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

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?

• How did you get $\dfrac{dV}{dt}=\pi2rh\dfrac{dr}{dt}+\pi r^{2}h\dfrac{dh}{dt}$ . Why did you put h at the end? Mar 2 '19 at 0:17

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:

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

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 $$\frac{dh}{dt}$$.