Answer to Question #97450 in Differential Geometry | Topology for Warsi

Question

find the arc length of r(t)=(e t sin t , e t cos t ,e t)

Answer 1

$x(t)=e^t\sin(t)$

$y(t)=e^t\cos(t)$

$z(t)=e^t$

$x'(t)=e^t\sin(t)+e^t\cos(t)$

$y'(t)=e^t\cos(t)-e^t\sin(t)$

$z'(t)=e^t$

arc length of $r(t) = \intop _a^b\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt=$

$=\intop _a^b\sqrt{(e^t\sin{t}+e^t\cos{t})^2+(e^t\cos{t}-e^t\sin{t})^2+(e^t)^2}dt=$

$=\intop _a^b\sqrt{e^{2t}\sin^{2}{t}+2e^{2t}\sin{t}\cos{t}+e^{2t}\cos^{2}{t}+e^{2t}\cos^{2}{t}-2e^{2t}\sin{t}\cos{t}+e^{2t}\sin^{2}{t}+e^{2t}}dt=$

$=\intop _a^b\sqrt{2e^{2t}\sin^{2}{t}+2e^{2t}\cos^{2}{t}+e^{2t}}dt=$

$=\intop _a^b\sqrt{2e^{2t}(\sin^{2}{t}+\cos^{2}{t})+e^{2t}}dt=$

$=\intop _a^b\sqrt{2e^{2t}+e^{2t}}dt=$

$=\intop _a^b\sqrt{3e^{2t}}dt=$

$=\sqrt{3}\intop _a^b{e^{t}}dt=$

$=\sqrt{3}{e^{t}}|\ _a^b=$

$=\sqrt{3}(e^{b}-e^{a})$