The height of a cylinder with a fixed radius of 4 cm is increasing at the rate of 2 cm/min. Find the rate of change of the volume of the cylinder (with respect to time) when the height is 14cm.

Question

compassionate · Answer

@dan815

xapproachesinfinity · Answer

well you v=pir^2h
take d/dt(v)
and plug in your numbers

xapproachesinfinity · Answer

$\Huge\frac{d}{dt}(V)=\pi r^2\frac{d}{dt}(h)$
r is fixed r=4

anonymous · Answer

@xapproachesinginity What do you do after you have got 16pi d/dt(h). Or is 16pi your answer

xapproachesinfinity · Answer

no you have h' as well it is given
$\large \frac{d}{dt}(h)=2 ~cm/min$

xapproachesinfinity · Answer

they said the height is increasing at a rate of 2 cm/min

anonymous · Answer

so it would be 8pi?

xapproachesinfinity · Answer

why?m 2*16pi

xapproachesinfinity · Answer

do you got it?

anonymous · Answer

so the answer would be 32pi

anonymous · Answer

@xapproachesinfinity

anonymous · Answer

@JFraser can u help me?

anonymous · Answer

@phi @paki @Zarkon @Destinymasha @agent0smith

jfraser · Answer

I don't remember this kind of calculus well enough to help, sorry