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diferentiation

Started by neep, Oct 08 2008 07:11 PM

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#1 neep

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Posted 08 October 2008 - 07:11 PM

need help with 2 questions from TeeJay Notes.
Here are the two questions
1. If P=(2m+3)^4 find dm/dt when m=1 given that dp/dt = 2
2. A spherical balloon is inflated so that its volume increases at a constant rate of 200cm^3/second. find the rate increase of the surface area of the balloon when the radius is 100cm.

Thanks for any help in a advance.

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#2 Marcus

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Posted 09 October 2008 - 03:36 PM

Question 1

For this question you need to know that

$\frac{dP}{dm}=\frac{dP}{dt} \times \frac{dt}{dm}$ you probably found this in the parametric equations topic
rearrange
$\frac{dm}{dt}=\frac{dP}{dt} \times \frac{dm}{dP}$
(you can kinda treat them as fractions...though they most definately are not)

now you find $\frac{dP}{dm}$
$\begin{align*} P&=(2m+3)^{4}\\ \frac{dP}{dm}&= 8(2m+3)^{3}\\ \end{align*}$

now $\frac{dm}{dP}$ = $\frac{1}{\frac{dP}{dm}}$
so $\frac{dm}{dP}=\frac{1}{8(2m+3)^{3}}$

substitute in m and we get $\frac{dm}{dP}=\frac{1}{1000}$

and solve
$\begin{align*} \frac{dm}{dt}&=\frac{dP}{dt} \times \frac{dm}{dP}\\ \frac{dm}{dt}&=2 \times \frac{1}{1000}\\ \frac{dm}{dt}&=\frac{1}{500}\\ \end{align*}$

Question 2 -- the way I did this is quite complex...so there is probably an easier way :S :S

now for this you need two equations:
equation for surface area $A=4\pi r^{2}$ and the equation for volume $V=\frac{4}{3} \pi r^{3}$

you may notice now if you differentiate volume you get surface area so $A=\frac{dV}{dr}$
we are also told that $\frac{dV}{dt}=200$ we could also work out $\frac{dA}{dr}=8 \pi r$ from surface area equation (which is 4 times circumference nice that they are all related

)

What we are being asked for is $\frac{dA}{dt}$

so we have to make some sort of equation out of all this so $\frac{dA}{dt}=\frac{dA}{dr} \times \frac{dr}{dV} \times \frac{dV}{dt}$ (notice the dr and dV 'cancel out')

so $\frac{dV}{dt}=200$
now r=100
$\begin{align*} \frac{dA}{dr}&=8 \pi r\\ \frac{dA}{dr}&=800 \pi \end{align*}$
and
$\begin{align*} \frac{dr}{dV}&=\frac{1}{\frac{dV}{dr}}\\ \frac{dr}{dV}&=\frac{1}{A}\\ \frac{dr}{dV}&=\frac{1}{4 \pi r^{2}}\\ \frac{dr}{dV}&=\frac{1}{40,000 \pi} \end{align*}$

then finally substitute them all in
$\begin{align*} \frac{dA}{dt}&=\frac{dA}{dr} \times \frac{dr}{dV} \times \frac{dV}{dt}\\ \frac{dA}{dt}&=(800 \pi)(\frac{1}{40,000 \pi})(200)\\ \frac{dA}{dt}&=\frac{160,000 \pi}{40,000 \pi}\\ \frac{dA}{dt}&=4 \end{align*}$

couple of REALLY horrible questions there...I hope I haven't made any mistakes...but point out anything you don't understand

=-=-=Marcus=-=-=

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