Edexcel A level Maths (9MA0) · Pure mathematics › Differentiation
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Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A stone is dropped into a pond, creating a circular ripple. The radius, \(r\) cm, of the ripple increases at a constant rate of 4 cm s−1.
(a) Find the rate at which the area enclosed by the ripple is increasing at the instant when the radius is 15 cm. Give your answer as an exact multiple of \(\pi\).[3]
(b) The ripple starts from a single point. Find, as an exact multiple of \(\pi\), the rate at which the area is increasing 8 seconds after the ripple starts.[2]
Show the answer and mark scheme
(a)Answer: \(120\pi\) cm2 s−1
M1 for \(\frac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r\)
M1 for using \(\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}\)
A1 for \(120\pi\) cm2 s−1
Worked solution: \(A = \pi r^{2}\), so \(\frac{\mathrm{d}A}{\mathrm{d}t} = 2\pi r \times \frac{\mathrm{d}r}{\mathrm{d}t} = 2\pi \times 15 \times 4 = 120\pi\) cm2 s−1.
(b)Answer: \(256\pi\) cm2 s−1
M1 for \(r = 4 \times 8 = 32\)
A1 for \(256\pi\) cm2 s−1
Worked solution: After 8 s, \(r = 32\), so \(\frac{\mathrm{d}A}{\mathrm{d}t} = 2\pi \times 32 \times 4 = 256\pi\) cm2 s−1.
Question 2Medium7 marks
A spherical balloon is inflated so that its volume increases at a constant rate of 50 cm3 s−1. [The volume of a sphere of radius \(r\) is \(\frac{4}{3}\pi r^3\) and its surface area is \(4\pi r^2\).]
(a) Show that, when the radius is \(r\) cm, \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{25}{2\pi r^2}\).[3]
(b) Find the rate at which the surface area of the balloon is increasing when the radius is 5 cm.[3]
(c) State one modelling assumption made in this question and explain why it may not be realistic.[1]
M1 for \(\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\)
M1 for the chain rule: \(\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}r}\)
A1* for \(\frac{25}{2\pi r^2}\)
Worked solution: \(V = \frac{4}{3}\pi r^3\), so \(\dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\). Then \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{\mathrm{d}V/\mathrm{d}t}{\mathrm{d}V/\mathrm{d}r} = \dfrac{50}{4\pi r^2} = \dfrac{25}{2\pi r^2}\).
(b)Answer: 20 cm2 s−1
M1 for \(\frac{\mathrm{d}S}{\mathrm{d}r} = 8\pi r\)
M1 for \(\frac{\mathrm{d}S}{\mathrm{d}t} = 8\pi r \times \frac{25}{2\pi r^2} = \frac{100}{r}\)
A1 for 20 (cm2 s−1)
Worked solution: \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = \dfrac{\mathrm{d}S}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 8\pi r \times \dfrac{25}{2\pi r^2} = \dfrac{100}{r}\). When \(r = 5\), this is 20 cm2 s−1.
(c)Answer: e.g. the balloon stays a perfect sphere — a real balloon is not spherical (especially near the neck); or air is pumped in at a constant rate — in practice the rate varies.
B1 for an assumption with a reason why it may be unrealistic, e.g. the balloon remains spherical (real balloons are elongated / have a neck), or the rate of inflation is constant (a person or pump will not keep an exactly constant rate)
Worked solution: The model assumes the balloon is always a perfect sphere. Real balloons are not exactly spherical, particularly near the neck, so the relation between \(V\) and \(r\) is only approximate.
Question 3Hard6 marks
A water trough is 200 cm long. Its cross-section is an isosceles triangle with its vertex at the bottom; the triangle is 30 cm wide at the top and 20 cm deep. Water flows into the empty trough at a constant rate of 450 cm3 s−1.
(a) Show that, when the depth of the water is \(h\) cm, the volume of water in the trough is \(150h^{2}\) cm3.[3]
(b) Find the rate at which the depth of the water is increasing at the instant when the depth is 5 cm.[3]
Show the answer and mark scheme
(a)Answer: Proof
M1 for using similar triangles to find the width of the water surface, \(1.5h\)
M1 for volume \(= \frac{1}{2} \times 1.5h \times h \times 200\)
A1* for \(V = 150h^{2}\) (cso)
Worked solution: By similar triangles, the width of the water surface at depth \(h\) is \(\frac{30}{20}h = 1.5h\). \(V = \frac{1}{2} \times 1.5h \times h \times 200 = 150h^{2}\).
(b)Answer: \(0.3\) cm s−1
M1 for \(\frac{\mathrm{d}V}{\mathrm{d}h} = 300h\)
M1 for using \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \div \frac{\mathrm{d}V}{\mathrm{d}h}\)
A1 for \(0.3\) cm s−1
Worked solution: \(\frac{\mathrm{d}V}{\mathrm{d}h} = 300h\), so \(\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{450}{300h} = \frac{450}{1500} = 0.3\) cm s−1.