1.07s Parametric and implicit differentiation

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]

The equation of a curve is \(2x^4 + xy^3 + y^4 = 10\).

1. Show that \(\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}\). [4]
2. Hence show that there are two points on the curve at which the tangent is parallel to the \(x\)-axis and find the coordinates of these points. [4]

The parametric equations of a curve are $$x = 2\sin\theta + \sin 2\theta, \quad y = 2\cos\theta + \cos 2\theta,$$ where \(0 < \theta < \pi\).

1. Obtain an expression for \(\frac{dy}{dx}\) in terms of \(\theta\). [3]
2. Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the \(y\)-axis. [4]

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]

Answer only one of the following two alternatives. EITHER The curve \(C\) is defined parametrically by $$x = 18t - t^2 \quad \text{and} \quad y = 8t^{\frac{1}{2}},$$ where \(0 < t \leqslant 4\).

1. Show that at all points of \(C\), $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = \frac{-3(9 + t)}{2t^2(9 - t)^3}.$$ [4]
2. Show that the mean value of \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) with respect to \(x\) over the interval \(0 < x \leqslant 56\) is \(\frac{3}{70}\). [4]
3. Find the area of the surface generated when \(C\) is rotated through \(2\pi\) radians about the \(x\)-axis, showing full working. [6]

OR Let \(I_n = \int_1^{\sqrt{2}} (x^2 - 1)^n \mathrm{d}x\).

1. Show that, for \(n \geqslant 1\), $$(2n + 1)I_n = \sqrt{2} - 2nI_{n-1}.$$ [5]
2. Using the substitution \(x = \sec \theta\), show that $$I_n = \int_0^{\frac{1}{4}\pi} \tan^{2n+1} \theta \sec \theta \, \mathrm{d}\theta.$$ [4]
3. Deduce the exact value of $$\int_0^{\frac{1}{4}\pi} \frac{\sin^7 \theta}{\cos^8 \theta} \, \mathrm{d}\theta.$$ [5]

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]

The curve \(C\) has parametric equations $$x = 3t + 2t^{-1} + at^3, \quad y = 4t - \frac{3}{2}t^{-1} + bt^3, \quad \text{for } 1 \leq t \leq 2,$$ where \(a\) and \(b\) are constants.

1. It is given that \(a = \frac{2}{3}\) and \(b = -\frac{1}{2}\). Show that \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \frac{25}{4}(t^2 + t^{-2})^2\) and find the exact length of \(C\). [6]
2. It is given instead that \(a = b = 0\). Find the value of \(\frac{d^2y}{dx^2}\) when \(t = 1\). [4]

The curve C has parametric equations $$x = \frac{5}{3}t^{\frac{3}{2}} - 2t^{\frac{1}{2}}, \quad y = 2t + 5, \quad \text{for } 0 < t \leq 3.$$

1. Find the exact length of C. [5]
2. Find the set of values of \(t\) for which \(\frac{d^2y}{dx^2} > 0\). [5]

The curve \(C\) has equation $$4y^2 + 4\ln(xy) = 1.$$

1. Show that, at the point \(\left(2, \frac{1}{2}\right)\) on \(C\), \(\frac{dy}{dx} = -\frac{1}{6}\). [3]
2. Find the value of \(\frac{d^2y}{dx^2}\) at the point \(\left(2, \frac{1}{2}\right)\). [4]

The curve \(C\) has parametric equations $$x = \frac{1}{2}e^{2t} - \frac{1}{3}t^3 - \frac{1}{2}, \quad y = 2e^t(t-1), \quad \text{for } 0 \leqslant t \leqslant 1.$$ Find the exact length of \(C\). [7]

The curve \(C\) is defined by the equation $$8x^3 - 3y^2 + 2xy = 9$$ Find an equation of the normal to \(C\) at the point \((2, 5)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [7]

\includegraphics{figure_3} Figure 3 shows a sketch of the curve \(C\) with parametric equations $$x = 1 + 3\tan t, \quad y = 2\cos 2t, \quad -\frac{\pi}{6} \leq t \leq \frac{\pi}{3}$$ The curve crosses the \(x\)-axis at point \(P\), as shown in Figure 3.

1. Find the equation of the tangent to \(C\) at \(P\), writing your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants to be found. [5]

The curve \(C\) has equation \(y = f(x)\), where \(f\) is a function with domain \(\left[k, 1 + 3\sqrt{3}\right]\)

1. Find the exact value of the constant \(k\). [1]
2. Find the range of \(f\). [2]

\includegraphics{figure_4} Figure 4 shows a sketch of the closed curve with equation $$(x + y)^3 + 10y^2 = 108x$$

1. Show that $$\frac{dy}{dx} = \frac{108 - 3(x + y)^2}{20y + 3(x + y)^2}$$ [5]

The curve is used to model the shape of a cycle track with both \(x\) and \(y\) measured in km. The points \(P\) and \(Q\) represent points that are furthest north and furthest south of the origin \(O\), as shown in Figure 4. Using the result given in part (a),

1. find how far the point \(Q\) is south of \(O\). Give your answer to the nearest 100 m. [4]

A curve has equation $$x^2 + 2xy - 3y^2 + 16 = 0.$$ Find the coordinates of the points on the curve where \(\frac{dy}{dx} = 0\). [7]

A curve has parametric equations $$x = 2\cot t, \quad y = 2\sin^2 t, \quad 0 < t \leq \frac{\pi}{2}.$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of the parameter \(t\). [4]
2. Find an equation of the tangent to the curve at the point where \(t = \frac{\pi}{4}\). [4]
3. Find a cartesian equation of the curve in the form \(y = f(x)\). State the domain on which the curve is defined. [4]

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

1. Find the value of \(\frac{dy}{dx}\) at the point on \(C\) with coordinates \((9, 5)\). [4]
2. Find a cartesian equation of the curve in the form $$y = \frac{ax + b}{cx + d}$$ where \(a\), \(b\), \(c\) and \(d\) are integers. [3]

The curve \(C\) has the equation $$\sin(\pi y) - y - x^2 y = -5, \quad x > 0$$

1. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5] The point \(P\) with coordinates \((2, 1)\) lies on \(C\). The tangent to \(C\) at \(P\) meets the \(x\)-axis at the point \(A\).
2. Find the exact value of the \(x\)-coordinate of \(A\). [4]

The curve \(C\) has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$

1. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5]
2. Find the coordinates of the points on \(C\) where \(\frac{dy}{dx} = 0\) (Solutions based entirely on graphical or numerical methods are not acceptable.) [6]

A curve \(C\) has parametric equations $$x = 4t + 3, \quad y = 4t + 8 + \frac{5}{2t}, \quad t \neq 0$$

1. Find the value of \(\frac{dy}{dx}\) at the point on \(C\) where \(t = 2\), giving your answer as a fraction in its simplest form. [3]
2. Show that the cartesian equation of the curve \(C\) can be written in the form $$y = \frac{x^2 + ax + b}{x - 3}, \quad x \neq 3$$ where \(a\) and \(b\) are integers to be determined. [3]

A curve has equation $$x^3 - 2xy - 4x + y^3 - 51 = 0.$$ Find an equation of the normal to the curve at the point \((4, 3)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [8]

\includegraphics{figure_1} The curve shown in Fig. 1 has parametric equations $$x = \cos t, \quad y = \sin 2t, \quad 0 \leq t < 2\pi.$$

1. Find an expression for \(\frac{dy}{dx}\) in terms of the parameter \(t\). [3]
2. Find the values of the parameter \(t\) at the points where \(\frac{dy}{dx} = 0\). [3]
3. Hence give the exact values of the coordinates of the points on the curve where the tangents are parallel to the \(x\)-axis. [2]
4. Show that a cartesian equation for the part of the curve where \(0 \leq t < \pi\) is $$y = 2x\sqrt{(1 - x^2)}.$$ [3]
5. Write down a cartesian equation for the part of the curve where \(\pi \leq t < 2\pi\). [1]

\includegraphics{figure_1} A table top, in the shape of a parallelogram, is made from two types of wood. The design is shown in Fig. 1. The area inside the ellipse is made from one type of wood, and the surrounding area is made from a second type of wood. The ellipse has parametric equations, $$x = 5 \cos \theta, \quad y = 4 \sin \theta, \quad 0 \leq \theta < 2\pi$$ The parallelogram consists of four line segments, which are tangents to the ellipse at the points where \(\theta = \alpha\), \(\theta = -\alpha\), \(\theta = \pi - \alpha\), \(\theta = -\pi + \alpha\).

1. Find an equation of the tangent to the ellipse at \((5 \cos \alpha, 4 \sin \alpha)\), and show that it can be written in the form $$5y \sin \alpha + 4x \cos \alpha = 20.$$ [4]
2. Find by integration the area enclosed by the ellipse. [4]
3. Hence show that the area enclosed between the ellipse and the parallelogram is $$\frac{80}{\sin 2\alpha} - 20\pi.$$ [4]
4. Given that \(0 < \alpha < \frac{\pi}{4}\), find the value of \(\alpha\) for which the areas of two types of wood are equal. [3]

The circle \(C\) has equation \(x^2 + y^2 - 8x - 16y - 209 = 0\).

1. Find the coordinates of the centre of \(C\) and the radius of \(C\). [3]

The point \(P(x, y)\) lies on \(C\).

1. Find, in terms of \(x\) and \(y\), the gradient of the tangent to \(C\) at \(P\). [3]
2. Hence or otherwise, find an equation of the tangent to \(C\) at the point \((21, 8)\). [2]

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 \(t = \frac{1}{2}\pi\).

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

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The rectangular hyperbola \(H\) has equation \(xy = 36\) The point \(P(4, 9)\) lies on \(H\)

1. Show, using calculus, that the normal to \(H\) at \(P\) has equation $$4x - 9y + 65 = 0$$ [4]

The normal to \(H\) at \(P\) crosses \(H\) again at the point \(Q\)

1. Determine an equation for the tangent to \(H\) at \(Q\), giving your answer in the form \(y = mx + c\) where \(m\) and \(c\) are rational constants. [5]