Understanding Line Integrals: Scalar vs Vector Fields

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The discussion centers on the differences between line integrals of scalar functions and vector fields, specifically why the integrals of a scalar function multiplied by the magnitude of the derivative do not equal the integral of the gradient of that function dotted with the derivative. The example provided illustrates this discrepancy using the function f(x,y) = xy and its gradient, leading to different results when evaluated along a specific curve. Participants express confusion over the equality stated in textbooks, as empirical examples consistently yield unequal results. The conversation highlights the need for a deeper understanding of the relationships between scalar fields, vector fields, and their integrals. Clarification on the underlying principles and potential misinterpretations of the formulas is sought.
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 F(r)= \nabla f then it is certainly true that \int_C \nabla F\cdot dr= \int_a^b \nabla F(r)\cdot r'(t)dt. But I don't know what you mean by "equation1 equal to equation2".
 
Given \nabla f = F(r)
\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b F(r(t)) \cdot r'(t)dt
Why?
 
cshum00 said:
Given \nabla f = F(r)
\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b F(r(t)) \cdot r'(t)dt
Why?
Why should \int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b \nabla f(r(t)) \cdot r'(t)dt?
 
Well, let's try some example:

Let f(x,y) = xy
F=\nabla f(x,y)=<y,x>

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

f(r(t))=t^2
F(r(t))=<t,t>

If \int_a^b f(r(t))|r'(t)|dt = \int_a^b F(r(t)) \cdot r'(t)dt
\int_0^1 t^2 \sqrt2dt = \int_0^1 <t,t> \cdot <1,1>dt
(\sqrt2/3) t^3 |_0^1 = \int_0^1 2t dt
\sqrt2/3 = t^2 |_0^1
\sqrt2/3 \neq 1
 
The only reason i have been asking this:
\int_a^b f(r(t)) |r'(t)|dt \neq \int_a^b \nabla f(r(t)) \cdot r'(t)dt
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 f(x,y)=f(x,y)
F = \nabla f(x,y)=<f_x, f_y>

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

\int_a^b f(r(t)) |r'(t)|dt = \int_a^b \nabla f(r(t)) \cdot r'(t)dt
\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
\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))dtSince limits and variable of integration are the same then
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))
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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