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

7 \includegraphics[max width=\textwidth, alt={}, center]{b9556031-871d-4dd3-9523-e3438a41339f-3_655_685_262_730} The diagram shows the curve \(y = x \mathrm { e } ^ { 2 x }\) and its minimum point \(M\).

1. Find the exact coordinates of \(M\).
2. Show that the curve intersects the line \(y = 20\) at the point whose \(x\)-coordinate is the root of the equation $$x = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x } \right)$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x _ { n } } \right)$$ with initial value \(x _ { 1 } = 1.3\), to calculate the root correct to 2 decimal places, giving the result of each iteration to 4 decimal places.

5 Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = 4\) in each of the following cases:

1. \(y = x \ln ( x - 3 )\),
2. \(y = \frac { x - 1 } { x + 1 }\).

6 The curve \(y = 4 x ^ { 2 } \ln x\) has one stationary point.

1. Find the coordinates of this stationary point, giving your answers correct to 3 decimal places.
2. Determine whether this point is a maximum or a minimum point.

6 \includegraphics[max width=\textwidth, alt={}, center]{48ab71ff-c37b-4e0b-b031-d99b0cf517a8-3_421_976_251_580} The diagram shows the curve \(y = \frac { \sin 2 x } { x + 2 }\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The \(x\)-coordinate of the maximum point \(M\) is denoted by \(\alpha\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that \(\alpha\) satisfies the equation \(\tan 2 x = 2 x + 4\).
2. Show by calculation that \(\alpha\) lies between 0.6 and 0.7 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 x _ { n } + 4 \right)\) to find the value of \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

2 Find the gradient of each of the following curves at the point for which \(x = 0\).

1. \(y = 3 \sin x + \tan 2 x\)
2. \(y = \frac { 6 } { 1 + \mathrm { e } ^ { 2 x } }\)

8 \includegraphics[max width=\textwidth, alt={}, center]{de8af872-9f77-4787-8e66-ed199405ca25-3_581_650_1272_744} The diagram shows the curve $$y = \tan x \cos 2 x , \text { for } 0 \leqslant x < \frac { 1 } { 2 } \pi$$ and its maximum point \(M\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 \cos ^ { 2 } x - \sec ^ { 2 } x - 2\).
2. Hence find the \(x\)-coordinate of \(M\), giving your answer correct to 2 decimal places.

5 The equation of a curve is \(y = 6 x \mathrm { e } ^ { \frac { 1 } { 3 } x }\). At the point on the curve with \(x\)-coordinate \(p\), the gradient of the curve is 40 .

1. Show that \(p = 3 \ln \left( \frac { 20 } { p + 3 } \right)\).
2. Show by calculation that \(3.3 < p < 3.5\).
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

3 The equation of a curve is \(y = x \sin 2 x\), where \(x\) is in radians. Find the equation of the tangent to the curve at the point where \(x = \frac { 1 } { 4 } \pi\).

9 \includegraphics[max width=\textwidth, alt={}, center]{a74e4ddf-d254-45f3-bd9a-adf7cd53b3a6-4_611_895_255_625} The diagram shows the curve \(y = \sqrt { } \left( \frac { 1 - x } { 1 + x } \right)\).

1. By first differentiating \(\frac { 1 - x } { 1 + x }\), obtain an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\). Hence show that the gradient of the normal to the curve at the point \(( x , y )\) is \(( 1 + x ) \sqrt { } \left( 1 - x ^ { 2 } \right)\).
2. The gradient of the normal to the curve has its maximum value at the point \(P\) shown in the diagram. Find, by differentiation, the \(x\)-coordinate of \(P\).

2 Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in each of the following cases:

1. \(y = \ln ( 1 + \sin 2 x )\),
2. \(y = \frac { \tan x } { x }\).

4 The curve with equation \(y = \frac { \mathrm { e } ^ { 2 x } } { 4 + \mathrm { e } ^ { 3 x } }\) has one stationary point. Find the exact values of the coordinates of this point.

6 The curve with equation \(y = x ^ { 2 } \cos \frac { 1 } { 2 } x\) has a stationary point at \(x = p\) in the interval \(0 < x < \pi\).

1. Show that \(p\) satisfies the equation \(\tan \frac { 1 } { 2 } p = \frac { 4 } { p }\).
2. Verify by calculation that \(p\) lies between 2 and 2.5.
3. Use the iterative formula \(p _ { n + 1 } = 2 \tan ^ { - 1 } \left( \frac { 4 } { p _ { n } } \right)\) to determine the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

5 A curve has equation \(y = \frac { 2 } { 3 } \ln \left( 1 + 3 \cos ^ { 2 } x \right)\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\tan x\).
2. Hence find the \(x\)-coordinate of the point on the curve where the gradient is - 1 . Give your answer correct to 3 significant figures.

5 The curve with equation \(y = \mathrm { e } ^ { - a x } \tan x\), where \(a\) is a positive constant, has only one point in the interval \(0 < x < \frac { 1 } { 2 } \pi\) at which the tangent is parallel to the \(x\)-axis. Find the value of \(a\) and state the exact value of the \(x\)-coordinate of this point.

10 \includegraphics[max width=\textwidth, alt={}, center]{dcfbe7af-c212-42b1-8a90-8e0418cf0ffd-16_330_689_264_726} The diagram shows the curve \(y = \sin ^ { 3 } x \sqrt { } ( \cos x )\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and its maximum point \(M\).

1. Using the substitution \(u = \cos x\), find by integration the exact area of the shaded region bounded by the curve and the \(x\)-axis.
2. Showing all your working, find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 \includegraphics[max width=\textwidth, alt={}, center]{8c533469-393c-4e4c-a6ec-eab1303741e7-3_480_901_973_621} The diagram shows the curve \(y = x ^ { 2 } e ^ { - \frac { 1 } { 2 } x }\).

1. Find the \(x\)-coordinate of \(M\), the maximum point of the curve.
2. Find the area of the shaded region enclosed by the curve, the \(x\)-axis and the line \(x = 1\), giving your answer in terms of e.

2 The equation of a curve is \(y = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). Show that the gradient of the curve at the point for which \(x = \ln 3\) is \(\frac { 9 } { 50 }\).

8 \includegraphics[max width=\textwidth, alt={}, center]{7fe27759-d014-4bc6-8391-342d9df8280e-3_397_750_255_699} The diagram shows the curve \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x ^ { 2 } } \sqrt { } \left( 1 + 2 x ^ { 2 } \right)\) for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. The sequence of values given by the iterative formula $$x _ { n + 1 } = \sqrt { } \left( \ln \left( 4 + 8 x _ { n } ^ { 2 } \right) \right) ,$$ with initial value \(x _ { 1 } = 2\), converges to a certain value \(\alpha\). State an equation satisfied by \(\alpha\) and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 0.5\).
3. Use the iterative formula to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8 \includegraphics[max width=\textwidth, alt={}, center]{346e8866-ca23-4ea6-81bf-bf62502a16d1-3_397_750_255_699} The diagram shows the curve \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x ^ { 2 } } \sqrt { } \left( 1 + 2 x ^ { 2 } \right)\) for \(x \geqslant 0\), and its maximum point \(M\).

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. The sequence of values given by the iterative formula $$x _ { n + 1 } = \sqrt { } \left( \ln \left( 4 + 8 x _ { n } ^ { 2 } \right) \right) ,$$ with initial value \(x _ { 1 } = 2\), converges to a certain value \(\alpha\). State an equation satisfied by \(\alpha\) and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 0.5\).
3. Use the iterative formula to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

4 The curve with equation \(y = \frac { 2 - \sin x } { \cos x }\) has one stationary point in the interval \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\).

1. Find the exact coordinates of this point.
2. Determine whether this point is a maximum or a minimum point.

9 \includegraphics[max width=\textwidth, alt={}, center]{21878d10-7f16-4dbb-86ef-65a7ba5eeafb-16_446_956_260_593} The diagram shows the curve \(y = \left( 1 + x ^ { 2 } \right) \mathrm { e } ^ { - \frac { 1 } { 2 } x }\) for \(x \geqslant 0\). The shaded region \(R\) is enclosed by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).

1. Find the exact values of the \(x\)-coordinates of the stationary points of the curve.
2. Show that the exact value of the area of \(R\) is \(18 - \frac { 42 } { \mathrm { e } }\).

2 The curve with equation \(y = \frac { \mathrm { e } ^ { - 2 x } } { 1 - x ^ { 2 } }\) has a stationary point in the interval \(- 1 < x < 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of this stationary point, giving the answer correct to 3 decimal places.

10 \includegraphics[max width=\textwidth, alt={}, center]{5b5ed7d1-028e-4f9a-ae9e-26071d0df678-18_449_787_262_678} The diagram shows the graph of \(y = \mathrm { e } ^ { \cos x } \sin ^ { 3 } x\) for \(0 \leqslant x \leqslant \pi\), and its maximum point \(M\). The shaded region \(R\) is bounded by the curve and the \(x\)-axis.

1. Find the \(x\)-coordinate of \(M\). Show all necessary working and give your answer correct to 2 decimal places.
2. By first using the substitution \(u = \cos x\), find the exact value of the area of \(R\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

5 The equation of a curve is \(y = \mathrm { e } ^ { - 2 x } \tan x\), for \(0 \leqslant x < \frac { 1 } { 2 } \pi\).

1. Obtain an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that it can be written in the form \(\mathrm { e } ^ { - 2 x } ( a + b \tan x ) ^ { 2 }\), where \(a\) and \(b\) are constants.
2. Explain why the gradient of the curve is never negative.
3. Find the value of \(x\) for which the gradient is least.

3 Find the exact coordinates of the stationary point of the curve with equation \(y = \frac { 3 x } { \ln x }\).