1.07m Tangents and normals: gradient and equations

8 \includegraphics[max width=\textwidth, alt={}, center]{8f815127-61b2-4a7f-8687-747950ea6597-3_693_1061_262_541} The diagram shows the curve \(y = x ^ { 2 } \mathrm { e } ^ { - x }\) and its maximum point \(M\).

1. Find the \(x\)-coordinate of \(M\).
2. Show that the tangent to the curve at the point where \(x = 1\) passes through the origin.
3. Use the trapezium rule, with two intervals, to estimate the value of $$\int _ { 1 } ^ { 3 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.

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

1. Find the coordinates of the two points on the curve where \(x = 1\).
2. Show by differentiation that at one of these points the tangent to the curve is parallel to the \(x\)-axis. Find the equation of the tangent to the curve at the other point, giving your answer in the form \(a x + b y + c = 0\).

8 \includegraphics[max width=\textwidth, alt={}, center]{dde12c57-5129-43ae-b385-9a8f21f51e49-3_566_787_255_680} The diagram shows the curve \(y = x \sin x\), for \(0 \leqslant x \leqslant \pi\). The point \(Q \left( \frac { 1 } { 2 } \pi , \frac { 1 } { 2 } \pi \right)\) lies on the curve.

1. Show that the normal to the curve at \(Q\) passes through the point \(( \pi , 0 )\).
2. Find \(\frac { \mathrm { d } } { \mathrm { d } x } ( \sin x - x \cos x )\).
3. Hence evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x\).

8 \includegraphics[max width=\textwidth, alt={}, center]{2aceb797-097c-499b-99b6-cce9f287cb51-3_566_787_255_680} The diagram shows the curve \(y = x \sin x\), for \(0 \leqslant x \leqslant \pi\). The point \(Q \left( \frac { 1 } { 2 } \pi , \frac { 1 } { 2 } \pi \right)\) lies on the curve.

1. Show that the normal to the curve at \(Q\) passes through the point \(( \pi , 0 )\).
2. Find \(\frac { \mathrm { d } } { \mathrm { d } x } ( \sin x - x \cos x )\).
3. Hence evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x\).

3 \includegraphics[max width=\textwidth, alt={}, center]{55794ceb-2d52-459c-8724-6a6a29ab159a-2_705_737_591_703} The diagram shows the part of the curve \(y = \frac { 1 } { 2 } \tan 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Find the \(x\)-coordinates of the points on this part of the curve at which the gradient is 4 .

7 The parametric equations of a curve are $$x = \mathrm { e } ^ { 3 t } , \quad y = t ^ { 2 } \mathrm { e } ^ { t } + 3$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ( t + 2 ) } { 3 \mathrm { e } ^ { 2 t } }\).
2. Show that the tangent to the curve at the point \(( 1,3 )\) is parallel to the \(x\)-axis.
3. Find the exact coordinates of the other point on the curve at which the tangent is parallel to the \(x\)-axis.

6 The parametric equations of a curve are $$x = 1 + 2 \sin ^ { 2 } \theta , \quad y = 4 \tan \theta$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sin \theta \cos ^ { 3 } \theta }\).
2. Find the equation of the tangent to the curve at the point where \(\theta = \frac { 1 } { 4 } \pi\), giving your answer in the form \(y = m x + c\).

5 The parametric equations of a curve are $$x = 1 + \sqrt { } t , \quad y = 3 \ln t$$

1. Find the exact value of the gradient of the curve at the point \(P\) where \(y = 6\).
2. Show that the tangent to the curve at \(P\) passes through the point \(( 1,0 )\).

3 A curve has equation $$3 \ln x + 6 x y + y ^ { 2 } = 16$$ Find the equation of the normal to the curve at the point \(( 1,2 )\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

5 A curve has parametric equations $$x = t + \ln ( t + 1 ) , \quad y = 3 t \mathrm { e } ^ { 2 t }$$

1. Find the equation of the tangent to the curve at the origin.
2. Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places. [4]

2 A curve has equation $$y = \frac { 3 x + 1 } { x - 5 }$$ Find the coordinates of the points on the curve at which the gradient is - 4 .

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 \(p\).

7 \includegraphics[max width=\textwidth, alt={}, center]{1990cbac-d96f-4484-be4b-67dab35b3147-10_647_519_260_813} For the curve shown in the diagram, the normal to the curve at the point \(P\) with coordinates \(( x , y )\) meets the \(x\)-axis at \(N\). The point \(M\) is the foot of the perpendicular from \(P\) to the \(x\)-axis. The curve is such that for all values of \(x\) in the interval \(0 \leqslant x < \frac { 1 } { 2 } \pi\), the area of triangle \(P M N\) is equal to \(\tan x\).

1. 1. Show that \(\frac { M N } { y } = \frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence show that \(x\) and \(y\) satisfy the differential equation \(\frac { 1 } { 2 } y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \tan x\).
2. Given that \(y = 1\) when \(x = 0\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).

5 The parametric equations of a curve are $$x = t \mathrm { e } ^ { 2 t } , \quad y = t ^ { 2 } + t + 3$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - 2 t }\).
2. Hence show that the normal to the curve, where \(t = - 1\), passes through the point \(\left( 0,3 - \frac { 1 } { \mathrm { e } ^ { 4 } } \right)\).

9 The equation of a curve is \(y \mathrm { e } ^ { 2 x } - y ^ { 2 } \mathrm { e } ^ { x } = 2\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y \mathrm { e } ^ { x } - y ^ { 2 } } { 2 y - \mathrm { e } ^ { x } }\).
2. Find the exact coordinates of the point on the curve where the tangent is parallel to the \(y\)-axis.

7 The equation of a curve is \(\ln ( x + y ) = x - 2 y\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x + y - 1 } { 2 ( x + y ) + 1 }\).
2. Find the coordinates of the point on the curve where the tangent is parallel to the \(x\)-axis. \(\quad\) [3]

7 The equation of a curve is \(y = \frac { x } { \cos ^ { 2 } x }\), for \(0 \leqslant x < \frac { 1 } { 2 } \pi\). At the point where \(x = a\), the tangent to the curve has gradient equal to 12 .

1. Show that \(a = \cos ^ { - 1 } \left( \sqrt [ 3 ] { \frac { \cos a + 2 a \sin a } { 12 } } \right)\).
2. Verify by calculation that \(a\) lies between 0.9 and 1 .
3. Use an iterative formula based on the equation in part (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

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 .

12. The curve with equation \(y = \mathrm { f } ( x ) , x > 0\), passes through the point \(P ( 4 , - 2 )\). Given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x \sqrt { x } - 10 x ^ { - \frac { 1 } { 2 } }$$

1. find the equation of the tangent to the curve at \(P\), writing your answer in the form \(y = m x + c\), where \(m\) and \(c\) are integers to be found.
2. Find \(\mathrm { f } ( x )\).

11. A curve has equation \(y = \mathrm { f } ( x )\), where $$f ^ { \prime \prime } ( x ) = \frac { 6 } { \sqrt { x ^ { 3 } } } + x \quad x > 0$$ The point \(P ( 4 , - 50 )\) lies on the curve.
Given that \(\mathrm { f } ^ { \prime } ( x ) = - 4\) at \(P\),

1. find the equation of the normal at \(P\), writing your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants,
   (3)
2. find \(\mathrm { f } ( x )\).
   (8)

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1. A curve has equation

$$y = 2 x ^ { 3 } - 5 x ^ { 2 } - \frac { 3 } { 2 x } + 7 \quad x > 0$$

1. Find, in simplest form, \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) The point \(P\) lies on the curve and has \(x\) coordinate \(\frac { 1 } { 2 }\)
2. Find an equation of the normal to the curve at \(P\), writing your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers to be found.

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6. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where \(x > 0\) Given that

• \(\mathrm { f } ^ { \prime } ( x ) = \frac { ( x + 3 ) ^ { 2 } } { x \sqrt { x } }\)
• the point \(P ( 4,20 )\) lies on \(C\)
• Find \(\mathrm { f } ( x )\), simplifying your answer.
  

10. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = ( 3 x + 20 ) ( x + 6 ) ( 2 x - 3 )$$

1. Use the given information to state the values of \(x\) for which $$f ( x ) > 0$$
2. Expand \(( 3 x + 20 ) ( x + 6 ) ( 2 x - 3 )\), writing your answer as a polynomial in simplest form. The straight line \(l\) is the tangent to \(C\) at the point where \(C\) cuts the \(y\)-axis.
   Given that \(l\) cuts \(C\) at the point \(P\), as shown in Figure 4,
3. find, using algebra, the \(x\) coordinate of \(P\) (Solutions based on calculator technology are not acceptable.)

1. A curve \(C\) has equation \(y = \mathrm { f } ( x ) , \quad x > 0\)

Given that

• \(\mathrm { f } ^ { \prime \prime } ( x ) = 4 x + \frac { 1 } { \sqrt { x } }\)
• the point \(P\) has \(x\) coordinate 4 and lies on \(C\)
• the tangent to \(C\) at \(P\) has equation \(y = 3 x + 4\)
  1. find an equation of the normal to \(C\) at \(P\)
  2. find \(\mathrm { f } ( x )\), writing your answer in simplest form.

1. The curve \(C\) has equation

$$y = \frac { 5 x ^ { 3 } - 8 } { 2 x ^ { 2 } } \quad x > 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) writing your answer in simplest form. The point \(P ( 2,4 )\) lies on \(C\).
2. Find an equation for the tangent to \(C\) at \(P\) writing your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.