1.03g Parametric equations: of curves and conversion to cartesian

5 The parametric equations of a curve are $$x = 1 + \sqrt { } t , \quad y = 3 \ln t$$

1. Find the exact value of the gradient of the curve at the point \(P\) where \(y = 6\).
2. Show that the tangent to the curve at \(P\) passes through the point \(( 1,0 )\).

7 \includegraphics[max width=\textwidth, alt={}, center]{250b4df9-2646-4246-bb6d-2be92bf29598-3_553_689_258_726} The parametric equations of a curve are $$x = 6 \sin ^ { 2 } t , \quad y = 2 \sin 2 t + 3 \cos 2 t$$ for \(0 \leqslant t < \pi\). The curve crosses the \(x\)-axis at points \(B\) and \(D\) and the stationary points are \(A\) and \(C\), as shown in the diagram.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 3 } \cot 2 t - 1\).
2. Find the values of \(t\) at \(A\) and \(C\), giving each answer correct to 3 decimal places.
3. Find the value of the gradient of the curve at \(B\).

6 \includegraphics[max width=\textwidth, alt={}, center]{7e100be2-9768-4fcd-b516-c714e53b0665-3_453_650_258_744} The diagram shows the curve with parametric equations $$x = 3 \cos t , \quad y = 2 \cos \left( t - \frac { 1 } { 6 } \pi \right)$$ for \(0 \leqslant t < 2 \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 } ( \sqrt { } 3 - \cot t )\).
2. Find the equation of the tangent to the curve at the point where the curve crosses the positive \(y\)-axis. Give the answer in the form \(y = m x + c\).

3 The parametric equations of a curve are $$x = ( t + 1 ) \mathrm { e } ^ { t } , \quad y = 6 ( t + 4 ) ^ { \frac { 1 } { 2 } }$$ Find the equation of the tangent to the curve when \(t = 0\), giving the answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers.

7 \includegraphics[max width=\textwidth, alt={}, center]{9bbcee46-c5b8-4836-a4b4-f317bf8b1c0a-3_533_698_735_717} The diagram shows the curve with parametric equations $$x = 4 \sin \theta , \quad y = 1 + 3 \cos \left( \theta + \frac { 1 } { 6 } \pi \right)$$ for \(0 \leqslant \theta < 2 \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be expressed in the form \(k ( 1 + ( \sqrt { } 3 ) \tan \theta )\) where the exact value of \(k\) is to be determined.
2. Find the equation of the normal to the curve at the point where the curve crosses the positive \(y\)-axis. Give your answer in the form \(y = m x + c\), where the constants \(m\) and \(c\) are exact.

6 The parametric equations of a curve are $$x = 2 \mathrm { e } ^ { 2 t } + 4 \mathrm { e } ^ { t } , \quad y = 5 t \mathrm { e } ^ { 2 t }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) and hence find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places.
2. Find the gradient of the normal to the curve at the point where the curve crosses the \(x\)-axis.

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

1. Find the equation of the tangent to the curve at the origin.
2. Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places. [4]

7 The parametric equations of a curve are $$x = 3 \sin 2 \theta , \quad y = 1 + 2 \tan 2 \theta$$ for \(0 \leqslant \theta < \frac { 1 } { 4 } \pi\).

1. Find the exact gradient of the curve at the point for which \(\theta = \frac { 1 } { 6 } \pi\).
2. Find the value of \(\theta\) at the point where the gradient of the curve is 2 , giving the value correct to 3 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{77672e56-a268-47b8-ab8b-cd84b4b3de4f-10_551_689_258_726} The parametric equations of a curve are $$x = 6 \sin ^ { 2 } t , \quad y = 2 \sin 2 t + 3 \cos 2 t$$ for \(0 \leqslant t < \pi\). The curve crosses the \(x\)-axis at points \(B\) and \(D\) and the stationary points are \(A\) and \(C\), as shown in the diagram.

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 3 } \cot 2 t - 1\).
2. Find the values of \(t\) at \(A\) and \(C\), giving each answer correct to 3 decimal places.
3. Find the value of the gradient of the curve at \(B\).

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

1. Show that the gradient of the curve is always positive.
2. Find the equation of the tangent to the curve at the point where it intersects the \(y\)-axis.

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 The parametric equations of a curve are \(x = \frac { 1 } { \cos t } , y = \ln \tan t\), where \(0 < t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \cos t } { \sin ^ { 2 } t }\).
2. Find the equation of the tangent to the curve at the point where \(y = 0\).

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

4 The parametric equations of a curve are $$x = 1 - \cos \theta , \quad y = \cos \theta - \frac { 1 } { 4 } \cos 2 \theta$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2 \sin ^ { 2 } \left( \frac { 1 } { 2 } \theta \right)\).

5 The parametric equations of a curve are $$x = t \mathrm { e } ^ { 2 t } , \quad y = t ^ { 2 } + t + 3$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - 2 t }\).
2. Hence show that the normal to the curve, where \(t = - 1\), passes through the point \(\left( 0,3 - \frac { 1 } { \mathrm { e } ^ { 4 } } \right)\).

3 The parametric equations of a curve are $$x = 3 - \cos 2 \theta , \quad y = 2 \theta + \sin 2 \theta$$ for \(0 < \theta < \frac { 1 } { 2 } \pi\).
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).

6 The parametric equations of a curve are $$x = \sqrt { t } + 3 , \quad y = \ln t$$ for \(t > 0\).

1. Obtain a simplified expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Hence find the exact coordinates of the point on the curve at which the gradient of the normal is - 2 .

2 The parametric equations of a curve are $$x = ( \ln t ) ^ { 2 } , \quad y = \mathrm { e } ^ { 2 - t ^ { 2 } }$$ for \(t > 0\).
Find the gradient of the curve at the point where \(t = \mathrm { e }\), simplifying your answer.

7 The parametric equations of a curve are $$x = 3 \sin 2 t , \quad y = \tan t + \cot t$$ for \(0 < t < \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 \sin ^ { 2 } 2 t }\). \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-10_2716_40_109_2009} \includegraphics[max width=\textwidth, alt={}, center]{6280ab81-0bdb-47b4-8651-bff1261a0adf-11_2723_33_99_22}
2. Find the equation of the normal to the curve at the point where \(t = \frac { 1 } { 4 } \pi\). Give your answer in the form \(p y + q x + r = 0\), where \(p , q\) and \(r\) are integers.

3 The curve \(C\) has parametric equations $$\mathrm { x } = \mathrm { e } ^ { \mathrm { t } } - \frac { 1 } { 3 } \mathrm { t } ^ { 3 } , \quad \mathrm { y } = 4 \mathrm { e } ^ { \frac { 1 } { 2 } \mathrm { t } } ( \mathrm { t } - 2 ) , \quad \text { for } 0 \leqslant t \leqslant 2$$ Find, in terms of e , the 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 \(\frac { d ^ { 2 } y } { d x ^ { 2 } } > 0\).

2 It is given that $$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = \cos ^ { - 1 } t \quad \text { for } 0 < t < 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }\). \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-05_2723_33_99_22}
2. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - t ^ { a } \left( 1 - t ^ { 2 } \right) ^ { b } \left( 2 - t ^ { 2 } \right)\), where \(a\) and \(b\) are constants to be determined.

5 The curve \(C\) has parametric equations $$x = \mathrm { e } ^ { t } - 4 t + 3 , \quad y = 8 \mathrm { e } ^ { \frac { 1 } { 2 } t } , \quad \text { for } 0 \leqslant t \leqslant 2 .$$

1. Find, in terms of e , the length of \(C\).
2. Find, in terms of \(\pi\) and e , the area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of a design for a scraper blade. The blade \(A O B C D A\) consists of an isosceles triangle \(C O D\) joined along its equal sides to sectors \(O B C\) and \(O D A\) of a circle with centre \(O\) and radius 8 cm . Angles \(A O D\) and \(B O C\) are equal. \(A O B\) is a straight line and is parallel to the line \(D C . D C\) has length 7 cm .

1. Show that the angle \(C O D\) is 0.906 radians, correct to 3 significant figures.
2. Find the perimeter of \(A O B C D A\), giving your answer to 3 significant figures.
3. Find the area of \(A O B C D A\), giving your answer to 3 significant figures.

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 \(k\).