Path Integral: James J. Callahan's Advanced Calculus - A Geometric View

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Page $22: 1.24$. Determine the work done by the force field F in moving a particle along the oriented curve $\overrightarrow{C}$, where: c. F = (y,x), $\overrightarrow{C}$: any path from $(5,2)$ to $(7,11)$.

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Do I have to pick up a third point, like (7,2), to construct a C1 and C2 and then integrate two part integrals and sum up them into one? I have no clue to solve it because I am confused with the question.

I try the second method which treats $y$ as a function of $x$. the slope of y is, $m= (11-2)/(7-5)=7/2$, then $(y-2)=m(x-5)$, so $$y=7x/2-35/2+2=7x/2 - 31/2$$

Let $x=t$, then $y= 7t/2-31/2$; $dx = 1dt$, $dy = 7/2dt$.

Then, $$\int_C{ F \cdot dx} = \int_5^7{ (7t/2-31/2, t) \cdot (1, 7/2) dt }$$

Is this way right to fix the problem?

I also put the context here to clarify. enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

I guess it ask me to fix Pdx+Qdy, as $$\mathbb{F=(P,Q)}$$, and $$\mathbb{X=(x,y)}, then $$\Int_C{F \cdot dX} = \Int_C{Pdx+Qdy}$$

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$\vec{F}=A(x,y)\hat{i}+B(x,y)\hat{j}$

Suppose $\frac{\partial B}{\partial x}=\frac{\partial A}{\partial y}$, i.e. the curl is 0. The integral is path invariant and there $\exists V(x,y)| \partial V/\partial x=A $ and $\partial V/\partial y=B$.

$\partial(\int A(x,y)dx+Q(y))/\partial y=B$

$\int \frac{\partial A}{\partial y} dx+Q'=B$

$Q= \int[B-\int \frac{\partial A}{\partial y}dx] dy$

$V=\int A(x,y) dx +\int[B-\int \frac{\partial A}{\partial y}dx] dy $

Suppose $A(x,y)=y, B(x,y)=x$.

Then $V= \int ydx +\int[x-\int 1 dx]dy=yx$.

$V(7,11)-V(5,2)= (11)(7)-(2)(5)=67$


Looks like they want you to use the line integral without the potential function.

$dW=\vec{F}\cdot d\vec{s}.$

$ \vec{F}=P(x,y,z)\hat{i}+Q(x,y,z)\hat{j}$

$d\vec{s}=dx\hat{i}+dy\hat{j}$

$\frac{d\vec{s}}{dt}= \frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$

$\vec{F}\cdot d\vec{s}=\vec{F}\cdot \frac{\vec{ds}}{dt} dt$

$\int \vec{F}\cdot d\vec{s} = \int \vec{F}\cdot \frac{d\vec{s}}{dt}dt=\int P(x,y)\frac{dx}{dt}+Q(x,y) \frac{dy}{dt} dt$

Let $x=f(t), y=g(t)$

Then $W=\int y\frac{dx}{dt} + x\frac{dy}{dt} dt=\int \frac{d(xy)}{dt}dt=xy|_{(5,2)}^{(7,11)}=77-10=67$

So same result as expected, including the same final calculations.

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@TurlocTheRed @coreyman317 Thanks for your help!

I think I got the right answer under such a context.

I am not good at latex editing on this website. So I post my photo here as an answer.

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