A2 June 2023 Paper 1 Q7

OCR ACurrent spec11 marksDifferentiation & Maclaurin Second Order Differentials

7 An engineer is modelling the motion of a particle \(P\) of mass 0.5 kg in a wind tunnel.

\(P\) is modelled as travelling in a straight line. The point \(O\) is a fixed point within the wind tunnel. The displacement of \(P\) from \(O\) at time \(t\) seconds is \(x\) metres, for \(t \geqslant 0\).

You are given that \(x \geqslant 0\) for all \(t \geqslant 0\) and that \(P\) does not reach the end of the wind tunnel.

If \(t \geqslant 0\), then \(P\) is subject to three forces which are modelled in the following way.

The first force has a magnitude of \(5(t + 1)\cosh t\) N and acts in the positive \(x\)-direction.
The second force has a magnitude of \(0.5x\) N and acts towards \(O\).
The third force has a magnitude of \(\left|\dfrac{\mathrm{d}x}{\mathrm{d}t}\right|\) N and acts in the direction of motion of the particle.

(a) The engineer applies the equation “\(F = ma\)” to the model of the motion of \(P\) and derives the following differential equation.\[5(t + 1)\cosh t - 0.5x + \dfrac{\mathrm{d}x}{\mathrm{d}t} = 0.5\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\]

(i) Explain the sign of the \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) term in the engineer’s differential equation. [1]

When \(t = 0\) the displacement of \(P\) is 6 m, and it is travelling towards \(O\) with a speed of \(5\,\mathrm{m\,s^{-1}}\).

(ii) Without attempting to solve the differential equation, find the acceleration of \(P\) when \(t = 0\). [2]

Let the particular solution to the differential equation in part (a) be a function f such that \(x = \mathrm{f}(t)\) for \(t \geqslant 0\).

The particular solution to the differential equation can be expressed as a Maclaurin series.

(b)

(i) Show that the Maclaurin series for \(\mathrm{f}(t)\) up to and including the term in \(t\) is \(6 - 5t\). [1]

(ii) Use your answer to part (a)(ii) to show that the term in \(t^2\) in the Maclaurin series for \(\mathrm{f}(t)\) is \(-3t^2\). [1]

(iii) By differentiating the differential equation in part (a) with respect to \(t\), show that the term in \(t^3\) in the Maclaurin series for \(\mathrm{f}(t)\) is \(0.5t^3\). [4]

You are given that the complete Maclaurin series for the function f is valid for all values of \(t \geqslant 0\).

After 0.25 seconds \(P\) has travelled 1.43 m towards the origin.

(c)

(i) By using the Maclaurin series for \(\mathrm{f}(t)\) up to and including the term in \(t^3\), evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 0.25\). [1]

(ii) Explain why it might not be sensible to use the Maclaurin series for \(\mathrm{f}(t)\) up to and including the term in \(t^3\) to evaluate the suitability of the model for determining the displacement of \(P\) from \(O\) when \(t = 10\). [1]

Mark scheme (a)

Scheme	Marks
(i) e.g. sign of \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) is +ve because the force is in the same direction of motion	B1
	[1]
(ii) \(\dfrac{1}{2}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = 5\cosh 0 - 3 - 5\) \(= -3\)	M1
\(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -6\ (\mathrm{m\,s^{-2}})\)	A1
	[2]

Notes

(a)(i)

B1: Convincingly shown

(corrected from the printed mark scheme: the printed answer says “sign of \(\frac{\mathrm{d}y}{\mathrm{d}t}\)”; the term in the equation is \(\frac{\mathrm{d}x}{\mathrm{d}t}\))

(a)(ii)

M1: Substitutes \(t = 0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \pm 5\)

A1: Or 6 towards \(O\)

Mark scheme (b)

Scheme	Marks
(i) Maclaurin: \(x = \mathrm{f}(0) + \mathrm{f}^{\prime}(0)t + \ldots\) When \(t = 0\), \((x = 6 =)\ \mathrm{f}(0) = 6\), \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}t} =\right) = \mathrm{f}^{\prime}(0) = -5\) \(\Rightarrow x = 6 - 5t \ldots\)	B1
	[1]
(ii) 3^rd term of Maclaurin is \(\mathrm{f}^{\prime\prime}(0)\dfrac{t^2}{2!}\) With (from (a)(ii)) \(\mathrm{f}^{\prime\prime}(0) = -6\) So 3^rd term is \((-6)\dfrac{t^2}{2!} = -3t^2\)	B1
	[1]
(iii) \(\Rightarrow 5\cosh t + 5(t + 1)\sinh t - 0.5\dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = \dfrac{1}{2}\dfrac{\mathrm{d}^3x}{\mathrm{d}t^3}\)	M1 A1
When \(t = 0\), \(\quad 5 + 0 + 2.5 - 6 = \dfrac{1}{2}\mathrm{f}^{\prime\prime\prime}(0)\) \(\Rightarrow \mathrm{f}^{\prime\prime\prime}(0) = 3\)	M1
\(\Rightarrow \text{4th term} = \mathrm{f}^{\prime\prime\prime}(0)\dfrac{t^3}{3!} = \dfrac{1}{2}t^3\)	A1
	[4]

Notes

(b)(i)

B1: First two terms of Maclaurin’s formula, possibly in generalised form, must be seen
AG Convincingly shown.

(b)(ii)

B1: AG Convincingly shown

(b)(iii)

M1: For \(\frac{\mathrm{d}}{\mathrm{d}t}(\cosh t) = \sinh t\) and product rule attempted

A1: Fully correct differentiation

M1: Substitute \(t = 0\)

A1: AG. Allow embedded answer.

Mark scheme (c)

Scheme	Marks
(i) \(x = 6 - 5t - 3t^2 + \dfrac{1}{2}t^3\) \(\Rightarrow\) When \(t = 0.25\), \(x = 6 - 1.25 - 0.1875 - 0.0078 \approx 4.570\ldots\) \(\Rightarrow\) Distance travelled \(6 - 4.570 \approx 1.430\ldots\) So suitable as value close	B1
	[1]
(ii) e.g. more terms may be required Higher terms may be large The candidate calculates the term in \(t^4\) \((11t^4/12)\) and indicates that this term is large for values of \(t \gt 1\).	B1
	[1]

Notes

(c)(i)

B1: Substitutes \(t = 0.25\) to obtain an approximation and correct conclusion

(c)(ii)

B1: Allow any correct explanation that explains/implies that for \(t \gt 1\) some of the higher power terms are large and so non-negligible.
\(\mathrm{f}(10) = 156\) is too large is not enough