1.03g Parametric equations: of curves and conversion to cartesian

6 The parametric equations of a curve are $$x = a \cos ^ { 3 } t , \quad y = a \sin ^ { 3 } t$$ where \(a\) is a positive constant and \(0 < t < \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Show that the equation of the tangent to the curve at the point with parameter \(t\) is $$x \sin t + y \cos t = a \sin t \cos t$$
3. Hence show that, if this tangent meets the \(x\)-axis at \(X\) and the \(y\)-axis at \(Y\), then the length of \(X Y\) is always equal to \(a\).

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

3 The parametric equations of a curve are $$x = \ln ( 2 t + 3 ) , \quad y = \frac { 3 t + 2 } { 2 t + 3 }$$ Find the gradient of the curve at the point where it crosses the \(y\)-axis.

10 \includegraphics[max width=\textwidth, alt={}, center]{3eefd6c1-924c-4b7e-8d17-a2942fb48234-3_515_508_1105_815} The diagram shows part of the curve with parametric equations $$x = 2 \ln ( t + 2 ) , \quad y = t ^ { 3 } + 2 t + 3$$

1. Find the gradient of the curve at the origin.
2. At the point \(P\) on the curve, the value of the parameter is \(p\). It is given that the gradient of the curve at \(P\) is \(\frac { 1 } { 2 }\).
   1. Show that \(p = \frac { 1 } { 3 p ^ { 2 } + 2 } - 2\).
   2. By first using an iterative formula based on the equation in part (a), determine the coordinates of the point \(P\). Give the result of each iteration to 5 decimal places and each coordinate of \(P\) correct to 2 decimal places.

4 The parametric equations of a curve are $$x = t + \cos t , \quad y = \ln ( 1 + \sin t )$$ where \(- \frac { 1 } { 2 } \pi < t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec t\).
2. Hence find the \(x\)-coordinates of the points on the curve at which the gradient is equal to 3 . Give your answers correct to 3 significant figures.

4 The parametric equations of a curve are $$x = \ln \cos \theta , \quad y = 3 \theta - \tan \theta ,$$ where \(0 \leqslant \theta < \frac { 1 } { 2 } \pi\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\tan \theta\).
2. Find the exact \(y\)-coordinate of the point on the curve at which the gradient of the normal is equal to 1 . \includegraphics[max width=\textwidth, alt={}, center]{b00cefad-7c3c-4672-b309-f19aafab8b01-08_378_689_260_726} The diagram shows a semicircle with centre \(O\), radius \(r\) and diameter \(A B\). The point \(P\) on its circumference is such that the area of the minor segment on \(A P\) is equal to half the area of the minor segment on \(B P\). The angle \(A O P\) is \(x\) radians.

2 The parametric equations of a curve are $$x = \frac { t } { 2 t + 3 } , \quad y = \mathrm { e } ^ { - 2 t }$$ Find the gradient of the curve at the point for which \(t = 0\).

2 The parametric equations of a curve are $$x = 3 \left( 1 + \sin ^ { 2 } t \right) , \quad y = 2 \cos ^ { 3 } t$$ Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer as far as possible.

8 \includegraphics[max width=\textwidth, alt={}, center]{6025cf1d-525e-4f12-9517-f20ef5fff2fa-3_698_1006_758_571} The diagram shows the curve with parametric equations $$x = \sin t + \cos t , \quad y = \sin ^ { 3 } t + \cos ^ { 3 } t$$ for \(\frac { 1 } { 4 } \pi < t < \frac { 5 } { 4 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3 \sin t \cos t\).
2. Find the gradient of the curve at the origin.
3. Find the values of \(t\) for which the gradient of the curve is 1 , giving your answers correct to 2 significant figures.

3 The parametric equations of a curve are $$x = \frac { 4 t } { 2 t + 3 } , \quad y = 2 \ln ( 2 t + 3 )$$

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer.
2. Find the gradient of the curve at the point for which \(x = 1\).

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { - t } \cos t , \quad y = \mathrm { e } ^ { - t } \sin t$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \tan \left( t - \frac { 1 } { 4 } \pi \right)\).

2 A curve is defined for \(0 < \theta < \frac { 1 } { 2 } \pi\) by the parametric equations $$x = \tan \theta , \quad y = 2 \cos ^ { 2 } \theta \sin \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 \cos ^ { 5 } \theta - 4 \cos ^ { 3 } \theta\).

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.

4 The parametric equations of a curve are $$x = 2 \sin \theta + \sin 2 \theta , \quad y = 2 \cos \theta + \cos 2 \theta$$ where \(0 < \theta < \pi\).

1. Obtain an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the \(y\)-axis.

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

6

1. Show that the curve with Cartesian equation $$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 36 \left( x ^ { 2 } - y ^ { 2 } \right)$$ has polar equation \(r ^ { 2 } = 36 \cos 2 \theta\).
   The curve \(C\) has polar equation \(r ^ { 2 } = 36 \cos 2 \theta\), for \(- \frac { 1 } { 4 } \pi \leqslant \theta \leqslant \frac { 1 } { 4 } \pi\).
2. Sketch \(C\) and state the maximum distance of a point on \(C\) from the pole.
3. Find the area of the region enclosed by \(C\).
4. Find the maximum distance of a point on \(C\) from the initial line, giving the answer in exact form.

5 The curve \(C\) has polar equation \(r = \operatorname { asec } ^ { 2 } \theta\), where \(a\) is a positive constant and \(0 \leqslant \theta \leqslant \frac { 1 } { 4 } \pi\).

1. Sketch \(C\), stating the polar coordinates of the point of intersection of \(C\) with the initial line and also with the half-line \(\theta = \frac { 1 } { 4 } \pi\).
2. Find the maximum distance of a point of \(C\) from the initial line.
3. Find the area of the region enclosed by \(C\), the initial line and the half-line \(\theta = \frac { 1 } { 4 } \pi\).
4. Find, in the form \(y = f ( x )\), the Cartesian equation of \(C\).

6

1. Show that the curve with Cartesian equation $$\left( x - \frac { 1 } { 2 } \right) ^ { 2 } + y ^ { 2 } = \frac { 1 } { 4 }$$ has polar equation \(r = \cos \theta\).
   The curves \(C _ { 1 }\) and \(C _ { 2 }\) have polar equations $$r = \cos \theta \quad \text { and } \quad r = \sin 2 \theta$$ respectively, where \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the pole and at another point \(P\).
2. Find the polar coordinates of \(P\).
3. In a single diagram sketch \(C _ { 1 }\) and \(C _ { 2 }\), clearly identifying each curve, and mark the point \(P\).
4. The region \(R\) is enclosed by \(C _ { 1 }\) and \(C _ { 2 }\) and includes the line \(O P\). Find, in exact form, the area of \(R\).

6 \includegraphics[max width=\textwidth, alt={}, center]{f5e0b088-73db-405b-a832-aa01d9fcba64-08_396_716_260_712} The diagram shows the curve with parametric equations $$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$ for \(0 \leqslant t \leqslant 2\). At the point \(P\) on the curve, the \(y\)-coordinate is 1 .

1. Show that the value of \(t\) at the point \(P\) satisfies the equation \(t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }\).
2. Use the iterative formula \(t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }\) with \(t _ { 1 } = 0.7\) to find the value of \(t\) at \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 2 significant figures.

6 \includegraphics[max width=\textwidth, alt={}, center]{0d15e5a1-d05f-48bc-8613-198804ff605c-08_396_716_260_712} The diagram shows the curve with parametric equations $$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$ for \(0 \leqslant t \leqslant 2\). At the point \(P\) on the curve, the \(y\)-coordinate is 1 .

1. Show that the value of \(t\) at the point \(P\) satisfies the equation \(t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }\).
2. Use the iterative formula \(t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }\) with \(t _ { 1 } = 0.7\) to find the value of \(t\) at \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
3. Find the gradient of the curve at \(P\), giving the answer correct to 2 significant figures.

7 The parametric equations of a curve are $$x = 2 t - \sin 2 t , \quad y = 5 t + \cos 2 t$$ for \(0 \leqslant t \leqslant \frac { 1 } { 2 } \pi\). At the point \(P\) on the curve, the gradient of the curve is 2 .

1. Show that the value of the parameter at \(P\) satisfies the equation \(2 \sin 2 t - 4 \cos 2 t = 1\).
2. By first expressing \(2 \sin 2 t - 4 \cos 2 t\) in the form \(R \sin ( 2 t - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), find the coordinates of \(P\). Give each coordinate correct to 3 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.

4 The parametric equations of a curve are $$x = 1 - \mathrm { e } ^ { - t } , \quad y = \mathrm { e } ^ { t } + \mathrm { e } ^ { - t }$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 t } - 1\).
2. Hence find the exact value of \(t\) at the point on the curve at which the gradient is 2 .

4 The parametric equations of a curve are $$x = 1 + \ln ( t - 2 ) , \quad y = t + \frac { 9 } { t } , \quad \text { for } t > 2$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( t ^ { 2 } - 9 \right) ( t - 2 ) } { t ^ { 2 } }\).
2. Find the coordinates of the only point on the curve at which the gradient is equal to 0 .

4 The parametric equations of a curve are $$x = \ln ( 1 - 2 t ) , \quad y = \frac { 2 } { t } , \quad \text { for } t < 0$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 - 2 t } { t ^ { 2 } }\).
2. Find the exact coordinates of the only point on the curve at which the gradient is 3 .

5 The parametric equations of a curve are $$x = \cos 2 \theta - \cos \theta , \quad y = 4 \sin ^ { 2 } \theta$$ for \(0 \leqslant \theta \leqslant \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 8 \cos \theta } { 1 - 4 \cos \theta }\).
2. Find the coordinates of the point on the curve at which the gradient is - 4 .