1.07s Parametric and implicit differentiation

A curve has the equation $$3x^2 + xy - 2y^2 + 25 = 0.$$ Find an equation for the normal to the curve at the point with coordinates \((1, 4)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [8]

\includegraphics{figure_2} Figure 2 shows the curve with parametric equations $$x = t + \sin t, \quad y = \sin t, \quad 0 \leq t \leq \pi.$$

1. Find \(\frac{dy}{dx}\) in terms of \(t\). [3]
2. Find, in exact form, the coordinates of the point where the tangent to the curve is parallel to the \(x\)-axis. [3]
3. Show that the region bounded by the curve and the \(x\)-axis has area 2. [6]

\includegraphics{figure_5} The diagram 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 \(a^2\). [5]

A curve has the equation $$x^2 - 4xy + 2y^2 = 1.$$

1. Find an expression for \(\frac{dy}{dx}\) in its simplest form in terms of \(x\) and \(y\). [4]
2. Show that the tangent to the curve at the point \(P(1, 2)\) has the equation $$3x - 2y + 1 = 0.$$ [3]

The tangent to the curve at the point \(Q\) is parallel to the tangent at \(P\).

1. Find the coordinates of \(Q\). [4]

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 \(y\). [4]
2. Find an equation for the normal to the curve at the point \((3, -2)\). [3]

A curve has parametric equations $$x = t^3 + 1, \quad y = \frac{2}{t}, \quad t \neq 0.$$

1. Find an equation for the normal to the curve at the point where \(t = 1\), giving your answer in the form \(y = mx + c\). [6]
2. Find a cartesian equation for the curve in the form \(y = f(x)\). [3]

A curve has the equation $$x^2 + 2xy^2 + y = 4.$$ Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). [5]

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

1. Verify that $$x = k\cos\theta, \quad y = \frac{k}{2}\sin\theta,$$ are parametric equations for the curve. [3]
2. Hence or otherwise show that \(\frac{dy}{dx} = -\frac{x}{4y}\). [3]
3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). [1]

\includegraphics{figure_8}

1. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1\), \(k = 3\) and \(k = 4\). [3]

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.

1. Explain why the path of the stream is modelled by the differential equation $$\frac{dy}{dx} = \frac{4y}{x}.$$ [2]
2. Solve this differential equation. Given that the path of the stream passes through the point (2, 1), show that its equation is \(y = \frac{x^4}{16}\). [6]

\includegraphics{figure_3} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2\theta, \quad y = 10 \sin \theta + 5 \sin 2\theta, \quad (0 \leqslant \theta < 2\pi),$$ where \(x\) and \(y\) are in metres.

1. Show that \(\frac{\text{d}y}{\text{d}x} = -\frac{\cos \theta + \cos 2\theta}{\sin \theta + \sin 2\theta}\). Verify that \(\frac{\text{d}y}{\text{d}x} = 0\) when \(\theta = \frac{1}{3}\pi\). Hence find the exact coordinates of the highest point A on the path of C. [6]
2. Express \(x^2 + y^2\) in terms of \(\theta\). Hence show that $$x^2 + y^2 = 125 + 100 \cos \theta.$$ [4]
3. Using this result, or otherwise, find the greatest and least distances of C from O. [2]

You are given that, at the point B on the path vertically above O, $$2 \cos^2 \theta + 2 \cos \theta - 1 = 0.$$

1. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. [4]

The arc of the curve with equation \(y = 4 - \ln(1-x^2)\) from \(x = 0\) to \(x = \frac{3}{4}\) has length \(s\).

1. Show that \(s = \int_0^{\frac{3}{4}} \frac{\sqrt{1+x^2}}{1-x^2} \, dx\). [4 marks]
2. Find the value of \(s\), giving your answer in the form \(p + \ln N\), where \(p\) is a rational number and \(N\) is an integer. [6 marks]

1. Prove that the derivative of \(\cos^{-1} x\) is \(-\frac{1}{\sqrt{1 - x^2}}\). [3]

A curve has equation \(y = \cos^{-1}(1 - x^2)\), for \(0 < x < \sqrt{2}\).

1. Find and simplify \(\frac{dy}{dx}\), and hence show that $$(2 - x^2)\frac{d^2y}{dx^2} = x\frac{dy}{dx}.$$ [5]

A curve has parametric equations $$x = t + a \sin t, \quad y = 1 - a \cos t,$$ where \(a\) is a positive constant.

1. Draw, on separate diagrams, sketches of the curve for \(-2\pi < t < 2\pi\) in the cases \(a = 1\), \(a = 2\) and \(a = 0.5\). By investigating other cases, state the value(s) of \(a\) for which the curve has
   1. loops,
   2. cusps. [7]
2. Suppose that the point P\((x, y)\) lies on the curve. Show that the point P\('(-x, y)\) also lies on the curve. What does this indicate about the symmetry of the curve? [3]
3. Find an expression in terms of \(a\) and \(t\) for the gradient of the curve. Hence find, in terms of \(a\), the coordinates of the turning points on the curve for \(-2\pi < t < 2\pi\) and \(a \neq 1\). [5]
4. In the case \(a = \frac{1}{2}\pi\), show that \(t = \frac{1}{3}\pi\) and \(t = \frac{5}{3}\pi\) give the same point. Find the angle at which the curve crosses itself at this point. [3]

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]

Find the coordinates of the stationary points of the curve with equation $$x^3 + y^3 - 3xy = 48$$ and determine their nature. [14]

1. Given that \(y = \ln [t + \sqrt{(1 + t^2)}]\), show that \(\frac{dy}{dt} = \frac{1}{\sqrt{(1+t^2)}}\). [3]

The curve \(C\) has parametric equations $$x = \frac{1}{\sqrt{(1+t^2)}}, \quad y = \ln [t + \sqrt{(1 + t^2)}], \quad t \in \mathbb{R}.$$ A student was asked to prove that, for \(t > 0\), the gradient of the tangent to \(C\) is negative. The attempted proof was as follows: $$y = \ln \left(t + \frac{1}{x}\right)$$ $$= \ln \left(\frac{tx + 1}{x}\right)$$ $$= \ln (tx + 1) - \ln x$$ $$\therefore \frac{dy}{dx} = \frac{t}{tx + 1} - \frac{1}{x}$$ $$= \frac{\frac{t}{x}}{t + \frac{1}{x}} - \frac{1}{x}$$ $$= \frac{t\sqrt{(1+t^2)}}{t + \sqrt{(1+t^2)}} - \sqrt{(1 + t^2)}$$ $$= -\frac{(1+t^2)}{t + \sqrt{(1+t^2)}}$$ As \((1 + t^2) > 0\), and \(t + \sqrt{(1 + t^2)} > 0\) for \(t > 0\), \(\frac{dy}{dx} < 0\) for \(t > 0\).

1. 1. Identify the error in this attempt.
   2. Give a correct version of the proof. [6]
2. Prove that \(\ln [-t + \sqrt{(1 + t^2)}] = -\ln [t + \sqrt{(1 + t^2)}]\). [3]
3. Deduce that \(C\) is symmetric about the \(x\)-axis and sketch the graph of \(C\). [3]

\includegraphics{figure_5} The diagram shows the curve \(C\) with parametric equations \(x = \frac{3}{t}\), \(y = t^2 e^{-2t}\), where \(t > 0\). The maximum point on \(C\) is denoted by \(P\).

1. Determine the exact coordinates of \(P\). [4] The shaded region \(R\) is enclosed by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 6\).
2. Show that the area of \(R\) is given by $$\int_a^b 3te^{-2t} dt,$$ where \(a\) and \(b\) are constants to be determined. [3]
3. Hence determine the exact area of \(R\). [5]

In this question you must show detailed reasoning. \includegraphics{figure_6} The diagram shows the curve with equation \(4xy = 2(x^2 + 4y^2) - 9x\).

1. Show that \(\frac{dy}{dx} = \frac{4x - 4y - 9}{4x - 16y}\). [3] At the point \(P\) on the curve the tangent to the curve is parallel to the \(y\)-axis and at the point \(Q\) on the curve the tangent to the curve is parallel to the \(x\)-axis.
2. Show that the distance \(PQ\) is \(k\sqrt{5}\), where \(k\) is a rational number to be determined. [8]

A circle \(C\) has equation \(x^2 + y^2 - 6x + 10y + k = 0\).

1. Find the set of possible values of \(k\). [2]
2. It is given that \(k = -46\). Determine the coordinates of the two points on \(C\) at which the gradient of the tangent is \(\frac{1}{2}\). [5]

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]

A curve has equation $$y^3e^{2x} + 2y - 16x = k$$ where \(k\) is a constant. The curve has a stationary point on the \(y\)-axis. Determine the value of \(k\) [7 marks]

A sculpture formed from a prism is fixed on a horizontal platform, as shown in the diagram. The shape of the cross-section of the sculpture can be modelled by the equation \(x^2 + 2xy + 2y^2 = 10\), where \(x\) and \(y\) are measured in metres. The \(x\) and \(y\) axes are horizontal and vertical respectively. \includegraphics{figure_12} Find the maximum vertical height above the platform of the sculpture. [8 marks]

Find the coordinates of the stationary point of the curve with equation \((x + y - 2)^2 = e^y - 1\) [7 marks]

The curve defined by the parametric equations $$x = t^2 \text{ and } y = 2t \quad -\sqrt{2} \leq t \leq \sqrt{2}$$ is shown in Figure 1 below. \includegraphics{figure_1}

1. Find a Cartesian equation of the curve in the form \(y^2 = f(x)\) [2 marks]
2. The point \(A\) lies on the curve where \(t = a\) The tangent to the curve at \(A\) is at an angle \(\theta\) to a line through \(A\) parallel to the \(x\)-axis. The point \(B\) has coordinates \((1, 0)\) The line \(AB\) is at an angle \(\phi\) to the \(x\)-axis. \includegraphics{figure_1_extended}
   1. By considering the gradient of the curve, show that $$\tan \theta = \frac{1}{a}$$ [3 marks]
   2. Find \(\tan \phi\) in terms of \(a\). [2 marks]
   3. Show that \(\tan 2\theta = \tan \phi\) [3 marks]