Finding Length of Curve with Y^3/15 + 5/4y

  • Thread starter Thread starter whatlifeforme
  • Start date Start date
  • Tags Tags
    Curve Length
Click For Summary

Homework Help Overview

The discussion revolves around finding the length of a curve defined by the equation x = y^3/15 + 5/4y over the interval from y = 3 to y = 5.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the correct formula for arc length and question the interpretation of the equation. There are discussions about the differentiation of the curve and the proper setup of the integral for calculating the length.

Discussion Status

Some participants have provided guidance on the correct formula for arc length, while others are attempting to clarify the expressions used in the calculations. There is an acknowledgment of a discrepancy in the expected answer, but no consensus has been reached regarding the correct approach.

Contextual Notes

There are indications of confusion regarding the notation used in the equation, particularly whether the term is 5/(4y) or (5/4)y. Additionally, there is a note about the original poster's struggle with the calculations leading to an incorrect answer.

whatlifeforme
Messages
218
Reaction score
0

Homework Statement


find the length of the curve.


Homework Equations


x=y^3/15 + 5/4y on 3<=y<=5


The Attempt at a Solution


(dy/dx)^2 = Y^4/25 - 1/2 + 25/16y^4

integral (3,5) y^2/5 + 5/4y^2

however, i got the wrong answer. the answer is 67/10.
 
Physics news on Phys.org
whatlifeforme said:

Homework Statement


find the length of the curve.

Homework Equations


x=y^3/15 + 5/4y on 3<=y<=5

Is that 5/(4y) or (5/4)y. If you don't use Latex, at least use parentheses.

The Attempt at a Solution


(dy/dx)^2 = Y^4/25 - 1/2 + 25/16y^4

That isn't (dy/dx)^2 although it may be (dx/dy)^2

integral (3,5) y^2/5 + 5/4y^2

however, i got the wrong answer. the answer is 67/10.

That isn't the right formula for arc length. You need$$
\sqrt{1 + \left(\frac {dx}{dy}\right) ^2}\, dy$$in the integrand.
 
ok so thus far this should be correct.

L = integral (3,5) sqrt(1 + y^4/25 - 1/2 + 25/16y^4)

this could be further simplified to: sqrt (y^2/5 + 5/4y^2)^2 --->

y^3/15 - 5/4y ] (3 to 5)

the answer is : 67/10 which I'm not getting.
 
edit: solved.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
20
Views
2K
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K