FP2 June 2009 Q8

EdexcelOld spec15 marksSecond Order Differentials

8. \[\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 5\frac{\mathrm{d}x}{\mathrm{d}t} + 6x = 2\mathrm{e}^{-t}\]

Given that \(x = 0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\) at \(t = 0\),

(a) find \(x\) in terms of \(t\). (8)

The solution to part (a) is used to represent the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds, where \(t > 0\), \(P\) is \(x\) metres from the origin \(O\).

(b) Show that the maximum distance between \(O\) and \(P\) is \(\dfrac{2\sqrt{3}}{9}\) m and justify that this distance is a maximum. (7)

Mark scheme (a)

Scheme	Marks
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 5\dfrac{\mathrm{d}x}{\mathrm{d}t} + 6x = 2\mathrm{e}^{-t},\quad x = 0,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2\) at \(t = 0\).
AE, \(m^2 + 5m + 6 = 0 \Rightarrow (m + 3)(m + 2) = 0\) \(\Rightarrow m = -3, -2.\)
So, \(x_{\text{CF}} = A\mathrm{e}^{-3t} + B\mathrm{e}^{-2t}\) \(A\mathrm{e}^{m_1 t} + B\mathrm{e}^{m_2 t}\), where \(m_1 \neq m_2\). \(A\mathrm{e}^{-3t} + B\mathrm{e}^{-2t}\)	M1 A1
\(\left\{x = k\mathrm{e}^{-t} \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = -k\mathrm{e}^{-t} \Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = k\mathrm{e}^{-t}\right\}\)
\(\Rightarrow k\mathrm{e}^{-t} + 5(-k\mathrm{e}^{-t}) + 6k\mathrm{e}^{-t} = 2\mathrm{e}^{-t} \Rightarrow 2k\mathrm{e}^{-t} = 2\mathrm{e}^{-t}\) \(\Rightarrow k = 1\) Substitutes \(k\mathrm{e}^{-t}\) into the differential equation given in the question. Finds \(k = 1\).	M1 A1
\(\left\{\text{So, } x_{\text{PI}} = \mathrm{e}^{-t}\right\}\)
So, \(x = A\mathrm{e}^{-3t} + B\mathrm{e}^{-2t} + \mathrm{e}^{-t}\) their \(x_{\text{CF}}\) + their \(x_{\text{PI}}\)	M1*
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -3A\mathrm{e}^{-3t} - 2B\mathrm{e}^{-2t} - \mathrm{e}^{-t}\) Finds \(\frac{\mathrm{d}x}{\mathrm{d}t}\) by differentiating their \(x_{\text{CF}}\) and their \(x_{\text{PI}}\)	dM1*
\(t = 0,\ x = 0 \Rightarrow \quad 0 = A + B + 1\) \(t = 0,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2 \Rightarrow \quad 2 = -3A - 2B - 1\) Applies \(t = 0,\ x = 0\) to \(x\) and \(t = 0,\ \frac{\mathrm{d}x}{\mathrm{d}t} = 2\) to \(\frac{\mathrm{d}x}{\mathrm{d}t}\) to form simultaneous equations.	ddM1*
\(\left\{\begin{array}{l}2A + 2B = -2 \\ -3A - 2B = 3\end{array}\right\}\) \(\Rightarrow A = -1,\ B = 0\)
So, \(x = -\mathrm{e}^{-3t} + \mathrm{e}^{-t}\) \(x = -\mathrm{e}^{-3t} + \mathrm{e}^{-t}\)	A1 cao
	(8)

Mark scheme (b)

Scheme	Marks
\(x = -\mathrm{e}^{-3t} + \mathrm{e}^{-t}\)
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3\mathrm{e}^{-3t} - \mathrm{e}^{-t} = 0\) Differentiates their \(x\) to give \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and puts \(\frac{\mathrm{d}x}{\mathrm{d}t}\) equal to 0.	M1
\(3 - \mathrm{e}^{2t} = 0\) \(\Rightarrow t = \frac{1}{2}\ln 3\) A credible attempt to solve. \(t = \frac{1}{2}\ln 3\) or \(t = \ln\sqrt{3}\) or awrt 0.55	dM1* A1
So, \(x = -\mathrm{e}^{-\frac{3}{2}\ln 3} + \mathrm{e}^{-\frac{1}{2}\ln 3} = -\mathrm{e}^{\ln 3^{-\frac{3}{2}}} + \mathrm{e}^{\ln 3^{-\frac{1}{2}}}\) \(x = -3^{-\frac{3}{2}} + 3^{-\frac{1}{2}}\) Substitutes their \(t\) back into \(x\) and an attempt to eliminate out the ln’s.	ddM1
\(= -\dfrac{1}{3\sqrt{3}} + \dfrac{1}{\sqrt{3}} = \dfrac{2}{3\sqrt{3}} = \underline{\dfrac{2\sqrt{3}}{9}}\) uses exact values to give \(\frac{2\sqrt{3}}{9}\)	A1 AG
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -9\mathrm{e}^{-3t} + \mathrm{e}^{-t}\) At \(t = \frac{1}{2}\ln 3\), \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -9\mathrm{e}^{-\frac{3}{2}\ln 3} + \mathrm{e}^{-\frac{1}{2}\ln 3}\) Finds \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\) and substitutes their \(t\) into \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\)	dM1*
\(= -9(3)^{-\frac{3}{2}} + 3^{-\frac{1}{2}} = -\dfrac{9}{3\sqrt{3}} + \dfrac{1}{\sqrt{3}} = -\dfrac{3}{\sqrt{3}} + \dfrac{1}{\sqrt{3}}\)
As \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\dfrac{9}{3\sqrt{3}} + \dfrac{1}{\sqrt{3}} = \left\{-\dfrac{2}{\sqrt{3}}\right\} < 0\) then \(x\) is maximum. \(-\frac{9}{3\sqrt{3}} + \frac{1}{\sqrt{3}} < 0\) and maximum conclusion.	A1
	(7)
	(15 marks)