Parametric equations

3 The parametric equations of a curve are $$x = 2 \theta + \sin 2 \theta , \quad y = 1 - \cos 2 \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \theta\).

3 The parametric equations of a curve are $$x = 2 \theta + \sin 2 \theta , \quad y = 1 - \cos 2 \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \theta\).

11. In this question you must show detailed reasoning. A curve has parametric equations $$x = \cos t - 3 t \text { and } y = 3 t - 4 \cos t - \sin 2 t , \text { for } 0 \leqslant t \leqslant \pi .$$ Show that the gradient of the curve is always negative.
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2. The curve \(C\) is described by the parametric equations $$x = 3 \cos t , \quad y = \cos 2 t , \quad 0 \leq t \leq \pi .$$

1. Find a cartesian equation of the curve \(C\).
2. Draw a sketch of the curve \(C\).

1 A curve is defined by the parametric equations $$x = \cos \theta \text { and } y = \sin \theta \quad \text { where } 0 \leq \theta \leq 2 \pi$$ Which of the options shown below is a Cartesian equation for this curve?
Circle your answer. $$\frac { y } { x } = \tan \theta \quad x ^ { 2 } + y ^ { 2 } = 1 \quad x ^ { 2 } - y ^ { 2 } = 1 \quad x ^ { 2 } y ^ { 2 } = 1$$

3. The curve \(C\) has parametric equations $$x = 3 + 2 \sin t \quad y = \frac { 6 } { 7 + \cos 2 t } \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$

1. Show that \(C\) has Cartesian equation $$y = \frac { 12 } { ( 7 - x ) ( 1 + x ) } \quad p \leqslant x \leqslant q$$ where \(p\) and \(q\) are constants to be found.
2. Hence, find a Cartesian equation for \(C\) in the form $$y = \frac { a } { x + b } + \frac { c } { x + d } \quad p \leqslant x \leqslant q$$ where \(a , b , c\) and \(d\) are constants.

4 A curve is defined by parametric equations $$x = \frac { 1 } { t } - 1 , y = \frac { 2 + t } { 1 + t }$$ Show that the cartesian equation of the curve is \(y = \frac { 3 + 2 x } { 2 + x }\).

4 A curve is defined by parametric equations $$x = \frac { 1 } { t } - 1 , y = \frac { 2 + t } { 1 + t }$$ Show that the cartesian equation of the curve is \(y = \frac { 3 + 2 x } { 2 + x }\).

10. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = \frac { 20 t } { 2 t + 1 } \quad y = t ( t - 4 ) , \quad t > 0$$ The curve cuts the \(x\)-axis at the point \(P\).

1. Find the \(x\) coordinate of \(P\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( t - A ) ( 2 t + 1 ) ^ { 2 } } { B }\) where \(A\) and \(B\) are constants to be found.
3. 1. Make \(t\) the subject of the formula $$x = \frac { 20 t } { 2 t + 1 }$$
   2. Hence find a cartesian equation of the curve \(C\). Write your answer in the form $$y = \mathrm { f } ( x ) , \quad 0 < x < k$$ where \(\mathrm { f } ( x )\) is a single fraction and \(k\) is a constant to be found.

3 The parametric equations of a curve are $$x = t + \ln ( t + 2 ) , \quad y = ( t - 1 ) \mathrm { e } ^ { - 2 t }$$ where \(t > - 2\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer.
2. Find the exact \(y\)-coordinate of the stationary point of the curve.

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

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the exact value of \(t\) at the stationary point.
3. Find the gradient of the curve at the point where it crosses the \(x\)-axis.
4. Express \(\sin 2 \theta ( 3 \sec \theta + 4 \operatorname { cosec } \theta )\) in the form \(a \sin \theta + b \cos \theta\), where \(a\) and \(b\) are integers.
5. Hence express \(\sin 2 \theta ( 3 \sec \theta + 4 \operatorname { cosec } \theta )\) in the form \(R \sin ( \theta + \alpha )\) where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
6. Using the result of part (ii), solve the equation \(\sin 2 \theta ( 3 \sec \theta + 4 \operatorname { cosec } \theta ) = 7\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

5 A line PQ is of length \(k\) (where \(k > 1\) ) and it passes through the point ( 1,0 ). PQ is inclined at angle \(\theta\) to the positive \(x\)-axis. The end Q moves along the \(y\)-axis. See Fig. 5. The end P traces out a locus. \begin{figure}[h] \end{figure}

1. Show that the locus of P may be expressed parametrically as follows. $$x = k \cos \theta \quad y = k \sin \theta - \tan \theta$$ You are now required to investigate curves with these parametric equations, where \(k\) may take any non-zero value and \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\).
2. Use your calculator to sketch the curve in each of the cases \(k = 2 , k = 1 , k = \frac { 1 } { 2 }\) and \(k = - 1\).
3. For what value(s) of \(k\) does the curve have
   (A) an asymptote (you should state what the asymptote is),
   (B) a cusp,
   (C) a loop?
4. For the case \(k = 2\), find the angle at which the curve crosses itself.
5. For the case \(k = 8\), find in an exact form the coordinates of the highest point on the loop.
6. Verify that the cartesian equation of the curve is $$y ^ { 2 } = \frac { ( x - 1 ) ^ { 2 } } { x ^ { 2 } } \left( k ^ { 2 } - x ^ { 2 } \right) .$$

3 A curve is defined parametrically by the equations $$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$ Find the gradient of the curve at the point where \(t = 2\).

1. A curve is defined by the parametric equations

$$x = t ^ { 3 } + 2 , \quad y = t ^ { 2 } - 1$$ Find the gradient of the curve at the point where \(t = - 2\)

11. The curve \(C\) has parametric equations $$x = \sin 2 \theta \quad y = \operatorname { cosec } ^ { 3 } \theta \quad 0 < \theta < \frac { \pi } { 2 }$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\)
2. Hence find the exact value of the gradient of the tangent to \(C\) at the point where \(y = 8\)
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3. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of a part of the curve \(C\) with parametric equations $$x = t ^ { 3 } , y = t ^ { 2 } .$$ The tangent at the point \(P ( 8,4 )\) cuts \(C\) at the point \(Q\) ．
Find the area of the shaded region between \(P Q\) and \(C\) ．

9. \begin{figure}[h] \end{figure} The curve \(C\) has parametric equations $$x = \ln ( t + 2 ) , \quad y = \frac { 4 } { t ^ { 2 } } \quad t > 0$$ The finite region \(R\), shown shaded in Figure 2, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = \ln 3\) and \(x = \ln 5\)

1. Show that the area of \(R\) is given by the integral $$\int _ { 1 } ^ { 3 } \frac { 4 } { t ^ { 2 } ( t + 2 ) } \mathrm { d } t$$
2. Hence find an exact value for the area of \(R\). Write your answer in the form ( \(a + \ln b\) ), where \(a\) and \(b\) are rational numbers.
3. Find a cartesian equation of the curve \(C\) in the form \(y = \mathrm { f } ( x )\).

3. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of a part of the curve \(C\) with parametric equations $$x = t ^ { 3 } , y = t ^ { 2 } .$$ The tangent at the point \(P ( 8,4 )\) cuts \(C\) at the point \(Q\) ．
Find the area of the shaded region between \(P Q\) and \(C\) ．

2 A curve \(C\) has parametric equations $$x = \mathrm { e } ^ { t } \cos t , \quad y = \mathrm { e } ^ { t } \sin t , \quad \text { for } 0 \leqslant t \leqslant \frac { 1 } { 2 } \pi$$ Find the arc length of \(C\).

5 The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 2 } { 3 } \mathrm { t } ^ { \frac { 3 } { 2 } } - 2 \mathrm { t } ^ { \frac { 1 } { 2 } } , \quad \mathrm { y } = 2 \mathrm { t } + 5 , \quad \text { for } 0 < t \leqslant 3$$

1. Find the exact length of \(C\).
2. Find the set of values of \(t\) for which \(\frac { d ^ { 2 } y } { d x ^ { 2 } } > 0\).

3 The curve \(C\) has parametric equations \(x = 2 t ^ { 3 } - 6 t , y = 6 t ^ { 2 }\).

1. Find the length of the arc of \(C\) for which \(0 \leqslant t \leqslant 1\).
2. Find the area of the surface generated when the arc of \(C\) for which \(0 \leqslant t \leqslant 1\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
3. Show that the equation of the normal to \(C\) at the point with parameter \(t\) is $$y = \frac { 1 } { 2 } \left( \frac { 1 } { t } - t \right) x + 2 t ^ { 2 } + t ^ { 4 } + 3$$
4. Find the cartesian equation of the envelope of the normals to \(C\).
5. The point \(\mathrm { P } ( 64 , a )\) is the centre of curvature corresponding to a point on \(C\). Find \(a\).

4 A curve is given by the parametric equations \(x = t ^ { 2 } , y = 3 t\) for all values of \(t\). Find the equation of the tangent to the curve at the point where \(t = - 2\).

4 A curve is given by the parametric equations \(x = t ^ { 2 } , y = 3 t\) for all values of \(t\). Find the equation of the tangent to the curve at the point where \(t = - 2\).

15. \begin{figure}[h] \end{figure} A table top, in the shape of a parallelogram, is made from two types of wood. The design is shown in Fig. 1. The area inside the ellipse is made from one type of wood, and the surrounding area is made from a second type of wood. The ellipse has parametric equations, $$x = 5 \cos \theta , \quad y = 4 \sin \theta , \quad 0 \leq \theta < 2 \pi$$ The parallelogram consists of four line segments, which are tangents to the ellipse at the points where \(\theta = \alpha , \theta = - \alpha , \theta = \pi - \alpha , \theta = - \pi + \alpha\).

1. Find an equation of the tangent to the ellipse at ( \(5 \cos \alpha , 4 \sin \alpha\) ), and show that it can be written in the form $$5 y \sin \alpha + 4 x \cos \alpha = 20$$
2. Find by integration the area enclosed by the ellipse.
3. Hence show that the area enclosed between the ellipse and the parallelogram is $$\frac { 80 } { \sin 2 \alpha } - 20 \pi$$
4. Given that \(0 < \alpha < \frac { \pi } { 4 }\), find the value of \(\alpha\) for which the areas of two types of wood are equal.
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13 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that \(t _ { 1 } t _ { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. \(\mathbf { C 1 . }\)

13 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that \(t _ { 1 } t _ { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. \(\mathbf { C 1 . }\)

7. \begin{figure}[h] \end{figure} Figure 4 shows a shape \(S ( \theta )\) made up of five line segments \(A B , B C , C D , D E\) and \(E A\) ． The lengths of the sides are \(A B = B C = 5 \mathrm {~cm} , C D = E A = 3 \mathrm {~cm}\) and \(D E = 7 \mathrm {~cm}\) ． Angle \(B A E =\) angle \(B C D = \theta\) radians． The length of each line segment always remains the same but the value of \(\theta\) can be varied so that different symmetrical shapes can be formed，with the added restriction that none of the line segments cross．
（a）Sketch \(S ( \pi )\) ，labelling the vertices clearly． The shape \(S ( \phi )\) is a trapezium．
（b）Sketch \(S ( \phi )\) and calculate the value of \(\phi\) ． The smallest possible value for \(\theta\) is \(\alpha\) ，where \(\alpha > 0\) ，and the largest possible value for \(\theta\) is \(\beta\) ， where \(\beta > \pi\) ．
（c）Show that \(\alpha = \arccos \left( \frac { 29 } { 40 } \right) \cdot \left[ \arccos ( x ) \right.\) is an alternative notation for \(\left. \cos ^ { - 1 } ( x ) \right]\)
（d）Find the value of \(\beta\) ． The area，in \(\mathrm { cm } ^ { 2 }\) ，of shape \(S ( \theta )\) is \(R ( \theta )\) ．
（e）Show that for \(\alpha \leqslant \theta < \pi\) $$R ( \theta ) = 15 \sin \theta + \frac { 7 } { 4 } \sqrt { 87 - 120 \cos \theta }$$ Given that this formula for \(R ( \theta )\) holds for \(\alpha \leqslant \theta \leqslant \beta\)
(f) show that \(R ( \theta )\) has only one stationary point and that this occurs when \(\theta = \frac { 2 \pi } { 3 }\)
(g) find the maximum and minimum values of \(R ( \theta )\). FOR STYLE, CLARITY AND PRESENTATION: 7 MARKS TOTAL FOR PAPER: 100 MARKS
END

2 It is given that $$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = t e ^ { t }$$

1. Show that \(\frac { d y } { d x } = - e ^ { t } \left( t ^ { 3 } + t ^ { 2 } \right)\).
2. Find \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) in terms of \(t\).

5 The curve \(C\) has parametric equations $$x = 3 t + 2 t ^ { - 1 } + a t ^ { 3 } , \quad y = 4 t - \frac { 3 } { 2 } t ^ { - 1 } + b t ^ { 3 } , \quad \text { for } 1 \leqslant t \leqslant 2$$ where \(a\) and \(b\) are constants.

1. It is given that \(a = \frac { 2 } { 3 }\) and \(b = - \frac { 1 } { 2 }\). Show that \(\left( \frac { d x } { d t } \right) ^ { 2 } + \left( \frac { d y } { d t } \right) ^ { 2 } = \frac { 25 } { 4 } \left( t ^ { 2 } + t ^ { - 2 } \right) ^ { 2 }\) and find the exact length of \(C\).
2. It is given instead that \(\mathrm { a } = \mathrm { b } = 0\). Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) when \(t = 1\).
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5 The curve \(C\) has parametric equations $$\mathrm { x } = \frac { 1 } { 2 } \mathrm { t } ^ { 2 } - \ln \mathrm { t } , \quad \mathrm { y } = 2 \mathrm { t } + 1 , \quad \text { for } \frac { 1 } { 2 } \leqslant t \leqslant 2$$

1. Find the exact length of \(C\).
2. Find \(\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dx } ^ { 2 } }\) in terms of \(t\), simplifying your answer.

6 A curve has parametric equations $$x = 9 t - \ln ( 9 t ) , \quad y = t ^ { 3 } - \ln \left( t ^ { 3 } \right)$$ Show that there is only one value of \(t\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) and state that value.

5 The parametric equations of a curve are $$x = 2 + 3 \sin \theta \text { and } y = 1 - 2 \cos \theta \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$$

1. Find the coordinates of the point on the curve where the gradient is \(\frac { 1 } { 2 }\).
2. Find the cartesian equation of the curve.

1 A family of curves is given by the parametric equations
\(x ( t ) = \cos ( t ) - \frac { \cos ( ( m + 1 ) t ) } { m + 1 }\) and \(y ( t ) = \sin ( t ) - \frac { \sin ( ( m + 1 ) t ) } { m + 1 }\)
where \(0 \leqslant t < 2 \pi\) and \(m\) is a positive integer.

1. 1. Sketch the curves in the cases \(m = 3 , m = 4\) and \(m = 5\) on separate axes in the Printed Answer Booklet.
   2. State one common feature of these three curves.
   3. State a feature for the case \(m = 4\) which is absent in the cases \(m = 3\) and \(m = 5\).
2. 1. Determine, in terms of \(m\), the values of \(t\) for which \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\) but \(\frac { \mathrm { d } y } { \mathrm {~d} t } \neq 0\).
   2. Describe the tangent to the curve at the points corresponding to such values of \(t\).
3. 1. Show that the curve lies between the circle centred at the origin with radius $$1 - \frac { 1 } { m + 1 }$$ and the circle centred at the origin with radius $$1 + \frac { 1 } { m + 1 }$$
   2. Hence, or otherwise, show that the area \(A\) bounded by the curve satisfies $$\frac { m ^ { 2 } \pi } { ( m + 1 ) ^ { 2 } } < A < \frac { ( m + 2 ) ^ { 2 } \pi } { ( m + 1 ) ^ { 2 } }$$
   3. Find the limit of the area bounded by the curve as \(m\) tends to infinity.
4. The arc length of a curve defined by parametric equations \(x ( t )\) and \(y ( t )\) between points corresponding to \(t = c\) and \(t = d\), where \(c < d\), is $$\int _ { c } ^ { d } \sqrt { \left( \frac { \mathrm {~d} x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } } \mathrm {~d} t$$ Use this to show that the length of the curve is independent of \(m\).

3．The curve \(C\) has parametric equations $$x = 15 t - t ^ { 3 } , \quad y = 3 - 2 t ^ { 2 }$$ Find the values of \(t\) at the points where the normal to \(C\) at \(( 14,1 )\) cuts \(C\) again．

4. A curve \(C\) has parametric equations $$x = \sin ^ { 2 } t , \quad y = 2 \tan t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). The tangent to \(C\) at the point where \(t = \frac { \pi } { 3 }\) cuts the \(x\)-axis at the point \(P\).
2. Find the \(x\)-coordinate of \(P\).
   \section*{LU}

3．The curve \(C\) has parametric equations $$x = 15 t - t ^ { 3 } , \quad y = 3 - 2 t ^ { 2 }$$ Find the values of \(t\) at the points where the normal to \(C\) at \(( 14,1 )\) cuts \(C\) again．

6 The curve \(C\) has parametric equations $$x = t ^ { 2 } , \quad y = \frac { 1 } { 4 } t ^ { 4 } - \ln t$$ for \(1 \leqslant t \leqslant 2\). Find the area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(y\)-axis.

1. A curve has parametric equations

$$x = 4 \mathrm { e } ^ { \frac { 1 } { 2 } t } \quad y = \mathrm { e } ^ { t } - t \quad 0 \leqslant t \leqslant 4$$ The curve is rotated through \(2 \pi\) radians about the \(x\)-axis.
Show that the area of the curved surface generated is $$\pi \left( \mathrm { e } ^ { 8 } + A \mathrm { e } ^ { 4 } + B \right)$$ where \(A\) and \(B\) are constants to be determined.

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with parametric equations $$x = \ln ( \sec \theta + \tan \theta ) - \sin \theta \quad y = \cos \theta \quad 0 \leqslant \theta \leqslant \frac { \pi } { 4 }$$ The curve \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis and is used to form a solid of revolution \(S\). Using calculus, show that the total surface area of \(S\) is given by $$\frac { \pi } { 2 } ( p + q \sqrt { 2 } )$$ where \(p\) and \(q\) are integers to be determined.

6 A curve has parametric equations $$x = \mathrm { e } ^ { 2 t } , \quad y = \frac { 2 t } { 1 + t }$$

1. Find the gradient of the curve at the point where \(t = 0\).
2. Find \(y\) in terms of \(x\).

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

1. Find the gradient of the curve at the point where \(t = 0\).
2. Find \(y\) in terms of \(x\).

2 Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u , y = u + \frac { 1 } { u } , \quad 1 \leqslant u \leqslant 10$$ The point A lies on the \(x\)-axis and AC is parallel to the \(y\)-axis. The tangent to the curve at C makes an angle \(\theta\) with AC, as shown. \begin{figure}[h] \end{figure}

1. Find the lengths \(\mathrm { OA } , \mathrm { OB }\) and AC .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\). Hence find the angle \(\theta\).
3. Show that the cartesian equation of the curve is \(y = \mathrm { e } ^ { \frac { 1 } { 5 } x } + \mathrm { e } ^ { - \frac { 1 } { 5 } x }\). An object is formed by rotating the region OACB through \(360 ^ { \circ }\) about \(\mathrm { O } x\).
4. Find the volume of the object.

4. The rectangular hyperbola \(H\) has parametric equations $$x = 4 t , \quad y = \frac { 4 } { t }$$ The straight line with equation \(3 y - 2 x = 10\) intersects \(H\) at the points \(A\) and \(B\). Given that the point \(A\) is above the \(x\)-axis,

1. find the coordinates of the point \(A\) and the coordinates of the point \(B\).
2. Find the coordinates of the midpoint of \(A B\).

6 A curve has parametric equations $$x = t ^ { 2 } - 6 t + 4 , \quad y = t - 3 .$$ Find

1. the coordinates of the point where the curve meets the \(x\)-axis,
2. the equation of the curve in cartesian form, giving your answer in a simple form without brackets,
3. the equation of the tangent to the curve at the point where \(t = 2\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

3. \begin{figure}[h] \end{figure} Figure 1 shows the curve \(C\) given by the parametric equations $$x = \frac { 5 } { \sqrt { 3 } } \sin t \quad y = 5 ( 1 - \cos t ) \quad 0 \leqslant t \leqslant 2 \pi$$ The circle with centre at the origin \(O\) and with radius \(\frac { 5 \sqrt { 2 } } { 2 }\) meets the curve \(C\) at the points \(A\) and \(B\) as shown in Figure 1.
（a）Determine the value of \(t\) at the point \(B\) ． The region \(R\) ，shown shaded in Figure 1，is bounded by the curve \(C\) and the circle．
（b）Determine the area of the region \(R\) ．

7. \begin{figure}[h] \end{figure} Figure 1 shows the curve with parametric equations $$x = t ^ { 3 } + 1 , \quad y = \frac { 2 } { t } , \quad t > 0 .$$ The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 9\).

1. Find the area of the shaded region.
2. Show that the volume of the solid formed when the shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis is \(12 \pi\).
3. Find a cartesian equation for the curve in the form \(y = \mathrm { f } ( x )\).
   7. continued

7. \begin{figure}[h] \end{figure} Figure 1 shows the curve with parametric equations $$x = t ^ { 3 } + 1 , \quad y = \frac { 2 } { t } , \quad t > 0 .$$ The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 9\).

1. Find the area of the shaded region.
2. Show that the volume of the solid formed when the shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis is \(12 \pi\).
3. Find a cartesian equation for the curve in the form \(y = \mathrm { f } ( x )\).
   7. continued

12. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with parametric equations $$x = \tan t , \quad y = 2 \sin ^ { 2 } t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The finite region \(S\), shown shaded in Figure 3, is bounded by the curve \(C\), the line \(x = \sqrt { 3 }\) and the \(x\)-axis. This shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.

1. Show that the volume of the solid of revolution formed is given by $$4 \pi \int _ { 0 } ^ { \frac { \pi } { 3 } } \left( \tan ^ { 2 } t - \sin ^ { 2 } t \right) \mathrm { d } t$$
2. Hence use integration to find the exact value for this volume.

4 A curve has parametric equations $$x = t ^ { 2 } + 3 t + 1 , \quad y = t ^ { 4 } + 1$$ The point \(P\) on the curve has parameter \(p\). It is given that the gradient of the curve at \(P\) is 4 .

1. Show that \(p = \sqrt [ 3 ] { } ( 2 p + 3 )\).
2. Verify by calculation that the value of \(p\) lies between 1.8 and 2.0.
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7
\includegraphics[max width=\textwidth, alt={}, center]{388d7076-636c-417d-84cb-e6e2a3e9a6a0-10_465_785_260_680} The diagram shows the curve with parametric equations $$x = 4 t + \mathrm { e } ^ { 2 t } , \quad y = 6 t \sin 2 t$$ for \(0 \leqslant t \leqslant 1\). The point \(P\) on the curve has parameter \(p\) and \(y\)-coordinate 3 .

1. Show that \(p = \frac { 1 } { 2 \sin 2 p }\).
2. Show by calculation that the value of \(p\) lies between 0.5 and 0.6 .
3. Use an iterative formula, based on the equation in part (a), to find the value of \(p\) correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures.
4. Find the gradient of the curve at \(P\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

6 A curve has parametric equations $$x = \frac { 1 } { ( 2 t + 1 ) ^ { 2 } } , \quad y = \sqrt { } ( t + 2 )$$ The point \(P\) on the curve has parameter \(p\) and it is given that the gradient of the curve at \(P\) is - 1 .

1. Show that \(p = ( p + 2 ) ^ { \frac { 1 } { 6 } } - \frac { 1 } { 2 }\).
2. Use an iterative process based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places.

2 Show that the curve, given by the parametric equations given below, represents a circle. $$x = 2 \cos \theta + 3 , y = 2 \sin \theta - 3$$ State the radius and centre of this circle.

4 The parametric equations of a curve are $$x = t ^ { 2 } + 1 , \quad y = 4 t + \ln ( 2 t - 1 )$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the equation of the normal to the curve at the point where \(t = 1\). Give your answer in the form \(a x + b y + c = 0\).

6. \begin{figure}[h] \end{figure} Figure 1 shows the curve \(C\) with parametric equations $$x = 2 \cos \theta - \cos 2 \theta , y = 2 \sin \theta - \sin 2 \theta , \quad 0 \leqslant \theta \leqslant \pi$$

1. Show that $$\left( \frac { \mathrm { d } x } { \mathrm {~d} \theta } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} \theta } \right) ^ { 2 } = 8 ( 1 - \cos \theta )$$ The curve \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Find the area of the surface generated, giving your answer in the form \(k \pi\), where \(k\) is a rational number.