Who has actually done DHA's spesh exam 2s?

[98.40_for_sure]

If yes, is it worth doing? im in the middle of one atm but i dunno if these questions are of any use and if time could better be spent

[kyzoo]

Yes do it. It requires a lot more concentration and mental acuity to do than VCAA, so doing it minimizes your chances of having a mental blank in the real exam.

[98.40_for_sure]

A multi choice question just took 5 minutes... :/
It was a REALLY nice question though.

If anyone wants to test their related rates knowledge, see below:

A sphere initially has a radius of 10cm. The surface area of the sphere is decreasing at a rate of 4cm^3 per second. The rate of increase of the volume of the sphere after 3 seconds is:

A: -2rt(100-(3/pi))
B: -6
C: 6
D: -26
E: -2rt(10-(3/pi))

[3Xamz]

Ahhh, that is a pretty nifty question 99.95 xD
Answer is B, I presume?

[98.40_for_sure]

Ahhh, that is a pretty nifty question 99.95 xD
Answer is B, I presume?

Nope, try again
It involves integrating as well as triple related rates

[3Xamz]

The surface area of the sphere is decreasing at a rate of 4cm^3 per second.

Surface area in terms of cm^3? :S

[98.40_for_sure]

MY BAD CM^2. but that shouldnt have affected you anyway

[3Xamz]

Yeah, but I keep getting -6
I don't see where we need to integrate?

[98.40_for_sure]

Post your working and lets see where you went wrong

[3Xamz]

Let Surface Area = A,
We want dV/dt at t = 3.

V = (4/3)*pi*r^3
dV/dr = 4*pi*r^2 = A
dA/dr = 8*pi*r

dV/dA = dV/dr * dr/dA = (4*pi*r^2) * (1/8pir) = r/2

We are given; dA/dt = -4cm^2/sec
dV/dt = dV/dA * dA/dt = r/2 * -4 = -2r.

SHIT! I subbed in r = 3, instead of t = 3 ><
Back to the drawing board...

[98.40_for_sure]

YES YES!! KEEP GOING!! almost there

ps. you use a drawing board?

[3Xamz]

Yeah, answer should be A (Y)

EDIT: Sorry didn't see your message, what's a drawing board?

DOUBLE EDIT: LOL figure of speech

[superflya]

YES YES!! KEEP GOING!! almost there

ps. you use a drawing board?

"back to the drawing board"- back to where u started