What is the arc-length parameterization for a given vector function?

[Sho Kano]

Homework Statement

Find the arc-length parameterization for r(t)=\left< { e }^{ 2t },{ e }^{ -2t },2\sqrt { 2 } t \right> ,t\ge 0

Homework Equations

s(t)=\int { \left| \dot { r } (t) \right| dt }

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?

 

[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.

 

[Sho Kano]

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.

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.

 

[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 } }

 

[Ray Vickson]

Homework Statement

Find the arc-length parameterization for r(t)=\left< { e }^{ 2t },{ e }^{ -2t },2\sqrt { 2 } t \right> ,t\ge 0

Homework Equations

s(t)=\int { \left| \dot { r } (t) \right| dt }

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?

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

 

[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?