The equation for length of a curve: what are the integral ends?

Click For Summary

Homework Help Overview

The problem involves finding the length of a curve defined by the parametric equation r(t) = over a specified interval. The endpoints of the curve are given as <0,0,-3> and <1,2,-3>, leading to questions about the appropriate limits for the integral used in the length calculation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of taking the derivative of the curve and applying the length formula. There is a question regarding the choice of integral limits, specifically whether using 0 to 1 is appropriate based on the curve's components.

Discussion Status

Some participants express uncertainty about their approach, while others affirm that the method seems reasonable. The discussion reflects a mix of instinctual reasoning and a desire for clarification on the underlying concepts.

Contextual Notes

There is mention of incomplete notes that may have contributed to the confusion regarding the integral setup and limits. Participants are navigating assumptions about the relationship between the parameter t and the curve's endpoints.

Cloudless
Messages
15
Reaction score
0

Homework Statement



The given curve is r(t) = <t2, 2t, -3>

Write an equation for the length of the curve from <0,0,-3> to <1, 2, -3>



2. The attempt at a solution

I take the derivative of r(t) for r'(t), then plug it into the length formula.

L = ∫ of √( (2t)2 + 22 )

For the Integral, I put from 0 to 1, because in the original equation, the x component t2 = 1 when t is 1, and the y component 2t = 2 when t is 1.

Am I doing something wrong? o_o
 
Physics news on Phys.org
Cloudless said:

Homework Statement



The given curve is r(t) = <t2, 2t, -3>

Write an equation for the length of the curve from <0,0,-3> to <1, 2, -3>



2. The attempt at a solution

I take the derivative of r(t) for r'(t), then plug it into the length formula.

L = ∫ of √( (2t)2 + 22 )

For the Integral, I put from 0 to 1, because in the original equation, the x component t2 = 1 when t is 1, and the y component 2t = 2 when t is 1.

Am I doing something wrong? o_o

Seems ok to me. Why do you think you are doing something wrong?
 
Ah ok. The notes never fully explained it so I was doing it by instinct mostly. xD

Thanks
 
dunno
 
Last edited:

Similar threads

Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K