7
2
- 5
\end{array} \right)$$
Given that
$$\overrightarrow { A B } = \left( \begin{array} { r }
- 2
4
3
\end{array} \right)$$
- find the coordinates of the point \(B\).
The point \(C\) has position vector
$$\overrightarrow { O C } = \left( \begin{array} { r }
a
5
- 1
\end{array} \right)$$
where \(a\) is a constant.
Given that \(\overrightarrow { O C }\) is perpendicular to \(\overrightarrow { B C }\) - find the possible values of \(a\).
- The curve \(C\) is defined by the equation
$$8 x ^ { 3 } - 3 y ^ { 2 } + 2 x y = 9$$
Find an equation of the normal to \(C\) at the point ( 2,5 ), giving your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers.
4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e583bf92-d6a9-4f1a-b3c8-372afa6e0a0e-08_558_542_258_749}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of a segment \(P Q R P\) of a circle with centre \(O\) and radius 5 cm .
Given that
- angle \(P O R\) is \(\theta\) radians
- \(\theta\) is increasing, from 0 to \(\pi\), at a constant rate of 0.1 radians per second
- the area of the segment \(P Q R P\) is \(A \mathrm {~cm} ^ { 2 }\)
- show that
$$\frac { \mathrm { d } A } { \mathrm {~d} \theta } = K ( 1 - \cos \theta )$$
where \(K\) is a constant to be found. - Find, in \(\mathrm { cm } ^ { 2 } \mathrm {~s} ^ { - 1 }\), the rate of increase of the area of the segment when \(\theta = \frac { \pi } { 3 }\)
5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e583bf92-d6a9-4f1a-b3c8-372afa6e0a0e-10_803_1086_248_493}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 2 shows a sketch of the curve defined by the parametric equations
$$x = t ^ { 2 } + 2 t \quad y = \frac { 2 } { t ( 3 - t ) } \quad a \leqslant t \leqslant b$$
where \(a\) and \(b\) are constants.
The ends of the curve lie on the line with equation \(y = 1\) - Find the value of \(a\) and the value of \(b\)
The region \(R\), shown shaded in Figure 2, is bounded by the curve and the line with equation \(y = 1\)
- Show that the area of region \(R\) is given by
$$M - k \int _ { a } ^ { b } \frac { t + 1 } { t ( 3 - t ) } \mathrm { d } t$$
where \(M\) and \(k\) are constants to be found.
- Write \(\frac { t + 1 } { t ( 3 - t ) }\) in partial fractions.
- Use algebraic integration to find the exact area of \(R\), giving your answer in simplest form.
- With respect to a fixed origin \(O\), the line \(l _ { 1 }\) is given by the equation
$$\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k } + \lambda ( 8 \mathbf { i } - \mathbf { j } + 4 \mathbf { k } )$$
where \(\lambda\) is a scalar parameter.
The point \(A\) lies on \(l _ { 1 }\)
Given that \(| \overrightarrow { O A } | = 5 \sqrt { 10 }\)- show that at \(A\) the parameter \(\lambda\) satisfies
$$81 \lambda ^ { 2 } + 52 \lambda - 220 = 0$$
Hence
- show that one possible position vector for \(A\) is \(- 15 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }\)
- find the other possible position vector for \(A\).
The line \(l _ { 2 }\) is parallel to \(l _ { 1 }\) and passes through \(O\).
Given that
- \(\overrightarrow { O A } = - 15 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }\)
- point \(B\) lies on \(l _ { 2 }\) where \(| \overrightarrow { O B } | = 4 \sqrt { 10 }\)
- find the area of triangle \(O A B\), giving your answer to one decimal place.
- The current, \(x\) amps, at time \(t\) seconds after a switch is closed in a particular electric circuit is modelled by the equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k - 3 x$$
where \(k\) is a constant.
Initially there is zero current in the circuit.- Solve the differential equation to find an equation, in terms of \(k\), for the current in the circuit at time \(t\) seconds.
Give your answer in the form \(x = \mathrm { f } ( t )\).
Given that in the long term the current in the circuit approaches 7 amps, - find the value of \(k\).
- Hence find the time in seconds it takes for the current to reach 5 amps, giving your answer to 2 significant figures.