Show dy/dx equals given expression

5 The variables \(x\) and \(y\) satisfy the relation \(\sin y = \tan x\), where \(- \frac { 1 } { 2 } \pi < y < \frac { 1 } { 2 } \pi\). Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \cos x \sqrt { } ( \cos 2 x ) } .$$

The curve \(C\) has equation $$13 x ^ { 2 } + 13 y ^ { 2 } - 10 x y = 52$$ Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) as a function of \(x\) and \(y\), simplifying your answer.
(6)

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 \(y\).

6 Fig. 6 shows the triangle OAP , where O is the origin and A is the point \(( 0,3 )\). The point \(\mathrm { P } ( x , 0 )\) moves on the positive \(x\)-axis. The point \(\mathrm { Q } ( 0 , y )\) moves between O and A in such a way that \(\mathrm { AQ } + \mathrm { AP } = 6\). \begin{figure}[h] \end{figure}

1. Write down the length AQ in terms of \(y\). Hence find AP in terms of \(y\), and show that $$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9 .$$
2. Use this result to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }\).
3. When \(x = 4\) and \(y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2\). Calculate \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) at this time. Section B (36 marks)

7 Given that \(x ^ { 2 } + x y + y ^ { 2 } = 12\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

6 A curve is defined by the equation \(( x + y ) ^ { 2 } = 4 x\). The point \(( 1,1 )\) lies on this curve.
By differentiating implicitly, show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x + y } - 1\).
Hence verify that the curve has a stationary point at \(( 1,1 )\).

2 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).

4 The equation of a curve is given by \(\mathrm { e } ^ { 2 y } = 1 + \sin x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find an expression for \(y\) in terms of \(x\), and differentiate it to verify the result in part (i).

1 Given that \(x ^ { 2 } + x y + y ^ { 2 } = 12\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

3 Fig. 6 shows the triangle OAP , where O is the origin and A is the point \(( 0,3 )\). The point \(\mathrm { P } ( x , 0 )\) moves on the positive \(x\)-axis. The point \(\mathrm { Q } ( 0 , y )\) moves between O and A in such a way that \(\mathrm { AQ } + \mathrm { AP } = 6\). \begin{figure}[h] \end{figure}

1. Write down the length AQ in terms of \(y\). Hence find AP in terms of \(y\), and show that $$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9$$
2. Use this result to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }\).
3. When \(x = 4\) and \(y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2\). Calculate \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) at this time.

6 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 \(y\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x _ { 2 } } }\). \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. (A) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

2 The variables \(x\) and \(y\) satisfy the equation \(x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }\). Both \(x\) and \(y\) are functions of \(t\).
2. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(x = 1 , y = 8\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\).

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

1. A curve has the equation

$$x ^ { 2 } ( 2 + y ) - y ^ { 2 } = 0$$ Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

2. A curve has the equation $$x ^ { 2 } + 2 x y ^ { 2 } + y = 4$$ Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) 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.

3．The curve \(C\) has equation $$x ^ { 2 } + y ^ { 2 } + f x y = g ^ { 2 }$$ where \(f\) and \(g\) are constants and \(g \neq 0\) ．
（a）Find an expression in terms of \(\alpha , \beta\) and \(f\) for the gradient of \(C\) at the point \(( \alpha , \beta )\) ． Given that \(f < 2\) and \(f \neq - 2\) and that the gradient of \(C\) at the point \(( \alpha , \beta )\) is 1 ，
（b）show that \(\alpha = - \beta = \frac { \pm g } { \sqrt { } ( 2 - f ) }\) ． Given that \(f = - 2\) ，
（c）sketch \(C\) ．

7 The variables \(x\) and \(y\) satisfy the equation \(x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }\). Both \(x\) and \(y\) are functions of \(t\).
2. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(x = 1 , y = 8\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\). Section B (36 marks)

5 The equation of a curve is given by \(\mathrm { e } ^ { 2 y } = 1 + \sin x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find an expression for \(y\) in terms of \(x\), and differentiate it to verify the result in part (i).

8 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\). \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. (A) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

8

1. Given that \(14 x ^ { 2 } - 7 x y + y ^ { 2 } = 2\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 28 x - 7 y } { 7 x - 2 y }\).
2. The points \(L\) and \(M\) on the curve \(14 x ^ { 2 } - 7 x y + y ^ { 2 } = 2\) each have \(x\)-coordinate 1 . The tangents to the curve at \(L\) and \(M\) meet at \(N\). Find the coordinates of \(N\).

3 Given that \(y \sin 2 x + \frac { 1 } { x } + y ^ { 2 } = 5\), find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

4

1. A curve is defined by the equation \(x ^ { 2 } - y ^ { 2 } = 8\).
   1. Show that at any point \(( p , q )\) on the curve, where \(q \neq 0\), the gradient of the curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { p } { q }\).
      (2 marks)
   2. Show that the tangents at the points \(( p , q )\) and \(( p , - q )\) intersect on the \(x\)-axis.
      (4 marks)
2. Show that \(x = t + \frac { 2 } { t } , y = t - \frac { 2 } { t }\) are parametric equations of the curve \(x ^ { 2 } - y ^ { 2 } = 8\).
   (2 marks)

6 The line \(l _ { 1 }\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 4 \\ - 5 \\ 3 \end{array} \right] + \lambda \left[ \begin{array} { r } - 1 \\ 3 \\ 1 \end{array} \right]\).
The line \(l _ { 2 }\) has equation \(\mathbf { r } = \left[ \begin{array} { r } 7 \\ - 8 \\ 6 \end{array} \right] + \mu \left[ \begin{array} { r } 2 \\ - 3 \\ 1 \end{array} \right]\).
The point \(P\) lies on \(l _ { 1 }\) where \(\lambda = - 1\). The point \(Q\) lies on \(l _ { 2 }\) where \(\mu = 2\).

1. Show that the vector \(\overrightarrow { P Q }\) is parallel to \(\left[ \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right]\).
2. The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(R ( 3 , b , c )\).
   1. Show that \(b = - 2\) and find the value of \(c\).
   2. The point \(S\) lies on a line through \(P\) that is parallel to \(l _ { 2 }\). The line \(R S\) is perpendicular to the line \(P Q\). \includegraphics[max width=\textwidth, alt={}, center]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-16_887_1159_1320_443} Find the coordinates of \(S\). \(7 \quad\) A curve has equation \(\cos 2 y + y \mathrm { e } ^ { 3 x } = 2 \pi\).
      The point \(A \left( \ln 2 , \frac { \pi } { 4 } \right)\) lies on this curve.