1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

4 Water flows into a bowl at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) (see Fig. 4). \begin{figure}[h] \end{figure} When the depth of water in the bowl is \(h \mathrm {~cm}\), the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \pi h ^ { 2 }\). Find the rate at which the depth is increasing at the instant in time when the depth is 5 cm .

6 When the gas in a balloon is kept at a constant temperature, the pressure \(P\) in atmospheres and the volume \(V \mathrm {~m} ^ { 3 }\) are related by the equation $$P = \frac { k } { V } ,$$ where \(k\) is a constant. [This is known as Boyle's Law.]
When the volume is \(100 \mathrm {~m} ^ { 3 }\), the pressure is 5 atmospheres, and the volume is increasing at a rate of \(10 \mathrm {~m} ^ { 3 }\) per second.

1. Show that \(k = 500\).
2. Find \(\frac { \mathrm { d } P } { \mathrm {~d} V }\) in terms of \(V\).
3. Find the rate at which the pressure is decreasing when \(V = 100\).

7 Fig. 4 shows a cone. The angle between the axis and the slant edge is \(30 ^ { \circ }\). Water is poured into the cone at a constant rate of \(2 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the radius of the water surface is \(r \mathrm {~cm}\) and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Write down the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
2. Show that \(V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }\), and find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
   [0pt] [You may assume that the volume of a cone of height \(h\) and radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
3. Use the results of parts (i) and (ii) to find the value of \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when \(r = 2\).

8 Fig. 4 is a diagram of a garden pond. \begin{figure}[h] \end{figure} The volume \(V \mathrm {~m} ^ { 3 }\) of water in the pond when the depth is \(h\) metres is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 3 - h ) .$$

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} h }\). Water is poured into the pond at the rate of \(0.02 \mathrm {~m} ^ { 3 }\) per minute.
2. Find the value of \(\frac { \mathrm { d } h } { \mathrm {~d} t }\) when \(h = 0.4\).

6 Fig. 8 shows part of the curve \(y = x \cos 3 x\). The curve crosses the \(x\)-axis at \(\mathrm { O } , \mathrm { P }\) and Q . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P and Q .
2. Find the exact gradient of the curve at the point P . Show also that the turning points of the curve occur when \(x \tan 3 x = \frac { 1 } { 3 }\).
3. Find the area of the region enclosed by the curve and the \(x\)-axis between O and P , giving your answer in exact form.

2 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\) \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. \(( A )\) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

7 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P - 1 } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{9296c786-a42a-4aa5-b326-39adbb544cbc-05_609_622_301_719} \captionsetup{labelformat=empty} \caption{Fig. 8}
      \end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where \(x\) and \(y\) are in metres.
      1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }\). Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 3 } \pi\). Hence find the exact coordinates of the highest point A on the path of C .
      2. Express \(x ^ { 2 } + y ^ { 2 }\) in terms of \(\theta\). Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
      3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
      4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. \section*{ADVANCED GCE UNIT MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) \section*{Paper B: Comprehension} \section*{THURSDAY 14 JUNE 2007} Afternoon
         Time: Up to 1 hour
         Additional materials:
         Rough paper
         MEI Examination Formulae and Tables (MF2) Candidate
         Name □
         Centre
         Number sufficient detail of the working to indicate that a correct method is being used. 1 This basic cycloid has parametric equations $$x = a \theta - a \sin \theta , \quad y = a - a \cos \theta$$
         \includegraphics[max width=\textwidth, alt={}]{9296c786-a42a-4aa5-b326-39adbb544cbc-10_307_1138_445_411}
         Find the coordinates of the points M and N , stating the value of \(\theta\) at each of them. Point M Point N 2 A sea wave has parametric equations (in suitable units) $$x = 7 \theta - 0.25 \sin \theta , \quad y = 0.25 \cos \theta$$ Find the wavelength and height of the wave.
         3 The graph below shows the profile of a wave.
         1. Assuming that it has parametric equations of the form given on line 68 , find the values of \(a\) and \(b\).
         2. Investigate whether the ratio of the trough length to the crest length is consistent with this shape. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-11_312_1141_623_415}
         3. \(\_\_\_\_\)
         4. \(\_\_\_\_\) 4 This diagram illustrates two wave shapes \(U\) and \(V\). They have the same wavelength and the same height. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-12_423_1552_356_205} One of the curves is a sine wave, the other is a curtate cycloid.
         5. State which is which, justifying your answer.
         6. \(\_\_\_\_\) The parametric equations for the curves are: $$x = a \theta , \quad y = b \cos \theta ,$$ and $$x = a \theta - b \sin \theta , \quad y = b \cos \theta .$$
         7. Show that the distance marked \(d\) on the diagram is equal to \(b\).
         8. Hence justify the statement in lines 109 to 111: "In such cases, the curtate cycloid and the sine curve with the same wavelength and height are very similar and so the sine curve is also a good model."
         9. \(\_\_\_\_\)
         10. \(\_\_\_\_\) 5 The diagram shows a curtate cycloid with scales given. Show that this curve could not be a scale drawing of the shape of a stable sea wave. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-13_289_1310_397_331}

5. (i) Given that $$x = \sec \frac { y } { 2 } , \quad 0 \leq y < \pi$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x \sqrt { x ^ { 2 } - 1 } }$$ (ii) Find an equation for the tangent to the curve \(y = \sqrt { 3 + 2 \cos x }\) at the point where \(x = \frac { \pi } { 3 }\).

8. (i) Find $$\int x ^ { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$ (ii) Using the substitution \(u = \sin t\), evaluate $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } 2 t \cos t \mathrm {~d} t$$

2 In a game of rugby, a kick is to be taken from a point P (see Fig. 7). P is a perpendicular distance \(y\) metres from the line TOA. Other distances and angles are as shown. \begin{figure}[h] \end{figure}

1. Show that \(\theta = \beta - \alpha\), and hence that \(\tan \theta = \frac { 6 y } { 160 + y ^ { 2 } }\). Calculate the angle \(\theta\) when \(y = 6\).
2. By differentiating implicitly, show that \(\frac { \mathrm { d } \theta } { \mathrm { d } y } = \frac { 6 \left( 160 - y ^ { 2 } \right) } { \left( 160 + y ^ { 2 } \right) ^ { 2 } } \cos ^ { 2 } \theta\).
3. Use this result to find the value of \(y\) that maximises the angle \(\theta\). Calculate this maximum value of \(\theta\). [You need not verify that this value is indeed a maximum.]

7.The variable \(y\) is defined by $$y = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \text { for } 0 < x < \frac { \pi } { 2 } .$$ A student was asked to prove that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = - 4 \cot 2 x .$$ The attempted proof was as follows: $$\begin{aligned} y & = \ln \left( \sec ^ { 2 } x + \operatorname { cosec } ^ { 2 } x \right) \\ & = \ln \left( \sec ^ { 2 } x \right) + \ln \left( \operatorname { cosec } ^ { 2 } x \right) \\ & = 2 \ln \sec x + 2 \ln \operatorname { cosec } x \\ \frac { \mathrm {~d} y } { \mathrm {~d} x } & = 2 \tan x - 2 \cot x \\ & = \frac { 2 \left( \sin ^ { 2 } x - \cos ^ { 2 } x \right) } { \sin x \cos x } \\ & = \frac { - 2 \cos 2 x } { \frac { 1 } { 2 } \sin 2 x } \\ & = - 4 \cot 2 x \end{aligned}$$

1. Identify the error in this attempt at a proof.
2. Give a correct version of the proof.
3. Find and simplify a general relationship between \(p\) and \(q\) ,where \(p\) and \(q\) are variables that depend on \(x\) ,such that the student would obtain the correct result when differentiating \(\ln ( p + q )\) with respect to \(x\) by the above incorrect method.
4. Given that \(p ( x ) = k \sec r x\) and \(q ( x ) = \operatorname { cosec } ^ { 2 } x\) ,where \(k\) and \(r\) are positive integers,find the values of \(k\) and \(r\) such that \(p\) and \(q\) satisfy the relationship found in part(c). \section*{END} Marks for presentation: 7
   TOTAL MARKS: 100

6.Figure 1 shows a sketch of part of the curve with equation \(y = x \sin ( \ln x ) , x \geqslant 1\) \begin{figure}[h] \end{figure} For \(x > 1\) ,the curve first crosses the \(x\)-axis at the point \(A\) .

1. Find the \(x\) coordinate of \(A\) .
2. Differentiate \(x \sin ( \ln x )\) and \(x \cos ( \ln x )\) with respect to \(x\) and hence find $$\int \sin ( \ln x ) \mathrm { d } x \text { and } \int \cos ( \ln x ) \mathrm { d } x$$
3. 1. Find \(\int x \sin ( \ln x ) \mathrm { d } x\) .
   2. Hence show that the area of the shaded region \(\boldsymbol { R }\) ,bounded by the curve and the \(x\)-axis between the points \(( 1,0 )\) and \(A\) ,is $$\frac { 1 } { 5 } \left( \mathrm { e } ^ { 2 \pi } + 1 \right)$$

1. $$\mathrm { f } ( x ) = x ^ { \left( x ^ { 2 } \right) } \quad x > 0$$ Use logarithms to find the \(x\) coordinate of the stationary point of the curve with equation \(y = \mathrm { f } ( x )\) .

4.Given that \(\mathrm { f } ( x ) = \mathrm { e } ^ { x ^ { 3 } - 2 x }\)

1. find \(\mathrm { f } ^ { \prime } ( x )\) The curves \(C _ { 1 }\) and \(C _ { 2 }\) are defined by the functions g and h respectively,where $$\begin{array} { l l } \mathrm { g } ( x ) = 8 x ^ { 3 } \mathrm { e } ^ { x ^ { 3 } - 2 x } & x \in \mathbb { R } , x > 0 \\ \mathrm {~h} ( x ) = \left( 3 x ^ { 5 } + 4 x \right) \mathrm { e } ^ { x ^ { 3 } - 2 x } & x \in \mathbb { R } , x > 0 \end{array}$$
2. Find the \(x\) coordinates of the points of intersection of \(C _ { 1 }\) and \(C _ { 2 }\) Given that \(C _ { 1 }\) lies above \(C _ { 2 }\) between these points of intersection,
3. find the area of the region bounded by the curves between these two points.
   Give your answer in the form \(A + B \mathrm { e } ^ { C }\) where \(A , B\) ,and \(C\) are exact real numbers to be found.

1. \hspace{0pt} [In this question you may assume the following formulae for the volume and curved] surface area of a cone of base radius \(r\) and height \(h\) and of a sphere of radius \(r\).

Cone: volume \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) and curved surface area \(S = \pi r \sqrt { h ^ { 2 } + r ^ { 2 } }\) Sphere: volume \(V = \frac { 4 \pi } { 3 } r ^ { 3 }\) and curved surface area \(S = 4 \pi r ^ { 2 }\) \includegraphics[max width=\textwidth, alt={}, center]{78ba3acc-4cca-4d15-8362-a27e425c5859-22_782_755_637_657} Figure 3
Figure 3 shows the design for a garden ornament.
The ornament is made of a hemisphere on top of a truncated cone.
The truncated cone has base radius \(2 r \mathrm {~cm}\), top radius \(r \mathrm {~cm}\) and height \(4 r \mathrm {~cm}\).
The hemisphere has radius \(R \mathrm {~cm}\).
Given that the volume of the ornament is \(2100 \pi \mathrm {~cm} ^ { 3 }\)

1. show that $$R ^ { 3 } = 3150 - 14 r ^ { 3 }$$
2. Find an expression involving \(\frac { \mathrm { d } R } { \mathrm {~d} r }\) in terms of \(r\) and/or \(R\). The base of the truncated cone of the ornament is fixed to the ground.
3. Show that the visible surface area of the ornament, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = ( 3 \sqrt { 17 } - 1 ) \pi r ^ { 2 } + 3 \pi R ^ { 2 }$$
4. Hence show that $$\frac { \mathrm { d } A } { \mathrm {~d} r } = \gamma \pi r - \frac { \delta \pi r ^ { 2 } } { R }$$ where \(\gamma\) and \(\delta\) are real numbers to be determined. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{78ba3acc-4cca-4d15-8362-a27e425c5859-23_705_803_625_630} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} Figure 4 shows a sketch of \(A\) against \(r\), for \(r \geqslant 0\) There is a local minimum at \(r = 0\) and a local maximum at the point \(M\). The overall minimum point is at the point \(N\), where the gradient of the curve is undefined.
5. 1. Determine the \(r\) coordinate of the point \(N\).
   2. Explain why, for the ornament, \(r\) must be less than this value.
6. Show that the \(r\) coordinate of the point \(M\) is $$\sqrt [ 3 ] { \frac { p ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } { 3 q ^ { 2 } + ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } }$$ where \(p\) and \(q\) are integers to be determined.

2. The curve \(C\) has equation \(y = x ^ { \sin x } , \quad x > 0\).

1. Find the equation of the tangent to \(C\) at the point where \(x = \frac { \pi } { 2 }\).
2. Prove that this tangent touches \(C\) at infinitely many points.

6. \includegraphics[max width=\textwidth, alt={}, center]{0214eebf-93f2-4338-9222-443000115225-4_346_1040_303_548} \section*{Figure 1} Figure 1 shows a sketch of the curve \(C _ { 1 }\) with equation $$y = \cos ( \cos x ) \sin x \quad \text { for } \quad 0 \leqslant x \leqslant \pi$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. Hence verify that the turning point is at \(x = \frac { \pi } { 2 }\) and find the \(y\) coordinate of this point.
3. Find the area of the region bounded by \(C _ { 1 }\) and the positive \(x\)-axis between \(x = 0\) and \(x = \pi\) Figure 2 shows a sketch of the curve \(C _ { 1 }\) and the curve \(C _ { 2 }\) with equation $$y = \sin ( \cos x ) \sin x \quad \text { for } \quad 0 \leqslant x \leqslant \pi$$ \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0214eebf-93f2-4338-9222-443000115225-4_519_1065_1631_484} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the origin and the point \(A ( a , b )\) ,where \(a < \pi\)
4. Find \(a\) and \(b\) ,giving \(b\) in a form not involving trigonometric functions.
5. Find the area of the shaded region between \(C _ { 1 }\) and \(C _ { 2 }\)

9. \begin{figure}[h] \end{figure} The population of a species of animal is being studied. The population \(P\), at time \(t\) years from the start of the study, is assumed to be $$P = \frac { 9000 \mathrm { e } ^ { k t } } { 3 \mathrm { e } ^ { k t } + 7 } , \quad t \geqslant 0$$ where \(k\) is a positive constant.
A sketch of the graph of \(P\) against \(t\) is shown in Figure 2 .
Use the given equation to

1. find the population at the start of the study,
2. find the value for the upper limit of the population. Given that \(P = 2500\) when \(t = 4\)
3. calculate the value of \(k\), giving your answer to 3 decimal places. Using this value for \(k\),
4. find, using \(\frac { \mathrm { d } P } { \mathrm {~d} t }\), the rate at which the population is increasing when \(t = 10\) Give your answer to the nearest integer.

13. \begin{figure}[h] \end{figure} Figure 4 shows a hemispherical bowl containing some water.
At \(t\) seconds, the height of the water is \(h \mathrm {~cm}\) and the volume of the water is \(V \mathrm {~cm} ^ { 3 }\), where $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 30 - h ) , \quad 0 < h \leqslant 10$$ The water is leaking from a hole in the bottom of the bowl. Given that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - \frac { 1 } { 10 } V\)

1. show that \(\frac { \mathrm { d } h } { \mathrm {~d} t } = - \frac { h ( 30 - h ) } { 30 ( 20 - h ) }\)
2. Write \(\frac { 30 ( 20 - h ) } { h ( 30 - h ) }\) in partial fraction form. Given that \(h = 10\) when \(t = 0\),
3. use your answers to parts (a) and (b) to find the time taken for the height of the water to fall to 5 cm . Give your answer in seconds to 2 decimal places.

4 For each of the following curves, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and determine the exact \(x\)-coordinate of the stationary point:

1. \(y = \left( 4 x ^ { 2 } + 1 \right) ^ { 5 }\),
2. \(y = \frac { x ^ { 2 } } { \ln x }\).

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 \(p = 9\).

3 A giant spherical balloon is being inflated in a theme park. The radius of the balloon is increasing at a rate of 12 cm per hour. Find the rate at which the surface area of the balloon is increasing at the instant when the radius is 150 cm . Give your answer in \(\mathrm { cm } ^ { 2 }\) per hour correct to 2 significant figures.
[0pt] [Surface area of sphere \(= 4 \pi r ^ { 2 }\).]

6 The curve with equation \(y = \frac { 3 x + 4 } { x ^ { 3 } - 4 x ^ { 2 } + 2 }\) has a stationary point at \(P\). It is given that \(P\) is close to the point with coordinates \(( 2.4 , - 1.6 )\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \sqrt [ 3 ] { \frac { 16 } { 3 } x + 1 }$$
2. By first using an iterative process based on the equation in part (i), find the coordinates of \(P\), giving each coordinate correct to 3 decimal places.

6 The volume, \(V \mathrm {~m} ^ { 3 }\), of liquid in a container is given by $$V = \left( 3 h ^ { 2 } + 4 \right) ^ { \frac { 3 } { 2 } } - 8 ,$$ where \(h \mathrm {~m}\) is the depth of the liquid.

1. Find the value of \(\frac { \mathrm { d } V } { \mathrm {~d} h }\) when \(h = 0.6\), giving your answer correct to 2 decimal places.
2. Liquid is leaking from the container. It is observed that, when the depth of the liquid is 0.6 m , the depth is decreasing at a rate of 0.015 m per hour. Find the rate at which the volume of liquid in the container is decreasing at the instant when the depth is 0.6 m .

3 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-2_435_472_932_794} The diagram shows a container in the form of a right circular cone. The angle between the axis and the slant height is \(\alpha\), where \(\alpha = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)\). Initially the container is empty, and then liquid is added at the rate of \(14 \mathrm {~cm} ^ { 3 }\) per minute. The depth of liquid in the container at time \(t\) minutes is \(x \mathrm {~cm}\).

1. Show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid in the container when the depth is \(x \mathrm {~cm}\) is given by $$V = \frac { 1 } { 12 } \pi x ^ { 3 } .$$ [The volume of a cone is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
2. Find the rate at which the depth of the liquid is increasing at the instant when the depth is 8 cm . Give your answer in cm per minute correct to 2 decimal places.