Find stationary points of parametric curve

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 .

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.

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.

5 \includegraphics[max width=\textwidth, alt={}, center]{77a45360-8e1d-4f4f-9830-075d832a14cf-08_334_895_258_625} The diagram shows the curve with parametric equations $$x = \tan \theta , \quad y = \cos ^ { 2 } \theta$$ for \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\).

1. Show that the gradient of the curve at the point with parameter \(\theta\) is \(- 2 \sin \theta \cos ^ { 3 } \theta\).
   The gradient of the curve has its maximum value at the point \(P\).
2. Find the exact value of the \(x\)-coordinate of \(P\).

13. \begin{figure}[h] \end{figure} The curve \(C\) shown in Figure 4 has parametric equations $$x = 1 + \sqrt { 3 } \tan \theta , \quad y = 5 \sec \theta , \quad - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$$ The curve \(C\) crosses the \(y\)-axis at \(A\) and has a minimum turning point at \(B\), as shown in Figure 4.

1. Find the exact coordinates of \(A\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \lambda \sin \theta\), giving the exact value of the constant \(\lambda\).
3. Find the coordinates of \(B\).
4. Show that the cartesian equation for the curve \(C\) can be written in the form $$y = k \sqrt { \left( x ^ { 2 } - 2 x + 4 \right) }$$ where \(k\) is a simplified surd to be found.

7 Fig. 7 shows the curve with parametric equations $$x = \cos \theta , y = \sin \theta - \frac { 1 } { 8 } \sin 2 \theta , 0 \leqslant \theta < 2 \pi$$ The curve crosses the \(x\)-axis at points \(\mathrm { A } ( 1,0 )\) and \(\mathrm { B } ( - 1,0 )\), and the positive \(y\)-axis at C . D is the maximum point of the curve, and E is the minimum point. The solid of revolution formed when this curve is rotated through \(360 ^ { \circ }\) about the \(x\)-axis is used to model the shape of an egg. \begin{figure}[h] \end{figure}

1. Show that, at the point \(\mathrm { A } , \theta = 0\). Write down the value of \(\theta\) at the point B , and find the coordinates of C .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence show that, at the point D, $$2 \cos ^ { 2 } \theta - 4 \cos \theta - 1 = 0 .$$
3. Solve this equation, and hence find the \(y\)-coordinate of D , giving your answer correct to 2 decimal places. The cartesian equation of the curve (for \(0 \leqslant \theta \leqslant \pi\) ) is $$y = \frac { 1 } { 4 } ( 4 - x ) \sqrt { 1 - x ^ { 2 } } .$$
4. Show that the volume of the solid of revolution of this curve about the \(x\)-axis is given by $$\frac { 1 } { 16 } \pi \int _ { - 1 } ^ { 1 } \left( 16 - 8 x - 15 x ^ { 2 } + 8 x ^ { 3 } - x ^ { 4 } \right) \mathrm { d } x .$$ Evaluate this integral.

2 Fig. 7a shows the curve with the parametric equations $$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve meets the \(x\)-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same. \begin{figure}[h] \end{figure}

1. State, with their coordinates, the points on the curve for which \(\theta = - \frac { \pi } { 2 } , \theta = 0\) and \(\theta = \frac { \pi } { 2 }\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac { \pi } { 2 }\), and verify that the two tangents to the curve at the origin meet at right angles.
3. Find the exact coordinates of the turning point Q . When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_324_389_1692_857} \captionsetup{labelformat=empty} \caption{Fig. 7b}
   \end{figure}
4. Express \(\sin ^ { 2 } \theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)\).
5. Find the volume of the paperweight shape.

6 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 carlesian equation \(\mathrm { y } ^ { 2 } - \mathrm { x } _ { 2 } = 4\).

1. Verify that $$\boldsymbol { x } = \boldsymbol { t } - - ^ { 1 } \quad \boldsymbol { t ^ { \prime } } \quad \boldsymbol { y } = \boldsymbol { t } + \frac { 1 } { \boldsymbol { t } ^ { \prime } }$$ are parametric equations of the curve.
   (u) Show lhat \(\left. \underset { d x } { \mathbf { d y } } = \frac { ( t - I ) ( r } { 12 + 1 } + 1 \right)\). Hence find the coordinates of the staionary points of the curve.

1 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.

4 Fig. 7a shows the curve with the parametric equations $$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve meets the \(x\)-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same. \begin{figure}[h] \end{figure}

1. State, with their coordinates, the points on the curve for which \(\theta = - \frac { \pi } { 2 } , \theta = 0\) and \(\theta = \frac { \pi } { 2 }\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac { \pi } { 2 }\), and verify that the two tangents to the curve at the origin meet at right angles.
3. Find the exact coordinates of the turning point Q . When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-4_324_389_1692_857} \captionsetup{labelformat=empty} \caption{Fig. 7b}
   \end{figure}
4. Express \(\sin ^ { 2 } \theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)\).
5. Find the volume of the paperweight shape.
6. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
7. Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.

9 A curve has parametric equations \(x = \frac { 1 } { t } - 1\) and \(y = 2 t + \frac { 1 } { t ^ { 2 } }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), simplifying your answer.
2. Find the coordinates of the stationary point and, by considering the gradient of the curve on either side of this point, determine its nature.
3. Find a cartesian equation of the curve.

12 The diagram shows the curve with parametric equations \(x = \sin 2 \theta + 2 , y = 2 \cos \theta + \cos 2 \theta\), for \(0 \leqslant \theta < 2 \pi\). \includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-10_771_673_397_239}

1. In this question you must show detailed reasoning. Determine the exact coordinates of all the stationary points on the curve.
2. Write down the equation of the line of symmetry of the curve.