iGCSE (2021 Edition)

# 11.22 Related rates of change

Lesson

In this chapter, we use the relation

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times\frac{\mathrm{d}u}{\mathrm{d}x}$dydx=dydu×dudx

This is the chain rule in the notation of Leibniz. We use the chain rule to differentiate functions that are a function of a function. In particular, given a function $y=f\left(u(x)\right)$y=f(u(x)), its derivative is given by the expression above.

It may be that we know how to express a variable $y$y as a function of $u$u. Thus, we can differentiate $y$y with respect to $u$u to obtain $\frac{\mathrm{d}y}{\mathrm{d}u}$dydu. We may also have an expression for $u$u as a function of $x$x and therefore we can find $\frac{\mathrm{d}u}{\mathrm{d}x}$dudx.

Putting these together, we obtain $\frac{\mathrm{d}y}{\mathrm{d}x}$dydx as the product of the two derivatives. That is, we find the rate of change of $y$y with respect to $x$x.

It can happen that we know how to write the derivatives $\frac{\mathrm{d}y}{\mathrm{d}t}$dydt and $\frac{\mathrm{d}x}{\mathrm{d}t}$dxdt but we require $\frac{\mathrm{d}y}{\mathrm{d}x}$dydx. To make this fit into the framework of the chain rule as described above, we need an additional fact:

If we have the rate of change of $y$y with respect to $x$x, then the rate of change of $x$x with respect to $y$y is its reciprocal.

In the present example, we have $\frac{\mathrm{d}x}{\mathrm{d}t}$dxdt and we need $\frac{\mathrm{d}t}{\mathrm{d}x}$dtdx in order to use the chain rule. So, we use the fact that $\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}$dtdx=1dxdt.

##### Example 1

A continuous supply of ink is seeping onto a porous plane surface so that a circular spot forms and grows over time. The area of the spot is increasing at the rate of $0.5\text{cm}^2$0.5cm2 per second. However, the radius of the spot is increasing at a rate that reduces as the spot grows. Find an expression for the time-rate of change of the radius and deduce the rate of change when the radius is $6\text{cm}$6cm.

The first thing to do in problems of this kind is to define some variables so that the problem can be expressed algebraically.

Let $A$A be the area of the spot, let $r$r be its radius and let $t$t be the elapsed time. The goal is to find $\frac{\mathrm{d}r}{\mathrm{d}t}$drdt when $r=6$r=6.

We are given that $\frac{\mathrm{d}A}{\mathrm{d}t}=0.5$dAdt=0.5, and we can make use of the fact that $A=\pi r^2$A=πr2 and hence, $\frac{\mathrm{d}A}{\mathrm{d}r}=2\pi r$dAdr=2πr.

The three derivatives combine to form $\frac{\mathrm{d}A}{\mathrm{d}r}.\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{\mathrm{d}A}{\mathrm{d}t}$dAdr.drdt=dAdt. That is, $2\pi r\times\frac{\mathrm{d}r}{\mathrm{d}t}=0.5$2πr×drdt=0.5. On rearranging, this is $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{0.5}{2\pi r}=\frac{1}{4\pi r}$drdt=0.52πr=14πr.

Therefore, when $r=6$r=6, we have $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{1}{24\pi}\approx0.01\text{cm}/\text{s}$drdt=124π0.01cm/s.

##### Example 2

The fuel supply to a certain rocket engine is regulated in such a way that the rocket travels with a constant acceleration of $20\text{m/}s^2$20m/s2 for $90$90 seconds after launch. What is the rate of change of velocity with respect to displacement when the velocity reaches $1000\text{km/s}$1000km/s?

Let $a$a be the acceleration, $v$v the velocity, $s$s the displacement, and let $t$t be the elapsed time. We are asked to find $\frac{\mathrm{d}v}{\mathrm{d}s}$dvds when $v=1000$v=1000.

Acceleration is the time rate of change of velocity. In symbols, $a=\frac{\mathrm{d}v}{\mathrm{d}t}$a=dvdt. But, according to the chain rule, this is $\frac{\mathrm{d}v}{\mathrm{d}s}.\frac{\mathrm{d}s}{\mathrm{d}t}$dvds.dsdt. We recall that velocity is the time rate of change of displacement, $v=\frac{\mathrm{d}s}{\mathrm{d}t}$v=dsdt. It follows that an alternative characterisation of acceleration is $a=v\frac{\mathrm{d}v}{\mathrm{d}s}$a=vdvds.

Therefore, $a=20=\frac{\mathrm{d}v}{\mathrm{d}s}\times1000$a=20=dvds×1000 and so, $\frac{\mathrm{d}v}{\mathrm{d}s}$dvds at $v=1000$v=1000 is $\frac{1}{50}\ \text{s}^{-1}$150 s1. This means that for a brief time the velocity increases by $\frac{1}{50}\ \text{m/s}$150 m/s in the space of one metre.

#### Worked Examples

##### Question 1

The volume $V$V of oxygen in a scuba diver’s oxygen cylinder is given by $V=\frac{22}{P}$V=22P, where $P$P is the pressure inside the tank.

1. Find the rate of change of $V$V with respect to $P$P.

2. During a dive, the pressure $P$P inside the cylinder increases at $0.5$0.5 units per second. Find the rate of change of the volume of oxygen when $P=2$P=2.

Let $t$t represent time in seconds.

##### Question 2

A spherical hot air balloon, whose volume and radius at time $t$t are $V$V m3 and $r$r m respectively, is filled with air at a rate of $4$4 m3/min.

At what rate is the radius of the balloon increasing when the radius is $2$2 m?

##### Question 3

A point moves along the curve $y=5x^3$y=5x3 in such a way that the $x$x-coordinate of the point increases by $\frac{1}{5}$15 units per second.

Let $t$t be the time at which the point reaches $\left(x,y\right)$(x,y).

Find the rate at which the $y$y-coordinate is changing with respect to time when $x=9$x=9.

### Outcomes

#### 0606C14.5C

Apply differentiation to connected rates of change.