Answer to Question #134207 in Calculus for Promise Omiponle

Question

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

Answer 1

$r(t) = (\displaystyle e^t\cos{t}, e^t\sin{t}, e^t), \hspace{0.2cm} \textsf{for}\hspace{0.1cm} 0\leq t \leq\ln(4)\\ r(t)\hspace{0.1cm}\textsf{is a parametric equation of}\hspace{0.1cm} x, y\hspace{0.1cm} \textsf{and}\hspace{0.1cm} z. \\ r(t) = (x(t), y(t), z(t)) \Rightarrow x(t) = e^t\cos{t}, y(t)=e^t\sin{t}, z(t)=e^t\\ \textsf{Length of the curve \textit{(s)}} = \int_{s_1}^{s_2} \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2 }\hspace{0.1cm} \mathrm{d}t\\ x(t) = e^t\cos{t}, y(t)=e^t\sin{t}, z(t)=e^t\\ \frac{\mathrm{d}x}{\mathrm{d}t} = e^t(\cos{t} -\sin{t}), \frac{\mathrm{d}y}{\mathrm{d}t} =e^t(\cos{t} + \sin{t}), \frac{\mathrm{d}z}{\mathrm{d}t} = e^t\\ s = \int_{0}^{\ln(4)}\sqrt{e^{2t}(\cos{t} -\sin{t})^2 + e^{2t}(\cos{t} -\sin{t})^2+ e^{2t}}\hspace{0.1cm} \mathrm{d}t\\ e^{2t}(\cos{t} -\sin{t})^2 + e^{2t}(\cos{t} -\sin{t})^2+ e^{2t} \\= e^{2t}((\cos{t} -\sin{t})^2 + (\cos{t} -\sin{t})^2+ 1) \\= e^{2t}(\cos^2{t} + \sin^2{t} - 2\cos{t}\sin{t} + cos^2{t} + \sin^2{t} + 2\cos{t}\sin{t} + 1)\\ = e^{2t}(1 + 1 + 1) = 3e^{2t}\\ \therefore s = \int_{0}^{\ln(4)} \sqrt{e^{2t}(1 + 1 + 1)}\hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\ln(4)} \sqrt{3}\sqrt{e^{2t}} \hspace{0.1cm} \mathrm{d}t\\ s = \int_{0}^{\ln(4)} \sqrt{3} e^t \hspace{0.1cm} \mathrm{d}t\\ s = \sqrt{3}e^t \vert_{0}^{\ln(4)} = \sqrt{3}(e^{\ln(4)} - e^{(0)})= \sqrt{3}(4 - 1) = 3\sqrt{3}\\ \therefore \textsf{The length of the curve}\hspace{0.1cm} r(t) = (e^t\cos{t}, e^t\sin{t}, e^t), \hspace{0.2cm} \textsf{for}\hspace{0.1cm} 0\leq t \leq \ln(4)\hspace{0.1cm}\\\textsf{is}\hspace{0.1cm} 3\sqrt{3}\hspace{0.1cm}\textsf{units} \approx 5.1962 \hspace{0.1cm}\textsf{units}$