Finding Length of a Curve: y2 = (x-1)3

Click For Summary

Homework Help Overview

The discussion revolves around finding the length of the curve defined by the equation y² = (x-1)³, specifically from the points (1,0) to (2,1). Participants are exploring the appropriate setup for the integral to calculate the arc length.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are considering whether to express the function in terms of x or y and are questioning the limits of integration based on the given points. There is confusion regarding the correct interpretation of the coordinates of the points provided in the problem statement.

Discussion Status

Some participants have offered guidance on using the form y = f(x) and pointed out the significance of the starting and ending points in determining the limits of integration. However, there remains uncertainty about the correct limits and the interpretation of the coordinates.

Contextual Notes

Participants are grappling with the definitions of the points (1,0) and (2,1) and how they relate to the function being analyzed. There is a noted discrepancy in understanding the values of x and y at these points, which is affecting their approach to the problem.

EEristavi
Messages
108
Reaction score
5

Homework Statement



I have to find length of the curve: y2 = (x-1)3 from (1,0) to (2,1)

Homework Equations



s = ∫ √(1 + (f '(x) )2 ) dx where we have integral from a to b

The Attempt at a Solution



I'm bit confused:
I'm thinking of writing function regarding x, f(x). However, I can also write for y, f(y).
Which is better? and what values should I take for definite integral ("limits")
 
Physics news on Phys.org
EEristavi said:

Homework Statement



I have to find length of the curve: y2 = (x-1)3 from (1,0) to (2,1)

Homework Equations



s = ∫ √(1 + (f '(x) )2 ) dx where we have integral from a to b

The Attempt at a Solution



I'm bit confused:
I'm thinking of writing function regarding x, f(x). However, I can also write for y, f(y).
Which is better? and what values should I take for definite integral ("limits")

Just try it in the form ##y=f(x)##. And you are given a starting point and an ending point. What does that tell you about limits on ##x##?
 
Dick said:
What does that tell you about limits on xxx?

well, from the problem statement x should change from 0 to 1, but its not correct (as I see from the solution), integral is taken from 1 to 2 (and I can't figure it out why)
 
EEristavi said:
well, from the problem statement x should change from 0 to 1, but its not correct (as I see from the solution), integral is taken from 1 to 2 (and I can't figure it out why)

You do know that, e.g., at the point ##(1,0)## the value of ##x## is 1, right?
 
Dick said:
You do know that, e.g., at the point (1,0)(1,0)(1,0) the value of xxx is 1, right?

OK, now I think, I need more clarification... I will write what is in my mind:

point (1, 0) means that y=1, x=0
but I see from function y = f (x), that if x = 0 => y = 0

So I guess, here is my problem of understanding...
 
EEristavi said:
OK, now I think, I need more clarification... I will write what is in my mind:

point (1, 0) means that y=1, x=0
but I see from function y = f (x), that if x = 0 => y = 0

So I guess, here is my problem of understanding...

Where I come from the point ##(1,0)## means ##x=1## and ##y=0##. That looks like the definition the problem is using also. I don't think I've ever heard of a different convention.
 
  • Like
Likes   Reactions: CWatters and EEristavi
Dick said:
Where I come from the point ##(1,0)## means ##x=1## and ##y=0##. That looks like the definition the problem is using also. I don't think I've ever heard of a different convention.
Ok, maybe it's been a while, since I've done some mathematics :D

Thank you :)
 

Similar threads

Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K