Connected Rates of Change – A Level Maths Year 2

1 Rates of Change With Respect to Time

A derivative with respect to time, $\dfrac{\mathrm{d}V}{\mathrm{d}t}$, is a rate: how fast $V$ is changing per second, per minute or per year. In many problems several quantities change together — as a balloon inflates, its radius, surface area and volume all grow — and we know the rate for one quantity but want it for another.

Connecting rates with the chain rule

$\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t}$

The middle link $\dfrac{\mathrm{d}V}{\mathrm{d}r}$ comes from geometry (a formula connecting $V$ and $r$). The given rate comes from the question. The chain rule glues them together. The "fractions" can be checked by imagining the $\mathrm{d}r$'s cancelling.

Signs carry meaning. A quantity that is decreasing has a negative rate of change. "Water leaks out at $5$ cm$^3$ s$^{-1}$" means $\dfrac{\mathrm{d}V}{\mathrm{d}t} = -5$.

2 Related Rates Problems

Write down the rate you are given and the rate you want, in $\dfrac{\mathrm{d}}{\mathrm{d}t}$ notation.
Find a formula linking the two quantities, and differentiate it.
Write the chain rule so that the "fractions" cancel to give the rate you want.
Substitute the values at the instant asked about — only after differentiating.

Worked Example 1 — A ripple

A stone dropped in a pond makes a circular ripple whose radius increases at $3$ cm s$^{-1}$. Find the rate at which the area is increasing when the radius is $10$ cm.

①Given $\dfrac{\mathrm{d}r}{\mathrm{d}t} = 3$; want $\dfrac{\mathrm{d}A}{\mathrm{d}t}$.

②$A = \pi r^2 \Rightarrow \dfrac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r$.

③$\dfrac{\mathrm{d}A}{\mathrm{d}t} = \dfrac{\mathrm{d}A}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 2\pi r \times 3 = 6\pi r$.

④When $r = 10$: $60\pi \approx 188$ cm$^2$ s$^{-1}$.

Worked Example 2 — Inflating a balloon

Air is pumped into a spherical balloon at $50$ cm$^3$ s$^{-1}$. When the radius is $5$ cm, find the rate of increase of (a) the radius (b) the surface area.

①Given $\dfrac{\mathrm{d}V}{\mathrm{d}t} = 50$. $\;V = \tfrac43\pi r^3 \Rightarrow \dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2 = 100\pi$ when $r = 5$.

②(a) $\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \div \dfrac{\mathrm{d}V}{\mathrm{d}r} = \dfrac{50}{100\pi} = \dfrac{1}{2\pi} \approx 0.159$ cm s$^{-1}$.

③(b) $S = 4\pi r^2 \Rightarrow \dfrac{\mathrm{d}S}{\mathrm{d}r} = 8\pi r = 40\pi$.

④$\dfrac{\mathrm{d}S}{\mathrm{d}t} = \dfrac{\mathrm{d}S}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 40\pi \times \dfrac{1}{2\pi} = 20$ cm$^2$ s$^{-1}$.

Dividing by $\dfrac{\mathrm{d}V}{\mathrm{d}r}$ is the same as multiplying by $\dfrac{\mathrm{d}r}{\mathrm{d}V}$ — this is $\dfrac{\mathrm{d}r}{\mathrm{d}V} = 1 \Big/ \dfrac{\mathrm{d}V}{\mathrm{d}r}$ from the rules page.

Worked Example 3 — Filling a cone

A container is an inverted cone of radius $8$ cm and height $20$ cm. Water is poured in at $30$ cm$^3$ s$^{-1}$. How fast is the depth rising when the water is $10$ cm deep?

①Similar triangles: $\dfrac{r}{h} = \dfrac{8}{20}$, so $r = \dfrac{2h}{5}$.

②$V = \tfrac13\pi r^2 h = \tfrac13\pi\left(\dfrac{4h^2}{25}\right)h = \dfrac{4\pi h^3}{75}$ — volume in terms of $h$ alone.

③$\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{4\pi h^2}{25} = 16\pi$ when $h = 10$.

④$\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{30}{16\pi} \approx 0.597$ cm s$^{-1}$.

Eliminate $r$ before differentiating. If you differentiate $\tfrac13\pi r^2 h$ treating $r$ as constant, you get the wrong answer — $r$ changes as the water rises.

Worked Example 4 — The sliding ladder

A $5$ m ladder leans against a vertical wall. The foot slides away from the wall at $0.4$ m s$^{-1}$. How fast is the top sliding down when the foot is $3$ m from the wall?

①Let the foot be $x$ m from the wall and the top $y$ m high. Pythagoras: $x^2 + y^2 = 25$.

②Differentiate with respect to $t$ (implicitly): $2x\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0$.

③When $x = 3$, $y = 4$, and $\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0.4$: $\;2(3)(0.4) + 2(4)\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0$.

④$\dfrac{\mathrm{d}y}{\mathrm{d}t} = -\dfrac{2.4}{8} = -0.3$. The top slides down at $0.3$ m s$^{-1}$.

This uses implicit differentiation, but with respect to $t$ instead of $x$. The negative sign confirms that $y$ is decreasing.

Worked Example 5 — A growing cube

The volume of a cube is increasing at $12$ cm$^3$ s$^{-1}$. Find the rate of increase of the surface area when the edge is $2$ cm.

①$V = x^3$, $\dfrac{\mathrm{d}V}{\mathrm{d}x} = 3x^2 = 12$ when $x = 2$, so $\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{12}{12} = 1$ cm s$^{-1}$.

②$S = 6x^2$, $\dfrac{\mathrm{d}S}{\mathrm{d}x} = 12x = 24$.

③$\dfrac{\mathrm{d}S}{\mathrm{d}t} = 24 \times 1 = 24$ cm$^2$ s$^{-1}$.

3 Constructing Simple Differential Equations

A differential equation is an equation involving a derivative. Many models are written in words about rates; your job is to translate them. Solving them is covered in Differential Equations.

Words	Mathematics
"the rate of change of $P$"	$\dfrac{\mathrm{d}P}{\mathrm{d}t}$
"is proportional to $P$"	$= kP$ (with $k$ a constant)
"decreases at a rate proportional to $m$"	$\dfrac{\mathrm{d}m}{\mathrm{d}t} = -km$, $\;k > 0$
"inversely proportional to $x$"	$= \dfrac{k}{x}$
"proportional to the square root of $h$"	$= k\sqrt{h}$

Worked Example 6 — Newton's law of cooling

A cup of tea at temperature $\theta$ °C cools in a room at $20$ °C. The rate at which the temperature falls is proportional to the difference between the temperature of the tea and the room. Write down a differential equation.

①Rate of change of temperature: $\dfrac{\mathrm{d}\theta}{\mathrm{d}t}$.

②Difference in temperature: $\theta - 20$.

③Falling, so negative: $\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 20)$, where $k > 0$ is a constant.

Worked Example 7 — A melting snowball

A spherical snowball melts so that its volume decreases at a rate proportional to its surface area. Show that its radius decreases at a constant rate.

①The model: $\dfrac{\mathrm{d}V}{\mathrm{d}t} = -kS = -4k\pi r^2$, $k > 0$.

②Chain rule: $\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 4\pi r^2\dfrac{\mathrm{d}r}{\mathrm{d}t}$.

③So $4\pi r^2\dfrac{\mathrm{d}r}{\mathrm{d}t} = -4k\pi r^2 \Rightarrow \dfrac{\mathrm{d}r}{\mathrm{d}t} = -k$, a constant. $\blacksquare$

4 Quick Reference

Chain rule

$\dfrac{\mathrm{d}A}{\mathrm{d}t} = \dfrac{\mathrm{d}A}{\mathrm{d}r}\times\dfrac{\mathrm{d}r}{\mathrm{d}t}$

Decreasing

Negative rate.

Order

Differentiate first, substitute values second.

Circle

$A = \pi r^2$, $C = 2\pi r$

Sphere

$V = \tfrac43\pi r^3$, $S = 4\pi r^2$

Cone

$V = \tfrac13\pi r^2 h$; use similar triangles to remove $r$.

Ladder

$x^2 + y^2 = L^2 \Rightarrow x\dot{x} + y\dot{y} = 0$

Proportional

"$\propto P$" $\to$ $kP$; include a minus sign for decrease.

5 Practice Questions

Question 1

The side of a square is increasing at $0.2$ cm s$^{-1}$. Find the rate of increase of the area when the side is $5$ cm.

▶ Show solution

$A = x^2$, $\dfrac{\mathrm{d}A}{\mathrm{d}x} = 2x = 10$.

$\dfrac{\mathrm{d}A}{\mathrm{d}t} = 10 \times 0.2 = 2$ cm$^2$ s$^{-1}$.

Question 2

$y = x^3 + 2x$ and $x$ is increasing at $3$ units per second. Find $\dfrac{\mathrm{d}y}{\mathrm{d}t}$ when $x = 1$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 + 2 = 5$ at $x = 1$.

$\dfrac{\mathrm{d}y}{\mathrm{d}t} = 5 \times 3 = 15$ units per second.

Question 3

The area of a circle increases at $20$ cm$^2$ s$^{-1}$. Find the rate of increase of the radius when $r = 4$ cm.

▶ Show solution

$\dfrac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r = 8\pi$.

$\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{20}{8\pi} = \dfrac{5}{2\pi} \approx 0.796$ cm s$^{-1}$.

Question 4

A radioactive substance decays so that its mass $m$ grams decreases at a rate proportional to the mass present. Write down a differential equation for $m$.

▶ Show solution

$\dfrac{\mathrm{d}m}{\mathrm{d}t} = -km$, where $k$ is a positive constant.

Question 5

In a school of $1000$ students, the rate at which a rumour spreads is proportional to the product of the number $N$ who have heard it and the number who have not. Write a differential equation for $N$.

▶ Show solution

Number who have not heard: $1000 - N$.

$\dfrac{\mathrm{d}N}{\mathrm{d}t} = kN(1000 - N)$, $k > 0$.

Question 6

The radius of a sphere increases at $0.1$ cm s$^{-1}$. Find the rate of increase of the volume when $r = 6$ cm.

▶ Show solution

$\dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2 = 144\pi$.

$\dfrac{\mathrm{d}V}{\mathrm{d}t} = 144\pi \times 0.1 = 14.4\pi \approx 45.2$ cm$^3$ s$^{-1}$.

Question 7

A $10$ m ladder leans against a wall. The top slides down at $0.5$ m s$^{-1}$. How fast is the foot moving away from the wall when the top is $6$ m above the ground?

▶ Show solution

$x^2 + y^2 = 100$; when $y = 6$, $x = 8$.

$2x\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0$ with $\dfrac{\mathrm{d}y}{\mathrm{d}t} = -0.5$:

$16\dfrac{\mathrm{d}x}{\mathrm{d}t} + 12(-0.5) = 0 \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{6}{16} = 0.375$ m s$^{-1}$.

Question 8

Water is poured at $5$ cm$^3$ s$^{-1}$ into an inverted cone whose semi-vertical angle is $30°$. Show that $V = \dfrac{\pi h^3}{9}$, and find the rate at which the depth is rising when $h = 3$ cm.

▶ Show solution

$r = h\tan 30° = \dfrac{h}{\sqrt3}$, so $V = \tfrac13\pi\left(\dfrac{h^2}{3}\right)h = \dfrac{\pi h^3}{9}$.

$\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{\pi h^2}{3} = 3\pi$ when $h = 3$.

$\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{5}{3\pi} \approx 0.531$ cm s$^{-1}$.

Question 9

An oil slick is circular and its area increases at a constant $6$ m$^2$ s$^{-1}$. Find the rate of increase of the circumference when the radius is $3$ m.

▶ Show solution

$\dfrac{\mathrm{d}A}{\mathrm{d}t} = 2\pi r\dfrac{\mathrm{d}r}{\mathrm{d}t} \Rightarrow 6 = 6\pi\dfrac{\mathrm{d}r}{\mathrm{d}t} \Rightarrow \dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{1}{\pi}$.

$C = 2\pi r \Rightarrow \dfrac{\mathrm{d}C}{\mathrm{d}t} = 2\pi \times \dfrac{1}{\pi} = 2$ m s$^{-1}$.

Question 10

A water trough is $2$ m long. Its cross-section is an inverted isosceles triangle, $0.6$ m wide at the top and $0.4$ m deep. Water flows in at $0.03$ m$^3$ per minute.

(a) Show that when the depth is $h$ m the volume is $V = 1.5h^2$ m$^3$.
(b) Find the rate at which the depth is rising when $h = 0.2$.
(c) Starting empty, how long does the trough take to fill?
(d) Find the rate at which the depth is rising just as the trough becomes full, and explain why it is smaller than in (b).

▶ Show solution

(a) By similar triangles the water surface has width $w = \dfrac{0.6}{0.4}h = 1.5h$. Cross-section area $= \tfrac12 \times 1.5h \times h = 0.75h^2$. Volume $= 0.75h^2 \times 2 = 1.5h^2$.

(b) $\dfrac{\mathrm{d}V}{\mathrm{d}h} = 3h = 0.6$. $\;\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{0.03}{0.6} = 0.05$ m per minute.

(c) Full volume $= 1.5 \times 0.4^2 = 0.24$ m$^3$; time $= 0.24 \div 0.03 = 8$ minutes.

(d) At $h = 0.4$: $\dfrac{\mathrm{d}V}{\mathrm{d}h} = 1.2$, so $\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{0.03}{1.2} = 0.025$ m per minute.

The trough is wider near the top, so the same volume of water spreads over a larger surface and raises the level less.

Connected Rates of Change · A Level Maths Year 2 · Created with MathJax