1.07m Tangents and normals: gradient and equations

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.

12

1. Calculate the gradient of the chord joining the points on the curve \(y = x ^ { 2 } - 7\) for which \(x = 3\) and \(x = 3.1\).
2. Given that \(\mathrm { f } ( x ) = x ^ { 2 } - 7\), find and simplify \(\frac { \mathrm { f } ( 3 + h ) - \mathrm { f } ( 3 ) } { h }\).
3. Use your result in part (ii) to find the gradient of \(y = x ^ { 2 } - 7\) at the point where \(x = 3\), showing your reasoning.
4. Find the equation of the tangent to the curve \(y = x ^ { 2 } - 7\) at the point where \(x = 3\).
5. This tangent crosses the \(x\)-axis at the point P . The curve crosses the positive \(x\)-axis at the point Q . Find the distance PQ , giving your answer correct to 3 decimal places.

4 Find the equation of the tangent to the curve \(y = x ^ { 3 } + 2 x - 7\) at the point where it cuts the \(y\) axis.

9 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 12 x + 9\). The curve passes through the point \(( 2 , - 2 )\).

1. Find the equation of the curve.
2. Show that the curve touches the \(x\)-axis at one point (A) and cuts it at another (B). State the coordinates of A and B.
3. The curve cuts the \(y\)-axis at C . Show that the tangent at C is perpendicular to the normal at B.

9. The curve \(C\) has the equation \(y = \mathrm { f } ( x )\) where $$f ^ { \prime } ( x ) = 1 + \frac { 2 } { \sqrt { x } } , \quad x > 0$$ The straight line \(l\) has the equation \(y = 2 x - 1\) and is a tangent to \(C\) at the point \(P\).

1. State the gradient of \(C\) at \(P\).
2. Find the \(x\)-coordinate of \(P\).
3. Find an equation for \(C\).
4. Show that \(C\) crosses the \(x\)-axis at the point \(( 1,0 )\) and at no other point.

9. \includegraphics[max width=\textwidth, alt={}, center]{33f9663f-26bb-445e-af6e-ca5ca927f7dd-3_638_757_1064_493} The diagram shows the curve with equation \(y = 5 + x - x ^ { 2 }\) and the normal to the curve at the point \(P ( 1,5 )\).

1. Find an equation for the normal to the curve at \(P\) in the form \(y = m x + c\).
2. Find the coordinates of the point \(Q\), where the normal to the curve at \(P\) intersects the curve again.
3. Show that the area of the shaded region bounded by the curve and the straight line \(P Q\) is \(\frac { 4 } { 3 }\).

9. The diagram shows the curve \(C\) with equation \(y = 3 x - 4 \sqrt { x } + 2\) and the tangent to \(C\) at the point \(A\). Given that \(A\) has \(x\)-coordinate 4,

1. show that the tangent to \(C\) at \(A\) has the equation \(y = 2 x - 2\). The shaded region is bounded by \(C\), the tangent to \(C\) at \(A\) and the \(y\)-axis.
2. Find the area of the shaded region.

2 A particle moves along a straight line containing a point O . Its displacement, \(x \mathrm {~m}\), from O at time \(t\) seconds is given by $$x = 12 t - t ^ { 3 } , \text { where } - 10 \leqslant t \leqslant 10$$ Find the values of \(x\) for which the velocity of the particle is zero.

9. The diagram shows the curve with equation \(y = 2 x - 3 \ln ( 2 x + 5 )\) and the normal to the curve at the point \(P ( - 2 , - 4 )\).

1. Find an equation for the normal to the curve at \(P\). The normal to the curve at \(P\) intersects the curve again at the point \(Q\) with \(x\)-coordinate \(q\).
2. Show that \(1 < q < 2\).
3. Show that \(q\) is a solution of the equation $$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
4. Use an iterative process based on the equation above with a starting value of 1.5 to find the value of \(q\) to 3 significant figures and justify the accuracy of your answer.

3. A curve has the equation \(x = y ^ { 2 } - 3 \ln 2 y\).

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { 2 y ^ { 2 } - 3 }$$
2. Find an equation for the tangent to the curve at the point where \(y = \frac { 1 } { 2 }\). Give your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers.

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 \(( 0 , - 20 )\).

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 Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = x ^ { 6 } + \sqrt { x }\).

1. Find the equation of the tangent to the curve \(y = x ^ { 4 }\) at the point where \(x = 2\). Give your answer in the form \(y = m x + c\).
2. Calculate the gradient of the chord joining the points on the curve \(y = x ^ { 4 }\) where \(x = 2\) and \(x = 2.1\).
3. (A) Expand \(( 2 + h ) ^ { 4 }\).
   (B) Simplify \(\frac { ( 2 + h ) ^ { 4 } - 2 ^ { 4 } } { h }\).
   (C) Show how your result in part (iii) (B) can be used to find the gradient of \(y = x ^ { 4 }\) at the point where \(x = 2\).
4. Calculate the gradient of the chord joining the points on the curve \(y = x ^ { 2 } - 7\) for which \(x = 3\) and \(x = 3.1\).
5. Given that \(\mathrm { f } ( x ) = x ^ { 2 } - 7\), find and simplify \(\frac { \mathrm { f } ( 3 + h ) - \mathrm { f } ( 3 ) } { h }\).
6. Use your result in part (ii) to find the gradient of \(y = x ^ { 2 } - 7\) at the point where \(x = 3\), showing your reasoning.
7. Find the equation of the tangent to the curve \(y = x ^ { 2 } - 7\) at the point where \(x = 3\).
8. This tangent crosses the \(x\)-axis at the point P . The curve crosses the positive \(x\)-axis at the point Q . Find the distance PQ , giving your answer correct to 3 decimal places.
9. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8ff8b67d-1489-4cb1-bcd2-b32db674e29f-3_651_770_242_737} \captionsetup{labelformat=empty} \caption{Fig. 12}
   \end{figure} Fig. 12 shows part of the curve \(y = x ^ { 4 }\) and the line \(y = 8 x\), which intersect at the origin and the point P .
   (A) Find the coordinates of P , and show that the area of triangle OPQ is 16 square units.
   (B) Find the area of the region bounded by the line and the curve.
10. You are given that \(\mathrm { f } ( x ) = x ^ { 4 }\).
    (A) Complete this identity for \(\mathrm { f } ( x + h )\). $$\mathrm { f } ( x + h ) = ( x + h ) ^ { 4 } = x ^ { 4 } + 4 x ^ { 3 } h + \ldots$$ (B) Simplify \(\frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
    (C) Find \(\lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\).
    (D) State what this limit represents.

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.

5 Differentiate \(4 x ^ { 2 } + \frac { 1 } { x }\) and hence find the \(x\)-coordinate of the stationary point of the curve \(y = 4 x ^ { 2 } + \frac { 1 } { x }\). \begin{figure}[h] \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find, in exact form, the range of values of \(x\) for which \(x ^ { 3 } - 6 x + 2\) is a decreasing function.
3. Find the equation of the tangent to the curve at the point \(( - 1,7 )\). Find also the coordinates of the point where this tangent crosses the curve again.

2 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x + 3\). The curve passes through the point ( 2,9 ).

1. Find the equation of the tangent to the curve at the point \(( 2,9 )\).
2. Find the equation of the curve and the coordinates of its points of intersection with the \(x\)-axis. Find also the coordinates of the minimum point of this curve.
3. Find the equation of the curve after it has been stretched parallel to the \(x\)-axis with scale factor \(\frac { 1 } { 2 }\). Write down the coordinates of the minimum point of the transformed curve. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-2_1020_940_244_679} \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 \(( 0 , - 20 )\).

1. Express \(\mathrm { f } ( x )\) in factorised form.
2. Show that the equation of the curve may be written as \(y = x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20\).
3. Use calculus to show that, correct to 1 decimal place, the \(x\)-coordinate of the minimum point on the curve is 0.4 . Find also the coordinates of the maximum point on the curve, giving your answers correct to 1 decimal place.
4. State, correct to 1 decimal place, the coordinates of the maximum point on the curve \(y = \mathrm { f } ( 2 x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-3_768_1023_223_598} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} Fig. 11 shows the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\).

1. Use calculus to find \(\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x\) and state what this represents.
2. Find the \(x\)-coordinates of the turning points of the curve \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\), giving your answers in surd form. Hence state the set of values of \(x\) for which \(y = x ^ { 3 } - 3 x ^ { 2 } - x + 3\) is a decreasing function.

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. 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 \(( 2 , - 4 )\).

3. The curve \(C\) has the equation \(y = 2 \mathrm { e } ^ { x } - 6 \ln x\) and passes through the point \(P\) with \(x\)-coordinate 1.

1. Find an equation for the tangent to \(C\) at \(P\). The tangent to \(C\) at \(P\) meets the coordinate axes at the points \(Q\) and \(R\).
2. Show that the area of triangle \(O Q R\), where \(O\) is the origin, is \(\frac { 9 } { 3 - \mathrm { e } }\).

A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).

1. Show that the tangent to the curve at the point where \(x = 1\) has the equation $$7 x - 4 y = 11$$ The curve has a stationary point with \(x\)-coordinate \(\alpha\).
2. Show that \(0.3 < \alpha < 0.4\)
3. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
4. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$ with \(x _ { 0 } = 0.35\), to find \(\alpha\) correct to 5 decimal places.
   You should show the result of each iteration.

1 Fig. 12 is a sketch of the curve \(y = 2 x ^ { 2 } - 11 x + 12\). \begin{figure}[h] \end{figure}

1. Show that the curve intersects the \(x\)-axis at \(( 4,0 )\) and find the coordinates of the other point of intersection of the curve and the \(x\)-axis.
2. Find the equation of the normal to the curve at the point \(( 4,0 )\). Show also that the area of the triangle bounded by this normal and the axes is 1.6 units \({ } ^ { 2 }\).
3. Find the area of the region bounded by the curve and the \(x\)-axis.

2 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 \(( 2 , - 4 )\).

3 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.

4 Fig. 10 shows a sketch of the graph of \(y = 7 x - x ^ { 2 } - 6\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the equation of the tangent to the curve at the point on the curve where \(x = 2\). Show that this tangent crosses the \(x\)-axis where \(x = \frac { 2 } { 3 }\).
2. Show that the curve crosses the \(x\)-axis where \(x = 1\) and find the \(x\)-coordinate of the other point of intersection of the curve with the \(x\)-axis.
3. Find \(\int _ { 1 } ^ { 2 } \left( 7 x - x ^ { 2 } - 6 \right) \mathrm { d } x\). Hence find the area of the region bounded by the curve, the tangent and the \(x\)-axis, shown shaded on Fig. 10. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{12e190fc-437f-499d-9c27-da49a7546755-3_643_1034_267_549} \captionsetup{labelformat=empty} \caption{Fig. 11}
   \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find, in exact form, the range of values of \(x\) for which \(x ^ { 3 } - 6 x + 2\) is a decreasing function.
3. Find the equation of the tangent to the curve at the point \(( - 1,7 )\). Find also the coordinates of the point where this tangent crosses the curve again.

A curve has the equation \(x = y \sqrt { 1 - 2 y }\).

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { 1 - 2 y } } { 1 - 3 y } .$$ The point \(A\) on the curve has \(y\)-coordinate - 1 .
2. Show that the equation of tangent to the curve at \(A\) can be written in the form $$\sqrt { 3 } x + p y + q = 0$$ where \(p\) and \(q\) are integers to be found.