Parametric curves and Cartesian conversion

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

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

8 A curve has parametric equations \(x = \frac { t } { 1 + t ^ { 3 } } , y = \frac { t ^ { 2 } } { 1 + t ^ { 3 } }\), where \(t \neq - 1\).

1. In this question you must show detailed reasoning. Determine the gradient of the curve at the point where \(t = 1\).
2. Verify that the cartesian equation of the curve is \(x ^ { 3 } + y ^ { 3 } = x y\).

3 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.

1 A curve is defined by the parametric equations $$x = \cos \theta \text { and } y = \sin \theta \quad \text { where } 0 \leq \theta \leq 2 \pi$$ Which of the options shown below is a Cartesian equation for this curve?
Circle your answer. $$\frac { y } { x } = \tan \theta \quad x ^ { 2 } + y ^ { 2 } = 1 \quad x ^ { 2 } - y ^ { 2 } = 1 \quad x ^ { 2 } y ^ { 2 } = 1$$

3. The curve \(C\) has parametric equations $$x = 3 + 2 \sin t \quad y = \frac { 6 } { 7 + \cos 2 t } \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$

1. Show that \(C\) has Cartesian equation $$y = \frac { 12 } { ( 7 - x ) ( 1 + x ) } \quad p \leqslant x \leqslant q$$ where \(p\) and \(q\) are constants to be found.
2. Hence, find a Cartesian equation for \(C\) in the form $$y = \frac { a } { x + b } + \frac { c } { x + d } \quad p \leqslant x \leqslant q$$ where \(a , b , c\) and \(d\) are constants.

13 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that \(t _ { 1 } t _ { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. \(\mathbf { C 1 . }\)

13 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that \(t _ { 1 } t _ { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. \(\mathbf { C 1 . }\)

7. \begin{figure}[h] \end{figure} Figure 4 shows a shape \(S ( \theta )\) made up of five line segments \(A B , B C , C D , D E\) and \(E A\) . The lengths of the sides are \(A B = B C = 5 \mathrm {~cm} , C D = E A = 3 \mathrm {~cm}\) and \(D E = 7 \mathrm {~cm}\) . Angle \(B A E =\) angle \(B C D = \theta\) radians. The length of each line segment always remains the same but the value of \(\theta\) can be varied so that different symmetrical shapes can be formed,with the added restriction that none of the line segments cross.

1. Sketch \(S ( \pi )\) ,labelling the vertices clearly. The shape \(S ( \phi )\) is a trapezium.
2. Sketch \(S ( \phi )\) and calculate the value of \(\phi\) . The smallest possible value for \(\theta\) is \(\alpha\) ,where \(\alpha > 0\) ,and the largest possible value for \(\theta\) is \(\beta\) , where \(\beta > \pi\) .
3. Show that \(\alpha = \arccos \left( \frac { 29 } { 40 } \right) \cdot \left[ \arccos ( x ) \right.\) is an alternative notation for \(\left. \cos ^ { - 1 } ( x ) \right]\)
4. Find the value of \(\beta\) . The area,in \(\mathrm { cm } ^ { 2 }\) ,of shape \(S ( \theta )\) is \(R ( \theta )\) .
5. Show that for \(\alpha \leqslant \theta < \pi\) $$R ( \theta ) = 15 \sin \theta + \frac { 7 } { 4 } \sqrt { 87 - 120 \cos \theta }$$ Given that this formula for \(R ( \theta )\) holds for \(\alpha \leqslant \theta \leqslant \beta\)
6. show that \(R ( \theta )\) has only one stationary point and that this occurs when \(\theta = \frac { 2 \pi } { 3 }\)
7. find the maximum and minimum values of \(R ( \theta )\). FOR STYLE, CLARITY AND PRESENTATION: 7 MARKS TOTAL FOR PAPER: 100 MARKS
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A curve is defined by the parametric equations $$x = t^3 + 2, \quad y = t^2 - 1$$

1. Find the gradient of the curve at the point where \(t = -2\) [4 marks]
2. Find a Cartesian equation of the curve. [2 marks]

2. A curve \(C\) has parametric equations $$x = \frac { 3 } { 2 } t - 5 , \quad y = 4 - \frac { 6 } { t } \quad t \neq 0$$

1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(t = 3\), giving your answer as a fraction in its simplest form.
2. Show that a cartesian equation of \(C\) can be expressed in the form $$y = \frac { a x + b } { x + 5 } \quad x \neq k$$ where \(a , b\) and \(k\) are integers to be found.

11. \begin{figure}[h] \end{figure} The curve \(C\) shown in Figure 3 has parametric equations $$x = 3 \cos t , \quad y = 9 \sin 2 t , \quad 0 \leqslant t \leqslant 2 \pi$$ The curve \(C\) meets the \(x\)-axis at the origin and at the points \(A\) and \(B\), as shown in Figure 3 .

1. Write down the coordinates of \(A\) and \(B\).
2. Find the values of \(t\) at which the curve passes through the origin.
3. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\), and hence find the gradient of the curve when \(t = \frac { \pi } { 6 }\)
4. Show that the cartesian equation for the curve \(C\) can be written in the form $$y ^ { 2 } = a x ^ { 2 } \left( b - x ^ { 2 } \right)$$ where \(a\) and \(b\) are integers to be determined.

1 A curve is defined by the parametric equations $$x = 1 + 2 t , \quad y = 1 - 4 t ^ { 2 }$$

1. 1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t }\).
      (2 marks)
   2. Hence find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find an equation of the normal to the curve at the point where \(t = 1\).
3. Find a cartesian equation of the curve.

The curve \(C\) has parametric equations $$x = 2\cos t, \quad y = \sqrt{3}\cos 2t, \quad 0 \leq t \leq \pi$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(t\). [2]

The point \(P\) lies on \(C\) where \(t = \frac{2\pi}{3}\) The line \(l\) is the normal to \(C\) at \(P\).

1. Show that an equation for \(l\) is $$2x - 2\sqrt{3}y - 1 = 0$$ [5]

The line \(l\) intersects the curve \(C\) again at the point \(Q\).

1. Find the exact coordinates of \(Q\). You must show clearly how you obtained your answers. [6]

The curve \(C_1\) has parametric equations \(x = 3p + 1\), \(y = 9p^2\). The curve \(C_2\) has parametric equations \(x = 4q\), \(y = 2q\). Find the Cartesian coordinates of the points of intersection of \(C_1\) and \(C_2\). [7]

1. The hyperbola \(H\) has Cartesian equation \(x y = 25\)

The parabola \(P\) has parametric equations \(x = 10 t ^ { 2 } , y = 20 t\) The hyperbola \(H\) intersects the parabola \(P\) at the point \(A\)

1. Use algebra to determine the coordinates of \(A\) The point \(B\) with coordinates \(( 10,20 )\) lies on \(P\)
2. Find an equation for the normal to \(P\) at \(B\) Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers to be determined.
3. Use algebra to determine, in simplest form, the exact coordinates of the points where this normal intersects the hyperbola \(H\) (6)

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 \(R\) .

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 \(x\).

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 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.

4

1. Using \(\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1\), show that \(\tan ^ { 2 } \theta + 1 \equiv \sec ^ { 2 } \theta\).
   A curve is given parametrically by $$x = a \sec \theta , \quad y = a \tan \theta$$ where \(a\) is a constant.
2. Find a Cartesian equation of the curve.
3. Determine an equation of the tangent to the curve at the point \(\theta = \frac { \pi } { 3 }\), giving your answer in exact form.
   [0pt] [5]

4

1. Using \(\sin ^ { 2 } \theta + \cos ^ { 2 } \theta \equiv 1\), show that \(\tan ^ { 2 } \theta + 1 \equiv \sec ^ { 2 } \theta\).
   A curve is given parametrically by $$x = a \sec \theta , \quad y = a \tan \theta$$ where \(a\) is a constant.
2. Find a Cartesian equation of the curve.
3. Determine an equation of the tangent to the curve at the point \(\theta = \frac { \pi } { 3 }\), giving your answer in exact form.
   [0pt] [5]

9. A curve has parametric equations $$x = \sec \theta + \tan \theta , \quad y = \operatorname { cosec } \theta + \cot \theta , \quad 0 < \theta < \frac { \pi } { 2 }$$

1. Show that \(x + \frac { 1 } { x } = 2 \sec \theta\). Given that \(y + \frac { 1 } { y } = 2 \operatorname { cosec } \theta\),
2. find a cartesian equation for the curve.
3. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} \theta } = \frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right)\).
4. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).

A curve has parametric equations $$x = \cos 2t, \quad y = \cosec t, \quad 0 < t < \frac{\pi}{2}.$$ The point \(P\) on the curve has \(x\)-coordinate \(\frac{1}{2}\).

1. Find the value of the parameter \(t\) at \(P\). [2]
2. Show that the tangent to the curve at \(P\) has the equation $$y = 2x + 1.$$ [5]

A curve has parametric equations $$x = \frac{t}{2-t}, \quad y = \frac{1}{1+t}, \quad -1 < t < 2.$$

1. Show that \(\frac{dy}{dx} = -\frac{1}{2}\left(\frac{2-t}{1+t}\right)^2\). [4]
2. Find an equation for the normal to the curve at the point where \(t = 1\). [3]
3. Show that the cartesian equation of the curve can be written in the form $$y = \frac{1+x}{1+3x}.$$ [4]

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.

A curve is defined parametrically by the equations $$x = \frac{3}{2}t^3 + 5$$ $$y = t^{\frac{3}{2}} \quad (t \geq 0)$$ Show that the arc length of the curve from \(t = 0\) to \(t = 2\) is equal to 26 units. [5 marks]

The curve \(C\) has parametric equations $$x = 15t - t^3, \quad y = 3 - 2t^2.$$ Find the values of \(t\) at the points where the normal to \(C\) at \((14, 1)\) cuts \(C\) again. [11]

2 Show that the curve, given by the parametric equations given below, represents a circle. $$x = 2 \cos \theta + 3 , y = 2 \sin \theta - 3$$ State the radius and centre of this circle.

\includegraphics{figure_2} Figure 2 shows a sketch of the curve defined by the parametric equations $$x = t^2 + 2t \quad y = \frac{2}{t(3-t)} \quad a \leq t \leq b$$ where \(a\) and \(b\) are constants. The ends of the curve lie on the line with equation \(y = 1\)

1. Find the value of \(a\) and the value of \(b\) [2]

The region \(R\), shown shaded in Figure 2, is bounded by the curve and the line with equation \(y = 1\)

1. Show that the area of region \(R\) is given by $$M - k \int_a^b \frac{t+1}{t(3-t)} dt$$ where \(M\) and \(k\) are constants to be found. [5]
2. 1. Write \(\frac{t+1}{t(3-t)}\) in partial fractions.
   2. Use algebraic integration to find the exact area of \(R\), giving your answer in simplest form. [6]

At time \(t\) hours after a high tide, the height, \(h\) metres, of the tide and the velocity, \(v\) knots, of the tidal flow can be modelled using the parametric equations $$v = 4 - \left(\frac{2t}{3} - 2\right)^2$$ $$h = 3 - 2\sqrt[3]{t - 3}$$ High tides and low tides occur alternately when the velocity of the tidal flow is zero. A high tide occurs at 2am.

1. 1. Use the model to find the height of this high tide. [1 mark]
   2. Find the time of the first low tide after 2am. [3 marks]
   3. Find the height of this low tide. [1 mark]
2. Use the model to find the height of the tide when it is flowing with maximum velocity. [3 marks]
3. Comment on the validity of the model. [2 marks]

Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u, \quad y = u + \frac{1}{u}, \quad 1 \leq u \leq 10.$$ 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. \includegraphics{figure_7}

1. Find the lengths OA, OB and AC. [5]
2. Find \(\frac{dy}{dx}\) in terms of \(u\). Hence find the angle \(\theta\). [6]
3. Show that the cartesian equation of the curve is \(y = e^{x/5} + e^{-x/5}\). [2]

An object is formed by rotating the region OACB through \(360°\) about Ox.

1. Find the volume of the object. [5]

1 The curve \(C\) is defined parametrically by $$x = 2 \cos ^ { 3 } t \quad \text { and } \quad y = 2 \sin ^ { 3 } t , \quad \text { for } 0 < t < \frac { 1 } { 2 } \pi .$$ Show that, at the point with parameter \(t\), $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 1 } { 6 } \sec ^ { 4 } t \operatorname { cosec } t$$

A curve \(C\) has parametric equations $$x = 1 - 3 t ^ { 2 } , \quad y = t \left( 1 - 3 t ^ { 2 } \right) , \quad \text { for } 0 \leqslant t \leqslant \frac { 1 } { \sqrt { 3 } }$$ Show that \(\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } = \left( 1 + 9 t ^ { 2 } \right) ^ { 2 }\). Hence find

1. the arc length of \(C\),
2. the surface area generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Use the fact that \(t = \frac { y } { x }\) to find a cartesian equation of \(C\). Hence show that the polar equation of \(C\) is \(r = \sec \theta \left( 1 - 3 \tan ^ { 2 } \theta \right)\), and state the domain of \(\theta\). Find the area of the region enclosed between \(C\) and the initial line. {www.cie.org.uk} after the live examination series. }

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.