How fast is the height of the water in a cylindrical tank increasing?

Question

A cylindrical tank with radius $5m$ is being filled with water at a rate of $3m^{3}/\min$. How fast is the height of the water increasing.

The radius $r=5m$ The rate of water is $\dfrac{dV}{dt}=3m^{3}/\min$ The height of the water in the cylinder is $h$

The volume of a cylinder is given by the formula

$V=\pi r^{2}h$

Differentiating both sides I have: $\dfrac{dV}{dt}=\pi2rh\dfrac{dr}{dt}+\pi r^{2}h\dfrac{dh}{dt}$

Is this step correct? Here I am stuck since I don't know to get $h$ for height. How should I proceed?

Answer 1

Each quantity is a function of time except for the radius $r$ because it does not change with time (well, technically speaking, you still could think of it as a function of time, but it would be a constant function then: $r(t)=5\ m$). All those quantities are related by this expression:

$$V(t)=\pi r^2 h(t)$$

This equality states that at any given time, the volume of the water in the tank as a function of time equals the height of the water in the tank as a function of time multiplied by $\pi r^2$. You know the rate at which the volume of the water in the tank is increasing and you know what $r$ is. You also know that $V(t)$ is equivalent to $\pi r^2 h(t)$. If they are equivalent, their derivatives must also be equivalent:

$$V'(t)=[\pi r^2 h(t)]'=\pi r^2 h'(t)$$

Now, just solve for $h'(t)$ which tells you exactly what the problem is asking you to find—how fast the height of the water in the tank increases:

$$h'(t)=\frac{V'(t)}{\pi r^2 }=\frac{3}{25\pi}\ m/min.$$

Answer 2

Notice that $r$ is fixed for your context.

$$\frac{dV}{dt}=\pi r^2 \frac{dh}{dt}$$

Just substitute in the values of $\frac{dV}{dt}$ and $r$ and you can solve for $\frac{dh}{dt}$.