VCE Specialist 3/4 Question Thread!

[scribble]

to sketch slope field graphs on the classpad, go to DiffEqGraph on the menu (its the fifth one down on the left on mine, but i don't remember if i moved mine around or not) and then just type your equation in where it says y'=
everything else is the same as regular Graph&Tab so you can zoom in and out or set the size of the window etc.

[saba.ay]

Could someone please help with the following two questions. I've attached the first from Essential Chapter 8 Review Question 14.

The second I can't attach because I don't have the ecopy of Chapter 9, but the question is Question 2 from Chapter 9 review: Extended Response p. 362.
With this question, I just need to know how to go about setting up the differential equation.
Please and thank you.

[lzxnl]

Hint for 14a. Rotate each line individually about the x axis, find the separate volumes and then add them up.
The equations of the lines are y=x/a and y=(x-1)/(a-1). For each line, the integral is pi*y^2 dx integrated over whatever x values they take, so [0,a] for the first line and [a,1] for the second line. Work out the integrals from that.

I wouldn't even bother doing that for the second question. Note that if you draw it out, you're revolving a straight line that intersects the axes. Therefore you get a cone that is easy to picture. Its radius is (1-k^2)^1/2 from the y-coordinate and its height is k. Its volume is hence 1/3*pi*(1-k^2)*k. Differentiate this, set the derivative to zero, prove that it's a local maximum...you get the drill.

[saba.ay]

Hint for 14a. Rotate each line individually about the x axis, find the separate volumes and then add them up.
The equations of the lines are y=x/a and y=(x-1)/(a-1). For each line, the integral is pi*y^2 dx integrated over whatever x values they take, so [0,a] for the first line and [a,1] for the second line. Work out the integrals from that.

I wouldn't even bother doing that for the second question. Note that if you draw it out, you're revolving a straight line that intersects the axes. Therefore you get a cone that is easy to picture. Its radius is (1-k^2)^1/2 from the y-coordinate and its height is k. Its volume is hence 1/3*pi*(1-k^2)*k. Differentiate this, set the derivative to zero, prove that it's a local maximum...you get the drill.

Thank you so much-breaking the triangle never occurred to me!

[e^1]

At the moment I'm trying to do the exercises ahead of the class. I got stuck on this question (Maths Quest).

If are 3-dimensional non-zero vectors.

Provide algebraically that

Thank you!

[lzxnl]

Firstly, all the norms are non-negative so squaring doesn't affect any inequality signs.
If we square the right hand side, we end up with (a+b).(a+b) = |a|^2 + |b|^2 + 2a.b = |a|^2 + |b|^2 | 2|a||b| cos angle
What's the left hand side squared? We end up with |a|^2 + |b|^2 + 2|a||b|. Note how each term is identical to the above squared expression except that (RHS)^2 has a cosine term as well. cos angle <= 1 so (a+b).(a+b) <= |a|^2 + |b|^2 + 2|a||b| = (LHS)^2
As (LHS)^2 >= (RHS)^2 and as LHS, RHS > 0, we can square root both sides without flipping any signs.
Hence (LHS) >= (RHS)

[SocialRhubarb]

^nliu's solution, now in a more easily digestible form.

[lzxnl]

Yeah I sometimes just ceebs with latex.

[Alwin]

Wow nliu, that's the first time I've seen the triangle inequality proved that way using cos.
For the sake for overkilling a question, this is the method I've always used:



Or, if the proof is only in R² or R³ (aka 2D or 3D) I just draw a triangle with two sides denoted by vectors a and b. The third side is a+b.
Clearly length of a+b ≤ length a +length  b

EDIT: nliu's method reminded me of another way lol - see spoiler

another random method for proof

[lzxnl]

It's how I've always proved the triangle inequality. I had not actually seen your way before.

[Jeggz]

It's how I've always proved the triangle inequality. I had not actually seen your way before.

Yeah me too, I've actually never seen your way either Alwin, interesting

[lzxnl]

It doesn't work for vectors though, and my proof only works for vectors.

[Nato]

need help with converting into polar form

[lzxnl]

If you plotted this number on an Argand diagram, you'd find that this number is in the first quadrant, which makes life easier.

Let 2sqrt 3 + 1i = r cos t + i*r sin t
Equating real and imaginary parts gives 2 sqrt 3 = r cos t (1)
and 1 = r sin t (2)
If we square both (1) and (2) and add them, the trig functions disappear and we get r^2 = 4*3 + 1 = 13, so r = sqrt(13)
Now that you have r, you can use either equation 1 or 2 to work out t, and there you have your number in polar form.

[pi]

need help with converting into polar form

Imagine it plotted on an argand diagram.

r = sqrt((2sqrt(3))^2 + 1^2) = sqrt(13) [simple Pythag]
theta = tan-1(y/x) = tan-1(1/2sqrt(3)) <- leave it as that if calc-free [simple trig]

So sqrt(13)tan-1(1/2sqrt(3))


edit: beaten