1.03g Parametric equations: of curves and conversion to cartesian

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve \(C\) with parametric equations $$x = 2 \sin t , \quad y = \ln ( \sec t ) , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The tangent to \(C\) at the point \(P\) ,where \(t = \frac { \pi } { 3 }\) ,cuts the \(x\)-axis at \(A\) .

1. Show that the \(x\)-coordinate of \(A\) is \(\frac { \sqrt { } 3 } { 3 } ( 3 - \ln 2 )\) . The shaded region \(R\) lies between \(C\) ,the positive \(x\)-axis and the tangent \(A P\) as shown in Figure 2 .
2. Show that the area of \(R\) is \(\sqrt { 3 } ( 1 + \ln 2 ) - 2 \ln ( 2 + \sqrt { 3 } ) - \frac { \sqrt { 3 } } { 6 } ( \ln 2 ) ^ { 2 }\) .

12. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with parametric equations $$x = 3 \sin t , \quad y = 2 \sin 2 t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$ The finite region \(S\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \frac { 3 } { 2 }\) The shaded region \(S\) 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 is given by $$k \int _ { 0 } ^ { a } \sin ^ { 2 } t \cos ^ { 3 } t \mathrm {~d} t$$ where \(k\) and \(a\) are constants to be given in terms of \(\pi\).
2. Use the substitution \(u = \sin t\), or otherwise, to find the exact value of this volume, giving your answer in the form \(\frac { p \pi } { q }\) where \(p\) and \(q\) are integers. (Solutions based entirely on graphical or numerical methods are not acceptable.)

1. A rectangular hyperbola \(H\) has equation \(x y = 25\)

The point \(P \left( 5 t , \frac { 5 } { t } \right) , t \neq 0\), is a general point on \(H\).

1. Show that the equation of the tangent to \(H\) at \(P\) is \(t ^ { 2 } y + x = 10 t\) The distinct points \(Q\) and \(R\) lie on \(H\). The tangent to \(H\) at the point \(Q\) and the tangent to \(H\) at the point \(R\) meet at the point \(( 15 , - 5 )\).
2. Find the coordinates of the points \(Q\) and \(R\).

6. The parabola \(C\) has Cartesian equation \(y ^ { 2 } = 8 x\) The point \(P \left( 2 p ^ { 2 } , 4 p \right)\) and the point \(Q \left( 2 q ^ { 2 } , 4 q \right)\), where \(p , q \neq 0 , p \neq q\), are points on \(C\).

1. Show that an equation of the normal to \(C\) at \(P\) is $$y + p x = 2 p ^ { 3 } + 4 p$$
2. Write down an equation of the normal to \(C\) at \(Q\) The normal to \(C\) at \(P\) and the normal to \(C\) at \(Q\) meet at the point \(N\)
3. Show that \(N\) has coordinates $$\left( 2 \left( p ^ { 2 } + p q + q ^ { 2 } + 2 \right) , - 2 p q ( p + q ) \right)$$ The line \(O N\), where \(O\) is the origin, is perpendicular to the line \(P Q\)
4. Find the value of \(( p + q ) ^ { 2 } - 3 p q\)

3 A curve has parametric equations \(x = a ( \theta + \sin \theta ) , y = a ( 1 - \cos \theta )\), for \(0 \leqslant \theta \leqslant \pi\), where \(a\) is a positive constant.

1. Show that the arc length \(s\) from the origin to a general point on the curve is given by \(s = 4 a \sin \frac { 1 } { 2 } \theta\).
2. Find the intrinsic equation of the curve giving \(s\) in terms of \(a\) and \(\psi\), where \(\tan \psi = \frac { \mathrm { d } y } { \mathrm {~d} x }\).
3. Hence, or otherwise, show that the radius of curvature at a point on the curve is \(4 a \cos \frac { 1 } { 2 } \theta\).
4. Find the coordinates of the centre of curvature corresponding to the point on the curve where \(\theta = \frac { 2 } { 3 } \pi\).
5. Find the area of the surface generated when the curve is rotated through \(2 \pi\) radians about the \(x\)-axis.

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 $$y ^ { 2 } = \frac { ( x - 1 ) ^ { 2 } } { x ^ { 2 } } \left( k ^ { 2 } - x ^ { 2 } \right) .$$

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t - 3 } , \quad y = 4 \ln t$$ where \(t > 0\). When \(t = a\) the gradient of the curve is 2 .

1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 2 } ( 3 - \ln a )\).
2. Verify by calculation that this equation has a root between 1 and 2 .
3. Use the iterative formula \(a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

7. \begin{figure}[h] \end{figure} A circular tower of radius 1 metre stands in a large horizontal field of grass.A goat is attached to one end of a rope and the other end of the rope is attached to a fixed point \(O\) at the base of the tower.The goat cannot enter the tower. Taking the point \(O\) as the origin( 0,0 ),the centre of the base of the tower is at the point \(T ( 0,1 )\) ,where the unit of length is the metre. The rope has length \(\pi\) metres and you may ignore the size of the goat.
The curve \(C\) shown in Figure 4 represents the edge of the region that the goat can reach.

1. Write down the equation of \(C\) for \(y < 0\) When the goat is at the point \(G ( x , y )\) ,with \(x > 0\) and \(y > 0\) ,as shown in Figure 4 ,the rope lies along \(O A G\) where \(O A\) is an arc of the circle with angle \(O T A = \theta\) radians and \(A G\) is a tangent to the circle at \(A\) .
2. With the aid of a suitable diagram show that $$\begin{aligned} & x = \sin \theta + ( \pi - \theta ) \cos \theta \\ & y = 1 - \cos \theta + ( \pi - \theta ) \sin \theta \end{aligned}$$
3. By considering \(\int y \frac { \mathrm {~d} x } { \mathrm {~d} \theta } \mathrm {~d} \theta\), show that the area, in the first quadrant, between \(C\), the positive \(x\)-axis and the positive \(y\)-axis can be expressed in the form $$\int _ { 0 } ^ { \pi } u \sin u \mathrm {~d} u + \int _ { 0 } ^ { \pi } u ^ { 2 } \sin ^ { 2 } u \mathrm {~d} u + \int _ { 0 } ^ { \pi } u \sin u \cos u \mathrm {~d} u$$
4. Show that \(\int _ { 0 } ^ { \pi } u ^ { 2 } \sin ^ { 2 } u \mathrm {~d} u = \frac { \pi ^ { 3 } } { 6 } + \int _ { 0 } ^ { \pi } u \sin u \cos u \mathrm {~d} u\)
5. Hence find the area of grass that can be reached by the goat.

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.

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.

4 A curve has parametric equations $$x = 2 + t ^ { 2 } , \quad y = 4 t$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the equation of the normal at the point where \(t = 4\), giving your answer in the form \(y = m x + c\).
3. Find a cartesian equation of the curve.

8 A curve is defined by the parametric equations $$x = \sin ^ { 2 } \theta , \quad y = 4 \sin \theta - \sin ^ { 3 } \theta ,$$ where \(- \frac { 1 } { 2 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 - 3 \sin ^ { 2 } \theta } { 2 \sin \theta }\).
2. Find the coordinates of the point on the curve at which the gradient is 2 .
3. Show that the curve has no stationary points.
4. Find a cartesian equation of the curve, giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

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.

5 A curve has parametric equations $$x = 2 t + t ^ { 2 } , \quad y = 2 t ^ { 2 } + t ^ { 3 }$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) and find the gradient of the curve at the point \(( 3 , - 9 )\).
2. By considering \(\frac { y } { x }\), find a cartesian equation of the curve, giving your answer in a form not involving fractions.

7 The parametric equations of a curve are \(x = \frac { t + 2 } { t + 1 } , y = \frac { 2 } { t + 3 }\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } > 0\).
2. Find the cartesian equation of the curve, giving your answer in a form not involving fractions.

8 A curve has parametric equations $$x = \frac { 1 } { t + 1 } , \quad y = t - 1 .$$ The line \(y = 3 x\) intersects the curve at two points.

1. Show that the value of \(t\) at one of these points is - 2 and find the value of \(t\) at the other point.
2. Find the equation of the normal to the curve at the point for which \(t = - 2\).
3. Find the value of \(t\) at the point where this normal meets the curve again.
4. Find a cartesian equation of the curve, giving your answer in the form \(y = \mathrm { f } ( x )\).

7 A curve has parametric equations $$x = 2 \sin t , \quad y = \cos 2 t + 2 \sin t$$ for \(- \frac { 1 } { 2 } \pi \leqslant t \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - 2 \sin t\) and hence find the coordinates of the stationary point.
2. Find the cartesian equation of the curve.
3. State the set of values that \(x\) can take and hence sketch the curve.

10

1. Express \(\frac { x + 8 } { x ( x + 2 ) }\) in partial fractions.
2. By first using division, express \(\frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }\) in the form \(P + \frac { Q } { x } + \frac { R } { x + 2 }\). A curve has parametric equations \(x = \frac { 2 t } { 1 - t } , y = 3 t + \frac { 4 } { t }\).
3. Show that the cartesian equation of the curve is \(y = \frac { 7 x ^ { 2 } + 16 x + 16 } { x ( x + 2 ) }\).
4. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\). Give your answer in the form \(L + M \ln 2 + N \ln 3\).

9 A curve has parametric equations \(x = 1 - \cos t , y = \sin t \sin 2 t\), for \(0 \leqslant t \leqslant \pi\).

1. Find the coordinates of the points where the curve meets the \(x\)-axis.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \cos 2 t + 2 \cos ^ { 2 } t\). Hence find, in an exact form, the coordinates of the stationary points.
3. Find the cartesian equation of the curve. Give your answer in the form \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a polynomial.
4. Sketch the curve.

8 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
3. (A) Show that \(y = x \cos \theta\).
   (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
   (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
4. Find the volume of the balloon.

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 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find \(t\) in terms of \(x\). Hence find the cartesian equation of the curve, giving your answer as simply as possible.

8 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h] \end{figure} In the following, all lengths are in metres.

1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
   You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).

3 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$

1. Find the exact value of the gradient of the curve at the point where \(\theta = \frac { 1 } { 6 } \pi\).
2. Show that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }\).

5 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).

1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{132ae754-bd4c-4819-80ef-4823ac2ead4f-02_545_853_1738_607} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.