Understanding Line Integrals: Scalar vs Vector Fields

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Discussion Overview

The discussion revolves around the concept of line integrals in the context of scalar and vector fields. Participants explore the relationship between line integrals of scalar functions and vector fields, questioning why certain equations that should theoretically be equal yield different results in practice.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants note that the line integral of a scalar function and the line integral of a vector field should theoretically yield the same results under certain conditions.
  • One participant expresses confusion about the equality of two specific line integral equations involving a scalar function and its gradient, questioning the validity of the book's assertion that they should be equal.
  • Another participant provides an example using a specific scalar function and its gradient, demonstrating that the two line integrals do not yield the same result when computed.
  • A participant attempts to derive a general proof but struggles with the relationship between the scalar function, its gradient, and the parametrization of the curve.
  • There is a mention of the right-hand side of the equation resembling the product rule of derivatives, indicating a potential connection that remains unexplored.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the equality of the two line integral expressions. There are multiple competing views, with some participants asserting that the integrals should be equal while others provide counterexamples that suggest otherwise.

Contextual Notes

Participants express uncertainty regarding the assumptions underlying the equations and the implications of the gradient in the context of line integrals. There are unresolved mathematical steps in the proposed proofs and examples.

cshum00
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Well i know that the line integral is
36b2dc0ff136923eb88a33c1f36d7ca8.png
given a scalar function f. equation1

But the line integral is also
cf0ea5b21fdf24a36e4b98844ccd673b.png
given a vector field F. equation2

So, given scalar function f and taking the gradient vector of it in order to turn it into a vector field F. Why is equation1 not equal to equation2?
 
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It's not clear to me what you are asking. If [itex]F(r)= \nabla f[/itex] then it is certainly true that [itex]\int_C \nabla F\cdot dr= \int_a^b \nabla F(r)\cdot r'(t)dt[/itex]. But I don't know what you mean by "equation1 equal to equation2".
 
Given [itex]\nabla f = F(r)[/itex]
[itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b F(r(t)) \cdot r'(t)dt[/itex]
Why?
 
cshum00 said:
Given [itex]\nabla f = F(r)[/itex]
[itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b F(r(t)) \cdot r'(t)dt[/itex]
Why?
Why should [itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b \nabla f(r(t)) \cdot r'(t)dt[/itex]?
 
Well, let's try some example:

Let [itex]f(x,y) = xy[/itex]
[itex]F=\nabla f(x,y)=<y,x>[/itex]

Curve C => [itex]y=x, 0<x<1[/itex]
Parametrize [itex]r(t)=<t,t>, 0<t<1[/itex]
[itex]r'(t)=<1,1>[/itex]
[itex]|r'(t)|=\sqrt2[/itex]

[itex]f(r(t))=t^2[/itex]
[itex]F(r(t))=<t,t>[/itex]

If [itex]\int_a^b f(r(t))|r'(t)|dt = \int_a^b F(r(t)) \cdot r'(t)dt[/itex]
[itex]\int_0^1 t^2 \sqrt2dt = \int_0^1 <t,t> \cdot <1,1>dt[/itex]
[itex](\sqrt2/3) t^3 |_0^1 = \int_0^1 2t dt[/itex]
[itex]\sqrt2/3 = t^2 |_0^1[/itex]
[itex]\sqrt2/3 \neq 1[/itex]
 
The only reason i have been asking this:
[itex]\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b \nabla f(r(t)) \cdot r'(t)dt[/itex]
is because the formula given on the book say that they should be equal. However, as i try many examples; i never get the same answer. Here is some general proof i have been trying but never got far.

Let [itex]f(x,y)=f(x,y)[/itex]
[itex]F = \nabla f(x,y)=<f_x, f_y>[/itex]

Let Parametrization of Curve C:
[itex]r(t) = <g(t), h(t)>[/itex]
[itex]r'(t) = <g'(t), h'(t)>[/itex]
[itex]|r'(t)| = [g'(t)^2 + h'^2(t)]^\frac{1}{2}[/itex]

[itex]\int_a^b f(r(t)) |r'(t)|dt = \int_a^b \nabla f(r(t)) \cdot r'(t)dt[/itex]
[itex]\int_a^b f(g(t), h(t))[g'(t)^2 + h'(t)^2]^\frac{1}{2}dt = \int_a^b <f(_x, f_y> \cdot <g'(t), h'(t)>dt[/itex]
[itex]\int_a^b f(g(t), h(t))[g'(t)^2 + h'(t)^2]^\frac{1}{2}dt = \int_a^b g'(t)f_x(g(t), h(t))+ h'(t)f_y(g(t), h(t))dt[/itex]Since limits and variable of integration are the same then
[itex]f(g(t), h(t))[g'(t)^2 + h'(t)^2]^\frac{1}{2} = g'(t)f_x(g(t), h(t)) + h'(t)f_y(g(t), h(t))[/itex]
But since i don't have any idea of what is going on inside those partial derivatives, i don't have a clue how to proceed.

Edit: In addition, the right hand side kind of looks like the product rule of a derivative.
 

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