Reparametrize the curve in terms of arc length

melifaro

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

R(t) is defined by x = e^t, y = [itex]\sqrt{2}[/itex]t, z = -e^-t

Arc length S = ∫ ||R'(t)||dt

||R'(t)||= sqrt{[itex]\dot{x}[/itex]² + [itex]\dot{y}[/itex]² + [itex]\dot{z}[/itex]²}


The attempt at a solution

Getting R'(t) ==> x = e^t, y = [itex]\sqrt{2}[/itex], z = e^-t

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

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

DivisionByZro

Recall that:

[tex] \frac{e^{t}-e^{-t}}{2} = sinh(t) [/tex]

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

SammyS

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 = e^t - e^-t

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

melifaro

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

S = 2 sinh(t)
sinh^-1([itex]\frac{S}{2}[/itex]) = t ?

So R(s) is defined by x = e^{sinh-1([itex]\frac{S}{2}[/itex])}, y = [itex]\sqrt{2}[/itex]sinh^-1([itex]\frac{S}{2}[/itex]), z = e^{-sinh-1([itex]\frac{S}{2}[/itex])}

EDIT

This is so simple and so smart. Thanks a lot!

SammyS

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.

DivisionByZro

You are close, but can still simplify it:

[tex] 2sinh(t) = S [/tex]
[tex] t = sin^{-1}(\frac{S}{2}) [/tex]
[tex] t= ln(\frac{1}{2}(S\pm\sqrt{4+s^{2}})) [/tex]

Edit: fixed

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

SammyS

It's because you were solving for e^t. That's always positive, as a function of a real variable.

melifaro

Thank you!