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

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The discussion centers on determining which integral represents total distance given a velocity function s'(t). The first integral, ∫_{a}^{b} √(1 + [s'(t)]²) dt, is identified as the arc length, not total distance. The second integral, ∫_{a}^{b} |s(t)| dt, is also deemed incorrect for calculating distance unless it is corrected to ∫_{a}^{b} |s'(t)| dt, which accurately represents total distance traveled. The distinction between distance and displacement is emphasized, with total distance calculated as the integral of the absolute value of the velocity function. Understanding these integrals is crucial for correctly interpreting motion in both one and higher dimensions.
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) \int_{a}^{b} \sqrt{1 + [s'(t)]^2} dt

(2) \int_{a}^{b} |s(t)| dt


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 f(x). If it happens that f(x) is a speed, you don't get the distance !
Let's take f(x) = 0, speed zero.
\int_a^{b} \sqrt{1+{|f(x)|}^2}\ dt= \int_a^{b} \sqrt{1+0}\ dt = b-a !

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

(2) \int_{a}^{b} |s(t)| dt

(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 f(x) = 0, speed zero.
\int_a^{b} \sqrt{1+{|f(x)|}^2}\ dt= \int_a^{b} \sqrt{1+0}\ dt = b-a !

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 \displaystyle \int_{t_1}^{\,t_2}\,|s'(t)|\,dt\,.

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

 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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