## Reparametrize the curve in terms of arc length

Reparametrize the curve R(t) in terms of arc length measured from the point where t = 0

R(t) is defined by x = et, y = $\sqrt{2}$t, z = -e-t

Arc length S = ∫ ||R'(t)||dt

||R'(t)||= sqrt{$\dot{x}$2 + $\dot{y}$2 + $\dot{z}$2}

The attempt at a solution

Getting R'(t) ==> x = et, y = $\sqrt{2}$, z = e-t

Then ||R'(t)|| = sqrt{e2t + 2 + e-2t} = et + e-t

S = ∫(et + e-t)dt from 0 to some t
So
S = et - e-t

This is the point where I get stuck. How can I transform this equation into form t = ... ?

I tried taking ln of the whole equation but it doesn't seem to work.

Help please!
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 Blog Entries: 1 Recall that: $$\frac{e^{t}-e^{-t}}{2} = sinh(t)$$ Where sinh(t) is the hyperbolic sine function. Can you take it from here?

Mentor
 Quote by melifaro Reparametrize the curve R(t) in terms of arc length measured from the point where t = 0 R(t) is defined by x = et, y = $\sqrt{2}$t, z = -e-t Arc length S = ∫ ||R'(t)||dt ||R'(t)||= sqrt{$\dot{x}$2 + $\dot{y}$2 + $\dot{z}$2} The attempt at a solution Getting R'(t) ==> x = et, y = $\sqrt{2}$, z = e-t Then ||R'(t)|| = sqrt{e2t + 2 + e-2t} = et + e-t S = ∫(et + e-t)dt from 0 to some t So S = et - e-t This is the point where I get stuck. How can I transform this equation into form t = ... ? I tried taking ln of the whole equation but it doesn't seem to work. Help please!
Hello melifaro. Welcome to PF !

What's the definition of the hyperbolic sine, sinh(x) ?

Added in Edit:

Of course using the sinh function will leave S as a function of arcsinh(t).

Alternatively take the equation
S = et - e-t
and multiply by et to get an equation that's quadratic in et. Solve for et. (Keep only the + solution from the ± result. Why can you do that?) Take the natural log of the result.

## Reparametrize the curve in terms of arc length

I'm not very familiar with hyperbolic functions but does that mean that

S = 2 sinh(t)
sinh-1($\frac{S}{2}$) = t ?

So R(s) is defined by x = esinh-1($\frac{S}{2}$), y = $\sqrt{2}$sinh-1($\frac{S}{2}$), z = e-sinh-1($\frac{S}{2}$)

EDIT

 Quote by SammyS Alternatively take the equation S = et - e-t and multiply by et to get an equation that's quadratic in et. Solve for et. (Keep only the + solution from the ± result. Why can you do that?) Take the natural log of the result.
This is so simple and so smart. Thanks a lot!

Mentor
 Quote by melifaro I'm not very familiar with hyperbolic functions but does that mean that S = 2 sinh(t) sinh-1($\frac{S}{2}$) = t ? So R(s) is defined by x = esinh-1($\frac{S}{2}$), y = $\sqrt{2}$sinh-1($\frac{S}{2}$), z = -e-sinh-1($\frac{S}{2}$) EDIT This is so simple and so smart. Thanks a lot!
The x & z components come out very nicely using the result from the quadratic equation.

BTW: You left the negative off of the result for z.
 Blog Entries: 1 You are close, but can still simplify it: $$2sinh(t) = S$$ $$t = sin^{-1}(\frac{S}{2})$$ $$t= ln(\frac{1}{2}(S\pm\sqrt{4+s^{2}}))$$ Edit: fixed

 Quote by SammyS [...]Keep only the + solution from the ± result. Why can you do that?[...]
Just to make sure, is it because t is always positive from the problem description?

Edit

 Quote by DivisionByZro You are close, but can still simplify it: $$2sinh(t) = S$$ $$t = sin^{-1}(\frac{S}{2})$$ $$t= ln(S\pm\sqrt{1+s^{2}})$$
Yes, thanks. As I said I am not familiar with hyperbolic functions so SammyS' solution better for me

Mentor
 Quote by melifaro Just to make sure, is it because t is always positive from the problem description?
It's because you were solving for et. That's always positive, as a function of a real variable.

 Quote by SammyS It's because you were solving for et. That's always positive, as a function of a real variable.
Thank you!

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