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.
You have found that
$$h'(t) = \frac 1{2 \pi}$$
There's no $t$ on the right hand side, which means that $\frac 1{2 \pi}$ is the change in height regardless of the value $t$. So even if we call the time at which the liquid fills a quarter of the container $t_0^*$, we will still get $h'(t_0^*) = \boxed{\frac 1{2 \pi}}$.
*Extra credit: if we wanted to find $t_0^*$, what equation would we need to solve?
Best Answer
Physicists obtain the chain rule (calculus) by extending a fraction and swapping the denominators. In your scenario, this would look like this: $$ \frac{dV}{dt} = \frac{dV}{dt} \cdot \frac{dh}{dh} = \frac{dV}{dh} \cdot \frac{dh}{dt} $$ You already know that $\frac{dV}{dt} = 150\frac{in^3}{sec}$, and we can derive $\frac{dV}{dh} = \pi r^2$. So, with the formula from above: $$ 150\frac{in^3}{sec} = \pi r^2 \cdot \frac{dh}{dt} $$ Solving for $\frac{dh}{dt}$ (which is the function we are looking for), we get $$\frac{dh}{dt} = \frac{150}{\pi r^2}\frac{in^3}{sec}$$ We see that this doesn't depend on the time $t$, which is expected, as volume and height are proportional (in a cylinder, the radius is the same at any hight).
As we don't know anything about the relationship of the hight of the cylinder to the hight in question (which doesn't have any influence on the result), the problem can't be solved. The information "hight of the cylinder is approximately ten times the radius" must be chosen to confuse you. We can't use a formula like $h=10r$, because that $h$ is the hight of the cylinder. If $h=10r$ at any hight $h$, the cylinder would be an inverted cone.