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

3. \begin{figure}[h] \end{figure} A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is \(h \mathrm {~m}\), the volume \(V \mathrm {~m} ^ { 3 }\) is given by $$V = \frac { 1 } { 12 } \pi \cdot h ^ { 2 } ( 3 - 4 h ) , \quad 0 \leqslant h \leqslant 0.25$$

1. Find, in terms of \(\pi , \frac { \mathrm { d } V } { \mathrm {~d} h }\) when \(h = 0.1\) Water flows into the bowl at a rate of \(\frac { \pi } { 800 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Find the rate of change of \(h\), in \(\mathrm { m } \mathrm { s } ^ { - 1 }\), when \(h = 0.1\)

2. \begin{figure}[h] \end{figure} Figure 1 shows a metal cube which is expanding uniformly as it is heated. At time \(t\) seconds, the length of each edge of the cube is \(x \mathrm {~cm}\), and the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(\frac { \mathrm { d } V } { \mathrm {~d} x } = 3 x ^ { 2 }\) Given that the volume, \(V \mathrm {~cm} ^ { 3 }\), increases at a constant rate of \(0.048 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\),
2. find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\), when \(x = 8\)
3. find the rate of increase of the total surface area of the cube, in \(\mathrm { cm } ^ { 2 } \mathrm {~s} ^ { - 1 }\), when \(x = 8\)

5. At time \(t\) seconds the radius of a sphere is \(r \mathrm {~cm}\), its volume is \(V \mathrm {~cm} ^ { 3 }\) and its surface area is \(S \mathrm {~cm} ^ { 2 }\). [You are given that \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) and that \(S = 4 \pi r ^ { 2 }\) ] The volume of the sphere is increasing uniformly at a constant rate of \(3 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).

1. Find \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when the radius of the sphere is 4 cm , giving your answer to 3 significant figures.
2. Find the rate at which the surface area of the sphere is increasing when the radius is 4 cm .

4. \begin{figure}[h] \end{figure} A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of \(30 ^ { \circ }\), as shown in Figure 1. The height of the container is 50 cm . When the depth of the water in the container is \(h \mathrm {~cm}\), the surface of the water has radius \(r \mathrm {~cm}\) and the volume of water is \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(V = \frac { 1 } { 9 } \pi h ^ { 3 }\) [0pt] [You may assume the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.] Given that the volume of water in the container increases at a constant rate of \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\),
2. find the rate of change of the depth of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 15\) Give your answer in its simplest form in terms of \(\pi\).

5. Given that $$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 3 y$$

1. show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 1 } { \left( 2 - x ^ { 2 } \right) } \left( 2 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \left( 1 - 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) - 5 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$ Given also that \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }\) at \(x = 0\)
2. obtain a series solution for \(y\) in ascending powers of \(x\) with simplified coefficients, up to and including the term in \(x ^ { 3 }\)

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of curve \(C\) with polar equation $$r = 3 \sin 2 \theta \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ The point \(P\) on \(C\) has polar coordinates \(( R , \phi )\). The tangent to \(C\) at \(P\) is perpendicular to the initial line.

1. Show that \(\tan \phi = \frac { 1 } { \sqrt { 2 } }\)
2. Determine the exact value of \(R\). The region \(S\), shown shaded in Figure 1, is bounded by \(C\) and the line \(O P\), where \(O\) is the pole.
3. Use calculus to show that the exact area of \(S\) is $$p \arctan \frac { 1 } { \sqrt { 2 } } + q \sqrt { 2 }$$ where \(p\) and \(q\) are constants to be determined. Solutions relying entirely on calculator technology are not acceptable.

5. $$y = \sqrt { 4 + \ln x } \quad x > \frac { 1 } { 2 }$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \frac { 9 + 2 \ln x } { 4 x ^ { 2 } ( 4 + \ln x ) ^ { \frac { 3 } { 2 } } }$$
2. Hence, or otherwise, determine the Taylor series expansion about \(x = 1\) for \(y\), in ascending powers of ( \(x - 1\) ), up to and including the term in \(( x - 1 ) ^ { 2 }\), giving each coefficient in simplest form.

4. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } - x$$

1. Show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = A y \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } + B \frac { \mathrm {~d} y } { \mathrm {~d} x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(A\) and \(B\) are integers to be determined. Given that \(y = 1\) at \(x = - 1\)
2. determine the Taylor series solution for \(y\), in ascending powers of \(( x + 1 )\) up to and including the term in \(( x + 1 ) ^ { 4 }\), giving each coefficient in simplest form.

8. \begin{figure}[h] \end{figure} The curve \(C\) shown in Figure 1 has polar equation $$r = 1 - \sin \theta \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point \(P\) lies on \(C\), such that the tangent to \(C\) at \(P\) is parallel to the initial line.

1. Use calculus to determine the polar coordinates of \(P\) The finite region \(R\), shown shaded in Figure 1, is bounded by
   $$\frac { 1 } { 32 } ( a \pi + b \sqrt { 3 } + c )$$ where \(a\), \(b\) and \(c\) are integers to be determined.

8. (a) Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 13 y = 0 , \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 0$$ (b) Hence find the general solution of the differential equation (I).

1. (a) Show that the substitution \(z = y ^ { - 2 }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 x y = x \mathrm { e } ^ { - x ^ { 2 } } y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 x z = - 2 x \mathrm { e } ^ { - x ^ { 2 } }$$ (b) Solve differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence find the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

7. \begin{figure}[h] \end{figure} The curve \(C\), shown in Figure 1, has polar equation $$r = 2 a ( 1 + \cos \theta ) \quad 0 \leqslant \theta \leqslant \pi$$ where \(a\) is a positive constant. The tangent to \(C\) at the point \(A\) is parallel to the initial line.

1. Determine the polar coordinates of \(A\). The point \(B\) on the curve has polar coordinates \(\quad a ( 2 + \sqrt { 3 } ) , \frac { \pi } { 6 }\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\) and the line \(A B\).
2. Use calculus to determine the exact area of the shaded region \(R\). Give your answer in the form $$\frac { a ^ { 2 } } { 4 } ( d \pi - e + f \sqrt { 3 } )$$ where \(d , e\) and \(f\) are integers.

1. The curve \(C\), with pole \(O\), has polar equation

$$r = 1 + \cos \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ At the point \(A\) on \(C\), the tangent to \(C\) is parallel to the initial line.

1. Find the polar coordinates of \(A\).
2. Find the finite area enclosed by the initial line, the line \(O A\) and the curve \(C\), giving your answer in the form \(a \pi + b \sqrt { 3 }\), where \(a\) and \(b\) are rational constants to be found.

6. \begin{figure}[h] \end{figure} The curve shown in Figure 1 has polar equation $$r = 4 a ( 1 + \cos \theta ) \quad 0 \leqslant \theta < \pi$$ where \(a\) is a positive constant.
The tangent to the curve at the point \(A\) is parallel to the initial line.

1. Show that the polar coordinates of \(A\) are \(\left( 6 a , \frac { \pi } { 3 } \right)\) The point \(B\) lies on the curve such that angle \(A O B = \frac { \pi } { 6 }\) The finite region \(R\), shown shaded in Figure 1, is bounded by the line \(A B\) and the curve.
2. Use calculus to determine the area of the shaded region \(R\), giving your answer in the form \(a ^ { 2 } ( n \pi + p \sqrt { 3 } + q )\), where \(n , p\) and \(q\) are integers.

2. $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y ^ { 3 } = 4$$

1. Show that $$x \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = a y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + \left( b y ^ { 2 } + c \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(a\), \(b\) and \(c\) are integers to be determined. Given that \(y = 1\) at \(x = 2\)
2. determine the Taylor series expansion for \(y\) in ascending powers of \(( x - 2 )\), up to and including the term in \(( x - 2 ) ^ { 3 }\), giving each coefficient in simplest form.

1. (a) Given that \(t = \ln x\), where \(x > 0\), show that

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$ (b) Hence show that the transformation \(t = \ln x\), where \(x > 0\), transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 y = 1 + 4 \ln x - 2 ( \ln x ) ^ { 2 }$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } - 2 y = 1 + 4 t - 2 t ^ { 2 }$$ (c) Solve differential equation (II) to determine \(y\) in terms of \(t\).
(d) Hence determine the general solution of differential equation (I).

3. \begin{figure}[h] \end{figure} A bowl with circular cross section and height 20 cm is shown in Figure 2.
The bowl is initially empty and water starts flowing into the bowl.
When the depth of water is \(h \mathrm {~cm}\), the volume of water in the bowl, \(V \mathrm {~cm} ^ { 3 }\), is modelled by the equation $$V = \frac { 1 } { 3 } h ^ { 2 } ( h + 4 ) \quad 0 \leqslant h \leqslant 20$$ Given that the water flows into the bowl at a constant rate of \(160 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\), find, according to the model,

1. the time taken to fill the bowl,
2. the rate of change of the depth of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 5\)

3. \begin{figure}[h] \end{figure} A tablet is dissolving in water.
The tablet is modelled as a cylinder, shown in Figure 1.
At \(t\) seconds after the tablet is dropped into the water, the radius of the tablet is \(x \mathrm {~mm}\) and the length of the tablet is \(3 x \mathrm {~mm}\). The cross-sectional area of the tablet is decreasing at a constant rate of \(0.5 \mathrm {~mm} ^ { 2 } \mathrm {~s} ^ { - 1 }\)

1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) when \(x = 7\)
2. Find, according to the model, the rate of decrease of the volume of the tablet when \(x = 4\)

2. \begin{figure}[h] \end{figure} Figure 1 shows a cube which is increasing in size.
At time \(t\) seconds,

• the length of each edge of the cube is \(x \mathrm {~cm}\)
• the surface area of the cube is \(S \mathrm {~cm} ^ { 2 }\)
• the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\)

Given that the surface area of the cube is increasing at a constant rate of \(4 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\)

1. show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }\) where \(k\) is a constant to be found,
2. show that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = V ^ { p }\) where \(p\) is a constant to be found.

3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in \(\mathrm { mg } l ^ { - 1 }\), at time \(t\) hours satisfies the differential equation $$2 x \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) ^ { 2 } = x ^ { 2 } - 3 x ^ { 4 } , \quad x > 0$$

1. Show that the substitution \(\mathrm { y } = \frac { 1 } { x ^ { 2 } }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + y = 3$$
2. Find the general solution of differential equation \(I\). Given that at time \(t = 0 , x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\),
3. find an expression for \(x\) in terms of \(t\),
4. write down the maximum value of \(x\) as \(t\) varies.

8. (a) Sketch the curve \(C\) with polar equation $$r = 5 + \sqrt { 3 } \cos \theta , \quad 0 \leq \theta \leq 2 \pi$$ (b) Find the polar coordinates of the points where the tangents to \(C\) are parallel to the initial line \(\theta = 0\). Give your answers to 3 significant figures where appropriate.
(c) Using integration, find the area enclosed by the curve \(C\), giving your answer in terms of \(\pi\).

7. (a) Show that the transformation \(z = y ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 2 z \tan x = 1$$ (b) Solve the differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence obtain the general solution of the differential equation (I).

3. $$f ( x ) = \ln ( 1 + \sin k x )$$ where \(k\) is a constant, \(x \in \mathbb { R }\) and \(- \frac { \pi } { 2 } < k x < \frac { 3 \pi } { 2 }\)

1. Find f \({ } ^ { \prime } ( x )\)
2. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { - k ^ { 2 } } { 1 + \sin k x }\)
3. Find the Maclaurin series of \(\mathrm { f } ( x )\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).

4. $$y = \ln \left( \frac { 1 } { 1 - 2 x } \right) , \quad | x | < \frac { 1 } { 2 }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\)
2. Hence, or otherwise, find the series expansion of \(\ln \left( \frac { 1 } { 1 - 2 x } \right)\) about \(x = 0\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\). Give each coefficient in its simplest form.
3. Use your expansion to find an approximate value for \(\ln \left( \frac { 3 } { 2 } \right)\), giving your answer
   to 3 decimal places.