contestada

Let C be a circle of radius 9 centered at (0,0), traversed counterclockwise. Use this curve to answer the questions below. (a) Let F = y i + x j Find a potential function. f(x,y) = xy Does the Fundamental Theorem of Line Integrals apply to F · dr C ? Does Green's Theorem apply to F · dr C ? (b) Let G = y x2 + y2 i − x x2 + y2 j . Find a potential function. g(x,y) = Where is the potential function not defined? Does the Fundamental Theorem of Line Integrals apply to G · dr C ? Does Green's Theorem apply to G · dr C ? (c) Let H = x x2 + y2 i + y x2 + y2 j . Find a potential function. h(x,y) = Where is the potential function not defined? Does the Fundamental Theorem of Line Integrals apply to H · dr C ? Does Green's Theorem apply to H · dr C ? Submit Answer Save Progress

Respuesta :

LammettHash

a. We're looking for a scalar function [tex]f(x,y)[/tex] such that [tex]\vec F(x,y)=\nabla f(x,y)[/tex]. This means

[tex]\dfrac{\partial f}{\partial x}=y[/tex]

[tex]\dfrac{\partial f}{\partial y}=x[/tex]

Integrate both sides of the first PDE with respect to [tex]x[/tex]:

[tex]\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=\int y\,\mathrm dx\implies f(x,y)=xy+g(y)[/tex]

Differentating both sides with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=C[/tex]

so that [tex]\boxed{f(x,y)=xy+C}[/tex]. A potential function exists, so the fundamental theorem does apply.

Green's theorem also applies because [tex]C[/tex] is a simple and smooth curve.

b. Now with (and I'm guessing as to what [tex]\vec G[/tex] is supposed to be)

[tex]\vec G(x,y)=\dfrac y{x^2+y^2}\,\vec\imath-\dfrac x{x^2+y^2}\,\vec\jmath[/tex]

we want to find [tex]g[/tex] such that

[tex]\dfrac{\partial g}{\partial x}=\dfrac y{x^2+y^2}[/tex]

[tex]\dfrac{\partial g}{\partial y}=-\dfrac x{x^2+y^2}[/tex]

Same procedure as in (a): integrating the first PDE wrt [tex]x[/tex] gives

[tex]g(x,y)=\tan^{-1}\dfrac xy+h(y)[/tex]

Differentiating wrt [tex]y[/tex] gives

[tex]-\dfrac x{x^2+y^2}=-\dfrac x{x^2+y^2}+\dfrac{\mathrm dh}{\mathrm dy}\implies h(y)=C[/tex]

so that [tex]\boxed{g(x,y)=\tan^{-1}\dfrac xy+C}[/tex], which is undefined whenever [tex]y=0[/tex], and the fundamental theorem applies, and Green's theorem also applies for the same reason as in (a).

c. Same as (b) with slight changes. Again, I'm assuming the same format for [tex]\vec H[/tex] as I did for [tex]\vec G[/tex], i.e.

[tex]\vec H(x,y)=\dfrac x{x^2+y^2}\,\vec\imath+\dfrac y{x^2+y^2}\,\vec\jmath[/tex]

Now

[tex]\dfrac{\partial h}{\partial x}=\dfrac x{x^2+y^2}\implies h(x,y)=\dfrac12\ln(x^2+y^2)+i(y)[/tex]

[tex]\dfrac{\partial h}{\partial x}=\dfrac y{x^2+y^2}=\dfrac y{x^2+y^2}+\dfrac{\mathrm di}{\mathrm dy}\implies i(y)=C[/tex]

[tex]\implies\boxed{h(x,y)=\dfrac12\ln(x^2+y^2)+C}[/tex]

which is undefined at the point (0, 0). Again, both the fundamental theorem and Green's theorem apply.