The question reads "Consider a circular cylinder of radius 1m and height 6m. We are filling the cylinder with oil at a rate of $0.5 m^3 s^{-1}$. Assume the cylinder is sitting on its base. How quickly is the height changing when the liquid fills a quarter of the container?"
My attempt at the solution:
$V = \pi r^2h$
$\frac{dV}{dt} = \pi 1^2 \frac{dh}{dt}$
Substituting $0.5m^3s^{-1}$ for $\frac{dV}{dt}$
$0.5 = \pi \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{0.5}{\pi}$
So I've arrived at a value for the rate of the increase of height, but I haven't used the fact that the question states that "the liquid fills a quarter of the container"
I'm not sure how to proceed from here to continue with the question.
Best Answer
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?