IGCSE · Practice questions · Connected rates of change and small increments

Differentiate, connect, estimate.

Six original Cambridge-style questions on connected rates of change and small increments: using \(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}\), writing a cone's volume in terms of its depth, giving decreasing rates a negative sign, and estimating changes with \(\delta y \approx \dfrac{dy}{dx}\,\delta x\).

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Formula, derivative, chain rule, units.

Formula: write the quantity in terms of ONE variable; for a cone use similar triangles to replace \(r\).
Differentiate, then substitute the value at that instant.
Connect: \(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}\); divide by \(\dfrac{dy}{dx}\) when \(\dfrac{dx}{dt}\) is the unknown.
Sign and units: decreasing means a negative rate; give units such as \(\text{cm}^{3}\,\text{s}^{-1}\).
Small increments: \(\delta y \approx \dfrac{dy}{dx}\,\delta x\) at the starting value; percentage change is \(\dfrac{\delta y}{y} \times 100\).

Recall

[2 marks]

The radius of a circle is increasing at \(0.2\,\text{cm}\,\text{s}^{-1}\). Find the rate of increase of the area of the circle when the radius is 10 cm.

\(A = \pi r^{2}\), so \(\dfrac{dA}{dr} = 2\pi r = 20\pi\) when \(r = 10\). ✓
\(\dfrac{dA}{dt} = 20\pi \times 0.2 = 4\pi \approx 12.6\,\text{cm}^{2}\,\text{s}^{-1}\). ✓

Calculation

[3 marks]

The variables \(x\) and \(y\) are related by \(y = \sqrt{x^{2} + 9}\). Given that \(x\) is increasing at 0.5 units per second, find the rate of increase of \(y\) when \(x = 4\).

\(y = (x^{2} + 9)^{\frac{1}{2}}\), so by the chain rule \(\dfrac{dy}{dx} = \dfrac{1}{2}(x^{2} + 9)^{-\frac{1}{2}} \times 2x = \dfrac{x}{\sqrt{x^{2} + 9}}\). ✓
At \(x = 4\): \(\dfrac{dy}{dx} = \dfrac{4}{\sqrt{25}} = \dfrac{4}{5}\). ✓
\(\dfrac{dy}{dt} = \dfrac{4}{5} \times 0.5 = 0.4\) units per second. ✓

Calculation

[3 marks]

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

\(y = 5x^{-2}\), so \(\dfrac{dy}{dx} = -10x^{-3} = -\dfrac{10}{x^{3}}\). ✓
At \(x = 2\): \(\dfrac{dy}{dx} = -\dfrac{10}{8} = -\dfrac{5}{4}\). ✓
\(\delta y \approx -\dfrac{5}{4}p = -1.25p\): since \(p\) is positive, \(y\) decreases by about \(1.25p\). ✓

Calculation

[4 marks]

The volume of a sphere is increasing at \(50\,\text{cm}^{3}\,\text{s}^{-1}\).

(a) Find the rate of increase of the radius when the radius is 5 cm.

(b) Find the rate of increase of the surface area, \(S = 4\pi r^{2}\), at the same instant.

(a) \(V = \frac{4}{3}\pi r^{3}\), \(\dfrac{dV}{dr} = 4\pi r^{2} = 100\pi\) when \(r = 5\). ✓
\(\dfrac{dr}{dt} = \dfrac{50}{100\pi} = \dfrac{1}{2\pi} \approx 0.159\,\text{cm}\,\text{s}^{-1}\). ✓
(b) \(\dfrac{dS}{dr} = 8\pi r = 40\pi\). ✓
\(\dfrac{dS}{dt} = 40\pi \times \dfrac{1}{2\pi} = 20\,\text{cm}^{2}\,\text{s}^{-1}\). ✓

Analysis

[5 marks]

A container is an inverted cone of height 30 cm and top radius 10 cm. Water leaks out through a small hole at the vertex at \(2\,\text{cm}^{3}\,\text{s}^{-1}\). When the depth of water is \(h\) cm, the volume is \(V\,\text{cm}^{3}\).

(a) Show that \(V = \dfrac{\pi h^{3}}{27}\).

(b) Find the rate of change of \(h\) when \(h = 9\).

(a) By similar triangles \(\dfrac{r}{h} = \dfrac{10}{30}\), so \(r = \dfrac{h}{3}\) and \(V = \dfrac{1}{3}\pi \left(\dfrac{h}{3}\right)^{2} h = \dfrac{\pi h^{3}}{27}\). ✓
(b) The volume is decreasing, so \(\dfrac{dV}{dt} = -2\). \(\dfrac{dV}{dh} = \dfrac{3\pi h^{2}}{27} = \dfrac{\pi h^{2}}{9} = 9\pi\) when \(h = 9\). ✓
\(\dfrac{dh}{dt} = \dfrac{-2}{9\pi} \approx -0.0707\,\text{cm}\,\text{s}^{-1}\): the depth falls at about \(0.0707\,\text{cm}\,\text{s}^{-1}\). ✓
(c) \(r = \dfrac{h}{3}\), so \(\dfrac{dr}{dt} = \dfrac{1}{3} \times \dfrac{dh}{dt} = -\dfrac{2}{27\pi} \approx -0.0236\,\text{cm}\,\text{s}^{-1}\). ✓

Analysis

[6 marks]

A closed box has a square base of side \(x\) cm and a height of \(2x\) cm.

(a) Show that its volume is \(V = 2x^{3}\) and its total surface area is \(S = 10x^{2}\).

(b) Find, in terms of \(p\), the approximate increase in \(V\) when \(x\) increases from 5 to \(5 + p\), where \(p\) is small.

(d) Find the approximate percentage increase in \(S\) when \(x\) increases by 1% from 5.

(a) \(V = x \times x \times 2x = 2x^{3}\). The two square faces give \(2x^{2}\) and the four side faces give \(4 \times x \times 2x = 8x^{2}\), so \(S = 10x^{2}\). ✓
(b) \(\dfrac{dV}{dx} = 6x^{2} = 150\) at \(x = 5\), so \(\delta V \approx 150p\). ✓
(c) \(150p = 6\), so \(p = 0.04\). ✓
(d) \(\dfrac{dS}{dx} = 20x = 100\) at \(x = 5\), and a 1% increase is \(\delta x = 0.05\), so \(\delta S \approx 100 \times 0.05 = 5\). ✓
At \(x = 5\), \(S = 250\), so the percentage increase is about \(\dfrac{5}{250} \times 100 = 2\%\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.