Finding arc length using integration

  • #1
Find the length of the positive arc of the curve [itex]y=cosh^{-1}(x)[/itex] (for which y≥0) between [itex]x=1[/itex] and [itex]x=\sqrt{5}[/itex].



My attempt: [itex]x=cosh(y) → \frac{dx}{dy} = sinh(y) → (\frac{dx}{dy})^{2}=sinh^{2}(y)[/itex], so [itex]ds=dy\sqrt{1+sinh^{2}(y)}[/itex], therefore the arc length is [itex]S=\int_{y=0}^{y=cosh^{-1}(\sqrt{5})} cosh(y) dy= 2[/itex]. Is this right? Even if it is, is there another method of doing it (e.g. parametric equations)?
 
Last edited by a moderator:

Answers and Replies

  • #2
761
13
No you have to integrate between 1 and sqrt(5).
 
  • #3
No you have to integrate between 1 and sqrt(5).

But those are the x limits. If I want to integrate with respect to dy, I need y limits. I could've used dy/dx instead of dx/dy, but I don't know how to differentiate [itex]y=cosh^{-1}(x)[/itex].
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,228
1,830
Find the length of the positive arc of the curve [itex]y=cosh^{-1}(x)[/itex] (for which y≥0) between [itex]x=1[/itex] and [itex]x=\sqrt{5}[/itex].



My attempt: [itex]x=cosh(y) → \frac{dx}{dy} = sinh(y) → (\frac{dx}{dy})^{2}=sinh^{2}(y)[/itex], so [itex]ds=dy\sqrt{1+sinh^{2}(y)}[/itex], therefore the arc length is [itex]S=\int_{y=0}^{y=cosh^{-1}(\sqrt{5})} cosh(y) dy= 2[/itex]. Is this right? Even if it is, is there another method of doing it (e.g. parametric equations)?
Looks fine to me.
 

Related Threads on Finding arc length using integration

  • Last Post
Replies
1
Views
6K
Replies
7
Views
8K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
6K
Replies
1
Views
674
  • Last Post
Replies
5
Views
1K
Replies
2
Views
9K
Replies
2
Views
919
Replies
3
Views
2K
Top