.

When I saw that x and y were alternating I instantly thought of sin and cos. I like this approach.
(2020) Year 11: Physics 3/4

digdog123

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9697 on: August 09, 2020, 08:53:11 pm »
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Hey guys , I was doing an external exam and for the solutions to one of the questions, it said this as an explanation (which should be attached below). I just wanted to know if this was a thing?? bc i tried it out with a few numbers, and maybe I did it wrong, but it didn't seem to be true. thanks in advance!
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schoolstudent115

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9698 on: August 09, 2020, 10:29:57 pm »
+6
Hey guys , I was doing an external exam and for the solutions to one of the questions, it said this as an explanation (which should be attached below). I just wanted to know if this was a thing?? bc i tried it out with a few numbers, and maybe I did it wrong, but it didn't seem to be true. thanks in advance!
This is true. Use De Moivre's theorem. .


However, notice that this is only for , since the general polar form for some is

The more general form for is (again using De Moivre's theorem)
(2020) Year 11: Physics 3/4

digdog123

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9699 on: August 10, 2020, 02:35:01 pm »
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This is true. Use De Moivre's theorem. .


However, notice that this is only for , since the general polar form for some is

The more general form for is (again using De Moivre's theorem)

Thanks so much!
« Last Edit: August 10, 2020, 02:38:05 pm by digdog123 »
beep boop

sogreatsosad

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9700 on: August 13, 2020, 07:37:52 pm »
-1
Hey, how do I solve this? A bit of an explanation too please :)


1729

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9701 on: August 13, 2020, 08:11:32 pm »
+2
Hey, how do I solve this? A bit of an explanation too please :)
To solve this problem all you need to know is that the question is asking you to find the smallest number n such that the expression is equal to some a*i where a is a real number. In my explanation I solved it by using the exponential form.

So to the solution:
You can turn the inner part (2-2sqrt(3)) into the form re^(itheta)
r = sqrt(a^2 + b^2)
a = 2 and b = -2sqrt(3) in this case so r = 4
factor out the 4 and you get 4*(1/2 - (sqrt(3)/2))
1/2 = cos(theta) since e^(i*theta) = cos(theta) + isin(theta)
therefore theta = pi/3
So our number is 4*e^(i pi/3)
Raise that to the nth power and you get 4*e^(i n pi/3)
This has to be equal to some imaginary number a*i where a is some real number
You can then divide both sides by 4 and take the natural log on both sides to get i*n pi/3 = ln(ia/4) = ln(i) + ln(a/4)  = i pi/2 + ln(a/4)
Then multiply both sides by 3/(i*pi) to get n = 3/2 + something times the natural log term.
Since the minimum value the natural log term can have is 0, the minimum value of n = 3/2

Let me know if anything was unclear in my explanation of the solution.
« Last Edit: August 13, 2020, 08:33:45 pm by 1729 »

sogreatsosad

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9702 on: August 13, 2020, 08:26:42 pm »
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Firstly, this question is worded pretty badly because 2 - 2sqrt(3)i is an imaginary number, n=1 would work as well. And like any other n less than that. Also, an imaginary number is just any number with i in it well; it's a complex number; i by itself is the imaginary number. Furthermore, in this equation, Im(z) give b where z = a +bi, is the imaginary part; bi is the imaginary number a is the real number the two of them make a complex number.

However to solve this problem all you need to know is that the question is asking you to find the smallest number n such that the expression is equal to some a*i where a is a real number. In my explanation I solved it by using the exponential form.

So to the solution:
You can turn the inner part (2-2sqrt(3)) into the form re^(itheta)
r = sqrt(a^2 + b^2)
a = 2 and b = -2sqrt(3) in this case so r = 4
factor out the 4 and you get 4*(1/2 - (sqrt(3)/2))
1/2 = cos(theta) since e^(i*theta) = cos(theta) + isin(theta)
therefore theta = pi/3
So our number is 4*e^(i pi/3)
Raise that to the nth power and you get 4*e^(i n pi/3)
This has to be equal to some imaginary number a*i where a is some real number
You can then divide both sides by 4 and take the natural log on both sides to get i*n pi/3 = ln(ia/4) = ln(i) + ln(a/4)  = i pi/2 + ln(a/4)
Then multiply both sides by 3/(i*pi) to get n = 3/2 + something times the natural log term.
Since the minimum value the natural log term can have is 0, the minimum value of n = 3/2

Let me know if anything was unclear in my explanation of the solution.

Hi, thanks for your time

Is there any way you could explain this using De Moivre's theorem? Thanks!

keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9703 on: August 13, 2020, 08:33:34 pm »
-1
However to solve this problem all you need to know is that the question is asking you to find the smallest number n such that the expression is equal to some a*i where a is a real number. In my explanation I solved it by using the exponential form.

So to the solution:
You can turn the inner part (2-2sqrt(3)) into the form re^(itheta)
r = sqrt(a^2 + b^2)
a = 2 and b = -2sqrt(3) in this case so r = 4
factor out the 4 and you get 4*(1/2 - (sqrt(3)/2))
1/2 = cos(theta) since e^(i*theta) = cos(theta) + isin(theta)
therefore theta = pi/3
So our number is 4*e^(i pi/3)
Raise that to the nth power and you get 4*e^(i n pi/3)
This has to be equal to some imaginary number a*i where a is some real number
You can then divide both sides by 4 and take the natural log on both sides to get i*n pi/3 = ln(ia/4) = ln(i) + ln(a/4)  = i pi/2 + ln(a/4)
Then multiply both sides by 3/(i*pi) to get n = 3/2 + something times the natural log term.
Since the minimum value the natural log term can have is 0, the minimum value of n = 3/2

Let me know if anything was unclear in my explanation of the solution.

Euler's formula is outside of the curriculum (I agree - it's stupid. It should be. But it's not.)

Hi, thanks for your time

Is there any way you could explain this using De Moivre's theorem? Thanks!

Asking in two places like this is a bit rude, don't you think?
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1729

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9704 on: August 13, 2020, 08:46:03 pm »
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Euler's formula is outside of the curriculum (I agree - it's stupid. It should be. But it's not.)
Oh my, the study designs everywhere. I thought, considering this is specialist 3/4 it would've been in the study design - I don't do VCE which is why I don't know. I'd never heard of this theorem before but I just looked it up. It's the same concept nearly.
Asking in two places like this is a bit rude, don't you think?
I didn't realise he asked in the specialist 1/2 thread as well but adding to the advice that keltingmeith did in the spesh 1/2.

You first wanna turn the complex number into the form cosx + isinx but you can't do that directly as you get cosx = 2 and sinx = -2sqrt(3) and those numbers are out of the ranges of the sine and cosine functions, cosx = 2 is impossible in the real numbers is what I mean. So you first factor out a 4 again because we wanna turn the numbers into things we know are the cosine/sine of a number. What I mean is the known numbers that come out of a sine or cosine function are  0, 1/2, 1/sqrt(2), sqrt(3)/2 or 1 yeah?
for the angles 0, 30, 45 ,60 and 90 so we wana turn that complex number into something like one of those numbers
factoring out a 4 does the trick since we get 4*(1/2 -sqrt(3)/2 i) both 1/2 and -sqrt(3)/2 are things we can work with. Now we can find the trigonometric form of the complex number setting cosx = 1/2 we get x = pi/3. Well I skipped a step our expression turns into 4^n * (cosx + isinx)^n which is what we want for De Moivre's theorem, and we find x = pi/3. Now you can plug it into the theorem to get 4^n * (cos(pi/3) + isin(pi/3))^n = 4^n * (cos(n pi/3) + isin(n pi/3)). Since we want this to be an imaginary number, the real part must be zero. Distribute the 4^n and you find that the real part of the number is 4^n * cos(n pi/3) which we want to be equal to 0. Since 4^n can't be 0, cos(n pi/3) must be zero. The smallest positive real n that makes cos(n pi/3) = 0 is n = 3/2 since we get cos(pi/2)