8
5
\end{array} \right) + \mu \left( \begin{array} { r }
5
- 4
- 3
\end{array} \right)$$
where \(\mu\) is a scalar parameter.
The point \(A\) is on \(l _ { 1 }\) where \(\mu = 2\).
- Write down the coordinates of \(A\).
The acute angle between \(O A\) and \(l _ { 1 }\) is \(\theta\), where \(O\) is the origin.
- Find the value of \(\cos \theta\).
The point \(B\) is such that \(\overrightarrow { O B } = 3 \overrightarrow { O A }\).
The line \(l _ { 2 }\) passes through the point \(B\) and is parallel to the line \(l _ { 1 }\). - Find a vector equation of \(l _ { 2 }\).
- Find the length of \(O B\), giving your answer as a simplified surd.
The point \(X\) lies on \(l _ { 2 }\). Given that the vector \(\overrightarrow { O X }\) is perpendicular to \(l _ { 2 }\),
- find the length of \(O X\), giving your answer to 3 significant figures.
5. The curve \(C\) has the equation
$$\sin ( \pi y ) - y - x ^ { 2 } y = - 5 , \quad x > 0$$ - Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
The point \(P\) with coordinates \(( 2,1 )\) lies on \(C\).
The tangent to \(C\) at \(P\) meets the \(x\)-axis at the point \(A\). - Find the exact value of the \(x\)-coordinate of \(A\).
6. (i) (a) Express \(\frac { 7 x } { ( x + 3 ) ( 2 x - 1 ) }\) in partial fractions. - Given that \(x > \frac { 1 } { 2 }\), find
$$\int \frac { 7 x } { ( x + 3 ) ( 2 x - 1 ) } d x$$
(ii) Using the substitution \(u ^ { 3 } = x\), or otherwise, find
$$\int \frac { 1 } { x + x ^ { \frac { 1 } { 3 } } } d x , \quad x > 0$$
7.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b0d21f0f-f5f6-4ca5-8e3e-98aee0d9db7a-11_703_1164_373_492}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 2 shows a sketch of part of the curve \(C\) with parametric equations
$$x = \tan \theta , \quad y = 1 + 2 \cos 2 \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
The curve \(C\) crosses the \(x\)-axis at \(( \sqrt { } 3,0 )\). The finite shaded region \(S\) shown in Figure 2 is bounded by \(C\), the line \(x = 1\) and the \(x\)-axis. This shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution. - Show that the volume of the solid of revolution formed is given by the integral
$$k \int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \left( 16 \cos ^ { 2 } \theta - 8 + \sec ^ { 2 } \theta \right) d \theta$$
where \(k\) is a constant.
- Hence, use integration to find the exact value for this volume.
8.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b0d21f0f-f5f6-4ca5-8e3e-98aee0d9db7a-13_869_545_312_811}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{figure}
Figure 3 shows a large vertical cylindrical tank containing a liquid. The radius of the circular cross-section of the tank is 40 cm . At time \(t\) minutes, the depth of liquid in the tank is \(h\) centimetres. The liquid leaks from a hole \(P\) at the bottom of the tank.
The liquid leaks from the tank at a rate of \(32 \pi \sqrt { } h \mathrm {~cm} ^ { 3 } \mathrm {~min} ^ { - 1 }\). - Show that at time \(t\) minutes, the height \(h \mathrm {~cm}\) of liquid in the tank satisfies the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = - 0.02 \sqrt { } h$$
- Find the time taken, to the nearest minute, for the depth of liquid in the tank to decrease from 100 cm to 50 cm .
\includegraphics[max width=\textwidth, alt={}]{b0d21f0f-f5f6-4ca5-8e3e-98aee0d9db7a-14_2639_1834_214_217}