SPS SPS FM Pure (SPS FM Pure) 2023 October

1. A curve is described by the equation

$$x ^ { 2 } + 4 x y + y ^ { 2 } + 27 = 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). A point \(Q\) lies on the curve.
   The tangent to the curve at \(Q\) is parallel to the \(y\)-axis.
   Given that the \(x\) coordinate of \(Q\) is negative,
2. use your answer to part (a) to find the coordinates of \(Q\).
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2. Given that \(x \geqslant 2\), find the general solution of the differential equation $$( 2 x - 3 ) ( x - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x - 1 ) y$$ [BLANK PAGE]

3. Liquid is pouring into a large vertical circular cylinder at a constant rate of \(1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is \(4000 \mathrm {~cm} ^ { 2 }\).

1. Show that at time \(t\) seconds, the height \(h \mathrm {~cm}\) of liquid in the cylinder satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h , \text { where } k \text { is a positive constant. }$$ When \(h = 25\), water is leaking out of the hole at \(400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Show that \(k = 0.02\)
3. Separate the variables of the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by $$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h$$ Using the substitution \(h = ( 20 - x ) ^ { 2 }\), or otherwise,
4. find the exact value of \(\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h\).
5. Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.
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4. The cubic equation \(x ^ { 3 } + 3 x ^ { 2 } + 2 = 0\) has roots \(\alpha , \beta\) and \(\gamma\).

1. Use the substitution \(x = \frac { 1 } { \sqrt { u } }\) to show that \(4 u ^ { 3 } + 12 u ^ { 2 } + 9 u - 1 = 0\).
2. Hence find the values of \(\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } }\) and \(\frac { 1 } { \alpha ^ { 2 } \beta ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } \alpha ^ { 2 } }\).
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5. (i) Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \left\{ ( r + 1 ) ^ { 3 } - r ^ { 3 } \right\} = ( n + 1 ) ^ { 3 } - 1$$ (ii) Show that \(( r + 1 ) ^ { 3 } - r ^ { 3 } \equiv 3 r ^ { 2 } + 3 r + 1\).
(iii) Use the results in parts (i) and (ii) and the standard result for \(\sum _ { r = 1 } ^ { n } r\) to show that $$3 \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 2 } n ( n + 1 ) ( 2 n + 1 )$$ [BLANK PAGE]
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