September 19, 2020, 10:01:00 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1321140 times) Tweet Share

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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9705 on: August 13, 2020, 08:53:18 pm »
0
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. 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.

Dude, it's in no study design all around the country that I've checked. Which actually annoys me, it's literally one extra equation, is commonly voted as one of the most important/beautiful formula in maths, would explain so much that we teach in specialist that currently has no explanation (or has complicated explanations), and then we wouldn't have to worry about the cis(theta) notation we use that nobody else in the world does.
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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9706 on: August 13, 2020, 08:55:25 pm »
+1
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. 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)

Firstly, I'd like to apologise. This is the first time I'm using something like Atarnotes, and I never considered that if everyone acted like me then this site would be a cluttered mess. I'm truly sorry about that, and will be more mindful in the future.

Secondly, thank you both for explaining this to me, it makes sense. If a question asked the smallest value to get a real number, then I'd simply ensure that sin(n*theta)=0, yes?
« Last Edit: August 13, 2020, 10:00:15 pm by sogreatsosad »

#### 1729

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9707 on: August 13, 2020, 08:59:56 pm »
0
Firstly, I'd like to apologise. This is the first time I'm using something like Atarnotes, and I never considered that if everyone acted like me then this site would be a cluttered mess. I'm truly sorry about that, and will be more mindful in the future.

Secondly, thank you both for explaining this to me, it makes sense. If a question asked the smallest value to get a real number, then I'd simply ensure that sin(n*theta)=0, yes?
Yep I'm pretty sure. If it says n is positive then you try to make the argument pi to get 0.

EDIT: Strangely you edited your message as soon as I posted, like exactly the same time not even kidding.
For example, (-2+sqrt2*i)^n, to find the smallest n where this is a real number, would this be done as I describe above? I got n=4/3
I'm pretty sure you'd have to factor out a 2 in this case but yeah when you find theta you set sin(n*theta) = 0
« Last Edit: August 13, 2020, 09:03:09 pm by 1729 »

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9708 on: August 14, 2020, 02:11:00 pm »
+1
Dude, it's in no study design all around the country that I've checked.

It's in HSC Extension 2.

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9709 on: August 14, 2020, 02:24:02 pm »
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It's in HSC Extension 2.

Welp, that's what I get for thinking they wouldn't change the syllabus in 5 years lol
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#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9710 on: August 20, 2020, 11:33:41 pm »
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does anyone know how to do this?

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9711 on: August 21, 2020, 12:11:00 am »
+1
does anyone know how to do this?

It may help to draw a diagram. Have you done this, and what else have you tried to solve this question?
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#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9712 on: August 21, 2020, 12:18:43 am »
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It may help to draw a diagram. Have you done this, and what else have you tried to solve this question?

I've drawn this but I am having trouble resolving the vectors

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9713 on: August 22, 2020, 02:36:46 pm »
+3

I've drawn this but I am having trouble resolving the vectors

Well, see attached. Notice that now you have two triangles you can work with - and remember that the sum in each direction is going to be 0. You'll also note that I've made an angle called t - and that the angle between the 10 vector and the 12 vector will be (180-60)+t degrees. Do you think you can take over from here?
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)

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#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9714 on: August 22, 2020, 03:24:32 pm »
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does anyone know how to do this?

I know this is not the way the questions asks this to be done, but P can be more easily found by drawing a triangle of forces and using the cosine rule, and then the other angles between the forces can be found using cosine / sine rules.

#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9715 on: August 22, 2020, 06:26:41 pm »
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Well, see attached. Notice that now you have two triangles you can work with - and remember that the sum in each direction is going to be 0. You'll also note that I've made an angle called t - and that the angle between the 10 vector and the 12 vector will be (180-60)+t degrees. Do you think you can take over from here?

Right. I see how it would be (180-50) + t = 130 + t

but from here Idk how to use the cosine rule since both t and p will be unknown

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9716 on: August 22, 2020, 06:49:30 pm »
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Right. I see how it would be (180-50) + t = 130 + t

but from here Idk how to use the cosine rule since both t and p will be unknown

TBH I'm not quite following KM's diagram, so let me try to explain how I would do this one by resolving into rectangular components.

First, I'd let the direction of i be parallel to the direction of the force of P newtons. So that force is Pi.

Second, I'd write the other two vectors in terms of their components. The 10 newtons force will be 10cos(50°)i + 10sin(50°)j, and the 12 newtons force will be 12cos(a)i + 12sin(a)j, where a is the angle measured anti-clockwise from the direction of the P newtons force. (There are other ways to resolve that 12 newtons force, by constructing some right-angled triangles which I think is what KM is suggesting, but this will do).

Then, since everything is in equilibrium, we have 10sin(50°) + 12sin(a) = 0 and P + 10cos(50°) + 12cos(a) = 0. Solve simultaneously for a and P. Note that this approach will give two solutions for a (since you'll get sin(a) = something negative), but only one of these corresponds to a positive solution for P.

Using this approach, the angle between the 12 newtons and 10 newtons forces will be a - 50.

If you use the cosine rule, then draw a triangle where the side lengths are 12, 10, P, and the angle between the sides of lengths 10 and P is (180 – 50)°, now use cosine rule to find P.

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9717 on: August 22, 2020, 06:50:46 pm »
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Right. I see how it would be (180-50) + t = 130 + t

but from here Idk how to use the cosine rule since both t and p will be unknown

You should try writing out some equations anyway, and seeing what happens. Notice what the hint tells you to do?
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#### Chocolatemilkshake

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9718 on: August 23, 2020, 03:08:12 pm »
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Hi would love some help with this question (volume of a solid when rotating about the y-axis). I just can't seem to figure it out.
It's attached below because I've got no idea how to write maths equations on AN
Thanks
VCE
2019 - Biology [50 + Premier's]
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#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9719 on: August 23, 2020, 03:37:58 pm »
+2
Hi would love some help with this question (volume of a solid when rotating about the y-axis). I just can't seem to figure it out.
It's attached below because I've got no idea how to write maths equations on AN
Thanks

First thing is rewrite the equation as $y^2 - x^2 = 1$, the graph of which is a hyperbola. So sketch that, along with the line x = 3, and find the region that is being rotated about the y-axis. You'll need to find a point of intersection between the line x = 3 and that hyperbola. That will help with finding the limits of the definite integral that you'll use to calculate the volume.

Next, the volume of the solid of revolution formed when you rotate a region about the y-axis is given by $\pi \int_{a}^{b} x^2dy.$, where $a$ and $b$ are the minimum and maximum y-values of the region being rotated about the y-axis. So rewrite the equation to get $x^2$ in terms of $y$, then calculate that definite integral.