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

4 The functions f and g are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = 2 x ^ { 3 } + 4 \quad \text { and } \quad \mathrm { g } ( x ) = \sqrt [ 3 ] { x - 10 }$$

1. Evaluate \(\mathrm { f } ^ { - 1 } ( - 50 )\).
2. Show that \(\operatorname { fg } ( x ) = 2 x - 16\).
3. Differentiate \(\operatorname { gf } ( x )\) with respect to \(x\).

8 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-4_616_1024_296_516} The diagram shows the curve \(y = \frac { 2 x + 4 } { x ^ { 2 } + 5 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the coordinates of the two stationary points.
2. The function g is defined for all real values of \(x\) by $$\mathrm { g } ( x ) = \left| \frac { 2 x + 4 } { x ^ { 2 } + 5 } \right| .$$
   1. Sketch the curve \(y = \mathrm { g } ( x )\) and state the range of g .
   2. It is given that the equation \(\mathrm { g } ( x ) = k\), where \(k\) is a constant, has exactly two distinct real roots. Write down the set of possible values of \(k\).

1 Find the equation of the tangent to the curve \(y = \frac { 5 x + 4 } { 3 x - 8 }\) at the point \(( 2 , - 7 )\).

3 The volume, \(V\) cubic metres, of water in a reservoir is given by $$V = 3 ( 2 + \sqrt { h } ) ^ { 6 } - 192 ,$$ where \(h\) metres is the depth of the water. Water is flowing into the reservoir at a constant rate of 150 cubic metres per hour. Find the rate at which the depth of water is increasing at the instant when the depth is 1.4 metres.

6 The curves \(C _ { 1 }\) and \(C _ { 2 }\) have equations $$y = \ln ( 4 x - 7 ) + 18 \quad \text { and } \quad y = a \left( x ^ { 2 } + b \right) ^ { \frac { 1 } { 2 } }$$ respectively, where \(a\) and \(b\) are positive constants. The point \(P\) lies on both curves and has \(x\)-coordinate 2 . It is given that the gradient of \(C _ { 1 }\) at \(P\) is equal to the gradient of \(C _ { 2 }\) at \(P\). Find the values of \(a\) and \(b\).

7 The variables \(x\) and \(y\) satisfy the equation \(x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }\). Both \(x\) and \(y\) are functions of \(t\).
2. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(x = 1 , y = 8\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\). Section B (36 marks)

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }\).
The curve has asymptotes \(x = 0\) and \(x = a\). \begin{figure}[h] \end{figure}

1. Find \(a\). Hence write down the domain of the function.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the coordinates of the turning point of the curve, and write down the range of the function. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
3. (A) Show algebraically that \(\mathrm { g } ( x )\) is an even function.
   (B) Show that \(\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )\).
   (C) Hence prove that the curve \(y = \mathrm { f } ( x )\) is symmetrical, and state its line of symmetry.

3

1. Given that \(y = \sqrt [ 3 ] { 1 + 3 x ^ { 2 } }\), use the chain rule to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
2. Given that \(y ^ { 3 } = 1 + 3 x ^ { 2 }\), use implicit differentiation to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Show that this result is equivalent to the result in part (i).

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 .

5 Given that \(y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }\).

4 Fig. 4 shows a cone with its axis vertical. The angle between the axis and the slant edge is \(45 ^ { \circ }\). Water is poured into the cone at a constant rate of \(5 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the height of the water surface above the vertex O of the cone is \(h \mathrm {~cm}\), and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure} Find \(V\) in terms of \(h\).
Hence find the rate at which the height of water is increasing when the height is 10 cm .
[0pt] [You are given that the volume \(V\) of a cone of height \(h\) and radius \(r\) is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ].

2 Given that \(y = \frac { \cos x } { 1 - \sin x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer.

3

1. Given that \(\mathrm { f } ( x ) = \mathrm { e } ^ { \sin x }\), find \(\mathrm { f } ^ { \prime } ( 0 )\) and \(\mathrm { f } ^ { \prime \prime } ( 0 )\).
2. Hence find the first three terms of the Maclaurin series for \(\mathrm { f } ( x )\).

6

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } + 1 } }\).
2. Given that \(y = \cosh \left( a \sinh ^ { - 1 } x \right)\), where \(a\) is a constant, show that $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } - a ^ { 2 } y = 0$$

5 It is given that \(y = \sin ^ { - 1 } 2 x\).

1. Using the derivative of \(\sin ^ { - 1 } x\) given in the List of Formulae (MF1), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that \(\left( 1 - 4 x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x }\).
3. Hence show that \(\left( 1 - 4 x ^ { 2 } \right) \frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } - 12 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0\).
4. Using your results from parts (i), (ii) and (iii), find the Maclaurin series for \(\sin ^ { - 1 } 2 x\) up to and including the term in \(x ^ { 3 }\).

10 It is given that \(x = t ^ { \frac { 1 } { 2 } }\), where \(x > 0\) and \(t > 0\), and \(y\) is a function of \(x\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x + \frac { 1 } { x } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 2 } y = 4 x ^ { 2 } \mathrm { e } ^ { - x ^ { 2 } }$$ reduces to the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { - t }$$
3. Find the general solution of ( \(*\) ), giving \(y\) in terms of \(x\).

9 The curve \(C\) has equation $$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$

1. Find the equations of the asymptotes of \(C\). \includegraphics[max width=\textwidth, alt={}, center]{9221f480-4af6-44be-a535-d2ceb0f8b5d2-14_61_1566_513_328}
2. Show that there is no point on \(C\) for which \(\frac { 1 } { 3 } < y < 3\).
3. Find the coordinates of the turning points of \(C\).
4. Sketch \(C\).

9 The curve \(C\) has equation $$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$

1. Find the equations of the asymptotes of \(C\). \includegraphics[max width=\textwidth, alt={}, center]{a0987277-06e9-451b-ae18-bb7de9e7661c-14_61_1566_513_328}
2. Show that there is no point on \(C\) for which \(\frac { 1 } { 3 } < y < 3\).
3. Find the coordinates of the turning points of \(C\).
4. Sketch \(C\).

9 The curve \(C\) has equation $$y = \frac { 3 x - 9 } { ( x - 2 ) ( x + 1 ) }$$

1. Find the equations of the asymptotes of \(C\). \includegraphics[max width=\textwidth, alt={}, center]{68e31138-756a-433a-bf42-0fdfadad091e-14_61_1566_513_328}
2. Show that there is no point on \(C\) for which \(\frac { 1 } { 3 } < y < 3\).
3. Find the coordinates of the turning points of \(C\).
4. Sketch \(C\).

3 Prove by induction that, for all \(n \geqslant 1\), $$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( \mathrm { e } ^ { x ^ { 2 } } \right) = \mathrm { P } _ { n } ( x ) \mathrm { e } ^ { x ^ { 2 } } ,$$ where \(\mathrm { P } _ { n } ( x )\) is a polynomial in \(x\) of degree \(n\) with the coefficient of \(x ^ { n }\) equal to \(2 ^ { n }\).

7

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$$
3. Find the general solution of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10\).
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 8\) when \(x = 1\).

7

1. Write down the expansions in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\) of:
   1. \(\cos x + \sin x\);
   2. \(\quad \ln ( 1 + 3 x )\).
2. It is given that \(y = \mathrm { e } ^ { \tan x }\).
   1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = ( 1 + \tan x ) ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }\).
   2. Find the value of \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) when \(x = 0\).
   3. Hence, by using Maclaurin's theorem, show that the first four terms in the expansion, in ascending powers of \(x\), of \(\mathrm { e } ^ { \tan x }\) are $$1 + x + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 }$$
3. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { \mathrm { e } ^ { \tan x } - ( \cos x + \sin x ) } { x \ln ( 1 + 3 x ) } \right]$$

8

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
3. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x\), given that \(y = \frac { 3 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 }\) when \(x = 1\).
   (5 marks)

7

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ giving your answer in the form \(y = \mathrm { f } ( t )\).
2. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }$$ (5 marks)
3. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ (2 marks)
4. Hence write down the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ (l mark)

6 It is given that \(y = ( 4 + \sin x ) ^ { \frac { 1 } { 2 } }\).

1. Express \(y \frac { \mathrm {~d} y } { \mathrm {~d} x }\) in terms of \(\cos x\).
2. Find the value of \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) when \(x = 0\).
3. Hence, by using Maclaurin's theorem, find the first four terms in the expansion, in ascending powers of \(x\), of \(( 4 + \sin x ) ^ { \frac { 1 } { 2 } }\).
   (2 marks)