1.07b Gradient as rate of change: dy/dx notation

3 The equation of a curve is \(y = ( x - 3 ) \sqrt { x + 1 } + 3\). The following points lie on the curve. Non-exact values are rounded to 4 decimal places. $$A ( 2 , k ) \quad B ( 2.9,2.8025 ) \quad C ( 2.99,2.9800 ) \quad D ( 2.999,2.9980 ) \quad E ( 3,3 )$$

1. Find \(k\), giving your answer correct to 4 decimal places.
2. Find the gradient of \(A E\), giving your answer correct to 4 decimal places.
   The gradients of \(B E , C E\) and \(D E\), rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997 respectively.
3. State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at the point \(E\).

1 The following points $$A ( 0,1 ) , \quad B ( 1,6 ) , \quad C ( 1.5,7.75 ) , \quad D ( 1.9,8.79 ) \quad \text { and } \quad E ( 2,9 )$$ lie on the curve \(y = \mathrm { f } ( x )\). The table below shows the gradients of the chords \(A E\) and \(B E\).

1. Complete the table to show the gradients of \(C E\) and \(D E\).
2. State what the values in the table indicate about the value of \(\mathrm { f } ^ { \prime } ( 2 )\).

3 The length, \(x\) metres, of a Green Anaconda snake which is \(t\) years old is given approximately by the formula $$x = 0.7 \sqrt { } ( 2 t - 1 ) ,$$ where \(1 \leqslant t \leqslant 10\). Using this formula, find

1. \(\frac { \mathrm { d } x } { \mathrm {~d} t }\),
2. the rate of growth of a Green Anaconda snake which is 5 years old.

10 A curve is defined for \(x > 0\) and is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x + \frac { 4 } { x ^ { 2 } }\). The point \(P ( 4,8 )\) lies on the curve.

1. Find the equation of the curve.
2. Show that the gradient of the curve has a minimum value when \(x = 2\) and state this minimum value.

8 \includegraphics[max width=\textwidth, alt={}, center]{02da6b6a-6db1-4bc3-ad4e-537e4f61dcac-3_365_663_1813_740} The inside lane of a school running track consists of two straight sections each of length \(x\) metres, and two semicircular sections each of radius \(r\) metres, as shown in the diagram. The straight sections are perpendicular to the diameters of the semicircular sections. The perimeter of the inside lane is 400 metres.

1. Show that the area, \(A \mathrm {~m} ^ { 2 }\), of the region enclosed by the inside lane is given by \(A = 400 r - \pi r ^ { 2 }\).
2. Given that \(x\) and \(r\) can vary, show that, when \(A\) has a stationary value, there are no straight sections in the track. Determine whether the stationary value is a maximum or a minimum. [5]

7 \includegraphics[max width=\textwidth, alt={}, center]{98ee8d3e-9aba-46a2-aa9c-b1e2093f393e-10_702_597_258_772} The diagram shows the curves \(y = 4 \cos \frac { 1 } { 2 } x\) and \(y = \frac { 1 } { 4 - x }\), for \(0 \leqslant x < 4\). When \(x = a\), the tangents to the curves are perpendicular.

1. Show that \(a = 4 - \sqrt { } \left( 2 \sin \frac { 1 } { 2 } a \right)\).
2. Verify by calculation that \(a\) lies between 2 and 3 .
3. Use an iterative formula based on the equation in part (i) to determine \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

1 The equation of a curve is \(y = \frac { 1 + x } { 1 + 2 x }\) for \(x > - \frac { 1 } { 2 }\). Show that the gradient of the curve is always negative.

4 A particle moves in a straight line. Its displacement \(t\) seconds after leaving the fixed point \(O\) is \(x\) metres, where \(x = \frac { 1 } { 2 } t ^ { 2 } + \frac { 1 } { 30 } t ^ { 3 }\). Find

1. the speed of the particle when \(t = 10\),
2. the value of \(t\) for which the acceleration of the particle is twice its initial acceleration.

3. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = x ^ { 2 } + 3 x - 2\) The point \(P ( 3,16 )\) lies on the curve.

1. Find the gradient of the tangent to the curve at \(P\). The point \(Q\) with \(x\) coordinate \(3 + h\) also lies on the curve.
2. Find, in terms of \(h\), the gradient of the line \(P Q\). Write your answer in simplest form.
3. Explain briefly the relationship between the answer to (b) and the answer to (a).

7. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation $$y = 2 \cos 3 x - 3 x + 4 \quad x > 0$$ where \(x\) is measured in radians. The curve crosses the \(x\)-axis at the point \(P\), as shown in Figure 3.
Given that the \(x\) coordinate of \(P\) is \(\alpha\),

1. show that \(\alpha\) lies between 0.8 and 0.9 The iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \arccos \left( 1.5 x _ { n } - 2 \right)$$ can be used to find an approximate value for \(\alpha\).
2. Using this iteration formula with \(x _ { 1 } = 0.8\) find, to 4 decimal places, the value of
   1. \(X _ { 2 }\)
   2. \(X _ { 5 }\) The point \(Q\) and the point \(R\) are local minimum points on the curve, as shown in Figure 3.
      Given that the \(x\) coordinates of \(Q\) and \(R\) are \(\beta\) and \(\lambda\) respectively, and that they are the two smallest values of \(x\) at which local minima occur,
3. find, using calculus, the exact value of \(\beta\) and the exact value of \(\lambda\).

6. (i) The curve \(C _ { 1 }\) has equation $$y = 3 \ln \left( x ^ { 2 } - 5 \right) - 4 x ^ { 2 } + 15 \quad x > \sqrt { 5 }$$ Show that \(C _ { 1 }\) has a stationary point at \(x = \frac { \sqrt { p } } { 2 }\) where \(p\) is a constant to be found.
(ii) A different curve \(C _ { 2 }\) has equation $$y = 4 x - 12 \sin ^ { 2 } x$$

1. Show that, for this curve, $$\frac { \mathrm { d } y } { \mathrm {~d} x } = A + B \sin 2 x$$ where \(A\) and \(B\) are constants to be found.
2. Hence, state the maximum gradient of this curve.

7. $$( 1 + 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x } = x + 4 y ^ { 2 }$$

1. Show that $$( 1 + 2 x ) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 1 + 2 ( 4 y - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$
2. Differentiate equation 1 with respect to \(x\) to obtain an equation involving $$\frac { \mathrm { d } ^ { 3 } } { \mathrm {~d} x ^ { 3 } } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x } , \quad x \text { and } y .$$ Given that \(y = \frac { 1 } { 2 }\) at \(x = 0\),
3. find a series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
   (6)(Total 11 marks)

4 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 \(V = 100\).

6

1. Find \(\int x \left( x ^ { 2 } + 2 \right) \mathrm { d } x\).
2. 1. Find \(\int \frac { 1 } { \sqrt { x } } \mathrm {~d} x\).
   2. The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { } x }\). Find the equation of the curve, given that it passes through the point \(( 4,0 )\).

1. At time \(t\) seconds \(( t \geqslant 0 )\), a particle \(P\) has position vector \(\mathbf { r }\) metres with respect to a fixed origin \(O\), where

$$\mathbf { r } = \left( t ^ { 3 } - \frac { 9 } { 2 } t ^ { 2 } - 24 t \right) \mathbf { i } + \left( - t ^ { 3 } + 3 t ^ { 2 } + 12 t \right) \mathbf { j }$$ At time \(T\) seconds, \(P\) is moving in a direction parallel to the vector \(\mathbf { - i } - \mathbf { j }\) Find

1. the value of \(T\),
2. the magnitude of the acceleration of \(P\) at the instant when \(t = T\).

3. At time \(t\) seconds \(( t \geqslant 0 )\) a particle \(P\) has position vector \(\mathbf { r }\) metres, with respect to a fixed origin \(O\), where
(b) the magnitude of the acceleration of \(P\) when \(t = 4\) $$\begin{aligned} & \qquad \mathbf { r } = \left( 16 t - 3 t ^ { 3 } \right) \mathbf { i } + \left( t ^ { 3 } - t ^ { 2 } + 2 \right) \mathbf { j } \\ & \text { Find } \\ & \text { (a) the velocity of } P \text { at the instant when it is moving parallel to the vector } \mathbf { j } \text {, } \end{aligned}$$ VILIV SIHI NI IIIIIM ION OC
VILV SIHI NI JAHAM ION OC
VJ4V SIHI NI JIIYM ION OC

2 Fig. 9 shows the line \(y = x\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)\). The line and the curve intersect at the origin and at the point \(\mathrm { P } ( a , a )\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { e } ^ { a } = 1 + 2 a\).
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac { 1 } { 2 } a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\).
3. Show that the inverse function of \(\mathrm { f } ( x )\) is \(\mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = \ln ( 1 + 2 x )\). Add a sketch of \(y = \mathrm { g } ( x )\) to the copy of Fig. 9.
4. Find the derivatives of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }\). Give a geometrical interpretation of this result.

3 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Write down the range of the function \(\mathrm { f } ( x )\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).

4 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \nless \leqslant 1\).
Fig. 6 also shows the curve \(y = \mathrm { g } ( x )\), where \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
P is the point on the curve \(y = \mathrm { f } ( x )\) with \(x\)-coordinate \(\frac { 1 } { 2 }\).

1. Find the \(y\)-coordinate of P , giving your answer in terms of \(\pi\). The point Q is the reflection of P in \(y = x\).
2. Find \(\mathrm { g } ( x )\) and its derivative \(\mathrm { g } ^ { \prime } ( x )\). Hence determine the exact gradient of the curve \(y = \mathrm { g } ( x )\) at the point Q . Write down the exact gradient of \(y = \mathrm { f } ( x )\) at the point P .

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\). The endpoints of the curve are \(\mathrm { P } ( - \pi , 1 )\) and \(\mathrm { Q } ( \pi , 3 )\), and \(\mathrm { f } ( x ) = a + \sin b x\), where \(a\) and \(b\) are constants. \begin{figure}[h] \end{figure}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac { 1 } { 2 }\).
2. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,2 ). Show that there is no point on the curve at which the gradient is greater than this.
3. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range. Write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 2,0 )\).
4. Find the area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point \(( 0,1 )\). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

8 The height of tide at the entrance to a harbour on a particular day may be modelled by the function \(h = 3 + 2 \sin 30 t + 1.5 \cos 30 t\) where \(h\) is measured in metres, \(t\) in hours after midnight and \(30 t\) is in degrees.
[0pt] [The values 2 and 1.5 represent the relative effects of the moon and sun respectively.]

1. Show that \(2 \sin 30 t + 1.5 \cos 30 t\) can be written in the form \(2.5 \sin ( 30 t + \alpha )\), where \(\alpha\) is to be determined.
2. Find the height of tide at high water and the first time that this occurs after midnight.
3. Find the range of tide during the day.
4. Sketch the graph of \(h\) against \(t\) for \(0 \leq t \leq 12\), indicating the maximum and minimum points.
5. A sailing boat may enter the harbour only if there is at least 2 metres of water. Find the times during this morning when it may enter the harbour.
6. From your graph estimate the time at which the water falling fastest and the rate at which it is falling.

1 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 $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ 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.