VCE Specialist 3/4 Question Thread!

[Homer]

having trouble with b

[Professor Polonsky]

Think of it as the distance between a line and a point, which will always be a perpendicular bisector of the line. There are a few ways to approach it from there, including dot product or simple geometry.

[507]

having trouble with b

Let the point where the injured hiker is be A, so
The rescue party's closest point is somewhere along so let's call this point B, where

The closest point is when is perpendicular to





Since they are perpendicular,



, k=0 is not applicable as dot product only applies to non-zero vectors.










So the distance is 1/5km or 200m, pretty sure you can use resolutes to solve this too.

[jono88]

Prove the identity 1/1-cosec(x) + 1/1+cosec(x) = -2tan^2(x)

How would i go about doing this question? Simplify numerators by cross multiplying? I end up getting 2/1-cosec^2(x)
Not sure what to do next :/

[Jeggz]

We know the formula , hence you can substitute this in.




This is equivalent to . Hope that made sense 

[zvezda]

Hey,
Just need some sort of confirmation on one of the questions in essentials: ex 8D q13.
My answer is different to the one at the back of the book, and I can't seem to figure out why.

[zvezda]

How would i go about answering q16 on exercise 8D?
Thanks

[b^3]

Hint: Translate the curve 4 units downwards, so that you would have the curve , and rotate about . This way you are forming the same solid of revolution with the same volume, but changing where you are taking the origin from.

[TimmyC]

Hey all, how would I factorise over C, z^3-3z^2+3z+7?
Would just like to know how I should approach this, thanks in advance!
Just putting it in spesh because it is infinitely more active :p

..do I have no idea because I have not learnt a thing about cubics?

[brightsky]

let P(z) = z^3-3z^2+3z+7
by inspection, (z+1) is a factor because P(-1) = -1 - 3 - 3 + 7 = 0
so P(z) = (z+1)(z^2 - 4z + 7) --> long divide or just deduce the coefficient of z in the quadratic factor
= (z+1)[(z-2)^2 +3]
= (z+1)[(z-2)^2 - (sqrt(3) i)^2]
= (z+1)(z-2-sqrt(3) i) (z-2+sqrt(3)i)

[TimmyC]

let P(z) = z^3-3z^2+3z+7
by inspection, (z+1) is a factor because P(-1) = -1 - 3 - 3 + 7 = 0
so P(z) = (z+1)(z^2 - 4z + 7) --> long divide or just deduce the coefficient of z in the quadratic factor
= (z+1)[(z-2)^2 +3]
= (z+1)[(z-2)^2 - (sqrt(3) i)^2]
= (z+1)(z-2-sqrt(3) i) (z-2+sqrt(3)i)

thanks mate!

[jono88]

Can someone help me anti-differentiate  integral (ax+b)(cx+d)^4
I let u=cx+d, du/dx=c     1/c du/dx = 1               x=(u-d)/c              not sure what to do next.

[Homer]

im stuck

[brightsky]

P, Q, R are collinear. this means that QR = 1/2 PQ.
r - q = 1/2 (q - p)
r = 1/2q - 1/2p + q = 3/2q - 1/2p
so A

[zvezda]

Just a quey with exam 2 2012 q1 part d,
When youre finding dy/dx, cant you work with numbers?
For example, when t=3, dx/dt=2 and dy/dt=4, therefore, dy/dx=4 x 1/2=2.
Is this legitimate?