Connected rates of change

In a nutshell

One of the many uses of the chain rule is to relate multiple rates of change. Recall that a derivative gives a rate of change.

Example 1

The volume $V$ of a cylinder with a base circle with radius $r$ (in centimetres) is given by the formula $V=\pi r^2h$ where $h$ is the cylinder's height (also in centimetres). Given that the radius of this cylinder grows by $2$ centimetres per second, find the rate of change of the volume when the radius is $5$ centimetres. Assume that the height stays constant, at $10\text{ cm}$ .

Start by giving the rate of change of the radius with differentiation notation:

$\frac{\text dr}{\text dt}=2$ 

where $t$ is the time passed in seconds. Next find the rate of change of the volume with respect to the radius. This can be found by differentiating the equation given for the volume:

$\begin{aligned}V&=\pi r^2h\\\frac{\text dV}{\text dr}&=2\pi rh\end{aligned}$ 

The question asks for the rate of change of the volume. Implicitly, this is with respect to time, thus you seek $\frac{\text dV}{\text dt}$ . The chain rule says that

$\frac{\text dV}{\text dt}=\frac{\text dV}{\text dr}\times\frac{\text dr}{\text dt}$ 

Thus

$\frac{\text dV}{\text dt}=2\pi rh\times 2=4\pi rh$ 

You have that $h=10$ and the question asks about when $r=5$ :

$\frac{\text dV}{\text dt}=4\pi \times 5\times10=200\pi\approx628.32$

Therefore, when the radius reaches $5$ centimetres, the rate of change of the volume of the cylinder is about $\underline{628\text{ cm}^3\text{s}^{-1}} (3{ s.f.})$ . In other words, the volume increases by nearly $630\text{ cm}^3$ per second at the point when the radius is $5\text{ cm}$ .

Differential equations

Equations involving rates of change are called differential equations.

Example 2

The rate of change of the population $P$ is directly proportional to the size of the population itself. Form a differential equation from this information.

You have that "the rate of change" refers to $\frac{\text d}{\text dt}$ , unless otherwise stated. Hence this information says

$\frac{\text dP}{\text dt}\propto P$ 

To turn this into an equation, you change the " $\propto$ " to " $=$ " and multiply one side by a constant $k$ :

$\underline{}$ $\underline{\frac{\text dP}{\text dt}=kP}$ 