Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 2 · Simple familiar

Question 15

[4 marks] Graphics calculator
The question

The volume of a sphere is increasing at a constant rate of 40 cm3 s−1.

a)

Show that the radius of the sphere is 4.924 cm, rounded to three decimal places, at the instant when the volume of the sphere is 500 cm3.

[1 mark] b)

Determine the rate at which the radius of the sphere is increasing at this instant.

[2 marks] c)

Determine the rate at which the surface area of the sphere is increasing at this instant.

[1 mark]
Watch the situation first

A steady volume, a slowing radius

The volume goes up by the same 40 cm3 every second. Watch the sphere grow to 500 cm3, then see how the slope of the volume curve links the two rates.

The volume grows by 40 cm³ every second, so the radius grows too At the instant V = 500 cm³, the radius is r ≈ 4.924 cm V = 4/3 πr³, so dV/dr = 4πr² is the slope of this curve Chain rule: dr/dt = dr/dV × dV/dt = 40 ÷ (4πr²) ≈ 0.131 cm s⁻¹ S = 4πr², so dS/dt = 8πr × dr/dt ≈ 16.2 cm² s⁻¹ r ≈ 4.924 cm S = 4πr² V = 0 cm³ V = 100 cm³ V = 200 cm³ V = 300 cm³ V = 400 cm³ V = 500 cm³ 0 1 2 3 4 5 6 0 200 400 600 800 1000 Radius r (cm) Volume V (cm³) 500 4.924 slope = dV/dr dV/dt = 40 cm³ s⁻¹ r = ∛(375 ÷ π) ≈ 4.924 cm dV/dr = 4πr² ≈ 304.6 dr/dt = 10 ÷ (πr²) ≈ 0.131 cm s⁻¹ dS/dt = 80 ÷ r ≈ 16.2 cm² s⁻¹

The sphere and the graph are drawn to scale. The volume rises steadily, but each extra 40 cm3 adds less to the radius as the sphere grows.

Now for the mathematics

Find the instant, then read off the rates

The volume always grows at 40 cm3 s−1. Change the volume and watch the radius, the slope of the curve and the two rates change with it. Then set the volume to the instant in the question.

r V Radius r (cm) Volume V (cm³) slope dV/dr
Volume curve Tangent at your volume
Part a)
Part b)
Part c)

This sphere is smaller than the one in the question, and its radius is growing faster, at about  cm s−1. Increase the volume to the instant in the question.

This sphere is bigger than the one in the question, and its radius is growing more slowly, at about  cm s−1. Decrease the volume to the instant in the question.

This is the instant in the question, $V=500$. The radius is 4.924 cm, it is growing at about 0.131 cm s−1, and the surface area is growing at about 16.2 cm2 s−1.

The volume rate stays at 40, but $\frac{dr}{dt}=\frac{10}{\pi r^2}$ gets smaller as $r$ grows. The steeper the tangent, the slower the radius grows.

Marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a)1 mark

Set the volume formula equal to 500 and solve for $r$. Evaluate it on the calculator, showing more decimal places than the target.

$$\begin{aligned}\frac{4}{3}\pi r^3&=500\\r&=\sqrt[3]{\frac{375}{\pi}}\\&=4.923725\ldots\\&\approx 4.924\text{ cm}\end{aligned}$$

Calculator use

Enter (375 ÷ π) ^ (1 ÷ 3). It gives 4.92373. Store it for parts b) and c): Ans → R (STO▸ R).

$r\approx 4.924$ cm, as required

Marker: correctly shows the required result

Step 2 · part b)1 mark

Differentiate the volume with respect to $r$.

$$V=\frac{4}{3}\pi r^3\ \Rightarrow\ \frac{dV}{dr}=4\pi r^2$$

Why?

You know $\dfrac{dV}{dt}=40$ and want $\dfrac{dr}{dt}$. $V$ is a function of $r$, so the chain rule links the two rates through $\dfrac{dV}{dr}$.

Marker: correctly determines $\dfrac{dV}{dr}$

Step 3 · part b)1 mark

Chain the rates, using $\dfrac{dr}{dV}=\dfrac{1}{4\pi r^2}$, and substitute the radius from part a).

$$\begin{aligned}\frac{dr}{dt}&=\frac{dr}{dV}\times\frac{dV}{dt}\\&=\frac{40}{4\pi(4.924)^2}\\&\approx 0.131\end{aligned}$$

Calculator use

Enter 40 ÷ (4 × π × R²). It gives 0.131299. Store it straight away for part c): Ans → D (STO▸ D).

$\dfrac{dr}{dt}\approx 0.131$ cm s−1

Marker: correctly determines the rate of increase of the radius

Step 4 · part c)1 mark

The formula sheet gives $S=4\pi r^2$. Differentiate it, then chain through $r$.

$$\begin{aligned}\frac{dS}{dt}&=8\pi r\times\frac{dr}{dt}\\&=8\pi(4.924)(0.1313)\\&\approx 16.2\end{aligned}$$

Calculator use

Enter 8 × π × R × D, using the stored values. It gives 16.2479. 80 ÷ R gives the same.

$\dfrac{dS}{dt}\approx 16.2$ cm2 s−1
Marking

$\dfrac{dS}{dt}=\dfrac{80}{r}$ is accepted. Using $\dfrac{dr}{dt}=0.131$ gives 16.2; this is accepted. Follow-through applies in part c) from an incorrect result in part b).

Marker: correctly determines the rate of increase of the surface area

Putting it all together

Each phrase of the question gave you something

“The volume of a sphere is increasing at a constant rate of 40 cm3 s−1”
The given rate, $\dfrac{dV}{dt}=40$.
“at the instant when the volume of the sphere is 500 cm3”
Substitute into $V=\dfrac{4}{3}\pi r^3$ to get $r=\sqrt[3]{\dfrac{375}{\pi}}$. Store it and use it in parts b) and c).
“Show that … 4.924 cm, rounded to three decimal places”
Show more decimal places than the target, $4.92373\ldots$, so the rounding is proved, not copied.
“the rate at which the radius of the sphere is increasing”
The wanted rate is $\dfrac{dr}{dt}$. Chain through $\dfrac{dV}{dr}=4\pi r^2$ to get $\dfrac{dr}{dt}=\dfrac{10}{\pi r^2}\approx 0.131$.
“the rate at which the surface area of the sphere is increasing”
Use $S=4\pi r^2$ from the formula sheet and chain through $r$: $\dfrac{dS}{dt}=8\pi r\times\dfrac{dr}{dt}=\dfrac{80}{r}\approx 16.2$.
Common errors

If I had to guess where you might lose marks…

  • Inverting the chain rule in b). Writing $\dfrac{dr}{dt}=\dfrac{dV}{dr}\div\dfrac{dV}{dt}=\dfrac{4\pi r^2}{40}$ gives 7.62 cm s−1. A radius growing faster than 7 cm each second, when the radius is under 5 cm, should ring alarm bells.
  • Using $S=4\pi r^2$ as the multiplier in c). Writing $\dfrac{dS}{dt}=4\pi r^2\times\dfrac{dr}{dt}$ gives exactly 40, which is just $\dfrac{dV}{dt}$ again, because $4\pi r^2$ is $\dfrac{dV}{dr}$, not $\dfrac{dS}{dr}$.
  • The wrong derivative of $S$. Using $\dfrac{dS}{dr}=4\pi r$ instead of $8\pi r$ gives $\dfrac{40}{r}\approx 8.12$ cm2 s−1, half the correct value.
  • Errors in the radius in a). Dropping $\pi$ gives $\sqrt[3]{375}=7.21$, and multiplying by $\dfrac{4}{3}$ instead of dividing gives $\sqrt[3]{\dfrac{2000}{3\pi}}=5.96$. In a “show that”, any value other than 4.924 tells you the algebra is wrong. Don’t force it.
  • Rounding too early. Rounding $\dfrac{dr}{dt}$ to 0.13 before part c) gives $8\pi(4.924)(0.13)=16.1$, which is wrong to 3 s.f. Rounding $r$ to 4.9 gives $\dfrac{dr}{dt}=0.133$ and $\dfrac{dS}{dt}=16.3$. Keep stored values and round only at the end.

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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