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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: ngRISING on November 09, 2009, 12:03:51 am

Title: Rates of Change . GerrySly is the best!
Post by: ngRISING on November 09, 2009, 12:03:51 am: how do i figure out which one is dV/dt etc etc? i'll post up the VCAA 09 MM Exam 1 Q. cause i got it wrong. got confused with the notation :S. i ended up with the answer 1200pie LOL. walked out and everyones like its pie/12. im like O_O . F**K.

so any last minute help to allow me to achieve a 25RAW will be appreciated.

(http://img156.imageshack.us/img156/2069/fmln.jpg)
Title: Re: Rates of Change
Post by: GerrySly on November 09, 2009, 12:14:01 am: $\frac{dV}{dt}=10$

$\frac{dV}{dr}=4\pi r$

$\begin{align*} \frac{dr}{dt}&=\frac{dr}{dV}\cdot \frac{dV}{dt}\\ &=\frac{1}{4\pi r}\cdot 10\\ &=\frac{5}{2\pi r} (mm^3/min)\\ \frac{dr}{dt}_{|r=30}&=\frac{5}{2\pi (30)}\\ &=\frac{1}{12\pi} mm/min \end{align*} $
Title: Re: Rates of Change
Post by: ngRISING on November 09, 2009, 12:16:33 am: im not asking how to solve it. TT explained that already. i was wondering how do i identify which goe son top & bottom :S
Title: Re: Rates of Change
Post by: kendraaaaa on November 09, 2009, 12:18:29 am: dy/dx

y = x^2

y goes on top, x on bottom.

so

V = 2pi*r

dV/dr
Title: Re: Rates of Change
Post by: GerrySly on November 09, 2009, 12:24:49 am: Well they are asking for the rate of change or radius with respect to time so $\frac{dr}{dt}$ then you expand using the chain rule.

$\frac{dr}{dt}&=\frac{dr}{}\cdot \frac{}{dt}$

Now you see the statement, "oil is being added at a rate of" that should immediately make you think that the volume is being increased so the rate $\frac{dV}{dt}$ . Now that you have that information you can fill in one of the empty parts in your chain rule expansion.

$\frac{dr}{dt}&=\frac{dr}{}\cdot \frac{dV}{dt}$

But since the chain rule states the two missing parts must be the same, the other empty slot must be dV as well

$\frac{dr}{dt}&=\frac{dr}{dV}\cdot \frac{dV}{dt}$

Then all you need to find is $\frac{dr}{dV}$ which you can see is a relationship between radius (r) and volume (V) and the only relationship we have is $V=2\pi r^2$ , so we can then see by differentiating and taking the reciprocal we get the required rate. Subbing that all in and then letting r=30 will get your answer.

Dunno how helpful that was cause I just went over how I approach those questions heh, good luck for tomorrow :)
Title: Re: Rates of Change
Post by: ngRISING on November 09, 2009, 12:31:37 am: Quote from: GerrySly on November 09, 2009, 12:24:49 am
Well they are asking for the rate of change or radius with respect to time so $\frac{dr}{dt}$ then you expand using the chain rule.

$\frac{dr}{dt}&=\frac{dr}{}\cdot \frac{}{dt}$

Now you see the statement, "oil is being added at a rate of" that should immediately make you think that the volume is being increased so the rate $\frac{dV}{dt}$ . Now that you have that information you can fill in one of the empty parts in your chain rule expansion.

$\frac{dr}{dt}&=\frac{dr}{}\cdot \frac{dV}{dt}$

But since the chain rule states the two missing parts must be the same, the other empty slot must be dV as well

$\frac{dr}{dt}&=\frac{dr}{dV}\cdot \frac{dV}{dt}$

Then all you need to find is $\frac{dr}{dV}$ which you can see is a relationship between radius (r) and volume (V) and the only relationship we have is $V=2\pi r^2$ , so we can then see by differentiating and taking the reciprocal we get the required rate. Subbing that all in and then letting r=30 will get your answer.

Dunno how helpful that was cause I just went over how I approach those questions heh, good luck for tomorrow :)

F**K YEAH. THANK YOU . LOTS OF LOVE ^^.
Title: Re: Rates of Change
Post by: ngRISING on November 09, 2009, 02:43:13 pm: OMG. THANK YOU AGAIN. I GOT THE Q RIGHT IN THE EXAM BECAUSE OF THIS. MANY THANKS ^^