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

6 \includegraphics[max width=\textwidth, alt={}, center]{d53a2d6b-4c5e-4bc6-8aa1-587e97c87920-2_371_839_1409_651} The diagram shows the part of the curve \(y = 3 \mathrm { e } ^ { - x } \sin 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\), and the stationary point \(M\).

1. Find the equation of the tangent to the curve at the origin.
2. Find the coordinates of \(M\), giving each coordinate correct to 3 decimal places.

5 The equation of a curve is \(y = \frac { \mathrm { e } ^ { 2 x } } { 4 x + 1 }\) and the point \(P\) on the curve has \(y\)-coordinate 10 .

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 1 } { 2 } \ln ( 40 x + 10 )\).
2. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 40 x _ { n } + 10 \right)\) with \(x _ { 1 } = 2.3\) to find the \(x\)-coordinate of \(P\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 3 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{d527d21f-0ab5-40fa-8cfd-ebfb4aba0a87-3_493_863_264_641} The diagram shows the part of the curve \(y = \sin ^ { 2 } x\) for \(0 \leqslant x \leqslant \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin 2 x\).
2. Hence find the \(x\)-coordinates of the points on the curve at which the gradient of the curve is 0.5 . [3]
3. By expressing \(\sin ^ { 2 } x\) in terms of \(\cos 2 x\), find the area of the region bounded by the curve and the \(x\)-axis between 0 and \(\pi\).

1 Find the gradient of the curve \(y = \ln ( 5 x + 1 )\) at the point where \(x = 4\).

4 A curve has equation \(y = \cos x \sin 2 x\).
Find the \(x\)-coordinate of the stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\), giving your answer correct to 3 significant figures.

4 The equation of a curve is \(y = x \tan ^ { - 1 } \left( \frac { 1 } { 2 } x \right)\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. The tangent to the curve at the point where \(x = 2\) meets the \(y\)-axis at the point with coordinates \(( 0 , p )\). Find \(p\).

8 The equation of a curve is \(y = e ^ { - 5 x } \tan ^ { 2 } x\) for \(- \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\).
Find the \(x\)-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.

10 \includegraphics[max width=\textwidth, alt={}, center]{c1fbc9ef-2dc6-43c3-bc58-179f683c9acf-18_471_686_276_717} The curve \(y = x \sqrt { \sin x }\) has one stationary point in the interval \(0 < x < \pi\), where \(x = a\) (see diagram).

1. Show that \(\tan a = - \frac { 1 } { 2 } a\).
2. Verify by calculation that \(a\) lies between 2 and 2.5.
3. Show that if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x _ { n + 1 } = \pi - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x _ { n } \right)\) converges, then it converges to \(a\), the root of the equation in part (a). [2]
4. Use the iterative formula given in part (c) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 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.

4 The equation of a curve is \(y = \cos ^ { 3 } x \sqrt { \sin x }\). It is given that the curve has one stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Find the \(x\)-coordinate of this stationary point, giving your answer correct to 3 significant figures.

4 The curve \(y = \mathrm { e } ^ { - 4 x } \tan x\) has two stationary points in the interval \(0 \leqslant x < \frac { 1 } { 2 } \pi\).

1. Obtain an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show it can be written in the form \(\sec ^ { 2 } x ( a + b \sin 2 x ) \mathrm { e } ^ { - 4 x }\), where \(a\) and \(b\) are constants.
2. Hence find the exact \(x\)-coordinates of the two stationary points.

5 \includegraphics[max width=\textwidth, alt={}, center]{72042f09-3495-42e9-bee9-96ec5ac0bf0c-06_352_643_274_744} The diagram shows the part of the curve \(y = x ^ { 2 } \cos 3 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 6 } \pi\), and its maximum point \(M\), where \(x = a\).

1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 3 } \tan ^ { - 1 } \left( \frac { 2 } { 3 a } \right)\).
2. Use an iterative formula based on the equation in (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

5 The equation of a curve is \(y = \frac { e ^ { \sin x } } { \cos ^ { 2 } x }\) for \(0 \leqslant x \leqslant 2 \pi\).
Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence find the \(x\)-coordinates of the stationary points of the curve.

11 The equation of a curve is \(y = \sqrt { \tan x }\), for \(0 \leqslant x < \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\tan x\), and verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = \frac { 1 } { 4 } \pi\).
   The value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is also 1 at another point on the curve where \(x = a\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{87be326f-f638-43e9-a654-b7b53d5141ef-18_605_492_1493_822}
2. Show that \(t ^ { 3 } + t ^ { 2 } + 3 t - 1 = 0\), where \(t = \tan a\).
3. Use the iterative formula $$a _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \left( 1 - \tan ^ { 2 } a _ { n } - \tan ^ { 3 } a _ { n } \right) \right)$$ to determine \(a\) correct to 2 decimal places, giving the result of each iteration to 4 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.

1 Find the exact coordinates of the points on the curve \(y = \frac { x ^ { 2 } } { 1 - 3 x }\) at which the gradient of the tangent is equal to 8 .

5 Find the exact coordinates of the stationary points of the curve \(y = \frac { \mathrm { e } ^ { 3 x ^ { 2 } - 1 } } { 1 - x ^ { 2 } }\).

10 \includegraphics[max width=\textwidth, alt={}, center]{a49b720b-f8d2-42ff-b147-5d676993aa4c-16_611_689_274_721} The diagram shows the curve \(y = x \cos 2 x\), for \(x \geqslant 0\).

1. Find the equation of the tangent to the curve at the point where \(x = \frac { 1 } { 2 } \pi\).
2. Find the exact area of the shaded region shown in the diagram, bounded by the curve and the \(x\)-axis.

3 A particle \(P\) travels from a point \(O\) along a straight line and comes to instantaneous rest at a point \(A\). The velocity of \(P\) at time \(t \mathrm {~s}\) after leaving \(O\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(v = 0.027 \left( 10 t ^ { 2 } - t ^ { 3 } \right)\). Find

1. the distance \(O A\),
2. the maximum velocity of \(P\) while moving from \(O\) to \(A\).

7 A car driver makes a journey in a straight line from \(A\) to \(B\), starting from rest. The speed of the car increases to a maximum, then decreases until the car is at rest at \(B\). The distance travelled by the car \(t\) seconds after leaving \(A\) is \(0.0000117 \left( 400 t ^ { 3 } - 3 t ^ { 4 } \right)\) metres.

1. Find the distance \(A B\).
2. Find the maximum speed of the car.
3. Find the acceleration of the car
   1. as it starts from \(A\),
   2. as it arrives at \(B\).
   3. Sketch the velocity-time graph for the journey.

8

1. Starting from the definitions of sech and tanh in terms of exponentials, prove that $$1 - \operatorname { sech } ^ { 2 } t = \tanh ^ { 2 } t$$ \includegraphics[max width=\textwidth, alt={}]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_77_1547_360_347} ......................................................................................................................................... ......................................................................................................................................... . ........................................................................................................................................ ........................................................................................................................................ ....................................................................................................................................... \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_72_1573_911_324} \includegraphics[max width=\textwidth, alt={}, center]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-14_67_1573_1005_324} The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 1 } { 2 } \tanh ^ { 2 } \mathrm { t } + \text { Insecht } , \quad \mathrm { y } = 1 + \tanh ^ { 4 } \mathrm { t } , \quad \text { for } t > 0$$
2. Show that \(\frac { d y } { d x } = - 4 \operatorname { sech } ^ { 2 } t\).
3. Find the coordinates of the point on \(C\) with \(\frac { d ^ { 2 } y } { d x ^ { 2 } } = - \frac { 9 } { 2 }\), giving your answer in the form \(( a + \ln b , c )\) where \(a , b\) and \(c\) are rational numbers.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

4 The curve \(C\) has equation $$4 y ^ { 3 } + ( x + y ) ^ { 6 } = 109 .$$

1. Show that, at the point \(( - 4,3 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 1 } { 17 }\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( - 4,3 )\).

3 It is given that $$\mathrm { x } = \sin ^ { - 1 } \mathrm { t } \quad \text { and } \quad \mathrm { y } = \mathrm { tcos } ^ { - 1 } \mathrm { t } , \quad \text { for } 0 \leqslant t < 1 .$$

1. Show that \(\frac { d y } { d x } = - t + \sqrt { 1 - t ^ { 2 } } \cos ^ { - 1 } t\).
2. Find \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) in terms of \(t\).

3 The curve \(C\) has equation $$x ^ { 3 } + 2 x y + 8 y ^ { 3 } = - 12$$

1. Show that, at the point \(( - 2 , - 1 )\) on \(C , \frac { \mathrm {~d} y } { \mathrm {~d} x } = - \frac { 1 } { 2 }\). \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-06_2714_37_143_2008}
2. Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(( - 2 , - 1 )\).

5 It is given that $$x = \sinh ^ { - 1 } t , \quad y = \cos ^ { - 1 } t$$ where \(- 1 < t < 1\).

1. By differentiating \(\cos y\) with respect to \(t\), show that \(\frac { d y } { d t } = - \frac { 1 } { \sqrt { 1 - t ^ { 2 } } }\).
2. Find \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) in terms of \(t\), simplifying your answer.

7 A particle \(P\) of mass 0.1 kg is projected vertically upwards from a point \(O\) with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Air resistance of magnitude \(0.1 v \mathrm {~N}\) opposes the motion, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(P\) at time \(t \mathrm {~s}\) after projection.

1. Show that, while \(P\) is moving upwards, \(\frac { 1 } { v + 10 } \frac { \mathrm {~d} v } { \mathrm {~d} t } = - 1\).
2. Hence find an expression for \(v\) in terms of \(t\), and explain why it is valid only for \(0 \leqslant t \leqslant \ln 3\).
3. Find the initial acceleration of \(P\).

7 A particle \(P\) of mass 0.3 kg is projected vertically upwards from the ground with an initial speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When \(P\) is at height \(x \mathrm {~m}\) above the ground, its upward speed is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that $$3 v - 90 \ln ( v + 30 ) + x = A ,$$ where \(A\) is a constant.

1. Differentiate this equation with respect to \(x\) and hence show that the acceleration of the particle is \(- \frac { 1 } { 3 } ( v + 30 ) \mathrm { m } \mathrm { s } ^ { - 2 }\).
2. Find, in terms of \(v\), the resisting force acting on the particle.
3. Find the time taken for \(P\) to reach its maximum height.