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 = $\sqrt{2}$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 = e^t, y = $\sqrt{2}$, 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:

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

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

 

[SammyS]

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 = $\sqrt{2}$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 = e^t, y = $\sqrt{2}$, 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!

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

So R(s) is defined by x = e^{sinh-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!

 

[SammyS]

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 = e^{sinh-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.

 

[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

 

[melifaro]

[...]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]

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

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

 

[melifaro]

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

Thank you!