9 Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20 ^ { \circ } \mathrm { C }\). At time \(t\) minutes later, the temperature of the liquid is \(\theta ^ { \circ } \mathrm { C }\).
- Explain how the information above leads to the differential equation
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 20 )$$
where \(k\) is a positive constant.
- The liquid is initially at a temperature of \(100 ^ { \circ } \mathrm { C }\). It takes 5 minutes for the liquid to cool from \(100 ^ { \circ } \mathrm { C }\) to \(68 ^ { \circ } \mathrm { C }\). Show that
$$\theta = 20 + 80 \mathrm { e } ^ { - \left( \frac { 1 } { 5 } \ln \frac { 5 } { 3 } \right) t }$$
- Calculate how much longer it takes for the liquid to cool by a further \(32 ^ { \circ } \mathrm { C }\).
1 Simplify \(\frac { x ^ { 3 } - 3 x ^ { 2 } } { x ^ { 2 } - 9 }\).
2 Given that \(\sin y = x y + x ^ { 2 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
3
- Find the quotient and the remainder when \(3 x ^ { 3 } - 2 x ^ { 2 } + x + 7\) is divided by \(x ^ { 2 } - 2 x + 5\).
- Hence, or otherwise, determine the values of the constants \(a\) and \(b\) such that, when \(3 x ^ { 3 } - 2 x ^ { 2 } + a x + b\) is divided by \(x ^ { 2 } - 2 x + 5\), there is no remainder.
4
- Use integration by parts to find \(\int x \sec ^ { 2 } x \mathrm {~d} x\).
- Hence find \(\int x \tan ^ { 2 } x \mathrm {~d} x\).
5 A curve is given parametrically by the equations \(x = t ^ { 2 } , y = 2 t\).
- Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), giving your answer in its simplest form.
- Show that the equation of the tangent to the curve at \(\left( p ^ { 2 } , 2 p \right)\) is
$$p y = x + p ^ { 2 } .$$
- Find the coordinates of the point where the tangent at \(( 9,6 )\) meets the tangent at \(( 25 , - 10 )\).
6
- Show that the substitution \(x = \sin ^ { 2 } \theta\) transforms \(\int \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\) to \(\int 2 \sin ^ { 2 } \theta \mathrm {~d} \theta\).
- Hence find \(\int _ { 0 } ^ { 1 } \sqrt { \frac { x } { 1 - x } } \mathrm {~d} x\).
7 The expression \(\frac { 11 + 8 x } { ( 2 - x ) ( 1 + x ) ^ { 2 } }\) is denoted by \(\mathrm { f } ( x )\).
- Express \(\mathrm { f } ( x )\) in the form \(\frac { A } { 2 - x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }\), where \(A , B\) and \(C\) are constants.
- Given that \(| x | < 1\), find the first 3 terms in the expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\).
8
- Solve the differential equation
$$\frac { d y } { d x } = \frac { 2 - x } { y - 3 }$$
giving the particular solution that satisfies the condition \(y = 4\) when \(x = 5\).
- Show that this particular solution can be expressed in the form
$$( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = k$$
where the values of the constants \(a , b\) and \(k\) are to be stated.
- Hence sketch the graph of the particular solution, indicating clearly its main features.
9 Two lines have vector equations
$$\mathbf { r } = \left( \begin{array} { r }
4
2
- 6
\end{array} \right) + t \left( \begin{array} { r }
- 8
1
- 2
\end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { r }
- 2
a
- 2
\end{array} \right) + s \left( \begin{array} { r }
- 9
2
- 5
\end{array} \right) ,$$
where \(a\) is a constant. - Calculate the acute angle between the lines.
- Given that these two lines intersect, find \(a\) and the point of intersection.
\section*{June 2006}
1 Find the gradient of the curve \(4 x ^ { 2 } + 2 x y + y ^ { 2 } = 12\) at the point \(( 1,2 )\).
2
- Expand \(( 1 - 3 x ) ^ { - 2 }\) in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\).
- Find the coefficient of \(x ^ { 2 }\) in the expansion of \(\frac { ( 1 + 2 x ) ^ { 2 } } { ( 1 - 3 x ) ^ { 2 } }\) in ascending powers of \(x\).
3
- Express \(\frac { 3 - 2 x } { x ( 3 - x ) }\) in partial fractions.
- Show that \(\int _ { 1 } ^ { 2 } \frac { 3 - 2 x } { x ( 3 - x ) } \mathrm { d } x = 0\).
- What does the result of part (ii) indicate about the graph of \(y = \frac { 3 - 2 x } { x ( 3 - x ) }\) between \(x = 1\) and \(x = 2\) ?
4 The position vectors of three points \(A , B\) and \(C\) relative to an origin \(O\) are given respectively by and
$$\begin{aligned}
& \overrightarrow { O A } = 7 \mathbf { i } + 3 \mathbf { j } - 3 \mathbf { k } ,
& \overrightarrow { O B } = 4 \mathbf { i } + 2 \mathbf { j } - 4 \mathbf { k }
& \overrightarrow { O C } = 5 \mathbf { i } + 4 \mathbf { j } - 5 \mathbf { k } .
\end{aligned}$$ - Find the angle between \(A B\) and \(A C\).
- Find the area of triangle \(A B C\).
5 A forest is burning so that, \(t\) hours after the start of the fire, the area burnt is \(A\) hectares. It is given that, at any instant, the rate at which this area is increasing is proportional to \(A ^ { 2 }\).
- Write down a differential equation which models this situation.
- After 1 hour, 1000 hectares have been burnt; after 2 hours, 2000 hectares have been burnt. Find after how many hours 3000 hectares have been burnt.
6
- Show that the substitution \(u = \mathrm { e } ^ { x } + 1\) transforms \(\int \frac { \mathrm { e } ^ { 2 x } } { \mathrm { e } ^ { x } + 1 } \mathrm {~d} x\) to \(\int \frac { u - 1 } { u } \mathrm {~d} u\).
- Hence show that \(\int _ { 0 } ^ { 1 } \frac { \mathrm { e } ^ { 2 x } } { \mathrm { e } ^ { x } + 1 } \mathrm {~d} x = \mathrm { e } - 1 - \ln \left( \frac { \mathrm { e } + 1 } { 2 } \right)\).
\section*{June 2006}
7 Two lines have vector equations
$$\mathbf { r } = \mathbf { i } - 2 \mathbf { j } + 4 \mathbf { k } + \lambda ( 3 \mathbf { i } + \mathbf { j } + a \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = - 8 \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \mu ( \mathbf { i } - 2 \mathbf { j } - \mathbf { k } ) ,$$
where \(a\) is a constant.
- Given that the lines are skew, find the value that \(a\) cannot take.
- Given instead that the lines intersect, find the point of intersection.
8
- Show that \(\int \cos ^ { 2 } 6 x \mathrm {~d} x = \frac { 1 } { 2 } x + \frac { 1 } { 24 } \sin 12 x + c\).
- Hence find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 12 } \pi } x \cos ^ { 2 } 6 x \mathrm {~d} x\).
9 A curve is given parametrically by the equations
$$x = 4 \cos t , \quad y = 3 \sin t$$
where \(0 \leqslant t \leqslant \frac { 1 } { 2 } \pi\).
- Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
- Show that the equation of the tangent at the point \(P\), where \(t = p\), is
$$3 x \cos p + 4 y \sin p = 12$$
- The tangent at \(P\) meets the \(x\)-axis at \(R\) and the \(y\)-axis at \(S\). \(O\) is the origin. Show that the area of triangle \(O R S\) is \(\frac { 12 } { \sin 2 p }\).
- Write down the least possible value of the area of triangle \(O R S\), and give the corresponding value of \(p\).
Jan 2007
1 It is given that
$$f ( x ) = \frac { x ^ { 2 } + 2 x - 24 } { x ^ { 2 } - 4 x } \quad \text { for } x \neq 0 , x \neq 4$$
Express \(\mathrm { f } ( x )\) in its simplest form.
2 Find the exact value of \(\int _ { 1 } ^ { 2 } x \ln x \mathrm {~d} x\).
3 The points \(A\) and \(B\) have position vectors \(\mathbf { a }\) and \(\mathbf { b }\) relative to an origin \(O\), where \(\mathbf { a } = 4 \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k }\) and \(\mathbf { b } = - 7 \mathbf { i } + 5 \mathbf { j } + 4 \mathbf { k }\). - Find the length of \(A B\).
- Use a scalar product to find angle \(O A B\).
4 Use the substitution \(u = 2 x - 5\) to show that \(\int _ { \frac { 5 } { 2 } } ^ { 3 } ( 4 x - 8 ) ( 2 x - 5 ) ^ { 7 } \mathrm {~d} x = \frac { 17 } { 72 }\).
- Expand \(( 1 - 3 x ) ^ { - \frac { 1 } { 3 } }\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
- Hence find the coefficient of \(x ^ { 3 }\) in the expansion of \(\left( 1 - 3 \left( x + x ^ { 3 } \right) \right) ^ { - \frac { 1 } { 3 } }\).
6
- Express \(\frac { 2 x + 1 } { ( x - 3 ) ^ { 2 } }\) in the form \(\frac { A } { x - 3 } + \frac { B } { ( x - 3 ) ^ { 2 } }\), where \(A\) and \(B\) are constants.
- Hence find the exact value of \(\int _ { 4 } ^ { 10 } \frac { 2 x + 1 } { ( x - 3 ) ^ { 2 } } \mathrm {~d} x\), giving your answer in the form \(a + b \ln c\), where \(a , b\) and \(c\) are integers.
7 The equation of a curve is \(2 x ^ { 2 } + x y + y ^ { 2 } = 14\). Show that there are two stationary points on the curve and find their coordinates.
8 The parametric equations of a curve are \(x = 2 t ^ { 2 } , y = 4 t\). Two points on the curve are \(P \left( 2 p ^ { 2 } , 4 p \right)\) and \(Q \left( 2 q ^ { 2 } , 4 q \right)\).
- Show that the gradient of the normal to the curve at \(P\) is \(- p\).
- Show that the gradient of the chord joining the points \(P\) and \(Q\) is \(\frac { 2 } { p + q }\).
- The chord \(P Q\) is the normal to the curve at \(P\). Show that \(p ^ { 2 } + p q + 2 = 0\).
- The normal at the point \(R ( 8,8 )\) meets the curve again at \(S\). The normal at \(S\) meets the curve again at \(T\). Find the coordinates of \(T\).
9
- Find the general solution of the differential equation
$$\frac { \sec ^ { 2 } y } { \cos ^ { 2 } ( 2 x ) } \frac { d y } { d x } = 2$$
- For the particular solution in which \(y = \frac { 1 } { 4 } \pi\) when \(x = 0\), find the value of \(y\) when \(x = \frac { 1 } { 6 } \pi\).