Question

a © lt (t) =<et, zee, 2t> a) compute the are arclength of the from tuoto tl. b) Reparemeterize the respect to *length. c) compute Ť, š, and B. F with d) Find formula for the curvature Evaluate it at t:1. e) Find the normal and osculating planes at t:1. f) Find the tangential and normal components of the acceleration t:1.

Answer 1

(a) We have

$\small \vec r(t)=\left \langle e^t,2e^{-t},2t \right \rangle$

So we have, differentiating with respect to t

$\small \vec r'(t)=\left \langle e^t,-2e^{-t},2 \right \rangle$

The norm of this tangent vector is given is

F(t) || = Ve2t + 4e-2t + 4

The arc length of this curve from t=0 to t=1 is given by

$\small \int_0^1 \| \vec{r}'(t)\|\, dt=\int _0^1\sqrt{e^{2t}+4e^{-2t}+4}\, dt$

Take the substituion u = -2t , so

$\small =\int _0^1\sqrt{e^{2t}+4e^{-2t}+4}\, dt=-\int _0^{-2}\frac{\sqrt{e^{-u}+4e^u+4}}{2}du$

$\small =\frac{1}{2} \int _{-2}^0\sqrt{e^{-u}+4e^u+4}\;du$

Again, taking the substitution $\small v=e^{-u}$ we get

$\small =-{1\over 2}\int _{e^2}^1\frac{v+2}{v\sqrt{v}}dv$

$\small ={1\over 2}\int^{e^2}_1\frac{v+2}{v\sqrt{v}}dv$

$\small ={1\over 2}\left [\int _1^{e^2}\frac{1}{\sqrt{v}}dv+\int _1^{e^2}\frac{2}{v\sqrt{v}}dv \right ]$

$\small ={1\over 2}\left [2e-2+2\left(-\frac{2}{e}+2\right) \right ]$

$\small =\frac{2e^2+2e-4}{2e}=e+1-2e^{-1}$

which is the required arc length.