What is the formula for finding arclength when given an integral?

  • Thread starter Thread starter JJ6
  • Start date Start date
  • Tags Tags
    Arclength
Click For Summary

Homework Help Overview

The problem involves finding the arclength of a curve defined by an integral of the square root of the cosine function, with the variable x constrained between -π/2 and π/2. The context is calculus, specifically relating to the application of the fundamental theorem of calculus in evaluating integrals.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to begin the problem, particularly due to the lack of a simple function for the integral of sqrt(cos(t)). Some participants discuss the application of the fundamental theorem of calculus to differentiate the integral, questioning the implications of constants in the differentiation process.

Discussion Status

Participants are exploring the differentiation of the integral and its implications for finding the arclength. There is a recognition of the fundamental theorem of calculus, and some guidance has been offered regarding the differentiation process, though no consensus has been reached on the interpretation of constants in this context.

Contextual Notes

The problem is constrained by the requirement to find the arclength based on an integral, and participants are navigating the complexities of differentiating an integral with variable limits.

JJ6
Messages
12
Reaction score
0

Homework Statement



Find the arclength of the curve given by y= integral from -pi/2 to x of sqrt(cost)dt. X is restricted between -pi/2 and pi/2.

Homework Equations



L = Integral from a to b of sqrt((dy/dx)^2 + 1)dx
L = Integral from a to b of sqrt((dy/dt)^2 + (dx/dt)^2)dt

The Attempt at a Solution



I'm not even sure how to start this problem since integral of sqrt(cost) has no simple function representing it. Can somebody please give me some direction?
 
Physics news on Phys.org
It is an application of the fondamental theorem of calculus.

[tex]\frac{d}{dx}\int_{x_0}^xf(t)dt=f(x)[/tex]
 
So does that mean that d/dx of the integral from -pi/2 to x of sqrt(cost)dt = sqrt(cosx)?

Does the -pi/2 just disappear because it is a constant?
 
Last edited:
JJ6 said:
So does that mean that d/dx of the integral from -pi/2 to x of sqrt(cost)dt = sqrt(cosx)?

Yep.

JJ6 said:
Does the -pi/2 just disappear because it is a constant?
Disapear from where?
 

Similar threads

Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K