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iGCSE (2021 Edition)

11.22 Related rates of change


In this chapter, we use the relation


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.









Apply differentiation to connected rates of change.

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