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

The length of a path over a curve in ℝ² is \displaystyle \int_{t_1}^{\,t_2}\,|s'(t)|\,dt\,.The area under the curve in ℝ² is \displaystyle \int_{t_1}^{\,t_2}\,s(t)\,|\,ds(t)|=\frac{1}{2}\,\int_{t_1}^{\,t_2}\,s(t)\,|s'(t)|\,dt\,.The path of a moving object is a curve in ℝ³, so the length of the path is \displaystyle \int_{t_1}^{\,t_
  • #1
flyingpig
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1

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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  • #2
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:
 
  • #3
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) ?
 
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  • #4
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]

 

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

What are integrals and why are they important in science?

Integrals are mathematical tools used to calculate the total area under a curve. They are important in science because they help us understand and analyze various physical phenomena, such as the motion of objects, the flow of fluids, and the distribution of electric charges.

How do you solve integrals?

The solution to an integral involves finding the antiderivative of the function being integrated. This can be done using various techniques, such as substitution, integration by parts, and trigonometric identities. The resulting antiderivative is then evaluated at the limits of integration to find the total area under the curve.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and gives a numerical value as the result. It represents the area under the curve between those limits. An indefinite integral, on the other hand, has no limits and gives a function as the result. It represents the general solution to the integral and can be used to find specific values by plugging in the appropriate limits.

Can you explain the concept of integration by parts?

Integration by parts is a technique used to solve integrals that involve products of functions. It involves breaking down the integral into two parts, one of which is differentiated and the other integrated. This allows us to turn a difficult integral into an easier one that can be solved using basic integration rules.

How are integrals used in real-world applications?

Integrals have a wide range of applications in various fields of science and engineering. They are used to calculate areas, volumes, and centers of mass, and to solve problems related to motion, work, and energy. In physics, they are used to analyze the behavior of particles and systems, while in economics, they are used to model demand and supply curves. They are also used in signal processing, image processing, and machine learning.

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