Theory & Formulas
1. Arc Length ($L$)
- Cartesian: $L = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
- Parametric: $L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
2. Surface Area of Revolution ($S$)
- Cartesian: $S = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
- Parametric: $S = 2\pi \int_{t_1}^{t_2} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
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In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 1Figure 1 shows a sketch of the curve with equation
$y = \cos 2x \qquad 0 \leqslant x \leqslant \frac{\pi}{4}$The curve is rotated through $2\pi$ radians about the $x$-axis.
(a) Show that the area of the curved surface generated is given by$S = 2\pi \int_{0}^{\frac{\pi}{4}} \cos 2x \sqrt{1 + 4\sin^2 2x} \, \mathrm{d}x$(b) Hence, using the substitution $2\sin 2x = \sinh \theta$, show that$S = \frac{\pi}{4} \left( \ln(a + \sqrt{b}) + a\sqrt{b} \right)$where $a$ and $b$ are integers to be determined.(a) $\begin{aligned} & y = \cos 2x \quad , \quad 0 \leqslant x \leqslant \frac{\pi}{4} \\ & \frac{dy}{dx} = -2 \sin 2x \implies \left(\frac{dy}{dx}\right)^2 = 4 \sin^2 2x \\ & S = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \\ \therefore\quad & S = 2\pi \int_{0}^{\frac{\pi}{4}} \cos 2x \sqrt{1 + 4\sin^2 2x} \, dx \end{aligned}$ (b) $\begin{aligned} & 2 \sin 2x = \sinh \theta \quad \Rightarrow \quad 4 \sin^2 2x = \sinh^2 \theta \\ & 4 \cos 2x \, dx = \cosh \theta \, d\theta \Rightarrow 2 \cos 2x \, dx = \frac{1}{2} \cosh \theta \, d\theta \\ & x = 0 \Rightarrow \sinh \theta = 0 \Rightarrow \theta = 0 \\ & x = \frac{\pi}{4} \Rightarrow \sinh \theta = 2 \Rightarrow \theta = \mathrm{arsinh}\, 2 \\ \therefore\quad& S = \frac{\pi}{2} \int_{0}^{\mathrm{arsinh}\, 2} \cosh \theta \sqrt{1 + \sinh^2 \theta} \, d\theta = \frac{\pi}{2} \int_{0}^{\mathrm{arsinh}\, 2} \cosh^2 \theta \, d\theta \\ & \cosh 2\theta = 2 \cosh^2 \theta - 1 \Rightarrow \cosh^2 \theta = \frac{1}{2} (1 + \cosh 2\theta) \\ \therefore\quad& S = \frac{\pi}{4} \int_{0}^{\mathrm{arsinh}\, 2} (1 + \cosh 2\theta) \, d\theta \\ & = \frac{\pi}{4} \left[ \theta + \frac{1}{2} \sinh 2\theta \right]_{0}^{\mathrm{arsinh}\, 2} \\ & = \frac{\pi}{4} \left[ \mathrm{arsinh}\, 2 + \frac{1}{2} \sinh (2 \, \mathrm{arsinh}\, 2) \right] \\ & = \frac{\pi}{4} \left[ \ln(2 + \sqrt{5}) + 2\sqrt{5} \right] \\ \therefore\quad & a = 2, b = 5 \end{aligned}$ -
Figure 2Figure 2 shows a sketch of the curve $C$ defined by the parametric equations
$x = (2t + 3)^{\frac{3}{2}} \qquad y = \frac{3}{2}t^2 + 3t + 6 \qquad -\frac{3}{2} \leqslant t \leqslant 3$(a) Show that $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = a(t + 2)^2$ where $a$ is an integer to be determined.Hence, using algebraic integration, determine:(b) the exact length of $C$.(c) the exact area of the surface generated when $C$ is rotated through $360^{\circ}$ about the $x$-axis, giving your answer in the form $k\pi$ where $k$ is a rational number.(a) $\begin{aligned} & \frac{dx}{dt} = \frac{3}{2}(2t+3)^{\frac{1}{2}} \cdot 2 = 3(2t+3)^{\frac{1}{2}} \\ & \frac{dy}{dt} = 3t + 3 = 3(t+1) \\ & \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 9(2t+3) + 9(t+1)^2 \\ & = 18t + 27 + 9(t^2 + 2t + 1) = 9t^2 + 36t + 36 \\ & = 9(t^2 + 4t + 4) = 9(t+2)^2 \quad \implies a = 9 \end{aligned}$ (b) $\begin{aligned} L &=\int_{-\frac{3}{2}}^{3}\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \\ &= \int_{-\frac{3}{2}}^{3} \sqrt{9(t+2)^2} \, dt = \int_{-\frac{3}{2}}^{3} 3(t+2) \, dt\\ & = 3 \left[ \frac{t^2}{2} + 2t \right]_{-\frac{3}{2}}^{3} = \frac{297}{8} \end{aligned}$ (c) $\begin{aligned} S & = 2\pi \int_{-\frac{3}{2}}^{3} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \\ & = 2\pi \int_{-\frac{3}{2}}^{3} \left( \frac{3}{2}t^2 + 3t + 6 \right) \cdot 3(t+2) \, dt \\ & = 9\pi \int_{-\frac{3}{2}}^{3} (t^2 + 2t + 4)(t+2) \, dt \\ & = 9\pi \int_{-\frac{3}{2}}^{3} (t^3 + 4t^2 + 8t + 8) \, dt \\ & = 9\pi \left[ \frac{t^4}{4} + \frac{4t^3}{3} + 4t^2 + 8t \right]_{-\frac{3}{2}}^{3} = \frac{70551\pi}{64} \end{aligned}$ -
Figure 3Figure 3 shows the curve with equation
$y = \ln \left( \tanh \frac{x}{2} \right) \qquad 1 \leqslant x \leqslant 2$(a) Show that the length, $s$, of the curve is given by$s = \int_{1}^{2} \coth x \, \mathrm{d}x$(b) Hence show that $s = \ln \left( e + \frac{1}{e} \right)$.(a) $\begin{aligned} & \frac{dy}{dx} = \frac{1}{\tanh \frac{x}{2}} \cdot \mathrm{sech}^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cosh \frac{x}{2}}{\sinh \frac{x}{2}} \cdot \frac{1}{\cosh^2 \frac{x}{2}} \cdot \frac{1}{2} \\ & = \frac{1}{2\sinh \frac{x}{2} \cosh \frac{x}{2}} = \frac{1}{\sinh x} = \mathrm{csch}\, x \\ & 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \mathrm{csch}^2 x = \mathrm{coth}^2 x \\ & s = \int_{1}^{2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_{1}^{2} \sqrt{\mathrm{coth}^2 x} \, dx = \int_{1}^{2} \mathrm{coth}\, x \, dx \end{aligned}$ (b) $\begin{aligned} s &= \int_{1}^{2} \frac{\cosh x}{\sinh x} \, dx = \bigg[ \ln|\sinh x| \bigg]_{1}^{2} \\ &= \ln(\sinh 2) - \ln(\sinh 1) = \ln\left( \frac{\sinh 2}{\sinh 1} \right) \\ &= \ln\left( \frac{2\sinh 1 \cosh 1}{\sinh 1} \right) = \ln(2\cosh 1) \\ &= \ln\left( 2 \cdot \frac{e^1 + e^{-1}}{2} \right) = \ln\left( e + \frac{1}{e} \right) \end{aligned}$ -
$y = \arccos(\mathrm{sech}\, x) \qquad x > 0$(a) Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{sech}\, x$.
Figure 4Figure 4 shows a sketch of part of the curve $C$ with equation $y = \mathrm{f}(x)$ where $\mathrm{f}(x) = \arccos(\mathrm{sech}\, x) + \coth x$ for $x > 0$. The point $P$ is a minimum turning point of $C$.
(b) Show that the $x$ coordinate of $P$ is $\ln(q + \sqrt{q})$ where $q = \frac{1}{2}(1 + \sqrt{k})$ and $k$ is an integer to be determined.(a) $\begin{aligned} & \text{Let } u = \mathrm{sech}\, x \Rightarrow \frac{du}{dx} = -\mathrm{sech}\, x \tanh x \\ & \frac{dy}{dx} = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} = \frac{-1}{\sqrt{1-\mathrm{sech}^2 x}} \cdot (-\mathrm{sech}\, x \tanh x) \\ & = \frac{\mathrm{sech}\, x \tanh x}{\sqrt{\tanh^2 x}} = \frac{\mathrm{sech}\, x \tanh x}{\tanh x} = \mathrm{sech}\, x \end{aligned}$ (b) $\begin{aligned} & \mathrm{f}(x) = \arccos(\mathrm{sech}\, x) + \coth x \\ & \mathrm{f}'(x) = \mathrm{sech}\, x - \mathrm{csch}^2 x \\ & \text{At min point } P, \mathrm{f}'(x) = 0 \implies \mathrm{sech}\, x = \mathrm{csch}^2 x \\ & \frac{1}{\cosh x} = \frac{1}{\sinh^2 x} \implies \sinh^2 x = \cosh x \implies \cosh^2 x - \cosh x - 1 = 0 \\ & \cosh x = \frac{1 \pm \sqrt{5}}{2} \implies \cosh x = \frac{1+\sqrt{5}}{2} \quad (\text{since } \cosh x \geqslant 1) \\ & x = \mathrm{arccosh}\left(\frac{1+\sqrt{5}}{2}\right) = \ln\left( \frac{1+\sqrt{5}}{2} + \sqrt{\left(\frac{1+\sqrt{5}}{2}\right)^2 - 1} \right) \\ & = \ln\left( \frac{1+\sqrt{5}}{2} + \sqrt{\frac{1+\sqrt{5}}{2}} \right) \quad \therefore q = \frac{1}{2}(1+\sqrt{5}), \ k = 5 \end{aligned}$ -
Figure 5Figure 5 shows a sketch of the curve $C$ with equation
$y = \frac{1}{2}(\tan x + \cot x) \qquad \frac{\pi}{6} \leqslant x \leqslant \frac{\pi}{3}$(a) Show that the length of $C$ is given by $\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\tan^2 x + \cot^2 x) \, \mathrm{d}x$.(b) Hence determine the exact length of $C$, giving your answer in simplest form.(a) $\begin{aligned} & \frac{dy}{dx} = \frac{1}{2}(\sec^2 x - \csc^2 x) = \frac{1}{2}\left[ (1 + \tan^2 x) - (1 + \cot^2 x) \right] = \frac{1}{2}(\tan^2 x - \cot^2 x) \\ & 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4}(\tan^2 x - \cot^2 x)^2 = \frac{4 + (\tan^4 x - 2\tan^2 x \cot^2 x + \cot^4 x)}{4} \\ & \text{Since } \tan x \cot x = 1 \implies 2\tan^2 x \cot^2 x = 2 \\ & = \frac{4 - 2 + \tan^4 x + \cot^4 x}{4} = \frac{(\tan^2 x + \cot^2 x)^2}{4} \\ & L = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{\frac{(\tan^2 x + \cot^2 x)^2}{4}} \, dx = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\tan^2 x + \cot^2 x) \, dx \end{aligned}$ (b) $\begin{aligned} L &= \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left[ (\sec^2 x - 1) + (\csc^2 x - 1) \right] \, dx = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\sec^2 x + \csc^2 x - 2) \, dx \\ &= \frac{1}{2} \left[ \tan x - \cot x - 2x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \\ &= \frac{1}{2} \left[ \left(\tan \frac{\pi}{3} - \cot \frac{\pi}{3} - \frac{2\pi}{3}\right) - \left(\tan \frac{\pi}{6} - \cot \frac{\pi}{6} - \frac{2\pi}{6}\right) \right] \\ &= \frac{1}{2} \left[ 2\sqrt{3} - \frac{2\sqrt{3}}{3} - \frac{\pi}{3} \right] = \frac{2\sqrt{3}}{3} - \frac{\pi}{6} \end{aligned}$ -
A curve has parametric equations
$x = 4e^{\frac{1}{2}t} \qquad y = e^t - t \qquad 0 \leqslant t \leqslant 4$The curve is rotated through $2\pi$ radians about the $x$-axis. Show that the area of the curved surface generated is $\pi(e^8 + Ae^4 + B)$ where $A$ and $B$ are constants to be determined.
$\begin{aligned} & \frac{dx}{dt} = 2e^{\frac{1}{2}t} \quad , \quad \frac{dy}{dt} = e^t - 1 \\ & \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 4e^t + (e^{2t} - 2e^t + 1) = e^{2t} + 2e^t + 1 = (e^t + 1)^2 \\ & S = 2\pi \int_{0}^{4} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = 2\pi \int_{0}^{4} (e^t - t)(e^t + 1) \, dt \\ & = 2\pi \int_{0}^{4} (e^{2t} + e^t - te^t - t) \, dt \\ & \text{Using parts for } \int te^t dt = te^t - e^t \\ & S = 2\pi \left[ \frac{1}{2}e^{2t} + e^t - (te^t - e^t) - \frac{1}{2}t^2 \right]_{0}^{4} = 2\pi \left[ \frac{1}{2}e^{2t} + 2e^t - te^t - \frac{1}{2}t^2 \right]_{0}^{4} \\ & = 2\pi \left[ \left(\frac{1}{2}e^8 + 2e^4 - 4e^4 - 8\right) - \left(\frac{1}{2} + 2 - 0 - 0\right) \right] \\ & = 2\pi \left( \frac{1}{2}e^8 - 2e^4 - \frac{21}{2} \right) = \pi(e^8 - 4e^4 - 21) \\ \therefore \quad& A = -4, B = -21 \end{aligned}$ -
Figure 6Figure 6 shows a sketch of the curve $C$ with parametric equations
$x = \ln(\sec \theta + \tan \theta) - \sin \theta \qquad y = \cos \theta \qquad 0 \leqslant \theta \leqslant \frac{\pi}{4}$The curve $C$ is rotated through $2\pi$ radians about the $x$-axis and is used to form a solid of revolution $S$. Using calculus, show that the total surface area of $S$ is given by $\frac{\pi}{2}\left(p + q\sqrt{2}\right)$ where $p$ and $q$ are integers to be determined.
$\begin{aligned} & \frac{dx}{d\theta} = \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta} - \cos \theta = \sec \theta - \cos \theta \\ & \frac{dy}{d\theta} = -\sin \theta \\ & \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = (\sec \theta - \cos \theta)^2 + \sin^2 \theta = \sec^2 \theta - 2 + \cos^2 \theta + \sin^2 \theta = \sec^2 \theta - 1 = \tan^2 \theta \\ & \text{Curved Area } = 2\pi \int_{0}^{\frac{\pi}{4}} \cos \theta \sqrt{\tan^2 \theta} \, d\theta = 2\pi \int_{0}^{\frac{\pi}{4}} \cos \theta \tan \theta \, d\theta = 2\pi \int_{0}^{\frac{\pi}{4}} \sin \theta \, d\theta \\ & = 2\pi \bigg[ -\cos \theta \bigg]_{0}^{\frac{\pi}{4}} = 2\pi \left( -\frac{1}{\sqrt{2}} - (-1) \right) = 2\pi - \pi\sqrt{2} \\ & \text{Area of Circular Ends:} \\ & \text{At } \theta = 0: y = 1 \Rightarrow A_1 = \pi(1)^2 = \pi \\ & \text{At } \theta = \frac{\pi}{4}: y = \frac{1}{\sqrt{2}} \Rightarrow A_2 = \pi\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2} \\ & \text{Total Area } S = (2\pi - \pi\sqrt{2}) + \pi + \frac{\pi}{2} = \frac{7\pi}{2} - \pi\sqrt{2} = \frac{\pi}{2} (7 - 2\sqrt{2})\\ \therefore \quad &p = 7, q = -2 \end{aligned}$
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