Related Rates Question With Cylinder?

Question

On a test we needed to solve the following question:

A right circular cylinder with a constant volume is decreasing in height at a rate of 0.2 in/sec. At the moment that the height is 4 inches and the radius is 3 inches, what is the rate at which the radius is decreasing?

After the test I still don't know the actual answer, however I am quite curious about what the answer should have been.

What I tried was the following.

Givens:

$V=(\pi)r^2(h)$
$V=(\pi)(3)^2(4)$
$V=36 \pi$

$r=3$
$h=4$

$dh/dt=-0.2$ in /sec

$dr/dt=?$

Work:

$V=(\pi)r^2(h)$
$36\pi=(\pi)r^2(dh/dt)+h(2\pi(r)(dr/dt))$
$36\pi=(\pi)(3)^2(-0.2)+4(2\pi(3)(dr/dt))$
$36\pi=9(\pi)(-0.2)+24(\pi)(4dr/dt)$
$36\pi=-1.8(\pi)+24(\pi)(4dr/dt)$
$37.8\pi=24(\pi)(4dr/dt)$
$37.8\pi/24(\pi)=(24(\pi)(4dr/dt))/24(\pi)$
$1.6 \pi=4(dr/dt)$
$dr/dr=0.4\pi$ in/sec

Could you tell me what I did was correct or did I make any mistakes? All help is greatly appreciated.

Answer 1

Not quite. You missed a key word in the problem. The first sentence tells you the cylinder is decreasing in height, but with a constant volume. If something is constant, then it is not changing. If it is not changing, its rate of change (i.e. derivative) is zero.

But wait, how can the cylinder be decreasing in height, yet remain constant in volume? This implies that the radius must be increasing, to balance out the lost volume from a decreasing height.

You are correct in differentiating $V=\pi r^2 h$ with respect to time. It's just implicit differentiation. So we have $$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} + 2\pi r h \frac{dr}{dt}.$$

However, at this point we must recognize that the problem just told us the volume is constant! If something is constant, it doesn't change, so its derivative is zero. That is,

$$0 = \pi r^2 \frac{dh}{dt} + 2\pi r h \frac{dr}{dt}.$$

Plugging in the given rate $dh/dt$, and evaluating at $r=3$ inches, and $h=4$ inches, we have

$$0 = \pi (3 \text{ in})^2 \left(\frac{-1 \text{ in}}{5 \text{ sec}}\right) + 2\pi (3 \text{ in}) (4 \text{ in}) \frac{dr}{dt}$$

$$0 = -\frac{9}{5}\pi\ \frac{\text{in}^3}{\text{sec}} + 24\pi \ \text{in}^2\frac{dr}{dt}.$$ $$\frac{9}{5}\pi\ \frac{\text{in}^3}{\text{sec}} = 24\pi \text{in}^2\frac{dr}{dt}$$

Now we divide by $24\pi\ \text{in}^2$. Notice how I deliberately involve the units in my calculations. It's not necessary, but it makes our work easier to check. For example, if the end result is not in the units we expect, then we know we make an error in our algebra. $$\frac{9}{120}\frac{\text{in}}{\text{sec}} = \frac{dr}{dt}.$$

And yes, our radius is indeed increasing, with the correct units.