1.07s Parametric and implicit differentiation

A curve \(C\) is defined by the parametric equations $$x = \frac{4 - e^{-6t}}{4}, \quad y = \frac{e^{3t}}{3t}, \quad t \neq 0$$

1. Find the exact value of \(\frac{dy}{dx}\) at the point on \(C\) where \(t = \frac{2}{3}\). [5 marks]
2. Show that \(x = \frac{4 - e^{-6t}}{4}\) can be rearranged into the form \(e^{3t} = \frac{e}{2\sqrt{(1-x)}}\). [2 marks]
3. Hence find the Cartesian equation of \(C\), giving your answer in the form $$y = \frac{e}{f(x)[1 - \ln(f(x))]}$$ [2 marks]

It is given that \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\).

1. By writing \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\) as \(2\tan\theta = 3x\), use implicit differentiation to show that $$\frac{d\theta}{dx} = \frac{k}{4 + 9x^2}$$, where \(k\) is an integer. [3 marks]
2. Hence solve the differential equation $$9y(4 + 9x^2)\frac{dy}{dx} = \cosec 3y$$ given that \(x = 0\) when \(y = \frac{\pi}{3}\). Give your answer in the form \(\mathbf{g}(y) = \mathbf{h}(x)\). [7 marks]

1. Use the identity for \(\cos(A + B)\) to prove that \(\cos 2A = 2\cos^2 A - 1\). [2]
2. Use the substitution \(x = 2\sqrt{2} \sin \theta\) to prove that $$\int_2^{\sqrt{6}} \sqrt{(8 - x^2)} \, dx = \frac{1}{3}(\pi + 3\sqrt{3} - 6).$$ [7]

A curve is given by the parametric equations $$x = \sec \theta, \quad y = \ln(1 + \cos 2\theta), \quad 0 \leq \theta < \frac{\pi}{2}.$$

1. Find an equation of the tangent to the curve at the point where \(\theta = \frac{\pi}{3}\). [5]

The curve \(C\) has equation \(5x^2 + 2xy - 3y^2 + 3 = 0\). The point \(P\) on the curve \(C\) has coordinates \((1, 2)\).

1. Find the gradient of the curve at \(P\). [5]
2. Find the equation of the normal to the curve \(C\) at \(P\), in the form \(y = ax + b\), where \(a\) and \(b\) are constants. [3]

A curve is given parametrically by the equations $$x = 5 \cos t, \quad y = -2 + 4 \sin t, \quad 0 \leq t < 2\pi.$$

1. Find the coordinates of all the points at which \(C\) intersects the coordinate axes, giving your answers in surd form where appropriate. [4]
2. Sketch the graph at \(C\). [2]

\(P\) is the point on \(C\) where \(t = \frac{1}{6}\pi\).

1. Show that the normal to \(C\) at \(P\) has equation $$8\sqrt{3}y = 10x - 25\sqrt{3}.$$ [4]

The equation of a curve is \(2x^2 + xy + y^2 = 14\). Show that there are two stationary points on the curve and find their coordinates. [8]

The parametric equations of a curve are \(x = 2t^2\), \(y = 4t\). Two points on the curve are \(P(2p^2, 4p)\) and \(Q(2q^2, 4q)\).

1. Show that the gradient of the normal to the curve at \(P\) is \(-p\). [2]
2. Show that the gradient of the chord joining the points \(P\) and \(Q\) is \(\frac{2}{p + q}\). [2]
3. The chord \(PQ\) is the normal to the curve at \(P\). Show that \(p^2 + pq + 2 = 0\). [2]
4. The normal at the point \(R(8, 8)\) meets the curve again at \(S\). The normal at \(S\) meets the curve again at \(T\). Find the coordinates of \(T\). [4]

The equation of a curve is \(xy^2 = 2x + 3y\).

1. Show that \(\frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}\). [5]
2. Show that the curve has no tangents which are parallel to the \(y\)-axis. [3]

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]

Find the gradient of the curve \(4x^2 + 2xy + y^2 = 12\) at the point \((1, 2)\). [4]

A curve is given parametrically by the equations $$x = 4\cos t, \quad y = 3\sin t,$$ where \(0 \leq t \leq \frac{1}{2}\pi\).

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]
2. Show that the equation of the tangent at the point \(P\), where \(t = p\), is $$3x\cos p + 4y\sin p = 12.$$ [3]
3. The tangent at \(P\) meets the \(x\)-axis at \(R\) and the \(y\)-axis at \(S\). \(O\) is the origin. Show that the area of triangle \(ORS\) is \(\frac{6}{\sin 2p}\). [3]
4. Write down the least possible value of the area of triangle \(ORS\), and give the corresponding value of \(p\). [3]

Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2t^2, y = 4t, \quad -\sqrt{2} < t < \sqrt{2}.$$ P\((2t^2, 4t)\) 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. \includegraphics{figure_8}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac{1}{t}\). [3]
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 \(\theta\). [8]

[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.

1. Show that the curve has cartesian equation \(y^2 = 8x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\). [7]

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]

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 \(\theta = \frac{1}{4}\pi\). [5]
2. Find \(y\) in terms of \(x\). [2]

Fig. 7a shows the curve with the parametric equations $$x = 2\cos\theta, \quad y = \sin 2\theta, \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.$$ 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. \includegraphics{figure_7a}

1. State, with their coordinates, the points on the curve for which \(\theta = -\frac{\pi}{2}\), \(\theta = 0\) and \(\theta = \frac{\pi}{2}\). [3]
2. Find \(\frac{dy}{dx}\) 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. [5]
3. Find the exact coordinates of the turning point Q. [3]

When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \includegraphics{figure_7b}

1. Express \(\sin^2\theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y^2 = x^2(1 - \frac{1}{4}x^2)\). [4]
2. Find the volume of the paperweight shape. [4]

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]

A curve has parametric equations \(x = e^{2t}, y = te^{2t}\).

1. Find \(\frac{dy}{dx}\) in terms of \(t\). Hence find the exact gradient of the curve at the point with parameter \(t = 1\). [4]
2. Find the cartesian equation of the curve in the form \(y = ax^b \ln x\), where \(a\) and \(b\) are constants to be determined. [3]

A curve has the equation $$x^2 + 3xy - 2y^2 + 17 = 0.$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5]
2. Find an equation for the normal to the curve at the point \((3, -2)\). [3]

A curve has parametric equations $$x = 3 \cos^2 t, \quad y = \sin 2t, \quad 0 \leq t < \pi.$$

1. Show that \(\frac{dy}{dx} = -\frac{2}{3} \cot 2t\). [4]
2. Find the coordinates of the points where the tangent to the curve is parallel to the \(x\)-axis. [3]
3. Show that the tangent to the curve at the point where \(t = \frac{\pi}{6}\) has the equation $$2x + 3\sqrt{3} y = 9.$$ [3]
4. Find a cartesian equation for the curve in the form \(y^2 = \text{f}(x)\). [4]

A curve has the equation $$2x^2 + xy - y^2 + 18 = 0.$$ Find the coordinates of the points where the tangent to the curve is parallel to the \(x\)-axis. [8]

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]

A curve has the equation $$x^2 - 3xy - y^2 = 12.$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5]
2. Find an equation for the tangent to the curve at the point \((2, -2)\). [3]

A curve has parametric equations $$x = t(t - 1), \quad y = \frac{4t}{1-t}, \quad t \neq 1.$$

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [4]

The point \(P\) on the curve has parameter \(t = -1\).

1. Show that the tangent to the curve at \(P\) has the equation $$x + 3y + 4 = 0.$$ [3]

The tangent to the curve at \(P\) meets the curve again at the point \(Q\).

1. Find the coordinates of \(Q\). [7]

A curve has the equation $$2 \sin 2x - \tan y = 0.$$

1. Find an expression for \(\frac{dy}{dx}\) in its simplest form in terms of \(x\) and \(y\). [5]
2. Show that the tangent to the curve at the point \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\) has the equation $$y = \frac{1}{2}x + \frac{\pi}{4}.$$ [3]

\includegraphics{figure_1} Figure 1 shows the curve with parametric equations $$x = a\sqrt{t}, \quad y = at(1-t), \quad t \geq 0,$$ where \(a\) is a positive constant.

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]

The curve meets the \(x\)-axis at the origin, \(O\), and at the point \(A\). The tangent to the curve at \(A\) meets the \(y\)-axis at the point \(B\) as shown.

1. Show that the area of triangle \(OAB\) is \(a^2\). [6]