$$\begin{gathered} \left(i\right) If \mathcal{Y}=\ln \left(\sin \mathcal{x}\right), find \frac{d\mathcal{Y}}{d\mathcal{x}} \\ \left(ii\right) Hence or otherwise find {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}cot\mathcal{x}d\mathcal{x} \\ Fnd {\int }_0^{\frac{\mathrm{\pi }}{12}}{\cos }^2\left(3\mathcal{x}\right)d\mathcal{x} \end{gathered}$$

$$\begin{gathered} The velocity of a particle is given by the equation v=\frac{1}{e^x} \\ If the initial displacement is x=0, find the equation for the displacement x, \\ in terms of t. \end{gathered}$$

$$\begin{gathered} A large industrial container is in the shape of a paraboloid, which is formed by rotating part \\ of the parabola y=\frac{1}{4}{x}^2 around the y-axis, as shown in the diagram. \\ Liquid is poured into the container at the rate of 2m^3 per minute. \end{gathered}$$

$$\begin{gathered} \left(i\right) Show that the volume V m^3 of liquid in the container when the depth of liquid is \\ h metres, is given by V= 2{\mathrm{\pi h}}^2. \\ \left(\mathrm{ii}\right) \mathrm{At} \mathrm{what} \mathrm{rate} \mathrm{is} \mathrm{the} \mathrm{height} \left(\mathrm{h}\right) \mathrm{of} \mathrm{the} \mathrm{liquid} \mathrm{rising} \mathrm{when} \mathrm{the} \mathrm{depth} \mathrm{is} 0.5 \mathrm{metres}? \\ \left(\mathrm{iii}\right) \mathrm{If} \mathrm{the} \mathrm{container} \mathrm{is} 3 \mathrm{metres} \mathrm{high}, \mathrm{how} \mathrm{long} \mathrm{will} \mathrm{it} \mathrm{take} \mathrm{to} \mathrm{fill} \mathrm{the} \mathrm{container}? \end{gathered}$$

$$\begin{gathered} \mathcal{y}=\ln \left(\sin \mathcal{x}\right), find \frac{d\mathcal{y}}{d\mathcal{x}} \\ hence , find {\int }_{\frac{\mathrm{\pi }}{ 4}}^{\frac{3\mathrm{\pi }}{4}}cot\mathcal{x} d\mathcal{x} \\ Find {\int }_0^{\frac{\mathrm{\pi }}{12}}{\cos }^2\left(3\mathcal{x}\right)d\mathcal{x} \end{gathered}$$

$$\begin{gathered} It is known that if \frac{d\mathcal{Y}}{d\mathcal{x}}=\mathcal{Y}, then \mathcal{Y}=e^{\mathcal{x}} is sa solution (s\mathrm{in}ce \frac{d\mathcal{Y}}{d\mathcal{x}}=e^{\mathcal{x}}=\mathcal{Y}) \\ If \frac{d\mathcal{Y}}{d\mathcal{x}}=\frac{1}{\mathcal{Y}} , find a solution for \mathcal{Y}. \end{gathered}$$

$$\begin{gathered} A particle move in a straight line and its position in meters at any time t \\ seconds is given by the equation : \mathcal{x}=5\cos \left(2t\right)-12\sin \left(2t\right) \\ Sow by differentiation that the motion is simple harmonic. \\ State the period of the motion \\ Express \mathcal{x} in the form R\cos (2t+\alpha ) \\ State the amplitude of the motion. \\ Find the general solution for all the times when the \\ particle is at the centre of the motion. \end{gathered}$$

$$\begin{gathered} \left(i\right) Sketch the graph of y=x^2-2x-4 \\ \left(ii\right) Hence sketch the graph of y=\frac{1}{\left|x^2-2x-4\right|} \end{gathered}$$

$$\begin{gathered} Find the co-ordinates of the point P which divides AB \\ externally in the ratio 2:3 where A is (1,-4) and B(6,9). \end{gathered}$$

$$\begin{gathered} The lenght of each edge of a cube is \mathcal{x} cm \\ Write expression for the surface area\left(A\right) and volume of \\ the cube and hence find\frac{dA}{d\mathcal{x}}and \frac{dV}{d\mathcal{x}} \\ The surface area of the cube increase ar a rate of 6c{m}^2/second. \\ Find the rate of change of the volume of the cube when the lenght of \\ each edge is 5 cm. \end{gathered}$$

$$\begin{gathered} A spherical bubble is expanding so that its volume is increa\sin g at the cons\tan t rate of 12m{m}^2 per second. \\ What is the rate of increase of the radius when the surface area is 500m{m}^2 \end{gathered}$$