1.07n Stationary points: find maxima, minima using derivatives

9

1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_506_812_676_653} \captionsetup{labelformat=empty} \caption{Figure 9.1}
   \end{figure} Using this model,
   (A) find the greatest height of the tunnel,
   (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_513_1256_1894_575} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
   Hence estimate the volume of earth removed when the tunnel is re-shaped.

9

1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_506_812_676_653} \captionsetup{labelformat=empty} \caption{Figure 9.1}
   \end{figure} Using this model,
   (A) find the greatest height of the tunnel,
   (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_513_1256_1894_575} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
   Hence estimate the volume of earth removed when the tunnel is re-shaped.

10 A curve has equation \(y = x ^ { 3 } - 6 x ^ { 2 } + 12\).

1. Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.
2. Find, in the form \(y = m x + c\), the equation of the normal to the curve at the point \(( 2 , - 4 )\).

6 A curve has gradient given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } - 6 x + 9\). Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
Show that the curve has a stationary point of inflection when \(x = 3\).

10 The equation of a curve is \(y = 7 + 6 x - x ^ { 2 }\).

1. Use calculus to find the coordinates of the turning point on this curve. Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.
2. Find \(\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x\), showing your working.
3. The curve and the line \(y = 12\) intersect at ( 1,12 ) and ( 5,12 ). Using your answer to part (ii), find the area of the finite region between the curve and the line \(y = 12\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{15b8f97b-c058-409f-907f-cb0a6102abc4-5_643_1034_331_513} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

10 \begin{figure}[h] \end{figure} Fig. 10 shows a solid cuboid with square base of side \(x \mathrm {~cm}\) and height \(h \mathrm {~cm}\). Its volume is \(120 \mathrm {~cm} ^ { 3 }\).

1. Find \(h\) in terms of \(x\). Hence show that the surface area, \(A \mathrm {~cm} ^ { 2 }\), of the cuboid is given by \(A = 2 x ^ { 2 } + \frac { 480 } { x }\).
2. Find \(\frac { \mathrm { d } A } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } A } { \mathrm {~d} x ^ { 2 } }\).
3. Hence find the value of \(x\) which gives the minimum surface area. Find also the value of the surface area in this case.

4 A curve has equation \(y = x + \frac { 1 } { x }\).
Use calculus to show that the curve has a turning point at \(x = 1\).
Show also that this point is a minimum.

9 The equation of a cubic curve is \(y = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the tangent to the curve when \(x = 3\) passes through the point \(( - 1 , - 41 )\).
2. Use calculus to find the coordinates of the turning points of the curve. You need not distinguish between the maximum and minimum.
3. Sketch the curve, given that the only real root of \(2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2 = 0\) is \(x = 0.2\) correct to 1 decimal place.

6 Use calculus to find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 6 x ^ { 2 } - 15 x\). Hence find the set of values of \(x\) for which \(x ^ { 3 } - 6 x ^ { 2 } - 15 x\) is an increasing function.

9 The equation of a curve is given by \(y = ( x - 1 ) ^ { 2 } ( x + 2 )\).

1. Write \(( x - 1 ) ^ { 2 } ( x + 2 )\) in the form \(x ^ { 3 } + p x ^ { 2 } + q x + r\) where \(p , q\) and \(r\) are to be determined.
2. Show that the curve \(y = ( x - 1 ) ^ { 2 } ( x + 2 )\) has a maximum point when \(x = - 1\) and find the coordinates of the minimum point.
3. Sketch the curve \(y = ( x - 1 ) ^ { 2 } ( x + 2 )\).
4. For what values of \(k\) does \(( x - 1 ) ^ { 2 } ( x + 2 ) = k\) have exactly one root.

11 The cross-section of a brick wall built on horizontal ground is given, for \(0 \leq x \leq 6\), by the following function $$\begin{array} { l l } 0 \leq x \leq 2 & y = 1 \\ 2 \leq x \leq 4 & y = - \frac { 1 } { 2 } x ^ { 2 } + 3 x - 3 \\ 4 \leq x \leq 6 & y = 1 \end{array}$$ Units are metres.

1. Show that the highest point on the wall is 1.5 metres above the ground.
2. Find the area of the cross-section of the wall.

12 Fig. 12 shows a window. The base and sides are parts of a rectangle with dimensions \(2 x\) metres horizontally by \(y\) metres vertically. The top is a semicircle of radius \(x\) metres. The perimeter of the window is 10 metres. \begin{figure}[h] \end{figure}

1. Express \(y\) as a function of \(x\).
2. Find the total area, \(A \mathrm {~m} ^ { 2 }\), in terms of \(x\) and \(y\). Use your answer to part (i) to show that this simplifies to $$A = 10 x - 2 x ^ { 2 } - \frac { 1 } { 2 } \pi x ^ { 2 }$$
3. Prove that for the maximum value of \(A\), \(y = x\) exactly.
   \section*{MEI STRUCTURED MATHEMATICS } \section*{CONCEPTS FOR ADVANCED MATHEMATICS, C2} \section*{Practice Paper C2-B
   Insert sheet for question 11} 11 Speed-time graph with the first two points plotted. \includegraphics[max width=\textwidth, alt={}, center]{73d1c02b-1b7b-426d-a171-c762597cfed4-5_768_1772_1389_205}

4 You are given that \(y = x ^ { 3 } - 12 x\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence find the coordinates of the turning points of the curve.

10 Fig. 10 shows the curve with equation \(y = x ^ { 2 } + \frac { 16 } { x }\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence calculate the coordinates of the stationary point on the curve.
3. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and explain why this confirms that he stationary point is a minimum.
4. Using the trapezium rule with 4 intervals, estimate the area between the curve and the \(x\) axis between \(x = 2\) and \(x = 4\).
5. State, giving a reason, whether this estimate of the area under-estimates or over-estimates the true area beneath the curve.

1

1. Use calculus to find, correct to 1 decimal place, the coordinates of the turning points of the curve \(y = x ^ { 3 } - 5 x\). [You need not determine the nature of the turning points.]
2. Find the coordinates of the points where the curve \(y = x ^ { 3 } - 5 x\) meets the axes and sketch the curve.
3. Find the equation of the tangent to the curve \(y = x ^ { 3 } - 5 x\) at the point \(( 1 , - 4 )\). Show that, where this tangent meets the curve again, the \(x\)-coordinate satisfies the equation $$x ^ { 3 } - 3 x + 2 = 0$$ Hence find the \(x\)-coordinate of the point where this tangent meets the curve again.

2 The equation of a cubic curve is \(y = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the tangent to the curve when \(x = 3\) passes through the point \(( - 1 , - 41 )\).
2. Use calculus to find the coordinates of the turning points of the curve. You need not distinguish between the maximum and minimum.
3. Sketch the curve, given that the only real root of \(2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2 = 0\) is \(x = 0.2\) correct to 1 decimal place. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b6ea89e3-a8a4-41a2-8ed5-eed6c2dfda7e-2_1017_935_285_638} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} Fig. 11 shows a sketch of the cubic curve \(y = \mathrm { f } ( x )\). The values of \(x\) where it crosses the \(x\)-axis are - 5 , - 2 and 2 , and it crosses the \(y\)-axis at \(( 0 , - 20 )\).

4

1. Differentiate \(x ^ { 3 } - 3 x ^ { 2 } - 9 x\). Hence find the \(x\)-coordinates of the stationary points on the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 9 x\), showing which is the maximum and which the minimum.
2. Find, in exact form, the coordinates of the points at which the curve crosses the \(x\)-axis.
3. Sketch the curve.

5 The equation of a curve is \(\quad y = 7 + 6 x - x ^ { 2 }\).

1. Use calculus to find the coordinates of the turning point on this curve. Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.
2. Find \(\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x\), showing your working.
3. The curve and the line \(y = 12\) intersect at \(( 1,12 )\) and \(( 5,12 )\). Using your answer to part (ii), find the area of the finite region between the curve and the line \(y = 12\).

8. The curve \(C\) has the equation \(y = \sqrt { x } + \mathrm { e } ^ { 1 - 4 x } , x \geq 0\).

1. Find an equation for the normal to the curve at the point \(\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)\). The curve \(C\) has a stationary point with \(x\)-coordinate \(\alpha\) where \(0.5 < \alpha < 1\).
2. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 4 } [ 1 + \ln ( 8 \sqrt { x } ) ]$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left[ 1 + \ln \left( 8 \sqrt { x _ { n } } \right) \right]$$ with \(x _ { 0 } = 1\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the value of \(x _ { 4 }\) to 3 decimal places.
4. Show that your value for \(x _ { 4 }\) is the value of \(\alpha\) correct to 3 decimal places.
5. Another attempt to find \(\alpha\) is made using the iterative formula $$x _ { n + 1 } = \frac { 1 } { 64 } \mathrm { e } ^ { 8 x _ { n } - 2 }$$ with \(x _ { 0 } = 1\). Describe the outcome of this attempt.

2 Fig. 9 shows a sketch of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - 22 x + 24\) and the line \(y = 6 x + 24\). \begin{figure}[h] \end{figure}

1. Differentiate \(y = x ^ { 3 } - 3 x ^ { 2 } - 22 x + 24\) and hence find the \(x\)-coordinates of the turning points of the curve. Give your answers to 2 decimal places.
2. You are given that the line and the curve intersect when \(x = 0\) and when \(x = - 4\). Find algebraically the \(x\)-coordinate of the other point of intersection.
3. Use calculus to find the area of the region bounded by the curve and the line \(y = 6 x + 24\) for \(- 4 \leqslant x \leqslant 0\), shown shaded on Fig. 9.

3

1. The standard formulae for the volume \(V\) and total surface area \(A\) of a solid cylinder of radius \(r\) and height \(h\) are $$V = \pi r ^ { 2 } h \quad \text { and } \quad A = 2 \pi r ^ { 2 } + 2 \pi r h .$$ Use these to show that, for a cylinder with \(A = 200\), $$V = 100 r - \pi r ^ { 3 }$$
2. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) and \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} r ^ { 2 } }\).
3. Use calculus to find the value of \(r\) that gives a maximum value for \(V\) and hence find this maximum value, giving your answers correct to 3 significant figures.

4

1. Differentiate \(x ^ { 3 } - 6 x ^ { 2 } - 15 x + 50\).
2. Hence find the \(x\)-coordinates of the stationary points on the curve \(y = x ^ { 3 } - 6 x ^ { 2 } - 15 x + 50\).

5 Use calculus to find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 6 x ^ { 2 } - 15 x\). Hence find the set of values of \(x\) for which \(x ^ { 3 } - 6 x ^ { 2 } - 15 x\) is an increasing function.

1 The equation of a cubic curve is \(y = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the tangent to the curve when \(x = 3\) passes through the point \(( - 1 , - 41 )\).
2. Use calculus to find the coordinates of the turning points of the curve. You need not distinguish between the maximum and minimum.
3. Sketch the curve, given that the only real root of \(2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2 = 0\) is \(x = 0.2\) correct to 1 decimal place.

2 A cubic curve has equation \(y = x ^ { 3 } - 3 x ^ { 2 } + 1\).

1. Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.
2. Show that the tangent to the curve at the point where \(x = - 1\) has gradient 9 . Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P . Show that the area of the triangle bounded by the normal at P and the \(x\) - and \(y\)-axes is 8 square units.