Reparametrize the curve in terms of arc length

  • Thread starter melifaro
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  • #1
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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 = [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 = et, y = [itex]\sqrt{2}[/itex], 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!
 

Answers and Replies

  • #2
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?
 
  • #3
SammyS
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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 = [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 = et, y = [itex]\sqrt{2}[/itex], 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.
 
Last edited:
  • #4
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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 = esinh-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

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:
  • #5
SammyS
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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 = esinh-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!
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
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
 
Last edited:
  • #7
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[...]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:

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

Yes, thanks. As I said I am not familiar with hyperbolic functions so SammyS' solution better for me
 
  • #8
SammyS
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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.
 
  • #9
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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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