# 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.

## Answers and Replies

Recall that:

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

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

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:
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

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!

Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

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:
[...]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

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

SammyS
Staff Emeritus