Related Rates of Change

A balloon inflates, a ladder slips, water rises in a cone, a shadow lengthens. In each case two quantities change together over time, and knowing the rate of one lets you deduce the rate of the other. The link is always the chain rule: if a geometric relationship ties the quantities, differentiating it with respect to time converts a known rate into the rate you want. Related rates is where the implicit differentiation of the previous lesson meets real modelling, and it is a perennial source of structured exam questions because it tests setup as much as calculus.

This lesson sets out a foolproof four-step method, works the classic scenarios (sphere, ladder, cone) with full mark schemes, and stresses the two places marks are lost: differentiating before substituting, and substituting numerical values too early. Master the discipline and these questions become almost mechanical.

Where this sits in AQA 7367

This is the related rates of change strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It applies the chain rule and implicit differentiation (previous lesson) to time-dependent geometric problems. Setting up and differentiating the relationship is AO1; modelling the scenario, choosing the right formula, and interpreting the answer (including sign and units) are the AO2/AO3 problem-solving marks — related rates is a classic AO3 "solve a problem in context" topic.

Core theory: the chain rule in time

If a quantity $A$ depends on $r$ , and $r$ depends on time $t$ , the chain rule links their rates:

$\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}.$

The middle factor $\tfrac{dA}{dr}$ comes from the geometric formula relating $A$ and $r$ ; the outer factors are the rates — one given, one sought. More generally, any chain of dependencies multiplies:

$\frac{dA}{dt} = \frac{dA}{dh}\cdot\frac{dh}{dr}\cdot\frac{dr}{dt},$

and when several variables are tied by one equation, differentiate the whole equation implicitly with respect to $t$ — every variable picks up a " $\cdot\,\tfrac{d(\cdot)}{dt}$ " factor, exactly as $y$ -terms picked up $\tfrac{dy}{dx}$ last lesson.

The four-step method

Identify every quantity, note which rate is given and which is required, with units. Draw a diagram for geometry.
Relate the quantities with a formula (area, volume, Pythagoras, trig). If it involves more than two variables, eliminate any that are not needed using a second relationship (e.g. $r = h\tan\theta$ for a cone).
Differentiate the relationship with respect to $t$ — implicitly, so each variable contributes its time-derivative.
Substitute the given numerical values only now, and solve for the unknown rate. Interpret the sign and state units.

Step 2 is where most marks are won or lost, so it rewards a moment's care. Ask: does my relationship involve exactly the variable whose rate I know and the variable whose rate I want? If a third, irrelevant variable appears (the surface radius $r$ in a cone problem where only the depth $h$ and volume $V$ are in play), you must remove it using a geometric constraint — usually a similar-triangles ratio such as $r = h\tan\theta$ or a fixed proportion. Skipping this leaves you differentiating an equation with an extra unknown rate $\tfrac{dr}{dt}$ you were never given, and the problem becomes unsolvable. The discipline of reducing to just two changing quantities before differentiating is the single most reliable predictor of a clean solution.

The single golden rule: differentiate first, substitute last. Plugging in a specific value (say $r=10$ ) before differentiating freezes the variable and destroys the relationship — the most damaging error in the topic.

It is worth dwelling on why "substitute last" is non-negotiable, because the error is so tempting. Suppose a sphere's volume is $V = \tfrac43\pi r^3$ and we want $\tfrac{dV}{dt}$ when $r=10$ . A student in a hurry writes $V = \tfrac43\pi(10)^3 = \tfrac{4000\pi}{3}$ and then differentiates — but $\tfrac{4000\pi}{3}$ is a constant, so its derivative is $0$ , giving the absurd answer $\tfrac{dV}{dt}=0$ . The point is that $r$ is changing; substituting its instantaneous value before differentiating treats a moving quantity as fixed. The correct order — differentiate $V=\tfrac43\pi r^3$ to get $\tfrac{dV}{dt} = 4\pi r^2\tfrac{dr}{dt}$ , then put $r=10$ — preserves the fact that $r$ varies right up until the final evaluation. Hold that example in mind and you will never substitute early again.

A second framing helps with setup. Every related-rates problem is really an instruction to apply $\tfrac{d}{dt}$ to an equation — and $\tfrac{d}{dt}$ obeys the same chain, product and implicit rules you met last lesson, just with $t$ as the independent variable. So " $A = \pi r^2$ " becomes " $\tfrac{dA}{dt} = 2\pi r\tfrac{dr}{dt}$ ", " $x^2+y^2=25$ " becomes " $2x\tfrac{dx}{dt}+2y\tfrac{dy}{dt}=0$ ", and " $V=\tfrac13\pi r^2 h$ " (if both $r$ and $h$ vary) becomes " $\tfrac{dV}{dt} = \tfrac13\pi(2rh\tfrac{dr}{dt} + r^2\tfrac{dh}{dt})$ " by the product rule. Seeing related rates as "implicit differentiation with respect to time" unifies the whole topic with the previous lesson and removes any sense of a new method to learn.

Worked examples (with mark scheme)

Example 1 — expanding circle

A circle's radius increases at $2\ \text{cm s}^{-1}$ . Find the rate of increase of its area when $r = 5\ \text{cm}$ .

Given $\tfrac{dr}{dt} = 2$ ; required $\tfrac{dA}{dt}$ at $r=5$ . Relate and differentiate:

$A = \pi r^2 \;\Rightarrow\; \frac{dA}{dr} = 2\pi r. \quad (\text{M1: differentiate the relationship})$

$\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt} = 2\pi r\cdot 2 = 4\pi r. \quad (\text{M1: chain rule})$

Substitute $r=5$ last:

$\frac{dA}{dt} = 4\pi(5) = 20\pi \approx 62.8\ \text{cm}^2\,\text{s}^{-1}. \quad (\text{A1})$

(M1 differentiate $A=\pi r^2$ ; M1 chain rule; A1 $20\pi$ with units. Note the substitution $r=5$ happens only at the end — leaving $\tfrac{dA}{dt} = 4\pi r$ general until then is the safe habit.)

There is a pleasing geometric reading of this answer. The result $\tfrac{dA}{dt} = 2\pi r\,\tfrac{dr}{dt}$ says the area grows by adding a thin annular ring of circumference $2\pi r$ and width $\tfrac{dr}{dt}\,dt$ in each instant — area added $=$ circumference $\times$ radial growth. Recognising that $\tfrac{dA}{dr} = 2\pi r$ is exactly the circumference is not a coincidence: it is the statement that the derivative of the area of a disc with respect to its radius is its perimeter, the two-dimensional shadow of the fact that $\tfrac{dV}{dr} = 4\pi r^2$ (the surface area) for a sphere. Holding this picture turns a formula into an intuition.

Example 2 — inflating sphere

A spherical balloon is inflated so its volume increases at $100\ \text{cm}^3\,\text{s}^{-1}$ . Find the rate of increase of the radius when $r = 10\ \text{cm}$ .

Given $\tfrac{dV}{dt} = 100$ ; required $\tfrac{dr}{dt}$ at $r=10$ :

$V = \frac{4}{3}\pi r^3 \;\Rightarrow\; \frac{dV}{dr} = 4\pi r^2. \quad (\text{M1})$

$\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \;\Rightarrow\; \frac{dr}{dt} = \frac{1}{4\pi r^2}\,\frac{dV}{dt}. \quad (\text{M1 rearrange for the required rate})$

Substitute $r=10,\ \tfrac{dV}{dt}=100$ :

$\frac{dr}{dt} = \frac{100}{4\pi(10)^2} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796\ \text{cm s}^{-1}. \quad (\text{A1})$

(M1 $\tfrac{dV}{dr}=4\pi r^2$ ; M1 rearrange to make $\tfrac{dr}{dt}$ the subject; A1 $\tfrac{1}{4\pi}$ . Rearranging symbolically before substituting keeps the arithmetic clean.)

Example 3 — the sliding ladder

A ladder $5\ \text{m}$ long leans against a vertical wall. Its foot slides away from the wall at $0.5\ \text{m s}^{-1}$ . Find the rate at which the top slides down when the foot is $3\ \text{m}$ from the wall.

Let $x$ = distance of the foot from the wall, $y$ = height of the top. By Pythagoras (the ladder length is constant):

$x^2 + y^2 = 25. \quad (\text{M1: relationship})$

Differentiate with respect to $t$ :

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0. \quad (\text{M1: differentiate w.r.t. } t)$

When $x=3$ , Pythagoras gives $y = \sqrt{25-9} = 4$ . With $\tfrac{dx}{dt}=0.5$ :

$2(3)(0.5) + 2(4)\frac{dy}{dt} = 0 \;\Rightarrow\; 3 + 8\frac{dy}{dt} = 0 \;\Rightarrow\; \frac{dy}{dt} = -\frac{3}{8}\ \text{m s}^{-1}. \quad (\text{A1; A1 interpret})$

The top slides down at $\tfrac{3}{8} = 0.375\ \text{m s}^{-1}$ ; the negative sign confirms $y$ is decreasing.

(M1 relationship; M1 differentiate w.r.t. $t$ ; A1 $-\tfrac38$ ; A1 for interpreting the sign and giving units. Finding $y=4$ from Pythagoras at the instant in question is essential — you cannot substitute $x=3$ without it.)

Example 4 — a shadow lengthening (similar triangles)

A street lamp is $6\ \text{m}$ high. A person $1.8\ \text{m}$ tall walks away from the lamp post at $1.5\ \text{m s}^{-1}$ . Find the rate at which the length of their shadow is increasing.

Let $d$ be the person's distance from the post and $s$ the shadow length. By similar triangles (the lamp, the person's head and the shadow tip are collinear):

$\frac{6}{d+s} = \frac{1.8}{s}. \quad (\text{M1: similar triangles})$

Cross-multiplying: $6s = 1.8(d+s) \Rightarrow 6s - 1.8s = 1.8d \Rightarrow 4.2s = 1.8d \Rightarrow s = \dfrac{1.8}{4.2}d = \dfrac{3}{7}d$ . (M1: eliminate to a relation between $s$ and $d$ )

Differentiate with respect to $t$ :

$\frac{ds}{dt} = \frac{3}{7}\frac{dd}{dt} = \frac{3}{7}(1.5) = \frac{4.5}{7} = \frac{9}{14} \approx 0.643\ \text{m s}^{-1}. \quad (\text{A1})$

The shadow lengthens at $\tfrac{9}{14}\ \text{m s}^{-1}$ — and notably this is independent of $d$ (the relation $s = \tfrac37 d$ is linear), so the shadow grows at a constant rate however far the person has walked.

(M1 set up the similar-triangle ratio; M1 rearrange to $s$ in terms of $d$ ; A1 the rate. A subtle point: the shadow tip moves at $\tfrac{dd}{dt}+\tfrac{ds}{dt} = 1.5 + \tfrac{9}{14} = \tfrac{30}{14} = \tfrac{15}{7}\ \text{m s}^{-1}$ , faster than the person — a result the Going further section generalises.)

Specimen-style exam question

(specimen-style — not from any past paper)

Water is poured into an inverted right circular cone of semi-vertical angle $30^\circ$ at a constant rate of $8\ \text{cm}^3\,\text{s}^{-1}$ . At time $t$ seconds the depth of water is $h$ cm. (a) Show that the volume of water is $V = \dfrac{\pi h^3}{9}$ . (b) Find the rate at which the depth is rising when $h = 6\ \text{cm}$ .

(a) For an inverted cone, the surface radius $r$ and depth $h$ satisfy $\tan 30^\circ = \dfrac{r}{h}$ , so $r = h\tan 30^\circ = \dfrac{h}{\sqrt3}$ . Then

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{h}{\sqrt3}\right)^2 h = \frac{1}{3}\pi\cdot\frac{h^2}{3}\cdot h = \frac{\pi h^3}{9}.$

(b) Differentiate with respect to $t$ : $\dfrac{dV}{dt} = \dfrac{\pi}{9}\cdot 3h^2\,\dfrac{dh}{dt} = \dfrac{\pi h^2}{3}\,\dfrac{dh}{dt}$ . Rearranging for the required rate and substituting $\tfrac{dV}{dt}=8,\ h=6$ :

$\frac{dh}{dt} = \frac{3}{\pi h^2}\,\frac{dV}{dt} = \frac{3\times 8}{\pi (6)^2} = \frac{24}{36\pi} = \frac{2}{3\pi} \approx 0.212\ \text{cm s}^{-1}.$

The decisive move in (a) is eliminating $r$ before differentiating, using the fixed ratio $r=h\tan\theta$ — leaving both $r$ and $h$ in the volume formula would force you to track two changing variables and an extra related rate $\tfrac{dr}{dt}$ . A top answer states this reduction explicitly: "since the cone's shape is fixed, $r$ and $h$ are proportional, so I express $V$ in $h$ alone." That single observation is what makes the differentiation a one-liner and is exactly the modelling insight the examiner rewards.

Related Rates of Change

Related Rates of Change

Where this sits in AQA 7367

Core theory: the chain rule in time

The four-step method

Worked examples (with mark scheme)

Example 1 — expanding circle

Example 2 — inflating sphere

Example 3 — the sliding ladder

Example 4 — a shadow lengthening (similar triangles)

Specimen-style exam question

Synoptic links

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