Answer to Question #134204 in Calculus for Promise Omiponle

Question

Find the length of the curve r(t) =<cost, sint, ln(cost)> for 0≤t≤pi/4.

Answer 1

$r(t) = (\displaystyle\cos{t}, \sin{t}, \ln(\cos{t})), \hspace{0.2cm} \textsf{for}\hspace{0.1cm} 0\leq t \leq \frac{\pi}{4}\\ r(t)\hspace{0.1cm}\textsf{is a parametric equation of}\hspace{0.1cm} x, y\hspace{0.1cm} \textsf{and}\hspace{0.1cm} z. \\\textsf{Also note that}\hspace{0.1cm} r(t): \mathbb{R} \rightarrow \mathbb{R}^3 \hspace{0.1cm} \\\textsf{is a vector valued function of a real variable}\\ \textsf{with independent scalar output variables} \hspace{0.1cm} x, y \hspace{0.1cm}\&\hspace{0.1cm} z\\ r(t) = (x(t), y(t), z(t)) \Rightarrow x(t) = \cos{t}, y(t)=\sin{t}, z(t)=\ln{\cos{t}}\\ \textsf{Length of the curve}\hspace{0.1cm} (s) = \int_{s_1}^{s_2} \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2 }\hspace{0.1cm} \mathrm{d}t\\ x(t) = \cos{t}, y(t)=\sin{t}, z(t)=\ln(\cos{t}) \\ \frac{\mathrm{d}x}{\mathrm{d}t} = -\sin{t}, \frac{\mathrm{d}y}{\mathrm{d}t} = \cos{t}, \frac{\mathrm{d}z}{\mathrm{d}t} = -\tan{t}\\ s = \int_{0}^{\frac{\pi}{4}} \sqrt{\sin^2{t} + \cos^2{t} + \tan^2{t}}\hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\frac{\pi}{4}} \sqrt{1 + \tan^2{t}}\hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\frac{\pi}{4}} \sqrt{\sec^2{t}} \hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\frac{\pi}{4}} \sec{t} \hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\frac{\pi}{4}} \sec{t}\left(\frac{\sec{t} + \tan{t}}{\sec{t} + \tan{t}}\right) \mathrm{d}t\\ s = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2{t} + \tan{t}\sec{t}}{\sec{t} + \tan{t}}\hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d}(\sec{t} + \tan{t})}{\sec{t} + \tan{t}}\\ s = \ln(\sec{t} + \tan{t})\vert_{0}^{\frac{\pi}{4}}\\ s = \ln\left(\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)\right) - \\\ln(\sec{0} + \tan{0})\\ s = \ln(\sqrt{2} + 1) - \ln(1 + 0) = \\\ln(\sqrt{2} + 1) - \ln(1) =\\ \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1)\\ \therefore \textsf{The length of the curve}\hspace{0.1cm} r(t) = (\cos{t}, \sin{t}, \ln{\cos{t}})), \hspace{0.2cm} \textsf{for}\hspace{0.1cm} 0\leq t \leq \frac{\pi}{4}\hspace{0.1cm}\\\textsf{is}\hspace{0.1cm} \ln(\sqrt{2} + 1)\hspace{0.1cm}\textsf{unit} \approx 0.8814 \hspace{0.1cm}\textsf{unit}$