AQA FP3 (Further Pure Mathematics 3) 2009 June

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { x ^ { 2 } + y + 1 }$$ and $$y ( 3 ) = 2$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), giving your answer to four decimal places.
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 3.2 )\), giving your answer to three decimal places.

2 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - y \tan x = 2 \sin x$$ given that \(y = 2\) when \(x = 0\).
(9 marks)

3 The diagram shows a sketch of a circle which passes through the origin \(O\).
\includegraphics[max width=\textwidth, alt={}, center]{13cfb9ca-9495-4b69-80c5-9fb7e8e0f957-3_423_451_356_794} The equation of the circle is \(( x - 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 25\) and \(O A\) is a diameter.

1. Find the cartesian coordinates of the point \(A\).
2. Using \(O\) as the pole and the positive \(x\)-axis as the initial line, the polar coordinates of \(A\) are \(( k , \alpha )\).
   1. Find the value of \(k\) and the value of \(\tan \alpha\).
   2. Find the polar equation of the circle \(( x - 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 25\), giving your answer in the form \(r = p \cos \theta + q \sin \theta\).

4 Evaluate the improper integral $$\int _ { 1 } ^ { \infty } \left( \frac { 1 } { x } - \frac { 4 } { 4 x + 1 } \right) \mathrm { d } x$$ showing the limiting process used and giving your answer in the form \(\ln k\), where \(k\) is a constant to be found.

5 It is given that \(y\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 8 \sin x + 4 \cos x$$

1. Find the value of the constant \(k\) for which \(y = k \sin x\) is a particular integral of the given differential equation.
2. Solve the differential equation, expressing \(y\) in terms of \(x\), given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4\) when \(x = 0\).
   (8 marks)

6 The function f is defined by $$\mathrm { f } ( x ) = ( 9 + \tan x ) ^ { \frac { 1 } { 2 } }$$

1. 1. Find \(f ^ { \prime \prime } ( x )\).
   2. By using Maclaurin's theorem, show that, for small values of \(x\), $$( 9 + \tan x ) ^ { \frac { 1 } { 2 } } \approx 3 + \frac { x } { 6 } - \frac { x ^ { 2 } } { 216 }$$
2. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { f ( x ) - 3 } { \sin 3 x } \right]$$

7 The diagram shows the curve \(C _ { 1 }\) with polar equation $$r = 1 + 6 \mathrm { e } ^ { - \frac { \theta } { \pi } } , \quad 0 \leqslant \theta \leqslant 2 \pi$$ \includegraphics[max width=\textwidth, alt={}, center]{13cfb9ca-9495-4b69-80c5-9fb7e8e0f957-4_300_513_1414_760}

1. Find, in terms of \(\pi\) and e , the area of the shaded region bounded by \(C _ { 1 }\) and the initial line.
2. The polar equation of a curve \(C _ { 2 }\) is $$r = \mathrm { e } ^ { \frac { \theta } { \pi } } , \quad 0 \leqslant \theta \leqslant 2 \pi$$ Sketch the curve \(C _ { 2 }\) and state the polar coordinates of the end-points of this curve.
3. The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the point \(P\). Find the polar coordinates of \(P\).

8

1. Given that \(x = t ^ { 2 }\), where \(t \geqslant 0\), and that \(y\) is a function of \(x\), show that:
   1. \(2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the substitution \(x = t ^ { 2 }\), where \(t \geqslant 0\), transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 3 y = 0$$ (2 marks)
3. Hence find the general solution of the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ giving your answer in the form \(y = \mathrm { g } ( x )\).