# Reparametrize the curve in terms of arc length

1. Jan 29, 2012

### 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.

2. Jan 29, 2012

### DivisionByZro

Recall that:

$$\frac{e^{t}-e^{-t}}{2} = sinh(t)$$

Where sinh(t) is the hyperbolic sine function. Can you take it from here?

3. Jan 29, 2012

### SammyS

Staff Emeritus
Hello melifaro. Welcome to PF !

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

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.

Last edited: Jan 29, 2012
4. Jan 29, 2012

### 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!

Last edited: Jan 29, 2012
5. Jan 29, 2012

### SammyS

Staff Emeritus
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.

6. Jan 29, 2012

### DivisionByZro

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

Last edited: Jan 29, 2012
7. Jan 29, 2012

### melifaro

Just to make sure, is it because t is always positive from the problem description?

Edit

Yes, thanks. As I said I am not familiar with hyperbolic functions so SammyS' solution better for me

8. Jan 29, 2012

### SammyS

Staff Emeritus
It's because you were solving for et. That's always positive, as a function of a real variable.

9. Jan 29, 2012

Thank you!