Home / IGCSE Add Maths 0606 / Calculus / Connected rates of change and small increments
Cambridge IGCSE · Topic 14.7
IGCSE Add Maths 0606 / Topic 14 / IGCSE

When one thing changes, everything linked to it changes too.

Blow up a balloon and its radius, surface area and volume all grow together. The chain rule links their rates: \(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}\). The same derivative also estimates the effect of a small change, \(\delta y \approx \dfrac{dy}{dx}\,\delta x\), without working out the new value in full.

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The key idea

If \(y\) depends on \(x\) and \(x\) changes with time \(t\), then \(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}\). Find \(\dfrac{dy}{dx}\) from the formula, put in the given value, then multiply by the known rate (or divide, if the unknown rate is \(\dfrac{dx}{dt}\)). A quantity that is decreasing has a NEGATIVE rate. For a small change \(\delta x\) in \(x\), the change in \(y\) is approximately \(\delta y \approx \dfrac{dy}{dx}\,\delta x\), with \(\dfrac{dy}{dx}\) taken at the starting value of \(x\).

R = 8 r H = 20 h similar triangles: r ÷ h = 8 ÷ 20, so r = 0.4h
Fig. 1: an inverted cone of height 20 cm and top radius 8 cm, partly filled with water to depth h. The water forms a smaller cone whose radius r and depth h are in the same ratio as the full cone, so r = 2h/5 and the volume of water can be written in terms of h alone
Section 01

Rates linked by the chain rule.

A rate of change is a derivative with respect to time: \(\dfrac{dr}{dt}\) is how fast the radius grows, \(\dfrac{dA}{dt}\) how fast the area grows. When \(A\) is a function of \(r\) and \(r\) changes with time, the chain rule from earlier in this topic connects the two rates: \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}\). The derivative \(\dfrac{dA}{dr}\) comes from the formula, and the question gives you one rate and asks for the other.

Worked example A spreading ripple
The radius of a circular ripple increases at \(0.5\,\text{cm}\,\text{s}^{-1}\). Find the rate at which its area is increasing when the radius is \(4\,\text{cm}\).
  1. \(A = \pi r^{2}\), so \(\dfrac{dA}{dr} = 2\pi r = 8\pi\) when \(r = 4\).
  2. \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt} = 8\pi \times 0.5 = 4\pi\).
  3. \(4\pi \approx 12.6\), in \(\text{cm}^{2}\,\text{s}^{-1}\).
Answer: \(4\pi \approx 12.6\,\text{cm}^{2}\,\text{s}^{-1}\)
Worked example Find the unknown rate by dividing
Air is pumped into a spherical balloon at \(100\,\text{cm}^{3}\,\text{s}^{-1}\). Find the rate at which the radius is increasing when \(r = 5\,\text{cm}\).
  1. \(V = \frac{4}{3}\pi r^{3}\), so \(\dfrac{dV}{dr} = 4\pi r^{2} = 100\pi\) when \(r = 5\).
  2. \(\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}\) gives \(100 = 100\pi \times \dfrac{dr}{dt}\).
  3. \(\dfrac{dr}{dt} = \dfrac{100}{100\pi} = \dfrac{1}{\pi} \approx 0.318\,\text{cm}\,\text{s}^{-1}\).
Answer: \(\dfrac{1}{\pi} \approx 0.318\,\text{cm}\,\text{s}^{-1}\)
Your turn
Each edge of a cube increases at \(0.2\,\text{cm}\,\text{s}^{-1}\). Find the rate of increase of the volume when the edge is \(3\,\text{cm}\).
Examiner trap

Substitute the value of \(r\) only AFTER differentiating. Putting \(r = 4\) into \(A = \pi r^{2}\) first gives the number \(16\pi\), whose derivative is 0. And check which rate is unknown: if you want \(\dfrac{dr}{dt}\), you divide by \(\dfrac{dV}{dr}\); multiplying gives \(10\,000\pi\), a radius growing faster than a jet.

Section 02

One variable first, then differentiate.

Sometimes the formula has two variables that both change. The volume of water in a cone, \(V = \frac{1}{3}\pi r^{2} h\), depends on the radius \(r\) of the surface and the depth \(h\). Before you can differentiate with respect to \(h\), use the shape to write \(r\) in terms of \(h\). In an inverted cone the water is a smaller cone, similar to the whole one, so \(r\) and \(h\) are in a fixed ratio (Fig. 1).

\[ \frac{r}{h} = \frac{8}{20} \;\Rightarrow\; r = \frac{2h}{5}, \qquad V = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^{2} h = \frac{4\pi h^{3}}{75} \]

Also watch the sign. A quantity that is decreasing (water leaking out, a balloon going down) has a negative rate, so write \(\dfrac{dV}{dt} = -6\), not \(6\). The answer then comes out negative, and you can say 'decreasing at \(\dots\)' with the positive number.

Worked example Filling a cone
Water is poured at \(10\,\text{cm}^{3}\,\text{s}^{-1}\) into the inverted cone of Fig. 1 (height 20 cm, top radius 8 cm). Find the rate at which the depth is increasing when the depth is 5 cm.
  1. Similar triangles: \(r = \dfrac{2h}{5}\), so \(V = \dfrac{4\pi h^{3}}{75}\).
  2. \(\dfrac{dV}{dh} = \dfrac{12\pi h^{2}}{75} = \dfrac{4\pi h^{2}}{25} = 4\pi\) when \(h = 5\).
  3. \(10 = 4\pi \times \dfrac{dh}{dt}\), so \(\dfrac{dh}{dt} = \dfrac{10}{4\pi} = \dfrac{5}{2\pi} \approx 0.796\,\text{cm}\,\text{s}^{-1}\).
Answer: \(\dfrac{5}{2\pi} \approx 0.796\,\text{cm}\,\text{s}^{-1}\)
Worked example A balloon going down
A spherical balloon loses air so that its volume decreases at \(6\,\text{cm}^{3}\,\text{s}^{-1}\). Find the rate of change of its surface area \(S = 4\pi r^{2}\) when \(r = 3\,\text{cm}\).
  1. Decreasing, so \(\dfrac{dV}{dt} = -6\). \(\dfrac{dV}{dr} = 4\pi r^{2} = 36\pi\), so \(\dfrac{dr}{dt} = \dfrac{-6}{36\pi} = -\dfrac{1}{6\pi}\).
  2. \(\dfrac{dS}{dr} = 8\pi r = 24\pi\).
  3. \(\dfrac{dS}{dt} = \dfrac{dS}{dr} \times \dfrac{dr}{dt} = 24\pi \times \left(-\dfrac{1}{6\pi}\right) = -4\).
Answer: \(\dfrac{dS}{dt} = -4\): the surface area is decreasing at \(4\,\text{cm}^{2}\,\text{s}^{-1}\)
Your turn
An inverted cone has its radius equal to its height at every level, so the water has \(r = h\). Water flows in at \(2\,\text{cm}^{3}\,\text{s}^{-1}\). Find \(\dfrac{dh}{dt}\) when \(h = 4\,\text{cm}\).
Examiner trap

Do not differentiate \(V = \frac{1}{3}\pi r^{2} h\) treating \(r\) as the fixed top radius 8. The water surface widens as the depth rises, so \(r\) changes too; replace it by \(\frac{2h}{5}\) BEFORE differentiating. And a rate for something decreasing must carry a minus sign, or every later rate has the wrong sign.

Stage 1 · Learn

Check your understanding

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The radius of a circle increases at \(3\,\text{cm}\,\text{s}^{-1}\). At what rate is the area increasing when \(r = 2\,\text{cm}\)?

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You know \(\dfrac{dV}{dt}\) and can find \(\dfrac{dV}{dr}\). Which expression gives \(\dfrac{dr}{dt}\)?

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Water drains from a tank at \(5\,\text{cm}^{3}\,\text{s}^{-1}\). What should you write for \(\dfrac{dV}{dt}\)?

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Given \(y = x^{2}\), use small increments to estimate the increase in \(y\) when \(x\) increases from 3 to 3.01.

Section 03

Small increments and approximations.

Near a point, a curve is almost straight, with gradient \(\dfrac{dy}{dx}\). So if \(x\) changes by a small amount \(\delta x\), the change in \(y\) is about the gradient times the change in \(x\). This gives a quick estimate of a change, a percentage change, or a value that is hard to work out directly.

\[ \delta y \approx \frac{dy}{dx} \times \delta x \qquad \text{with } \frac{dy}{dx} \text{ evaluated at the starting value of } x \]

Worked example A change in terms of p
Given that \(y = 2x^{3}\), find the approximate change in \(y\) when \(x\) increases from 2 to \(2 + p\), where \(p\) is small. Hence find the approximate percentage change in \(y\) when \(x\) increases by 1% from 2.
  1. \(\dfrac{dy}{dx} = 6x^{2} = 24\) when \(x = 2\), and \(\delta x = p\).
  2. \(\delta y \approx 24p\).
  3. A 1% increase from 2 is \(p = 0.02\), so \(\delta y \approx 24 \times 0.02 = 0.48\).
  4. At \(x = 2\), \(y = 16\); percentage change \(\approx \dfrac{0.48}{16} \times 100 = 3\%\).
Answer: \(\delta y \approx 24p\); about a 3% increase
Worked example Estimating a square root
Use \(y = \sqrt{x}\) and small increments to find an approximate value of \(\sqrt{4.02}\).
  1. Start from a value you know: \(x = 4\), \(y = 2\), and \(\delta x = 0.02\).
  2. \(\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} = \dfrac{1}{4}\) when \(x = 4\).
  3. \(\delta y \approx \dfrac{1}{4} \times 0.02 = 0.005\), so \(\sqrt{4.02} \approx 2 + 0.005 = 2.005\).
  4. Check: a calculator gives \(2.004994\dots\), so the estimate is correct to 4 decimal places.
Answer: \(\sqrt{4.02} \approx 2.005\)
Your turn
The radius of a circle increases from 5 cm to 5.02 cm. Use small increments to find the approximate increase in its area, in terms of \(\pi\).
Examiner trap

\(\delta y\) is the CHANGE in \(y\), not its new value: for \(y = 2x^{3}\) the new value is about \(16 + 24p\), and the change is \(24p\). Use \(\dfrac{dy}{dx}\) at the STARTING value (\(x = 2\), not \(2 + p\)), and remember the result is an approximation, so write \(\approx\).

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The volume of a cube is decreasing at \(6\,\text{cm}^{3}\,\text{s}^{-1}\). Find the rate of change of the edge length \(x\) when \(x = 2\,\text{cm}\).

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Given that \(y = 2x^{3} - 5x\), find, in terms of \(p\), the approximate change in \(y\) when \(x\) increases from 2 to \(2 + p\), where \(p\) is small.

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An inverted cone of height 12 cm and top radius 4 cm is filled with water at \(3\,\text{cm}^{3}\,\text{s}^{-1}\). Find the rate at which the depth \(h\) is rising when \(h = 6\,\text{cm}\).

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The radius of a sphere increases by 2%. Using small increments, find the approximate percentage increase in its volume.

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Rates connected, changes estimated.

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Six original Cambridge-style questions
Connected rates for circles, spheres, cubes and a draining cone, the sign of a decreasing rate, and small increments for approximate and percentage changes.