1.03g Parametric equations: of curves and conversion to cartesian

1 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$ \(\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P , and PR is the line through P parallel to the \(x\)-axis. Q is the point \(( 2,0 )\). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \begin{figure}[h] \end{figure}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac { 1 } { t }\).
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2 \theta\), and that angle TPQ is equal to \(\theta\).
   [0pt] [The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.
3. Show that the curve has cartesian equation \(y ^ { 2 } = 8 x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\).

3 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.

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.

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.

3 A curve has parametric equations $$x = 2 \sin \theta , \quad y = \cos 2 \theta$$

1. Find the exact coordinates and the gradient of the curve at the point with parameter \(\theta = \frac { 1 } { 3 } \pi\).
2. Find \(y\) in terms of \(x\).

4 The parametric equations of a curve are $$x = \cos 2 \theta , \quad y = \sin \theta \cos \theta \quad \text { for } 0 \leqslant \theta < \pi$$ Show that the cartesian equation of the curve is \(x ^ { 2 } + 4 y ^ { 2 } = 1\).
Sketch the curve.

5 Part of the track of a roller-coaster is modelled by a curve with the parametric equations $$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi .$$ This is shown in Fig. 8. B is a minimum point, and BC is vertical. \begin{figure}[h] \end{figure}

1. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
3. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2$$
4. Express \(\cos \theta - 4 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\). Hence solve the equation \(\cos \theta - 4 \sin \theta = 2\) for \(0 \leqslant \theta \leqslant 2 \pi\).

6 A curve has parametric equations $$x = a t ^ { 3 } , \quad y = \frac { a } { 1 + t ^ { 2 } }$$ where \(a\) is a constant.
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 2 } { 3 t \left( 1 + t ^ { 2 } \right) ^ { 2 } }\).
Hence find the gradient of the curve at the point \(\left( a , \frac { 1 } { 2 } a \right)\).

7 A curve has parametric equations \(x = 1 + u ^ { 2 } , y = 2 u ^ { 3 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\).
2. Hence find the gradient of the curve at the point with coordinates \(( 5,16 )\).

8 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 \(y = \frac { 3 + 2 x } { 2 + x }\).

2 Fig. 6 shows the arch ABCD of a bridge. \begin{figure}[h] \end{figure} The section from \(B\) to \(C\) is part of the curve \(O B C E\) with parametric equations $$x = a ( \theta - \sin \theta ) , y = a ( 1 - \cos \theta ) \text { for } 0 \leqslant \theta \leqslant 2 \pi \text {, }$$ where \(a\) is a constant.

1. Find, in terms of \(a\),
   (A) the length of the straight line OE,
   (B) the maximum height of the arch.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). The straight line sections AB and CD are inclined at \(30 ^ { \circ }\) to the horizontal, and are tangents to the curve at B and C respectively. BC is parallel to the \(x\)-axis. BF is parallel to the \(y\)-axis.
3. Show that at the point B the parameter \(\theta\) satisfies the equation $$\sin \theta = \frac { 1 } { \sqrt { 3 } } ( 1 \quad \cos \theta ) .$$ Verify that \(\theta = \frac { 2 } { 3 } \pi\) is a solution of this equation.
   Hence show that \(\mathrm { BF } = \frac { 3 } { 2 } a\), and find OF in terms of \(a\), giving your answer exactly.
4. Find BC and AF in terms of \(a\). Given that the straight line distance AD is 20 metres, calculate the value of \(a\).

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.

4 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 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.

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 \(x\).

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.

2 A curve has equation $$x ^ { 2 } + 4 y ^ { 2 } = k ^ { 2 } ,$$ where \(k\) is a positive constant.

1. Verify that $$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$ are parametric equations for the curve.
2. Hence or otherwise show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }\).
3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-2_658_1070_861_567} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
4. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1 , k = 3\) and \(k = 4\). On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.
5. Explain why the path of the stream is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
6. Solve this differential equation. Given that the path of the stream passes through the point \(( 2,1 )\), show that its equation is \(y = \frac { x ^ { 4 } } { 16 }\).

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 \(t = 2\).

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.

4 Part of the track of a roller-coaster is modelled by a curve with the parametric equations $$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$ This is shown in Fig. 8. B is a minimum point, and BC is vertical. \begin{figure}[h] \end{figure}

1. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
3. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2$$
4. Express \(\cos \theta - 4 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\). Hence solve the equation \(\cos \theta - 4 \sin \theta = 2\) for \(0 \leqslant \theta \leqslant 2 \pi\).

1 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]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.

7 Given that \(x = 2 \sec \theta\) and \(y = 3 \tan \theta\), show that \(\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1\).

5 A curve has parametric equations \(x = \frac { t ^ { 2 } } { 1 + t ^ { 2 } } , y = t ^ { 3 } - \lambda t\), where \(\lambda\) is a constant.

1. Use your calculator to obtain a sketch of the curve in each of the cases $$\lambda = - 1 , \quad \lambda = 0 \quad \text { and } \quad \lambda = 1 .$$ Name any special features of these curves.
2. By considering the value of \(x\) when \(t\) is large, write down the equation of the asymptote. For the remainder of this question, assume that \(\lambda\) is positive.
3. Find, in terms of \(\lambda\), the coordinates of the point where the curve intersects itself.
4. Show that the two points on the curve where the tangent is parallel to the \(x\)-axis have coordinates $$\left( \frac { \lambda } { 3 + \lambda } , \pm \sqrt { \frac { 4 \lambda ^ { 3 } } { 27 } } \right)$$ Fig. 5 shows a curve which intersects itself at the point ( 2,0 ) and has asymptote \(x = 8\). The stationary points A and B have \(y\)-coordinates 2 and - 2 . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{43b4c7ed-3556-4d87-8aef-0111fe747982-4_791_609_1482_769} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
5. For the curve sketched in Fig. 5, find parametric equations of the form \(x = \frac { a t ^ { 2 } } { 1 + t ^ { 2 } } , y = b \left( t ^ { 3 } - \lambda t \right)\), where \(a , \lambda\) and \(b\) are to be determined.

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.

1. 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.
2. Determine the area of the region \(R\) .

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 \(C\) .