Multi-part: volume and related rates

10

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4782d612-0ec1-418e-8ef3-c871dce82b44-16_451_442_269_888} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} Fig. 1 shows part of the curve \(y = x ^ { 2 } - 1\) and the line \(y = h\), where \(h\) is a constant.
   1. The shaded region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis. Show that the volume of revolution, \(V\), is given by \(V = \pi \left( \frac { 1 } { 2 } h ^ { 2 } + h \right)\).
   2. Find, showing all necessary working, the area of the shaded region when \(h = 3\).
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4782d612-0ec1-418e-8ef3-c871dce82b44-17_257_408_1126_904} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} Fig. 2 shows a cross-section of a bowl containing water. When the height of the water level is \(h \mathrm {~cm}\), the volume, \(V \mathrm {~cm} ^ { 3 }\), of water is given by \(V = \pi \left( \frac { 1 } { 2 } h ^ { 2 } + h \right)\). Water is poured into the bowl at a constant rate of \(2 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), at which the height of the water level is increasing when the height of the water level is 3 cm .

3 \includegraphics[max width=\textwidth, alt={}, center]{f759ce41-708e-4fe7-80b9-adc2be2972ac-04_489_465_258_840} The diagram shows a water container in the form of an inverted pyramid, which is such that when the height of the water level is \(h \mathrm {~cm}\) the surface of the water is a square of side \(\frac { 1 } { 2 } h \mathrm {~cm}\).

1. Express the volume of water in the container in terms of \(h\).
   [0pt] [The volume of a pyramid having a base area \(A\) and vertical height \(h\) is \(\frac { 1 } { 3 } A h\).]
   Water is steadily dripping into the container at a constant rate of \(20 \mathrm {~cm} ^ { 3 }\) per minute.
2. Find the rate, in cm per minute, at which the water level is rising when the height of the water level is 10 cm . \includegraphics[max width=\textwidth, alt={}, center]{f759ce41-708e-4fe7-80b9-adc2be2972ac-06_403_773_258_685} In the diagram, \(A B = A C = 8 \mathrm {~cm}\) and angle \(C A B = \frac { 2 } { 7 } \pi\) radians. The circular \(\operatorname { arc } B C\) has centre \(A\), the circular arc \(C D\) has centre \(B\) and \(A B D\) is a straight line.

8 The equation of a curve is \(y = \frac { 6 } { 5 - 2 x }\).

1. Calculate the gradient of the curve at the point where \(x = 1\).
2. A point with coordinates \(( x , y )\) moves along the curve in such a way that the rate of increase of \(y\) has a constant value of 0.02 units per second. Find the rate of increase of \(x\) when \(x = 1\).
3. The region between the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 1\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Show that the volume obtained is \(\frac { 12 } { 5 } \pi\).

9 \includegraphics[max width=\textwidth, alt={}, center]{d5f66324-e1fc-40e1-98e7-625187e24d3d-4_584_670_881_740} The diagram shows part of the curve \(y = \frac { 8 } { x } + 2 x\) and three points \(A , B\) and \(C\) on the curve with \(x\)-coordinates 1, 2 and 5 respectively.

1. A point \(P\) moves along the curve in such a way that its \(x\)-coordinate increases at a constant rate of 0.04 units per second. Find the rate at which the \(y\)-coordinate of \(P\) is changing as \(P\) passes through \(A\).
2. Find the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

9 \includegraphics[max width=\textwidth, alt={}, center]{ebfdf170-99c6-4785-b9d7-201c3425b4c9-4_556_720_676_715} The diagram shows the curve with equation \(y = 2 \ln ( x - 1 )\). The point \(P\) has coordinates ( \(0 , p\) ). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 0 , y = 0\) and \(y = p\). The units on the axes are centimetres. The region \(R\) is rotated completely about the \(\boldsymbol { y }\)-axis to form a solid.

1. Show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the solid is given by $$V = \pi \left( \mathrm { e } ^ { p } + 4 \mathrm { e } ^ { \frac { 1 } { 2 } p } + p - 5 \right) .$$
2. It is given that the point \(P\) is moving in the positive direction along the \(y\)-axis at a constant rate of \(0.2 \mathrm {~cm} \mathrm {~min} ^ { - 1 }\). Find the rate at which the volume of the solid is increasing at the instant when \(p = 4\), giving your answer correct to 2 significant figures.

8 \includegraphics[max width=\textwidth, alt={}, center]{c940af95-e291-402a-856c-9090d13163d5-4_538_702_264_719} The diagram shows the curve with equation $$y = \frac { 6 } { \sqrt { x } } - 3$$ The point \(P\) has coordinates \(( 0 , p )\). The shaded region is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = p\). The shaded region is rotated completely about the \(y\)-axis to form a solid of volume \(V\).

1. Show that \(V = 16 \pi \left( 1 - \frac { 27 } { ( p + 3 ) ^ { 3 } } \right)\).
2. It is given that \(P\) is moving along the \(y\)-axis in such a way that, at time \(t\), the variables \(p\) and \(t\) are related by $$\frac { \mathrm { d } p } { \mathrm {~d} t } = \frac { 1 } { 3 } p + 1 .$$ Find the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\) at the instant when \(p = 9\).