Connected rates of change and forming differential equations - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 16 exam-style questions on Connected rates of change and forming differential equations for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A lichen grows on a rock as a circular patch. At time \(t\) years after it is first measured, the patch has radius \(r\) mm and area \(A\) mm\(^2\).
In a model, the rate of increase of the area of the patch is proportional to the circumference of the patch.
Show that, according to the model, the radius of the patch increases at a constant rate.

Model answer

Model: \(\dfrac{dA}{dt} = k \times 2\pi r\), where \(k\) is a positive constant
\(A = \pi r^2\), so \(\dfrac{dA}{dr} = 2\pi r\) and \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt} = 2\pi r\dfrac{dr}{dt}\)
So \(2\pi r\dfrac{dr}{dt} = 2\pi kr\)
\(\dfrac{dr}{dt} = k\), which is a constant, so the radius increases at a constant rate.

Explanation

Translate the words into \(\dfrac{dA}{dt} = k \times 2\pi r\), then connect \(\dfrac{dA}{dt}\) to \(\dfrac{dr}{dt}\) with the chain rule, using \(\dfrac{dA}{dr} = 2\pi r\). The factor \(2\pi r\) cancels, leaving \(\dfrac{dr}{dt} = k\). A complete answer states the conclusion in words; a common error is to omit the constant of proportionality.

Mark scheme

1 mark
B1 Forms the model: \(\dfrac{dA}{dt} = k \times 2\pi r\), where \(k\) is a positive constant. An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a show-that question, and restating the result earns nothing. Accept \(\dfrac{dA}{dt} = kC\) with \(C = 2\pi r\) used, or \(\dfrac{dA}{dt} = \lambda r\).
1 mark
M1 Uses \(A = \pi r^2\) and the chain rule: \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt} = 2\pi r\dfrac{dr}{dt}\)
Dividing \(\dfrac{dA}{dt}\) by \(\dfrac{dA}{dr}\) to find \(\dfrac{dr}{dt}\) is equivalent.
1 mark
A1* \(2\pi r\dfrac{dr}{dt} = 2\pi kr\), so \(\dfrac{dr}{dt} = k\), a constant: the radius increases at a constant rate cso
The conclusion is needed. \(r ≠ 0\) may be assumed.

Question 2

4 marks

Figure 1 shows a cross-section of a straight slipway in a harbour. The slipway runs from the point \(B\) on the horizontal harbour floor up to the quay, and it is inclined at 10° to the horizontal.
The depth of the water above the harbour floor is \(h\) metres at time \(t\) hours after 6 am, where
\[h = 3 + 1.6\sin\left(\dfrac{\pi t}{6}\right)\]
The water surface meets the slipway at the point \(W\), and \(BW = s\) metres.
Find the rate at which the point \(W\) is moving up the slipway at 8 am. Give your answer in metres per hour to 3 significant figures.

Figure 1: a cross-section of the slipway, not drawn to scale. A horizontal line is the harbour floor. A straight slipway rises from the point B on the floor, at an angle marked 10 degrees to the floor, up to the quay on the right. A dashed horizontal line shows the water surface, which meets the slipway at the point W. The length of slipway from B to W is marked s m, and the vertical depth of the water from W down to the floor is marked h m.

Model answer

From the right-angled triangle with hypotenuse \(BW\): \(h = s\sin 10^\circ\), so \(s = \dfrac{h}{\sin 10^\circ}\)
\(\dfrac{dh}{dt} = 1.6 \times \dfrac{\pi}{6}\cos\left(\dfrac{\pi t}{6}\right)\)
\(\dfrac{ds}{dt} = \dfrac{ds}{dh} \times \dfrac{dh}{dt} = \dfrac{1}{\sin 10^\circ} \times \dfrac{1.6\pi}{6}\cos\left(\dfrac{\pi t}{6}\right)\)
At 8 am, \(t = 2\): \(\cos\dfrac{\pi}{3} = \dfrac{1}{2}\), so \(\dfrac{dh}{dt} = \dfrac{1.6\pi}{12} = 0.4189\)
\(\dfrac{ds}{dt} = \dfrac{0.4189}{\sin 10^\circ} = 2.41\) metres per hour (3 s.f.)

Explanation

The depth \(h\) is the vertical side of a right-angled triangle whose hypotenuse is \(BW\), so \(s = \dfrac{h}{\sin 10^\circ}\) and \(\dfrac{ds}{dt} = \dfrac{1}{\sin 10^\circ}\dfrac{dh}{dt}\). Differentiating the tide model in radians gives \(\dfrac{dh}{dt} = \dfrac{1.6\pi}{6}\cos\left(\dfrac{\pi t}{6}\right)\), which is about 0.419 m per hour at 8 am (\(t = 2\)). Common errors are using \(t = 8\), leaving out the factor \(\dfrac{\pi}{6}\) and using the wrong trigonometric ratio.

Mark scheme

1 mark
B1 \(h = s\sin 10^\circ\), i.e. \(s = \dfrac{h}{\sin 10^\circ}\); or a correct final answer of 2.41 metres per hour with no working, which scores 2 marks (special case)
An answer of 2.41 (metres per hour) with no working scores 2 marks (special case: the B1 and the final A1). Accept awrt 2.41, i.e. [2.405, 2.415). Answers without working may not gain full credit: for full marks \(\dfrac{dh}{dt}\) and the chain rule must be seen. Using \(\cos 10^\circ\) or \(\tan 10^\circ\) is B0.
1 mark
M1 Differentiates \(h\): \(\dfrac{dh}{dt} = k\cos\left(\dfrac{\pi t}{6}\right)\) (here \(k = 1.6 \times \dfrac{\pi}{6}\))
Must be working in radians. The value \(k = \dfrac{1.6\pi}{6} = \dfrac{4\pi}{15}\) is exact. \(\dfrac{dh}{dt} = 1.6\cos\left(\dfrac{\pi t}{6}\right)\) (no \(\dfrac{\pi}{6}\)) can score this mark but not the final mark.
1 mark
M1 Uses \(\dfrac{ds}{dt} = \dfrac{1}{\sin 10^\circ} \times \dfrac{dh}{dt}\) (chain rule with their relation) at \(t = 2\)
\(t = 2\) for 8 am; using \(t = 8\) is M0.
1 mark
A1 2.41 metres per hour (awrt 2.41, i.e. [2.405, 2.415))
Depends on both M marks. Units are not required, but penalise wrong units.

Question 3

6 marks

Long waves approach a beach over a sea bed that slopes uniformly. At a point \(x\) metres from the shoreline, the depth of the water is \(h\) metres, where \(h = 0.02x\).
In a model, a wave travels directly towards the shoreline, and its speed is \(v\) m s\(^{-1}\), where
\[v = \sqrt{9.8h}\]
Show that, according to the model, the speed of the wave decreases at a constant rate, and find this rate.
Hence find the time taken for the wave to travel from the point where the depth is 4 m to the point where the depth is 1 m. Give your answer in seconds to 3 significant figures.

Model answer

The wave moves towards the shoreline, so \(\dfrac{dx}{dt} = -v\); and \(h = 0.02x\), so \(\dfrac{dh}{dx} = 0.02\)
\(v = \sqrt{9.8}\,h^{\frac{1}{2}}\), so \(\dfrac{dv}{dh} = \dfrac{1}{2}\sqrt{9.8}\,h^{-\frac{1}{2}} = \dfrac{9.8}{2\sqrt{9.8h}} = \dfrac{4.9}{v}\)
\(\dfrac{dv}{dt} = \dfrac{dv}{dh} \times \dfrac{dh}{dx} \times \dfrac{dx}{dt} = \dfrac{4.9}{v} \times 0.02 \times (-v) = -0.098\)
This does not depend on \(v\), \(h\) or \(x\): the speed of the wave decreases at a constant rate of 0.098 m s\(^{-2}\).
Depth 4 m: \(v_1 = \sqrt{9.8 \times 4} = \sqrt{39.2}\); depth 1 m: \(v_2 = \sqrt{9.8}\)
Time \(= \dfrac{v_1 - v_2}{0.098} = \dfrac{\sqrt{39.2} - \sqrt{9.8}}{0.098} = \dfrac{\sqrt{9.8}}{0.098} = 31.9\) seconds (3 s.f.)

Explanation

Three rates are linked by the chain rule: \(\dfrac{dv}{dt} = \dfrac{dv}{dh} \times \dfrac{dh}{dx} \times \dfrac{dx}{dt}\), where \(\dfrac{dx}{dt} = -v\) because the wave moves towards the shoreline. Since \(\dfrac{dv}{dh} = \dfrac{4.9}{v}\), the factor \(v\) cancels and \(\dfrac{dv}{dt} = -0.098\), a constant, so the time to slow from \(\sqrt{39.2}\) to \(\sqrt{9.8}\) m s\(^{-1}\) is \(\dfrac{\sqrt{9.8}}{0.098} \approx 31.9\) s. Common errors are taking \(\dfrac{dx}{dt}\) as a constant, missing the factor 0.02 and using the depths as values of \(x\).

Mark scheme

1 mark
B1 \(\dfrac{dx}{dt} = -v\) (the wave moves towards the shoreline, so \(x\) decreases), or \(\dfrac{dh}{dt} = -0.02v\). An answer with no working scores 0 marks for the show-that; correct answers of 0.098 m s\(^{-2}\) and 31.9 seconds with no working score 1 mark (special case).
An answer with no working scores 0 marks for the show-that part: the cancelling must be seen. Special case: correct answers of 0.098 m s\(^{-2}\) and 31.9 s with no working score 1 mark (the final A1); accept awrt 31.9, i.e. [31.85, 31.95). Condone \(\dfrac{dx}{dt} = v\) if the decrease is dealt with correctly later.
1 mark
M1 Differentiates \(v = \sqrt{9.8}\,h^{\frac{1}{2}}\): \(\dfrac{dv}{dh} = \dfrac{1}{2}\sqrt{9.8}\,h^{-\frac{1}{2}}\) \(\left(= \dfrac{4.9}{v}\right)\) (a power of \(h\) reduced by 1)
The constants 9.8 and 4.9 are exact. Alternative: \(v^2 = 9.8h = 0.196x\) differentiated with respect to \(t\), \(2v\dfrac{dv}{dt} = 0.196\dfrac{dx}{dt}\), earns this mark and the next.
1 mark
M1 (dM1) Uses the chain rule \(\dfrac{dv}{dt} = \dfrac{dv}{dh} \times \dfrac{dh}{dx} \times \dfrac{dx}{dt}\) (or \(\dfrac{dv}{dh} \times \dfrac{dh}{dt}\)) with \(\dfrac{dh}{dx} = 0.02\)
Depends on the M mark just before. The value 0.02 is exact.
1 mark
A1 \(\dfrac{dv}{dt} = \dfrac{4.9}{v} \times 0.02 \times (-v) = -0.098\): \(v\) cancels, so the speed decreases at a constant rate of 0.098 m s\(^{-2}\) cso
0.098 is exact. The cancelling of \(v\) (or of \(\sqrt{h}\) or \(\sqrt{x}\)) and a conclusion that the rate is constant are both needed. Accept \(-0.098\) as the rate of change with "decreases" stated. Units are not required.
1 mark
M1 Uses the constant rate: time \(= \dfrac{v_1 - v_2}{0.098}\) with \(v_1 = \sqrt{9.8 \times 4}\) and \(v_2 = \sqrt{9.8 \times 1}\) (or uses \(v = u + at\) with \(a = -0.098\))
Uses their constant rate. Alternative: \(\dfrac{dt}{dx} = -\dfrac{1}{v}\), so the time is \(\displaystyle\int_{50}^{200} \dfrac{dx}{\sqrt{0.196x}}\) (\(x = 200\) and \(x = 50\) for depths 4 m and 1 m), which gives the same value. Using the depths 4 and 1 as the values of \(x\) is M0.
1 mark
A1 31.9 seconds (awrt 31.9, i.e. [31.85, 31.95))
From \(\dfrac{\sqrt{39.2} - \sqrt{9.8}}{0.098} = \dfrac{\sqrt{9.8}}{0.098} = 31.94\ldots\) Units are not required, but penalise wrong units.

Question 4

10 marks

Figure 1 shows the plan of a straight path in a park and a floodlight at the point \(F\), which is 30 m from the path. The point \(N\) on the path is the point nearest to \(F\).
A cyclist rides along the path, away from \(N\), at a constant speed of 4 m s\(^{-1}\). At time \(t\) seconds the cyclist is at the point \(C\), where \(NC = x\) metres and \(FC = s\) metres.
In a model, the intensity of the light from the floodlight that reaches the cyclist is \(I\) lux, where
\[I = \dfrac{54\,000}{s^2}\]
Show that
\[\dfrac{dI}{dt} = -\dfrac{432\,000x}{\left(900 + x^2\right)^2}\]
Find the rate at which the intensity is decreasing at the instant when the cyclist is 40 m from \(N\). Give your answer in lux per second to 3 significant figures.
Find, according to the model, the exact distance of the cyclist from \(N\) at the instant when the intensity is decreasing most rapidly, and find the exact value of this greatest rate of decrease.

Figure 1: a plan view. A horizontal straight line is the path. The floodlight F is above the path, and N is the point on the path directly below F, with a right angle marked at N; FN is marked 30 m. A line joins F to the point C on the path to the right of N. NC is marked x m and FC is marked s m.

Model answer

In the right-angled triangle \(FNC\): \(s^2 = 30^2 + x^2 = 900 + x^2\)
So \(I = \dfrac{54\,000}{900 + x^2} = 54\,000\left(900 + x^2\right)^{-1}\)
\(\dfrac{dI}{dx} = -54\,000\left(900 + x^2\right)^{-2} \times 2x = -\dfrac{108\,000x}{\left(900 + x^2\right)^2}\)
\(\dfrac{dx}{dt} = 4\), so \(\dfrac{dI}{dt} = \dfrac{dI}{dx} \times \dfrac{dx}{dt} = -\dfrac{108\,000x}{\left(900 + x^2\right)^2} \times 4 = -\dfrac{432\,000x}{\left(900 + x^2\right)^2}\)
When \(x = 40\): \(\dfrac{dI}{dt} = -\dfrac{432\,000 \times 40}{2500^2} = -2.7648\)
The intensity is decreasing at 2.76 lux per second (3 s.f.)
The rate of decrease is \(432\,000\,\mathrm{g}(x)\), where \(\mathrm{g}(x) = \dfrac{x}{\left(900 + x^2\right)^2}\)
\(\mathrm{g}'(x) = \dfrac{\left(900 + x^2\right)^2 - x \times 2\left(900 + x^2\right) \times 2x}{\left(900 + x^2\right)^4} = \dfrac{900 - 3x^2}{\left(900 + x^2\right)^3}\)
\(\mathrm{g}'(x) = 0\): \(3x^2 = 900\), \(x^2 = 300\), so \(x = 10\sqrt{3}\) (as \(x > 0\))
(\(\mathrm{g}'(x) > 0\) for \(0 < x < 10\sqrt{3}\) and \(\mathrm{g}'(x) < 0\) for \(x > 10\sqrt{3}\), so this gives the greatest rate)
Greatest rate of decrease \(= \dfrac{432\,000 \times 10\sqrt{3}}{\left(900 + 300\right)^2} = \dfrac{4\,320\,000\sqrt{3}}{1\,440\,000} = 3\sqrt{3}\) lux per second

Explanation

Pythagoras gives \(s^2 = 900 + x^2\), so \(I\) can be written in terms of \(x\), and the chain rule with \(\dfrac{dx}{dt} = 4\) gives the printed result; at \(x = 40\) the intensity falls at 2.7648 lux per second. The intensity decreases most rapidly where the rate \(\dfrac{432\,000x}{\left(900 + x^2\right)^2}\) is greatest, so this expression (not \(I\)) is differentiated and its derivative set to zero, giving \(x = 10\sqrt{3}\) and a greatest rate of \(3\sqrt{3}\) lux per second. Common errors are setting \(\dfrac{dI}{dt} = 0\) instead of maximising the rate, and giving decimals when exact values are asked for.

Mark scheme

1 mark
B1 \(s^2 = 900 + x^2\) (Pythagoras in triangle \(FNC\)), so \(I = \dfrac{54\,000}{900 + x^2}\). An answer with no working scores 0 marks for the show-that; correct answers of 2.76, \(10\sqrt{3}\) m and \(3\sqrt{3}\) lux per second with no working score 3 marks (special case).
An answer with no working scores 0 marks for the show-that, and restating the given result earns nothing. Special case: correct answers of 2.76 lux per second, \(x = 10\sqrt{3}\) m and \(3\sqrt{3}\) lux per second with no working score 3 marks (the A1 for 2.76 and the final two A1 marks). Accept awrt 2.76, i.e. [2.755, 2.765), for the rate when \(x = 40\).
1 mark
M1 Differentiates with respect to \(x\) using the chain rule: \(\dfrac{dI}{dx} = -54\,000 \times 2x\left(900 + x^2\right)^{-2}\) (allow \(kx\left(900 + x^2\right)^{-2}\))
Alternative: \(\dfrac{dI}{ds} = -108\,000s^{-3}\) together with \(\dfrac{ds}{dt} = \dfrac{x}{s}\dfrac{dx}{dt}\) from differentiating \(s^2 = 900 + x^2\) with respect to \(t\); this earns the same marks.
1 mark
M1 (dM1) Uses \(\dfrac{dI}{dt} = \dfrac{dI}{dx} \times \dfrac{dx}{dt}\) with \(\dfrac{dx}{dt} = 4\)
Depends on the M mark just before. Dividing by 4 instead of multiplying is M0.
1 mark
A1* \(\dfrac{dI}{dt} = -\dfrac{108\,000x}{\left(900 + x^2\right)^2} \times 4 = -\dfrac{432\,000x}{\left(900 + x^2\right)^2}\) cso
No errors. A correct line before the given answer must be seen, e.g. \(-108\,000x\left(900 + x^2\right)^{-2} \times 4\). The final line must include \(\dfrac{dI}{dt} =\).
1 mark
M1 Substitutes \(x = 40\) into the given \(\dfrac{dI}{dt}\) (with \(900 + 40^2 = 2500\))
May use the given result even if the show-that was not completed. Substituting \(t = 40\), or \(s = 40\) for \(x\), is M0.
1 mark
A1 2.76 lux per second (awrt 2.76, i.e. [2.755, 2.765))
The exact value is 2.7648. Accept \(-2.76\) described as a rate of change. Units are not required, but penalise wrong units.
1 mark
M1 Differentiates \(\dfrac{x}{\left(900 + x^2\right)^2}\) (or \(x\left(900 + x^2\right)^{-2}\)) with the quotient or product rule and sets the derivative equal to zero
The rate itself must be maximised: differentiating \(I\), or setting \(\dfrac{dI}{dt} = 0\) (giving \(x = 0\)), is M0. Allow the constant 432 000 to be kept or dropped.
1 mark
A1 \(900 - 3x^2 = 0\) oe, e.g. \(\left(900 + x^2\right)^2 - 4x^2\left(900 + x^2\right) = 0\)
A correct equation from a correct derivative.
1 mark
A1 \(x = 10\sqrt{3}\) m
Depends on the M mark at point 7. Exact: accept \(\sqrt{300}\); awrt 17.3, i.e. [17.25, 17.35), alone is A0. Ignore \(x = -10\sqrt{3}\) if it is rejected. A check that this gives the greatest rate is not required.
1 mark
A1 Greatest rate of decrease \(= 3\sqrt{3}\) lux per second
Depends on the M mark at point 7. From \(\dfrac{432\,000 \times 10\sqrt{3}}{1200^2}\). Exact: accept \(\sqrt{27}\); awrt 5.20, i.e. [5.195, 5.205), alone is A0.

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