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:
(2×3 + 3) dr/dt = 9/2
dr/dt = 1/2
Put values into equation:
dV/dt = 6π(3) (1/2)
dV/dt = 9π
Best Answer
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,
$$\frac{dV}{dt} = \frac{ 9\pi }{\frac 13 + \frac 23}=9\pi \>(m^3/hr)$$