VCE Specialist 3/4 Question Thread!

[psyxwar]

Can someone please explain to me the answers for 4 c, d and e on the 2007 VCAA Spesh exam 2? Please? There are no worked solutions and I dont understand the method behind doing these type of Qs, thanks in advance.

4.
c) let the k component of r(t) equal 0 and find t and sub it back into r(t)
d) find the minimum of |r(t)| (for example using fMin)
e) |r(a)-r(0)| where a is the time it lands

[M_BONG]

Hi guys

Can someone explain what the following means:

When the complex number z = x + yi is
a) Rotated by pi/2/ anticlockwise about the y-axis
b) Rotated by pi/2 clockwise about the y-axis
c) Rotated by 3pi/2 anticlockwise about the x-axis
d) Rotated by 3pi/2 clockwise about the x-axis

These seem to always pop up in the VCAA multiple choice, and not the trial exam/textbook. Someone explain? Is it different from it is rotated about, say, the x-axis?

Cheers!

[lzxnl]

Hi guys

Can someone explain what the following means:

When the complex number z = x + yi is
a) Rotated by pi/2/ anticlockwise about the y-axis
b) Rotated by pi/2 clockwise about the y-axis
c) Rotated by 3pi/2 anticlockwise about the x-axis
d) Rotated by 3pi/2 clockwise about the x-axis

These seem to always pop up in the VCAA multiple choice, and not the trial exam/textbook. Someone explain? Is it different from it is rotated about, say, the x-axis?

Cheers!

Rotating something about an axis!? What the...that's in 3D, not 2. Do you mean 'from' the axes?

[M_BONG]

Rotating something about an axis!? What the...that's in 3D, not 2. Do you mean 'from' the axes?

Yup! Apologies for the poor wording (I was doing it off the top of my head)

Something like this: (in the attachments)

What does it mean to rotate something about the origin?

[lzxnl]

Yup! Apologies for the poor wording (I was doing it off the top of my head)

Something like this: (in the attachments)

What does it mean to rotate something about the origin?

In that case...imagine you stick a large pole through the origin perpendicular to the xy plane. Rotating something about the origin means rotating about this large pole, if you want.

General rule: if you're rotating counterclockwise by theta radians, you are multiplying the complex number by cis theta
So you're rotating clockwise by pi/4 radians, you multiply the complex number by cis -pi/4. Negative as clockwise = -anticlockwise

[M_BONG]

Oh ok, cheers.

Also, with regards to implicit differentiation:

Eg. find dy/dx for (x^2 + y^2)^2

Do you have to expand the brackets or can you do it normally using chain rule?
I don't get the same results when I do it by chain rule.

This is what I do:

2 (x^2 + y^2) x ( 2x + 2y dy/dx) which is different from the answer as they first expanded the bracket and differentiated each term separately... so I can't apply normal Methods Chain rule techniques when implicitly differentiating?

[lzxnl]

Oh ok, cheers.

Also, with regards to implicit differentiation:

Eg. find dy/dx for (x^2 + y^2)^2

Do you have to expand the brackets or can you do it normally using chain rule?
I don't get the same results when I do it by chain rule.

This is what I do:

2 (x^2 + y^2) x ( 2x + 2y dy/dx) which is different from the answer as they first expanded the bracket and differentiated each term separately... so I can't apply normal Methods Chain rule techniques when implicitly differentiating?

I'm assuming you mean (x^2 + y^2)^2 = c where c is a constant
You can't differentiate, say, x + y. You can only differentiate a function or a relation

So for (x^2 + y^2)^2 = c, I'll do it both ways and I'll hopefully be able to show they're the same.
x^4 + 2x^2 y^2 + y^4 = c
4x^3 + 4x y^2 + 4x^2 y dy/dx + 4y^3 dy/dx = 0
dy/dx = -(x^3 + xy^2)/(x^2 y + y^3) = -x(x^2 + y^2)/y(x^2 + y^2) = -x/y

(x^2 + y^2)^2 = c
2(x^2 + y^2)*(2x + 2y dy/dx) = 0
Now, assume c > 0 as if c = 0, x^2 + y^2 = 0 means x and y are both zero only, which is rather trivial. If c is not 0, then x^2 + y^2 is not zero. Hence I can divide both sides by that.
2x + 2y dy/dx = 0
dy/dx = -x/y

There's nothing wrong with doing either...I think what you've forgotten is that you have to differentiate an equation. Even when we say 'differentiate x^3', we mean 'find dy/dx if y = x^3'
It's the rate of change between two variables and these variables have to have a known relationship before you can find their relative rates of change.

[keltingmeith]

Oh ok, cheers.

Also, with regards to implicit differentiation:

Eg. find dy/dx for (x^2 + y^2)^2

Do you have to expand the brackets or can you do it normally using chain rule?
I don't get the same results when I do it by chain rule.

This is what I do:

2 (x^2 + y^2) x ( 2x + 2y dy/dx) which is different from the answer as they first expanded the bracket and differentiated each term separately... so I can't apply normal Methods Chain rule techniques when implicitly differentiating?

There's actually a little trick with this one. As |zxn|, we're assuming that this is equal to some constant c, since implict differentiation is used on full equations (since that's the whole point of implict differentiation - differentiating implict functions):



Which matches both answers given by |zxn|.

For future reference, if you apply a differentiation rule correctly, you can use it in EVERY circumstance - including when/if you learn about partial differentiation at university.

[M_BONG]

I'm assuming you mean (x^2 + y^2)^2 = c where c is a constant
You can't differentiate, say, x + y. You can only differentiate a function or a relation

So for (x^2 + y^2)^2 = c, I'll do it both ways and I'll hopefully be able to show they're the same.
x^4 + 2x^2 y^2 + y^4 = c
4x^3 + 4x y^2 + 4x^2 y dy/dx + 4y^3 dy/dx = 0
dy/dx = -(x^3 + xy^2)/(x^2 y + y^3) = -x(x^2 + y^2)/y(x^2 + y^2) = -x/y

(x^2 + y^2)^2 = c
2(x^2 + y^2)*(2x + 2y dy/dx) = 0
Now, assume c > 0 as if c = 0, x^2 + y^2 = 0 means x and y are both zero only, which is rather trivial. If c is not 0, then x^2 + y^2 is not zero. Hence I can divide both sides by that.
2x + 2y dy/dx = 0
dy/dx = -x/y

There's nothing wrong with doing either...I think what you've forgotten is that you have to differentiate an equation. Even when we say 'differentiate x^3', we mean 'find dy/dx if y = x^3'
It's the rate of change between two variables and these variables have to have a known relationship before you can find their relative rates of change.

Cheers!
Made some algebraic mistake and assumed I had to expand things and can't do it the "normal" chain rule way, which was stupid of me.

Thanks!

[devilsadvocate]

Just a very quick question; when labelling forces like normal reaction, do we have to draw the arrow from the centre of the object (e.g. a box), or can we start the arrow from the edge of the object? Also for friction, do we need to draw it along the surface? Or do I not have to worry about it? Thanks in advance.

[keltingmeith]

Just a very quick question; when labelling forces like normal reaction, do we have to draw the arrow from the centre of the object (e.g. a box), or can we start the arrow from the edge of the object? Also for friction, do we need to draw it along the surface? Or do I not have to worry about it? Thanks in advance.

Generally, from the centre of the object is a good rule of thumb - implies more "correctness" in what you're saying, but that's more a physics thing than a specialist thing and shouldn't really matter. In terms of friction, it's acting on the object, so I'd do that from the centre as well - and draw it parallel to the surface supplying friction.

[Sayf44]

Can there ever be input/output mixing problems (differential equations) in exam 1?

[nhmn0301]

Can there ever be input/output mixing problems (differential equations) in exam 1?

There is one in my Tech Free SAC.

[psyxwar]

Can there ever be input/output mixing problems (differential equations) in exam 1?

Well why not? Anything is fair game really

[keltingmeith]

Anything from the study design is fair game for exam 1 and exam 2, unless otherwise stated in said design (but I don't think anything is exempted for specialist)

EDIT: LEL three in a row, whoops. .__. I think the point is clear now, at least.