17
\end{array} \right) + \lambda \left( \begin{array} { c }
- 2
1
- 4
\end{array} \right) \quad l _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { c }
- 5
11
p
\end{array} \right) + \mu \left( \begin{array} { l }
q
2
2
\end{array} \right)$$
where \(\lambda\) and \(\mu\) are parameters and \(p\) and \(q\) are constants. Given that \(l _ { 1 }\) and \(l _ { 2 }\) are perpendicular,
- show that \(q = - 3\).
Given further that \(l _ { 1 }\) and \(l _ { 2 }\) intersect, find
- the value of \(p\),
- the coordinates of the point of intersection.
The point \(A\) lies on \(l _ { 1 }\) and has position vector \(\left( \begin{array} { c } 9
3
13 \end{array} \right)\). The point \(C\) lies on \(l _ { 2 }\).
Given that a circle, with centre \(C\), cuts the line \(l _ { 1 }\) at the points \(A\) and \(B\), - find the position vector of \(B\).
5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-09_696_686_196_626}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm , as shown in Figure 2. Water is flowing into the container. When the height of water is \(h \mathrm {~cm}\), the surface of the water has radius \(r \mathrm {~cm}\) and the volume of water is \(V \mathrm {~cm} ^ { 3 }\). - Show that \(V = \frac { 4 \pi h ^ { 3 } } { 27 }\).
[0pt]
[The volume \(V\) of a right circular cone with vertical height \(h\) and base radius \(r\) is given by the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]
Water flows into the container at a rate of \(8 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). - Find, in terms of \(\pi\), the rate of change of \(h\) when \(h = 12\).
6. (a) Find \(\int \tan ^ { 2 } x \mathrm {~d} x\).
- Use integration by parts to find \(\int \frac { 1 } { x ^ { 3 } } \ln x \mathrm {~d} x\).
- Use the substitution \(u = 1 + e ^ { x }\) to show that
$$\int \frac { \mathrm { e } ^ { 3 x } } { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - \mathrm { e } ^ { x } + \ln \left( 1 + \mathrm { e } ^ { x } \right) + k$$
where \(k\) is a constant.
7.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-13_511_714_237_612}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{figure}
The curve \(C\) shown in Figure 3 has parametric equations
$$x = t ^ { 3 } - 8 t , \quad y = t ^ { 2 }$$
where \(t\) is a parameter. Given that the point \(A\) has parameter \(t = - 1\), - find the coordinates of \(A\).
The line \(l\) is the tangent to \(C\) at \(A\).
- Show that an equation for \(l\) is \(2 x - 5 y - 9 = 0\).
The line \(l\) also intersects the curve at the point \(B\).
- Find the coordinates of \(B\).