Wow I am stumped What is the difference between these two integrals?

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Homework Help Overview

The discussion revolves around understanding the differences between two integrals related to a velocity function, s'(t), and their implications for total distance and displacement. Participants are exploring concepts in calculus, particularly in the context of integrals and their geometric interpretations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to differentiate between the two integrals: one representing arc length and the other potentially representing area or displacement. There are questions about the validity of each integral in calculating distance.

Discussion Status

The discussion is active, with participants offering differing viewpoints on the nature of the integrals. Some have provided clarifications regarding the definitions and contexts of the integrals, while others are questioning the assumptions made about their applications.

Contextual Notes

There is mention of potential typos in the integrals being discussed, as well as concerns about the interpretation of velocity and distance in one-dimensional versus multi-dimensional contexts. Participants are also considering the implications of zero velocity on distance calculations.

flyingpig
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Homework Statement



Suppose s'(t) is a velocity function, then which of the integral will give you the total distance?

(1) [tex]\int_{a}^{b} \sqrt{1 + [s'(t)]^2} dt[/tex]

(2) [tex]\int_{a}^{b} |s(t)| dt[/tex]


The Attempt at a Solution



No clue at all...

the first is arc length, so it is like summing up the distances into one. The second one is absolute value of area of total displacement. They both are??
 
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What ? :confused::smile::smile:

Neither of them gives you a distance...
The first one is the length of a curve [itex]f(x)[/itex]. If it happens that [itex]f(x)[/itex] is a speed, you don't get the distance !
Let's take [itex]f(x) = 0[/itex], speed zero.
[tex]\int_a^{b} \sqrt{1+{|f(x)|}^2}\ dt= \int_a^{b} \sqrt{1+0}\ dt = b-a[/tex] !

Your speed is zero... but you travel some distance (b-a) ?

The second one, you basically multiply a length by a time... and the you integer (you sum). What do you get ? For sure you don't get a distance...
Under some circumstances you get an area...but... mmmm...
What was it for ? :cool:
 
flyingpig said:
(1) [tex]\int_{a}^{b} \sqrt{1 + [s'(t)]^2} dt[/tex]

(2) [tex]\int_{a}^{b} |s(t)| dt[/tex]

(1) is definitely correct for ℝ². However, if (2) is a typo and you actually meant

ab |s'(t)| dt,

then this is by far a better answer than (1), and I'll show you why.

Let s(t) be a parametric equation defined as (x(t),y(t)). Then the arc length integral is:

ab ds = ∫ab |s'(t)| dt

(Note that ds = |s'(t)| dt)

= ∫ab √( [dx/dt]² + [dy/dt]² ) dt

= ∫ab √( [dx]² + [dy]² )

= ∫ab √( [1]² + [dy/dx]² ) dx

= ∫ab √( 1 + [dy/dx]² ) dx

There isn't a difference in ℝ². However, in ℝn, you generally want to use:

ab |s'(t)| dt = ∫ab √( [dx1/dt]² +... + [dxn/dt]² ) dt

Btw Quinzio, your formula is just plain wrong.

Quinzio said:
Let's take [itex]f(x) = 0[/itex], speed zero.
[tex]\int_a^{b} \sqrt{1+{|f(x)|}^2}\ dt= \int_a^{b} \sqrt{1+0}\ dt = b-a[/tex] !

Your speed is zero... but you travel some distance (b-a) ?
 
Last edited:
If s'(t) is the velocity function in one dimension, then:
The total distance traveled in the time interval from t1 to t2 is [itex]\displaystyle \int_{t_1}^{\,t_2}\,|s'(t)|\,dt\,.[/itex]

The displacement during the same interval is [itex]\displaystyle \int_{t_1}^{\,t_2}\,s'(t)\,dt=s(t_2)-s(t_1)\,.[/itex]

 

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