Connected rates of change and forming differential equations - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 17 exam-style questions on Connected rates of change and forming differential equations for Edexcel A-Level Maths (Paper 1). 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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17 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 café sells cups of hot chocolate. The weekly demand is \(D\) cups when the price of a cup is £\(p\).

A model assumes that the rate of decrease of \(D\) with respect to \(p\) is proportional to \(D\).

When the price of a cup is £2.50, the weekly demand is 400 cups and the demand is decreasing at a rate of 120 cups per week for each £1 increase in the price.

Find a differential equation linking \(D\) and \(p\), giving the value of any constant in your equation.

Model answer

Rate of decrease of \(D\) with respect to \(p\) is proportional to \(D\):
\(\dfrac{dD}{dp} = -kD\), \(k > 0\)
When \(D = 400\), \(\dfrac{dD}{dp} = -120\): \(-120 = -400k\), so \(k = 0.3\)
\(\dfrac{dD}{dp} = -0.3D\)

Explanation

The rate of change of demand with respect to price is \(\dfrac{dD}{dp}\), and a rate of decrease means a negative sign with a positive constant, so \(\dfrac{dD}{dp} = -kD\). The data give \(-120 = -400k\), so \(k = 0.3\); the price of £2.50 is not needed. Marks are lost for giving only the value of \(k\) instead of the complete equation, and for using \(t\) instead of \(p\).

Mark scheme

1 mark
B1 \(\dfrac{dD}{dp} = -kD\) (\(k > 0\)) or \(\dfrac{dD}{dp} = kD\) with \(k\) to be found; or a correct final answer of \(\dfrac{dD}{dp} = -0.3D\) with no working
An answer of \(\dfrac{dD}{dp} = -0.3D\) with no working scores 3 marks. The derivative must be \(\dfrac{dD}{dp}\), not \(\dfrac{dD}{dt}\).
1 mark
M1 Substitutes \(D = 400\) and a rate of \(\pm 120\) into their equation to find their constant
E.g. \(-120 = -k \times 400\). Substituting \(p = 2.5\) for \(D\) is M0.
1 mark
A1 \(\dfrac{dD}{dp} = -0.3D\) oe, e.g. \(\dfrac{dD}{dp} = -\dfrac{3D}{10}\)
The complete equation is needed: just \(k = 0.3\) is A0 (examiners report this). \(\dfrac{dD}{dp} = 0.3D\) (wrong sign) is A0; \(\dfrac{dD}{dp} = kD\) with \(k = -0.3\) written into the equation is fine.

Question 2

4 marks

A raindrop falls through a cloud and grows as water vapour condenses on it. Its mass is \(m\) milligrams at time \(t\) minutes.

A model assumes that the rate of increase of \(m\) is proportional to \(m^{\frac{2}{3}}\).

When \(m = 8\), the mass of the raindrop is increasing at 0.6 milligrams per minute.

Use the model to find the rate at which the mass of the raindrop is increasing when \(m = 27\).

Model answer

\(\dfrac{dm}{dt} = km^{\frac{2}{3}}\)
When \(m = 8\): \(0.6 = k \times 8^{\frac{2}{3}} = 4k\), so \(k = 0.15\)
\(\dfrac{dm}{dt} = 0.15m^{\frac{2}{3}}\)
When \(m = 27\): \(\dfrac{dm}{dt} = 0.15 \times 27^{\frac{2}{3}} = 0.15 \times 9 = 1.35\)
The mass of the raindrop is increasing at 1.35 milligrams per minute.

Explanation

Write the model as \(\dfrac{dm}{dt} = km^{\frac{2}{3}}\) and use the given rate at \(m = 8\) to find \(k = 0.15\); the model then gives the rate at any other mass. (A power of \(\dfrac{2}{3}\) is natural here: the water condenses on the surface of the drop, and the surface area of drops of the same shape is proportional to \(m^{\frac{2}{3}}\).) A common error is to leave out the constant \(k\), or to evaluate \(8^{\frac{2}{3}}\) wrongly.

Mark scheme

1 mark
B1 \(\dfrac{dm}{dt} = km^{\frac{2}{3}}\); or a correct final answer of 1.35 milligrams per minute with no working, which scores 2 marks (special case)
An answer of 1.35 (milligrams per minute) with no working scores 2 marks (special case: the B1 and the final A1). Answers without working may not gain full credit: for full marks the equation \(\dfrac{dm}{dt} = km^{\frac{2}{3}}\) and \(k = 0.15\) (or the ratio method) must be seen. The constant of proportionality must be present; \(\dfrac{dm}{dt} = m^{\frac{2}{3}}\) is B0 (examiners report missing constants as a common error).
1 mark
M1 Substitutes \(m = 8\) and \(\dfrac{dm}{dt} = 0.6\) into their equation to find \(k\)
\(8^{\frac{2}{3}} = 4\).
1 mark
A1 \(k = 0.15\)
Or \(\dfrac{dm}{dt} = 0.15m^{\frac{2}{3}}\).
1 mark
A1 1.35 milligrams per minute
From \(0.15 \times 27^{\frac{2}{3}} = 0.15 \times 9\). Units are not required, but wrong units lose this mark. Alternative: the ratio method, \(0.6 \times \left(\dfrac{27}{8}\right)^{\frac{2}{3}} = 0.6 \times \dfrac{9}{4}\), scores M1 A1 A1 (with the B1 for the stated proportionality).

Question 3

6 marks

A cyclist freewheels down a long straight hill. He passes the top of the hill at time \(t = 0\). When he has travelled \(x\) metres from the top of the hill, his speed is \(v\) m s\(^{-1}\), where, in a model,
\[v^2 = 16 + 0.3x - 0.0001x^2, \quad 0 \leqslant x \leqslant 2500\]
At time \(t\) seconds, \(\dfrac{dx}{dt} = v\).
By differentiating the equation of the model with respect to \(t\), show that the acceleration of the cyclist, \(\dfrac{dv}{dt}\), is \((0.15 - 0.0001x)\) m s\(^{-2}\).
Hence find the greatest speed of the cyclist according to the model, giving your answer in m s\(^{-1}\) to 3 significant figures.

Model answer

Differentiating \(v^2 = 16 + 0.3x - 0.0001x^2\) with respect to \(t\):
\(2v\dfrac{dv}{dt} = (0.3 - 0.0002x)\dfrac{dx}{dt} = (0.3 - 0.0002x)v\)
\(\dfrac{dv}{dt} = \dfrac{0.3 - 0.0002x}{2} = 0.15 - 0.0001x\) as required
The speed is greatest when \(\dfrac{dv}{dt} = 0\): \(0.15 - 0.0001x = 0\), so \(x = 1500\)
\(v^2 = 16 + 0.3(1500) - 0.0001(1500)^2 = 16 + 450 - 225 = 241\)
Greatest speed \(= \sqrt{241} = 15.5\) m s\(^{-1}\) (3 s.f.)

Explanation

Differentiate both sides with respect to \(t\): \(v^2\) gives \(2v\dfrac{dv}{dt}\) and, by the chain rule, the right-hand side gives \((0.3 - 0.0002x)\dfrac{dx}{dt}\); replacing \(\dfrac{dx}{dt}\) by \(v\) and dividing by \(2v\) leaves \(0.15 - 0.0001x\). The speed is greatest where the acceleration is zero, at \(x = 1500\), which gives \(v = \sqrt{241} \approx 15.5\) m s\(^{-1}\). Common errors are differentiating \(v^2\) as \(2v\) without the factor \(\dfrac{dv}{dt}\), and giving \(v^2 = 241\) as the speed.

Mark scheme

1 mark
M1 Differentiates \(v^2\) implicitly with respect to \(t\), \(2v\dfrac{dv}{dt}\), and the right-hand side using the chain rule, \((0.3 - 0.0002x)\dfrac{dx}{dt}\) (allow one slip). An answer with no working scores 0 marks for the show-that; a correct greatest speed of 15.5 m s\(^{-1}\) with no working scores 1 mark (special case: the final A1).
An answer with no working scores 0 marks for the show-that: its three marks need the working. Special case: a greatest speed of awrt 15.5 m s\(^{-1}\), i.e. [15.45, 15.55), with no working scores 1 mark (the final A1). Alternative: differentiates with respect to \(x\), \(2v\dfrac{dv}{dx} = 0.3 - 0.0002x\), and uses \(\dfrac{dv}{dt} = \dfrac{dv}{dx} \times \dfrac{dx}{dt}\).
1 mark
M1 (dM1) Replaces \(\dfrac{dx}{dt}\) by \(v\) and divides by \(2v\)
Depends on the first M mark.
1 mark
A1* \(\dfrac{dv}{dt} = \dfrac{(0.3 - 0.0002x)v}{2v} = 0.15 - 0.0001x\) cso
Given answer: needs \(2v\dfrac{dv}{dt} = (0.3 - 0.0002x)v\) oe seen, and no errors.
1 mark
M1 Sets the acceleration equal to zero, \(0.15 - 0.0001x = 0\), to find \(x = 1500\) (or uses any valid method for the \(x\) that maximises \(v^2\))
Alternatives: \(\dfrac{d(v^2)}{dx} = 0.3 - 0.0002x = 0\), or completing the square, \(v^2 = 241 - 0.0001(x - 1500)^2\).
1 mark
M1 (dM1) Substitutes their \(x\) into \(v^2 = 16 + 0.3x - 0.0001x^2\) and takes the square root
Depends on the previous M mark. The exact value is \(v^2 = 241\); giving 241 as the speed is dM1A0.
1 mark
A1 Greatest speed awrt 15.5 m s\(^{-1}\), i.e. [15.45, 15.55)
\(\sqrt{241}\) is acceptable. A justification that it is a maximum is not required (the acceleration changes from positive to negative at \(x = 1500\)).

Question 4

8 marks

In a chemical reaction, two substances \(A\) and \(B\) combine to form a substance \(C\). At the start of the reaction there are 12 grams of \(A\), 9 grams of \(B\) and no \(C\). Each gram of \(C\) is formed from 0.5 grams of \(A\) and 0.5 grams of \(B\).
At time \(t\) minutes after the start of the reaction, \(x\) grams of \(C\) have formed. In a model, the rate at which \(C\) forms is proportional to the product of the masses of \(A\) and \(B\) that remain.
Show that
\[\dfrac{dx}{dt} = \lambda(24 - x)(18 - x)\]
where \(\lambda\) is a positive constant.
At the start of the reaction, \(C\) forms at a rate of 2.16 grams per minute.
Find the value of \(\lambda\).
Using algebra, find the mass of \(C\) at the instant when \(C\) is forming at half the rate at which it formed at the start of the reaction.
State, with a reason, the greatest mass of \(C\) that can form according to the model.

Model answer

Mass of \(A\) remaining \(= 12 - 0.5x\); mass of \(B\) remaining \(= 9 - 0.5x\)
\(\dfrac{dx}{dt} = k(12 - 0.5x)(9 - 0.5x) = k \times \dfrac{1}{2}(24 - x) \times \dfrac{1}{2}(18 - x) = \dfrac{k}{4}(24 - x)(18 - x)\)
So \(\dfrac{dx}{dt} = \lambda(24 - x)(18 - x)\), where \(\lambda = \dfrac{k}{4}\) is a positive constant
At the start, \(x = 0\): \(2.16 = \lambda \times 24 \times 18 = 432\lambda\), so \(\lambda = 0.005\)
Half the initial rate: \(0.005(24 - x)(18 - x) = 1.08\)
\((24 - x)(18 - x) = 216 \Rightarrow x^2 - 42x + 432 = 216 \Rightarrow x^2 - 42x + 216 = 0\)
\((x - 6)(x - 36) = 0\), so \(x = 6\) or \(x = 36\)
\(x = 36\) is impossible, as \(x \leqslant 18\) (there are only 9 g of \(B\)), so the mass of \(C\) is 6 grams
The greatest mass of \(C\) is 18 grams: when \(x = 18\) all of \(B\) has been used, so the rate \(\lambda(24 - x)(18 - x)\) is zero and no more \(C\) can form.

Explanation

The rate is \(k\) times the product of the remaining masses \((12 - 0.5x)\) and \((9 - 0.5x)\); taking a factor \(\dfrac{1}{2}\) out of each bracket gives \(\lambda(24 - x)(18 - x)\) with \(\lambda = \dfrac{k}{4}\). The initial rate gives \(\lambda = 0.005\), and half that rate leads to \(x^2 - 42x + 216 = 0\), so \(x = 6\) (36 is rejected), while \(B\) runs out when \(x = 18\). Common errors are using 12 and 9 instead of the remaining masses, and not rejecting the root 36, which would need more \(B\) than there is.

Mark scheme

1 mark
B1 The masses remaining are \((12 - 0.5x)\) grams of \(A\) and \((9 - 0.5x)\) grams of \(B\). Correct answers \(\lambda = 0.005\), \(x = 6\) and 18 grams with no working score 2 marks (the B1 for \(\lambda\) and the final B1).
Correct answers \(\lambda = 0.005\), \(x = 6\) and 18 grams with no working score 2 marks: the B1 for \(\lambda\) and the final B1, which needs a reason. The show-that needs working, and so does \(x = 6\), as the question says "Using algebra".
1 mark
M1 Writes \(\dfrac{dx}{dt} = k(12 - 0.5x)(9 - 0.5x)\) for a constant \(k\) (rate proportional to the product of their remaining masses)
Using the initial masses 12 and 9 (not the remaining masses) is M0.
1 mark
A1* \(\dfrac{dx}{dt} = \dfrac{k}{4}(24 - x)(18 - x) = \lambda(24 - x)(18 - x)\), where \(\lambda = \dfrac{k}{4}\) cso
Given answer: needs the factor \(\dfrac{1}{2}\) taken out of each bracket (or equivalent algebra) and \(\lambda\) identified as a positive constant.
1 mark
B1 \(\lambda = 0.005\), from \(2.16 = \lambda \times 24 \times 18 = 432\lambda\)
Accept \(\dfrac{1}{200}\).
1 mark
M1 Sets \(\lambda(24 - x)(18 - x) = 1.08\) with their \(\lambda\) (or \((24 - x)(18 - x) = 216\)) and expands to a 3TQ: \(x^2 - 42x + 216 = 0\)
Half the initial rate is 1.08 grams per minute; \((24 - x)(18 - x) = \dfrac{432}{2}\) needs no value of \(\lambda\).
1 mark
M1 (dM1) Solves their 3TQ by factorising, the formula or completing the square: \((x - 6)(x - 36) = 0\)
Depends on the previous M mark.
1 mark
A1 \(x = 6\) only (6 grams of \(C\)), rejecting \(x = 36\) because \(x\) cannot exceed 18
Both roots found and 36 rejected with a reason, e.g. "there is not enough \(B\)", "\(x \leqslant 18\)" or "the masses remaining would be negative".
1 mark
B1 18 grams, because all of the 9 g of \(B\) has then been used (\(9 - 0.5x = 0\)), so the rate of formation becomes zero
Needs 18 g and a reason referring to \(B\) running out or the rate \(\lambda(24 - x)(18 - x)\) being zero at \(x = 18\). Accept "\(x\) approaches 18" oe. 24 g is B0 (that is when \(A\) would run out).

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