September 21, 2020, 01:29:06 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1321559 times) Tweet 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.

Use a = dv/dt = -y’i + x’j = -xi - yj = -(xi + yj) = -r. Instead, no dy/dx involved.

And there’s no dy/dx in that solution, for one thing, it’s 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!
beep boop

#### 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. $\frac{1}{z}=z^{-1}$.
$z^{-1}= (cis(\theta))^{-1} = cis(-\theta)$

However, notice that this is only for $r=1$, since the general polar form for some $z$ is $r*cis(\theta)$

The more general form for $z^{-1}$ is (again using De Moivre's theorem) $(r*cis(\theta))^{-1} = \frac{1}{r} * cis(-\theta)$
(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 »
0
This is true. Use De Moivre's theorem. $\frac{1}{z}=z^{-1}$.
$z^{-1}= (cis(\theta))^{-1} = cis(-\theta)$

However, notice that this is only for $r=1$, since the general polar form for some $z$ is $r*cis(\theta)$

The more general form for $z^{-1}$ is (again using De Moivre's theorem) $(r*cis(\theta))^{-1} = \frac{1}{r} * cis(-\theta)$

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

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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 »

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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.

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.)

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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Something in VCE, shit was too long ago to remember

Try out my study score calculator, request your subjects, and help give feedback if you've already completed VCE!

#### 1729

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9704 on: August 13, 2020, 08:46:03 pm »
0
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)