Question

A water tanks has a diameter of 6m and depth of 5m. If the water is 3m deep and is rising at 5m/h, at what rate is the volume changing?

Answer 1

Because we are only given rate of change in height, you would need to know the relationship between r and h, so you can make some substitutions so \(V\) would only contain one variable; \(h\)

We are given that r = 3m when h = 5m. then we can find the relationahip using proportion:

So we have \(\dfrac{r}{3} = \dfrac{h}{5}\Longrightarrow r = \dfrac{3}{5}h\)

Now, it is known that \(V = \dfrac{1}{3}\pi r^2h\)

Substitute \(r\) with \(\dfrac{3}{5}h\) and we have \(V = \dfrac{1}{3}\pi\left(\dfrac{3}{5}h\right)^2h = \dfrac{3}{25}\pi h^3\)

Now we can take derivative of \(V\) with respect to \(t\) (time):

\(\dfrac{dV}{dt} = \dfrac{9}{25}\pi h^2 \dfrac{dh}{dt}\)

Now you are given \(h = 3~m\) and \(\dfrac{dh}{dt}=5~m/h\) (Positive because it is rising)

So you can plug in these values and solve for \(\dfrac{dV}{dt}\)

Does that help?