Rate of change of volume of cylinder with respect to surface area The surface area of a cylinder is increasing at a rate of 9π m^2/hr. 
The height of the cylinder is fixed at 3 meters. 
At a certain instant, the surface area is 36π m^2. 
What is the rate of change of the volume of the cylinder at the instant (in cubic meters per hour)
My daughter got stuck and asked me for help. I spent an hour trying to figure it out and I'm stuck too! Thanks!
Here is the answer, thanks to Rishi. Looks like there are a few other ways to solve. Thanks to everyone!!
Formulas:
S = 2πrh + 2πr^2
V = πr^2h
Steps:
dS/dt = 9π     <-- given
d/dt (2πrh + 2πr^2) = 9π
dr/dt (2πh + 4πr) = 9π
(2r + 3) dr/dt = 9/2    <-- divide by 2π
Equation:
dV/dt = 2πrh dr/dt = 6πr dr/dt
Solve for r at the instant:
6πr + 2πr^2 = 36π
r = 3
Use r to find dr/dt in above equation:
(2x3 + 3) dr/dt = 9/2
dr/dt = 1/2
Put values into equation:
dV/dt = 6π(3) (1/2)
dV/dt = 9π
 A: From the formulas,
$$V = πr^2h$$
$$S = 2πrh + 2πr^2$$
Eliminate $r$ to get the relationship between the surface $S$ and the volume $V$,
$$S = 2\sqrt{\pi h V}+\frac 2h V$$
Then, take the derivatives,
$$\frac{dS}{dt} = \sqrt{\frac{\pi h}{V}} \frac{dV}{dt}+\frac 2h \frac{dV}{dt}= \left(\frac 1r + \frac 2h \right)\frac{dV}{dt}$$
Thus, the rate of change for the volume is simply,
$$\frac{dV}{dt} = \frac{ 1 }{\frac 1r + \frac 2h}\frac{dS}{dt}$$
where the right-hand-side are all known, i.e. $h=3m$, $\frac{dS}{dt}=9\pi m^2/h$ and $r$ is solved from the given surface area at that instant $36\pi = 6\pi r +2\pi r^2$, or
$$r^2+3r-18=0$$
which yields $r=3m$. As a result, you should get the volume change rate,
$$\frac{dV}{dt} = \frac{ 9\pi }{\frac 13 + \frac 23}=9\pi \>(m^3/hr)$$
A: Surface area of a cylinder S is $2\pi rh+2\pi r^2$
Now $${ds \over dt}=9\pi$$
or $$\frac{d}{dt}(2\pi rh+2\pi r^2)=9\pi$$
or $${dr \over dt}.(2\pi h+4\pi r)=9\pi$$ 
or $$(2r+3){dr \over dt}={9 \over 2}$$
Since  $$v=\pi r^2h$$
$${dv \over dt}=2\pi rh {dr \over dt}=6\pi r{dr \over dt}$$
Now solve for r since $6\pi r+ 2\pi r^2=36 \pi$ and find ${dr \over dt} $ from above equation put values in last equation and finish it.
A: HINT
We have that


*

*area of tha base: $A(t)=\pi R^2(t)$

*lateral surface area: $S(t)=2\pi R(t)H=6\pi R(t)$

*total surface area: $S_{tot}(t)=2\pi R^2(t)+6\pi R(t)\implies S'_{tot}(t)=4\pi R'(t)R(t)+6\pi R'(t)$
then consider


*

*volume at time t: $V(t)=\pi  R^2(t)\cdot H \implies V'(t)=6\pi R'(r)R(t)$
A: $$\frac {dv}{dt}=\frac{dv}{ds}\times \frac {ds}{dt}$$ 
You have $\frac {ds}{dt}$
You need to find a formula for $v$ in terms of $s$ and find $\frac {dv}{ds}$ to finish the problem. 
