Find the length of the curve between two endpoints

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Problem:

Find the length of the curve $\vec r(t) = \sqrt 2t \hat i + \sqrt 2t \hat j + (1 - t^2) \hat k$ from $(0, 0, 1)$ to $(\sqrt 2, \sqrt 2, 0)$

Solution Attempt:

First find the arc length function for this vector by using $s(t) = \int ||r'(t)|| dt$

Where $r'(t) = \sqrt 2 \hat i + \sqrt 2 \hat j - 2t \hat k$ and $||r'(t)|| = 2\sqrt{1 + t^2}$

So $s(t) = \int||r'(t)||dt = \sec x \tan x + \log(\sec x + \tan x)$

This is where I get stuck. How do I evaluate this for $t$ given that the endpoints are $(0, 0, 1)$ to $(\sqrt 2, \sqrt 2, 0)$?

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1 Answer

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Presumably you made the substitution $t=\tan x$. Then $\sec x=\sqrt{1+t^2}$, so

$$s(t)=t\sqrt{1+t^2}+\ln\left(t+\sqrt{1+t^2}\right)\;.$$

By inspection $\vec r(0)=\langle 0,0,1\rangle$ and $\vec r(1)=\langle\sqrt2,\sqrt2,0\rangle$, so you just want

$$\left[t\sqrt{1+t^2}+\ln\left(t+\sqrt{1+t^2}\right)\right]_0^1\;.$$

Alternatively, you could avoid substituting back by noting that $\tan 0=0$ and $\tan\frac{\pi}4=1$, so that you want

$$\left[\sec x\tan x+\ln\left(\sec x+\tan x\right)\right]_0^{\pi/4}\;.$$

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