Implicit equations and differentiation

13 Determine the coordinates of the turning points on the curve with equation $$y ^ { 2 } + x y + x ^ { 2 } - x = 1 .$$

5 The equation of a curve is \(3 x ^ { 2 } + 2 x y + y ^ { 2 } = 6\). It is given that there are two points on the curve where the tangent is parallel to the \(x\)-axis.

1. Show by differentiation that, at these points, \(y = - 3 x\).
2. Hence find the coordinates of the two points.

5 \includegraphics[max width=\textwidth, alt={}, center]{7c125770-1ded-4763-8453-b07ef43e83e9-2_446_601_1969_772} The diagram shows the curve with equation $$x ^ { 3 } + x y ^ { 2 } + a y ^ { 2 } - 3 a x ^ { 2 } = 0$$ where \(a\) is a positive constant. The maximum point on the curve is \(M\). Find the \(x\)-coordinate of \(M\) in terms of \(a\).

1. By differentiating \(\frac { 1 } { \cos x }\), show that the derivative of \(\sec x\) is \(\sec x \tan x\). Hence show that if \(y = \ln ( \sec x + \tan x )\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec x\).
2. Using the substitution \(x = ( \sqrt { } 3 ) \tan \theta\), find the exact value of $$\int _ { 1 } ^ { 3 } \frac { 1 } { \sqrt { \left( 3 + x ^ { 2 } \right) } } \mathrm { d } x$$ expressing your answer as a single logarithm.

1 Find the gradient of the curve \(4 x ^ { 2 } + 2 x y + y ^ { 2 } = 12\) at the point \(( 1,2 )\).

4 A curve has equation \(y ^ { 2 } = 5 x - 4\).
Find the gradient of the curve at the points where \(x = 8\).

5 The curve \(C\) has equation $$3 x ^ { 2 } - 5 x y + \mathrm { e } ^ { 2 y - 4 } + 6 = 0$$ The point \(P\) with coordinates \(( 1,2 )\) lies on \(C\). The tangent to \(C\) at \(P\) meets the \(y\)-axis at the point \(A\) and the normal to \(C\) at \(P\) meets the \(y\)-axis at the point \(B\). Find the exact area of triangle \(A B P\).

3 Find the equation of the normal to the curve $$x ^ { 2 } \ln y + 2 x + 5 y = 11$$ at the point \(( 3,1 )\).

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 ).

1. The curve \(C\) has equation

$$( x + y ) ^ { 3 } = 3 x ^ { 2 } - 3 y - 2$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). The point \(P ( 1,0 )\) lies on \(C\).
2. Show that the normal to \(C\) at \(P\) has equation $$y = - 2 x + 2$$
3. Prove that the normal to \(C\) at \(P\) does not meet \(C\) again. You should use algebra for your proof and make your reasoning clear.

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5 A curve has equation \(x ^ { 2 } + 2 y ^ { 2 } + 5 x + 6 y = 10\). Find the equation of the tangent to the curve at the point \(( 2 , - 1 )\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
[0pt] [6]

4 A curve has equation $$3 x ^ { 2 } - y ^ { 2 } - 4 \ln ( 2 y + 3 ) = 26$$ Find the equation of the tangent to the curve at the point \(( 3 , - 1 )\).

2 Given that \(\sin y = x y + x ^ { 2 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

5 A curve is defined implicitly by the equation $$y ^ { 3 } = 2 x y + x ^ { 2 }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 ( x + y ) } { 3 y ^ { 2 } - 2 x }\).
2. Hence write down \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(x\) and \(y\).

1. \hspace{0pt} [In this question you may assume the following formulae for the volume and curved] surface area of a cone of base radius \(r\) and height \(h\) and of a sphere of radius \(r\).

Cone: volume \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) and curved surface area \(S = \pi r \sqrt { h ^ { 2 } + r ^ { 2 } }\) Sphere: volume \(V = \frac { 4 \pi } { 3 } r ^ { 3 }\) and curved surface area \(S = 4 \pi r ^ { 2 }\) \includegraphics[max width=\textwidth, alt={}, center]{78ba3acc-4cca-4d15-8362-a27e425c5859-22_782_755_637_657} Figure 3
Figure 3 shows the design for a garden ornament.
The ornament is made of a hemisphere on top of a truncated cone.
The truncated cone has base radius \(2 r \mathrm {~cm}\), top radius \(r \mathrm {~cm}\) and height \(4 r \mathrm {~cm}\).
The hemisphere has radius \(R \mathrm {~cm}\).
Given that the volume of the ornament is \(2100 \pi \mathrm {~cm} ^ { 3 }\)

1. show that $$R ^ { 3 } = 3150 - 14 r ^ { 3 }$$
2. Find an expression involving \(\frac { \mathrm { d } R } { \mathrm {~d} r }\) in terms of \(r\) and/or \(R\). The base of the truncated cone of the ornament is fixed to the ground.
3. Show that the visible surface area of the ornament, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = ( 3 \sqrt { 17 } - 1 ) \pi r ^ { 2 } + 3 \pi R ^ { 2 }$$
4. Hence show that $$\frac { \mathrm { d } A } { \mathrm {~d} r } = \gamma \pi r - \frac { \delta \pi r ^ { 2 } } { R }$$ where \(\gamma\) and \(\delta\) are real numbers to be determined. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{78ba3acc-4cca-4d15-8362-a27e425c5859-23_705_803_625_630} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} Figure 4 shows a sketch of \(A\) against \(r\), for \(r \geqslant 0\) There is a local minimum at \(r = 0\) and a local maximum at the point \(M\). The overall minimum point is at the point \(N\), where the gradient of the curve is undefined.
5. 1. Determine the \(r\) coordinate of the point \(N\).
   2. Explain why, for the ornament, \(r\) must be less than this value.
6. Show that the \(r\) coordinate of the point \(M\) is $$\sqrt [ 3 ] { \frac { p ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } { 3 q ^ { 2 } + ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } }$$ where \(p\) and \(q\) are integers to be determined.

4 A curve \(C\) has equation \(x ^ { 3 } - 3 x y + y ^ { 2 } = 4\). Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(( 0,2 )\) of \(C\).

2 A curve has equation $$( x + 1 ) y + y ^ { 2 } = 2$$

1. Show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 2 } { 3 }\) at the point \(( 0 , - 2 )\).
2. Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( 0 , - 2 )\).

10 The curve \(C\) has equation $$y = x ^ { 2 } + \lambda \sin ( x + y ) ,$$ where \(\lambda\) is a constant, and passes through the point \(A \left( \frac { 1 } { 4 } \pi , \frac { 1 } { 4 } \pi \right)\). Show that \(C\) has no tangent which is parallel to the \(y\)-axis. Show that, at \(A\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 - \frac { 1 } { 64 } \pi ( 4 - \pi ) ( \pi + 2 ) ^ { 2 }$$

7 The equation of a curve is \(x ^ { 3 } + 3 x y ^ { 2 } - y ^ { 3 } = 5\).

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

7 The equation of a curve is $$y ^ { 3 } + 4 x y = 16$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 4 y } { 3 y ^ { 2 } + 4 x }\).
2. Show that the curve has no stationary points.
3. Find the coordinates of the point on the curve where the tangent is parallel to the \(y\)-axis.

6 In this question you must show detailed reasoning. The diagram shows the curve with equation \(4 x y = 2 \left( x ^ { 2 } + 4 y ^ { 2 } \right) - 9 x\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x - 4 y - 9 } { 4 x - 16 y }\). At the point \(P\) on the curve the tangent to the curve is parallel to the \(y\)-axis and at the point \(Q\) on the curve the tangent to the curve is parallel to the \(x\)-axis.
2. Show that the distance \(P Q\) is \(k \sqrt { 5 }\), where \(k\) is a rational number to be determined.

2 The line \(4 y = x + c\), where \(c\) is a constant, is a tangent to the curve \(y ^ { 2 } = x + 3\) at the point \(P\) on the curve.

1. Find the value of \(c\).
2. Find the coordinates of \(P\).

7 A curve has equation \(2 x ^ { 3 } + 6 x y - 3 y ^ { 2 } = 2\).
Show that there are no points on this curve where the tangent is parallel to \(y = x\).

1 A curve has implicit equation \(y ^ { 2 } + 2 x \ln y = x ^ { 2 }\).
Verify that the point \(( 1,1 )\) lies on the curve, and find the gradient of the curve at this point.

2. A curve has the equation $$x ^ { 3 } + 2 x y - y ^ { 2 } + 24 = 0$$ Show that the normal to the curve at the point \(( 2 , - 4 )\) has the equation \(y = 3 x - 10\).

4. The curve \(C\) has equation $$4 x ^ { 2 } - y ^ { 3 } - 4 x y + 2 ^ { y } = 0$$ The point \(P\) with coordinates \(( - 2,4 )\) lies on \(C\).

1. Find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(P\). The normal to \(C\) at \(P\) meets the \(y\)-axis at the point \(A\).
2. Find the \(y\) coordinate of \(A\), giving your answer in the form \(p + q \ln 2\), where \(p\) and \(q\) are constants to be determined.
   (3)

5 The curve with equation $$6 \mathrm { e } ^ { 2 x } + k \mathrm { e } ^ { y } + \mathrm { e } ^ { 2 y } = c$$ where \(k\) and \(c\) are constants, passes through the point \(P\) with coordinates \(( \ln 3 , \ln 2 )\).

1. Show that \(58 + 2 k = c\).
2. Given also that the gradient of the curve at \(P\) is - 6 , find the values of \(k\) and \(c\).

7 The curve with equation \(\mathrm { e } ^ { 2 x } - 18 x + y ^ { 3 } + y = 11\) has a stationary point at \(( p , q )\).

1. Find the exact value of \(p\).
2. Show that \(q = \sqrt [ 3 ] { 2 + 18 \ln 3 - q }\).
3. Show by calculation that the value of \(q\) lies between 2.5 and 3.0.
4. Use an iterative formula, based on the equation in (b), to find the value of \(q\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

10. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C\) with equation $$x = y \mathrm { e } ^ { 2 y } \quad y \in \mathbb { R }$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( 1 + 2 y ) }$$ Given that the straight line with equation \(x = k\), where \(k\) is a constant, cuts \(C\) at exactly two points,
2. find the range of possible values for \(k\).

7. Given that $$x = 6 \sin ^ { 2 } 2 y \quad 0 < y < \frac { \pi } { 4 }$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { A \sqrt { \left( B x - x ^ { 2 } \right) } }$$ where \(A\) and \(B\) are integers to be found.
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6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 } { \sqrt { } ( 4 x + a ) }\), where \(a\) is a constant. The point \(P ( 2,14 )\) lies on the curve and the normal to the curve at \(P\) is \(3 y + x = 5\).

1. Show that \(a = 8\).
2. Find the equation of the curve.