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

Question

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 $(7,11)$.

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.

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}$$

Answer 1

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

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

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

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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.

Answer 2

@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.