Parametric differentiation

4 A curve is given by the parametric equations \(x = t ^ { 2 } , y = 3 t\) for all values of \(t\). Find the equation of the tangent to the curve at the point where \(t = - 2\).

4 A curve is given by the parametric equations \(x = t ^ { 2 } , y = 3 t\) for all values of \(t\). Find the equation of the tangent to the curve at the point where \(t = - 2\).

9. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\) with parametric equations $$x = \sec t \quad y = \sqrt { 3 } \tan \left( t + \frac { \pi } { 3 } \right) \quad \frac { \pi } { 6 } < t < \frac { \pi } { 2 }$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\)
2. Find an equation for the tangent to \(C\) at the point where \(t = \frac { \pi } { 3 }\) Give your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants.
3. Show that all points on \(C\) satisfy the equation $$y = \frac { A x ^ { 2 } + B \sqrt { 3 x ^ { 2 } - 3 } } { 4 - 3 x ^ { 2 } }$$ where \(A\) and \(B\) are constants to be found.

A curve is defined by the parametric equations $$x = t^3 + 2, \quad y = t^2 - 1$$ Find the gradient of the curve at the point where \(t = -2\) [2]

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 \(t = 0\).

The curve \(C\) has parametric equations $$x = \sin 2\theta \quad y = \cos\text{ec}^3 \theta \quad 0 < \theta < \frac{\pi}{2}$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(\theta\) [3]
2. Hence find the exact value of the gradient of the tangent to \(C\) at the point where \(y = 8\) [3]

A curve has parametric equations $$x = at^3, \quad y = \frac{a}{1+t^2},$$ where \(a\) is a constant. Show that \(\frac{dy}{dx} = \frac{-2}{3t(1+t^2)^2}\). Hence find the gradient of the curve at the point \((a, \frac{1}{2}a)\). [7]

5 \includegraphics[max width=\textwidth, alt={}, center]{83d0697c-b133-47da-a745-dfdafa7dbf10-08_663_433_260_854} The diagram shows the curve with parametric equations $$x = \ln ( 2 t + 3 ) , \quad y = \frac { 2 t - 3 } { 2 t + 3 } .$$ The curve crosses the \(y\)-axis at the point \(A\) and the \(x\)-axis at the point \(B\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { 2 t + 3 }\).
2. Find the gradient of the curve at \(A\).
3. Find the gradient of the curve at \(B\).

5 \includegraphics[max width=\textwidth, alt={}, center]{3966e088-0a2f-434a-94fc-40765cd157a7-06_376_848_269_644} The diagram shows the curve with parametric equations $$x = 4 \mathrm { e } ^ { 2 t } , \quad y = 5 \mathrm { e } ^ { - t } \cos 2 t$$ for \(- \frac { 1 } { 4 } \pi \leqslant t \leqslant \frac { 1 } { 4 } \pi\). The curve has a maximum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
2. Find the coordinates of \(M\), giving each coordinate correct to 3 significant figures.

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

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

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

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

\includegraphics{figure_4} Figure 4 shows a sketch of part of the curve \(C\) with parametric equations $$x = 3\tan\theta, \quad y = 4\cos^2\theta, \quad 0 \leq \theta < \frac{\pi}{2}$$ The point \(P\) lies on \(C\) and has coordinates \((3, 2)\). The line \(l\) is the normal to \(C\) at \(P\). The normal cuts the \(x\)-axis at the point \(Q\).

1. [(a)] Find the \(x\) coordinate of the point \(Q\). \hfill [6]

The finite region \(S\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis, the \(y\)-axis and the line \(l\). This shaded region is rotated \(2\pi\) radians about the \(x\)-axis to form a solid of revolution.

1. [(b)] Find the exact value of the volume of the solid of revolution, giving your answer in the form \(p\pi + q\pi^2\), where \(p\) and \(q\) are rational numbers to be determined. [You may use the formula \(V = \frac{1}{3}\pi r^2 h\) for the volume of a cone.] \hfill [9] \end{enumerate} \end{enumerate}

The parametric equations of a curve are $$x = e^{-t}\cos t, \quad y = e^{-t}\sin t.$$ Show that \(\frac{dy}{dx} = \tan(t - \frac{1}{4}\pi)\). [6]

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 \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).

In this question you must show detailed reasoning. A curve has parametric equations $$x = \cos t - 3t \text{ and } y = 3t - 4\cos t - \sin 2t, \text{ for } 0 \leqslant t \leqslant \pi.$$ Show that the gradient of the curve is always negative. [7]

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 \(t\), $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{1}{6}\sec^4 t \cosec t.$$ [4]

7 The curve \(C\) has parametric equations $$x = \sin t , \quad y = \sin 2 t , \quad \text { for } 0 \leqslant t \leqslant \pi .$$ Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) in terms of \(t\). Hence, or otherwise, find the coordinates of the stationary points on \(C\) and determine their nature.

The curve \(C\) has parametric equations $$x = \frac{1}{2}t^2 - \ln t, \quad y = 2t + 1, \quad \text{for } \frac{1}{2} \leqslant t \leqslant 2.$$

1. Find the exact length of \(C\). [5]
2. Find \(\frac{d^2y}{dx^2}\) in terms of \(t\), simplifying your answer. [4]

6 A curve has parametric equations $$x = 9 t - \ln ( 9 t ) , \quad y = t ^ { 3 } - \ln \left( t ^ { 3 } \right)$$ Show that there is only one value of \(t\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) and state that value.

5

1. By differentiating \(\frac { 1 } { \cos \theta }\), show that if \(y = \sec \theta\) then \(\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \sec \theta \tan \theta\).
2. The parametric equations of a curve are $$x = 1 + \tan \theta , \quad y = \sec \theta$$ for \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\). Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin \theta\).
3. Find the coordinates of the point on the curve at which the gradient of the curve is \(\frac { 1 } { 2 }\).

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.

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 $$x = 1 + \ln ( t - 2 ) , \quad y = t + \frac { 9 } { t } , \quad \text { for } t > 2$$

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 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.

A curve is given parametrically by the equations $$x = t^2, \quad y = \frac{1}{t}.$$

1. Find \(\frac{dy}{dx}\) in terms of \(t\), giving your answer in its simplest form. [3]
2. Show that the equation of the tangent at the point \(P\left(4, -\frac{1}{4}\right)\) is \(x - 16y = 12\). [3]
3. Find the value of the parameter at the point where the tangent at \(P\) meets the curve again. [4]

8 A curve has parametric equations $$x = \frac { 1 } { t + 1 } , \quad y = t - 1 .$$ The line \(y = 3 x\) intersects the curve at two points.

1. Show that the value of \(t\) at one of these points is - 2 and find the value of \(t\) at the other point.
2. Find the equation of the normal to the curve at the point for which \(t = - 2\).
3. Find the value of \(t\) at the point where this normal meets the curve again.
4. Find a cartesian equation of the curve, giving your answer in the form \(y = \mathrm { f } ( x )\).

11. The curve \(C\) has parametric equations $$x = 10 \cos 2 t , \quad y = 6 \sin t , \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$ The point \(A\) with coordinates \(( 5,3 )\) lies on \(C\).

1. Find the value of \(t\) at the point \(A\).
2. Show that an equation of the normal to \(C\) at \(A\) is $$3 y = 10 x - 41$$ The normal to \(C\) at \(A\) cuts \(C\) again at the point \(B\).
3. Find the exact coordinates of \(B\).

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

1. Find the gradient of the curve at the point for which \(\theta = 1\) radian.
2. Find the value of \(\sin \theta\) at the point on the curve where the tangent is parallel to the \(y\)-axis.

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation $$x = 2 y ^ { 2 } + 5 y - 6$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\). The point \(P\) lies on the curve and is shown in Figure 1.
   Given that the tangent to the curve at \(P\) is parallel to the \(y\)-axis,
2. find the coordinates of \(P\).

1 A family of curves is given by the parametric equations \(x ( t ) = \cos ( t ) - \frac { \cos ( ( m + 1 ) t ) } { m + 1 }\) and \(y ( t ) = \sin ( t ) - \frac { \sin ( ( m + 1 ) t ) } { m + 1 }\) where \(0 \leqslant t < 2 \pi\) and \(m\) is a positive integer.

1. 1. Sketch the curves in the cases \(m = 3 , m = 4\) and \(m = 5\) on separate axes in the Printed Answer Booklet.
   2. State one common feature of these three curves.
   3. State a feature for the case \(m = 4\) which is absent in the cases \(m = 3\) and \(m = 5\).
2. 1. Determine, in terms of \(m\), the values of \(t\) for which \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\) but \(\frac { \mathrm { d } y } { \mathrm {~d} t } \neq 0\).
   2. Describe the tangent to the curve at the points corresponding to such values of \(t\).
3. 1. Show that the curve lies between the circle centred at the origin with radius $$1 - \frac { 1 } { m + 1 }$$ and the circle centred at the origin with radius $$1 + \frac { 1 } { m + 1 }$$
   2. Hence, or otherwise, show that the area \(A\) bounded by the curve satisfies $$\frac { m ^ { 2 } \pi } { ( m + 1 ) ^ { 2 } } < A < \frac { ( m + 2 ) ^ { 2 } \pi } { ( m + 1 ) ^ { 2 } }$$
   3. Find the limit of the area bounded by the curve as \(m\) tends to infinity.
4. The arc length of a curve defined by parametric equations \(x ( t )\) and \(y ( t )\) between points corresponding to \(t = c\) and \(t = d\), where \(c < d\), is $$\int _ { c } ^ { d } \sqrt { \left( \frac { \mathrm {~d} x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 } } \mathrm {~d} t$$ Use this to show that the length of the curve is independent of \(m\).

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.

4 The parametric equations of a curve are $$x = 4 \sin \theta , \quad y = 3 - 2 \cos 2 \theta$$ where \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\). Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\), simplifying your answer as far as possible.

5 A line PQ is of length \(k\) (where \(k > 1\) ) and it passes through the point ( 1,0 ). PQ is inclined at angle \(\theta\) to the positive \(x\)-axis. The end Q moves along the \(y\)-axis. See Fig. 5. The end P traces out a locus. \begin{figure}[h] \end{figure}

1. Show that the locus of P may be expressed parametrically as follows. $$x = k \cos \theta \quad y = k \sin \theta - \tan \theta$$ You are now required to investigate curves with these parametric equations, where \(k\) may take any non-zero value and \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\).
2. Use your calculator to sketch the curve in each of the cases \(k = 2 , k = 1 , k = \frac { 1 } { 2 }\) and \(k = - 1\).
3. For what value(s) of \(k\) does the curve have
   (A) an asymptote (you should state what the asymptote is),
   (B) a cusp,
   (C) a loop?
4. For the case \(k = 2\), find the angle at which the curve crosses itself.
5. For the case \(k = 8\), find in an exact form the coordinates of the highest point on the loop.
6. Verify that the cartesian equation of the curve is $$y ^ { 2 } = \frac { ( x - 1 ) ^ { 2 } } { x ^ { 2 } } \left( k ^ { 2 } - x ^ { 2 } \right) .$$

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.

12 The parametric equations of a curve are given by \(x = 2 \cos \theta\) and \(y = 3 \sin \theta\) for \(0 \leq \theta < 2 \pi\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). The tangents to the curve at the points P and Q pass through the point \(( 2,6 )\).
2. Show that the values of \(\theta\) at the points P and Q satisfy the equation \(2 \sin \theta + \cos \theta = 1\).
3. Find the values of \(\theta\) at the points \(P\) and \(Q\).

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.

7 A curve is given parametrically by \(x = t ^ { 2 } + 1 , y = t ^ { 3 } - 2 t\) where \(t\) is any real number.

1. Show that the equation of the normal to the curve at the point where \(t = 2\) can be written in the form \(2 x + 5 y = 30\).
2. Show that this normal does not meet the curve again.

7 A curve has parametric equations $$x = \frac { 2 t + 3 } { t + 2 } , \quad y = t ^ { 2 } + a t + 1$$ where \(a\) is a constant. It is given that, at the point \(P\) on the curve, the gradient is 1 .

1. Show that the value of \(t\) at \(P\) satisfies the equation $$2 t ^ { 3 } + ( a + 8 ) t ^ { 2 } + ( 4 a + 8 ) t + 4 a - 1 = 0$$
2. It is given that \(( t + 1 )\) is a factor of $$2 t ^ { 3 } + ( a + 8 ) t ^ { 2 } + ( 4 a + 8 ) t + 4 a - 1$$ Find the value of \(a\).
3. Hence show that \(P\) is the only point on the curve at which the gradient is 1 .
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.