Homework Help: Arc-Length Parameterization

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1. Sep 15, 2016

Sho Kano

1. The problem statement, all variables and given/known data
Find the arc-length parameterization for $r(t)=\left< { e }^{ 2t },{ e }^{ -2t },2\sqrt { 2 } t \right> ,t\ge 0$

2. Relevant equations
$s(t)=\int { \left| \dot { r } (t) \right| dt }$

3. The attempt at a solution
$\dot { r } (t)=\left< { 2e }^{ 2t },-2{ e }^{ -2t },2\sqrt { 2 } \right> \\ \left| \dot { r } (t) \right| =\sqrt { 4{ e }^{ 4t }+4{ e }^{ -4t }+8 } \\ =2\sqrt { { e }^{ 4t }+{ e }^{ -4t }+2 } \\ =2\sqrt { { e }^{ 4t }+\frac { 1 }{ { e }^{ 4t } } +2 } \\ =2\sqrt { \frac { { e }^{ 8t }+{ 2e }^{ 4t }+1 }{ { e }^{ 4t } } } =2\sqrt { \frac { \left( { e }^{ 4t }+1 \right) \left( { e }^{ 4t }+1 \right) }{ { e }^{ 4t } } } \\ =\frac { 2 }{ { e }^{ 2t } } \left( { e }^{ 4t }+1 \right) \\ ={ 2e }^{ 2t }+{ 2e }^{ -2t }\\ s(t)-s(0)=\int _{ 0 }^{ t }{ { 2e }^{ 2t }+{ 2e }^{ -2t }dt } =0$ It can't equal 0...where did I go wrong?

2. Sep 15, 2016

andrewkirk

It looks correct to me until the very last step where you say the integral is zero. The integral is not zero. It couldn't be, because both terms of the integrand are positive.

Note that $2e^{2t}+2e^{-2t}$ is $2\cosh 2t$. Have a look at the graph of the cosh function and it will be immediately apparent that its definite integral from 0 to $t$ is positive.

3. Sep 15, 2016

Sho Kano

Shoot, I made a mistake in the integration. It should be $s(t)=2\int { { e }^{ 2t }+{ e }^{ -2t } } =2\left( \frac { { e }^{ 2t } }{ 2 } -\frac { { e }^{ -2t } }{ 2 } \right)$. That's obviously not zero.

4. Sep 15, 2016

Sho Kano

What's the next step? Taking the natural log doesn't seem like it will work out, because I'd be left with ${ e }^{ 4t }-1$ inside the log. $s={ e }^{ 2t }-{ e }^{ -2t }=\frac { { e }^{ 4t }-1 }{ { e }^{ 2t } }$

5. Sep 15, 2016

Ray Vickson

You could also put $|\dot{r}(t)| = \sqrt{8} \sqrt{1+\cosh(4t)}$.

6. Sep 15, 2016

Sho Kano

OK, this problem was a pain
$s={ e }^{ 2t }-{ e }^{ -2t }=2sinh(2t)\\ \frac { s }{ 2 } =sinh(2t)\\ arcsinh\left( \frac { s }{ 2 } \right) =2t=ln\left[ \frac { s }{ 2 } +\sqrt { 1+\frac { { s }^{ 2 } }{ 2 } } \right] \\ r(t(s))$. So is it unsolvable without using hyperbolic functions?