1.07m Tangents and normals: gradient and equations

5 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.

5. A curve has the equation \(y = \sqrt { 3 x + 11 }\). The point \(P\) on the curve has \(x\)-coordinate 3 .

1. Show that the tangent to the curve at \(P\) has the equation $$3 x - 4 \sqrt { 5 } y + 31 = 0$$ The normal to the curve at \(P\) crosses the \(y\)-axis at \(Q\).
2. Find the \(y\)-coordinate of \(Q\) in the form \(k \sqrt { 5 }\).

1. $$f ( x ) = \frac { 4 x - 1 } { 2 x + 1 }$$ Find an equation for the tangent to the curve \(y = \mathrm { f } ( x )\) at the point where \(x = - 2\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

5. \(\mathrm { f } ( x ) = 5 + \mathrm { e } ^ { 2 x - 3 } , x \in \mathbb { R }\).

1. State the range of f .
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain.
3. Solve the equation \(\mathrm { f } ( x ) = 7\).
4. Find an equation for the tangent to the curve \(y = \mathrm { f } ( x )\) at the point where \(y = 7\).

7. The curve with equation \(y = x ^ { \frac { 5 } { 2 } } \ln \frac { x } { 4 } , x > 0\) crosses the \(x\)-axis at the point \(P\).

1. Write down the coordinates of \(P\). The normal to the curve at \(P\) crosses the \(y\)-axis at the point \(Q\).
2. Find the area of triangle \(O P Q\) where \(O\) is the origin. The curve has a stationary point at \(R\).
3. Find the \(x\)-coordinate of \(R\) in exact form.

8 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-4_787_742_276_719} The diagram shows part of the curve \(y = \ln \left( 5 - x ^ { 2 } \right)\) which meets the \(x\)-axis at the point \(P\) with coordinates \(( 2,0 )\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(P Q\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\).
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures.
3. Deduce an approximation to the area of the region \(B\).

1 Find the equation of the tangent to the curve \(y = \frac { 2 x + 1 } { 3 x - 1 }\) at the point \(\left( 1 , \frac { 3 } { 2 } \right)\), giving your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers.

1 Find the equation of the tangent to the curve \(y = \sqrt { 4 x + 1 }\) at the point ( 2,3 ).

3 Find, in the form \(y = m x + c\), the equation of the tangent to the curve $$y = x ^ { 2 } \ln x$$ at the point with \(x\)-coordinate e.

8 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_476_608_287_756} The diagram shows the curve \(y = ( \ln x ) ^ { 2 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. The point \(P\) on the curve is the point at which the gradient takes its maximum value. Show that the tangent at \(P\) passes through the point \(( 0 , - 1 )\).

8 A curve has equation \(y = ( x + 2 ) \mathrm { e } ^ { - x }\).

1. Find the coordinates of the points where the curve cuts the axes.
2. Find the coordinates of the stationary point, S , on the curve.
3. By evaluating \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at S , determine whether the stationary point is a maximum or a minimum.
4. Sketch the curve in the domain \(- 3 < x < 3\).
5. Find where the normal to the curve at the point \(( 0,2 )\) cuts the curve again.
6. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).

5 The equation of a circle is \(x ^ { 2 } + y ^ { 2 } = 25\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { y }\).
2. Hence find the equation of the normal to the circle at the point ( 3,4 ).

A curve has the equation \(y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , x \neq 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
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2. Show that the curve has a stationary point in the interval [1.3,1.4]. The point \(A\) on the curve has \(x\)-coordinate 2 .
3. Show that the tangent to the curve at \(A\) passes through the origin. The tangent to the curve at \(A\) intersects the curve again at the point \(B\).
   The \(x\)-coordinate of \(B\) is to be estimated using the iterative formula $$x _ { n + 1 } = - \frac { 2 } { 3 } \sqrt { 3 + 3 x _ { n } \mathrm { e } ^ { x _ { n } - 2 } }$$ with \(x _ { 0 } = - 1\).
4. Find \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 7 significant figures and hence state the \(x\)-coordinate of \(B\) to 5 significant figures.

4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

7 A curve is defined by the equation \(\sin 2 x + \cos y = \sqrt { 3 }\).

1. Verify that the point \(\mathrm { P } \left( \frac { 1 } { 6 } \pi , \frac { 1 } { 6 } \pi \right)\) lies on the curve.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at the point P .

2 Fig. 7 shows the curve defined implicitly by the equation $$y ^ { 2 } + y = x ^ { 9 } + 2 x$$ together with the line \(x = 2\). \begin{figure}[h] \end{figure} Not to scale Find the coordinates of the points of intersection of the line and the curve.
Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.

1 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).

1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-1_628_935_728_657} \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{figure}
2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
   (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
   (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).

7 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.

8 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 \(( 1,0 )\).

1 Fig. 8 shows a sketch of part of the curve \(y = x \sin 2 x\), where \(x\) is in radians.
The curve crosses the \(x\)-axis at the point P . The tangent to the curve at P crosses the \(y\)-axis at Q . \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that the \(x\)-coordinates of the turning points of the curve satisfy the equation \(\tan 2 x + 2 x = 0\).
2. Find, in terms of \(\pi\), the \(x\)-coordinate of the point P . Show that the tangent PQ has equation \(2 \pi x + 2 y = \pi ^ { 2 }\).
   Find the exact coordinates of Q.
3. Show that the exact value of the area shaded in Fig. 8 is \(\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)\).

6 The equation of a curve is \(x ^ { 2 } + 3 x y + 4 y ^ { 2 } = 58\). Find the equation of the normal at the point \(( 2,3 )\) on the curve, giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

6 \includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-3_766_611_251_703} The diagram shows the curve with parametric equations $$x = a \sin \theta , \quad y = a \theta \cos \theta$$ where \(a\) is a positive constant and \(- \pi \leqslant \theta \leqslant \pi\). The curve meets the positive \(y\)-axis at \(A\) and the positive \(x\)-axis at \(B\).

1. Write down the value of \(\theta\) corresponding to the origin, and state the coordinates of \(A\) and \(B\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - \theta \tan \theta\), and hence find the equation of the tangent to the curve at the origin.

5.

1. Use the derivatives of \(\sin x\) and \(\cos x\) to prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \tan x ) = \sec ^ { 2 } x$$ The tangent to the curve \(y = 2 x \tan x\) at the point where \(x = \frac { \pi } { 4 }\) meets the \(y\)-axis at the point \(P\).
2. Find the \(y\)-coordinate of \(P\) in the form \(k \pi ^ { 2 }\) where \(k\) is a rational constant.

6. A curve has parametric equations $$x = 3 \cos ^ { 2 } t , \quad y = \sin 2 t , \quad 0 \leq t < \pi$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 2 } { 3 } \cot 2 t\).
2. Find the coordinates of the points where the tangent to the curve is parallel to the \(x\)-axis.
3. Show that the tangent to the curve at the point where \(t = \frac { \pi } { 6 }\) has the equation $$2 x + 3 \sqrt { 3 } y = 9$$
4. Find a cartesian equation for the curve in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

1. A curve has the equation

$$3 x ^ { 2 } - 2 x + x y + y ^ { 2 } - 11 = 0$$ The point \(P\) on the curve has coordinates \(( - 1,3 )\).

1. Show that the normal to the curve at \(P\) has the equation \(y = 2 - x\).
2. Find the coordinates of the point where the normal to the curve at \(P\) meets the curve again.