Multi-part: volume and related rates

CAIE P1 2017 March Q3

5 marks Standard +0.3

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 $\frac { 1 } { 2 } h \mathrm {~cm}$.

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.
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.

CAIE P1 2006 November Q8

10 marks Standard +0.3

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

Calculate the gradient of the curve at the point where $x = 1$.
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 $x = 1$.
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 $\frac { 12 } { 5 } \pi$.

CAIE P1 2013 November Q9

10 marks Standard +0.3

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.

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 $A$.
Find the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $x$-axis.

OCR C3 2006 June Q9

13 marks Standard +0.8

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.

Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the solid is given by $$V = \pi \left( \mathrm { e } ^ { p } + 4 \mathrm { e } ^ { \frac { 1 } { 2 } p } + p - 5 \right) .$$
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.

OCR C3 2009 January Q8

10 marks Standard +0.8

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 $V$.

Show that $V = 16 \pi \left( 1 - \frac { 27 } { ( p + 3 ) ^ { 3 } } \right)$.
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 $p = 9$.

CAIE P1 2017 June Q10

11 marks Standard +0.3

\includegraphics{figure_1} 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 $360°$ about the $y$-axis. Show that the volume of revolution, $V$, is given by $V = \pi\left(\frac{1}{2}h^2 + h\right)$. [3]
2. Find, showing all necessary working, the area of the shaded region when $h = 3$. [4]
\includegraphics{figure_2} Fig. 2 shows a cross-section of a bowl containing water. When the height of the water level is $h$ cm, the volume, $V$ cm$^3$, of water is given by $V = \pi\left(\frac{1}{4}h^2 + h\right)$. Water is poured into the bowl at a constant rate of 2 cm$^3$ s$^{-1}$. Find the rate, in cm s$^{-1}$, at which the height of the water level is increasing when the height of the water level is 3 cm. [4]

OCR C3 Q9

13 marks Challenging +1.2

\includegraphics{figure_9} 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 $y$-axis to form a solid.

Show that the volume, $V \text{ cm}^3$, of the solid is given by $$V = \pi(e^p + 4e^{\frac{p}{2}} + p - 5).$$ [8]
It is given that the point $P$ is moving in the positive direction along the $y$-axis at a constant rate of $0.2 \text{ cm min}^{-1}$. 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. [5]