Question

arclength f(t) = ((e^t)cos(t), (e^t)sin(t), 2) for t in [0, 2pi]

Answer 1

\[x(t)=e^t\cos{t}\quad,\quad y(t)=e^t\sin{t}\quad,\quad z(t)=2\]\[x'(t)=e^t(\cos{t}-\sin{t})\quad,\quad y'(t)=e^t(\sin{t}+\cos{t})\quad,\quad z'(t)=0\]\[l=\int\limits_0^{2\pi}\sqrt{(x')^2+(y')^2+(z')^2}dt=\]\[=\int\limits_0^{2\pi}\sqrt{e^{2t}(\cos{t}-\sin{t})^2+e^{2t}(\sin{t}+\cos{t})^2}dt=\]\[=\int\limits_0^{2\pi}e^t\sqrt{(\cos{t}-\sin{t})^2+(\sin{t}+\cos{t})^2}dt=\sqrt{2}\int\limits_0^{2\pi}e^tdt=\]\[=\sqrt{2}(e^{2\pi}-1)\]