• Support PF! Buy your school textbooks, materials and every day products Here!

Using dy/dx to find arc length of a parametric equation

  • #1
59
2

Homework Statement


I have attached a picture of the problem in the attachments

I need help on the last section, (part d)

Homework Equations


(1)##∫√( (dx/dt)^2+(dy/dt)^2)dt##
(2)##∫√( 1+(dy/dx)^2)dx##[/B]

The Attempt at a Solution


In order to get the answer we just need to find the arc-length of the position curve which could be done using the first equation. However, instead of using the first equation I divided ##dy/dt## by ##dx/dt## to find ##dy/dx##. Using dy/dx I plugged it into the second equation and got a different answer. Why is that so? Shouldn't they both yield the same answer? For both equations I used from 2 to 4 as the limits of integration.

Thanks[/B]
 

Attachments

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,017
Using dy/dx I plugged it into the second equation and got a different answer
I don't see what you do (not clearvoyant). Please post your attempt in full.
For both equations I used from 2 to 4 as the limits of integration.
And why is that ?
 
  • #3
59
2
I don't see what you do (not clearvoyant). Please post your attempt in full.
And why is that ?
I plugged this into a calculator and got 5.975
 

Attachments

  • #4
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,017
I plugged this into a calculator and got 5.975
Sure. But x starts at 1 and ends at what you (?) found .

And you have an integrand that is a function of t, not of x. So you either need to express dx as a function of t (and then you are back again to the solution manual), or express the integrand as a function of x, which I think is pretty tough.
 
  • #5
59
2
Sure. But x starts at 1 and ends at what you (?) found .

And you have an integrand that is a function of t, not of x. So you either need to express dx as a function of t (and then you are back again to the solution manual), or express the integrand as a function of x, which I think is pretty tough.
Wouldn't the limit of integration actually be from 1 to 1.252 (answer from part b), I tried that and I got 0.439. Also, what do you mean by expressing the integrand as a function of x? Isn't it already as a function of x? I just simply plugged it into the formula ##∫√(1+(dy/dx)^2)dx##
 
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,039
3,017
I tried
Can't be: you dont have ##t(x)## yet, so you can't integrate. Your result is simply the result of the wrong integration. And "I" got 0.441251 !?
Isn't it already as a function of x?
It certainly is not: dividing two functions of ##t## does not change the ##t## into an ##x## !
 
  • #7
59
2
Can't be: you dont have ##t(x)## yet, so you can't integrate. Your result is simply the result of the wrong integration. And "I" got 0.441251 !?
It certainly is not: dividing two functions of ##t## does not change the ##t## into an ##x## !
Ok, I see what you mean now, the ##dy/dx## is still in terms of t which is why the the integrand is incorrect. Thank you
 
  • Like
Reactions: BvU

Related Threads on Using dy/dx to find arc length of a parametric equation

Replies
9
Views
3K
  • Last Post
Replies
3
Views
975
Replies
6
Views
734
  • Last Post
Replies
2
Views
2K
Replies
4
Views
4K
Replies
2
Views
6K
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
11
Views
8K
Replies
2
Views
2K
Top