Question

1 pt) let f(x,y)=−yi+xjx2+y2 and let c be the circle r(t)=(cost)i+(sint)j, 0≤t≤2π.

Answer 1

What are $p$ and $q$ supposed to be? Can't answer a/b without that information.

I'll come back to part (c) in a moment. If we can show $\mathbf f$ is conservative, this part will be a breeze.


For part (d), to show whether $\mathbf f$ is conservative, we have to show that there is a scalar function $f$ such that $\nabla f=\mathbf f$. It appears that you've written

$\mathbf f=-y\,\mathbf i+x\,\mathbf j$

I'm not sure what to make of the $x^2+y^2$ that follows. $\nabla f=\mathbf f$ means that

$\dfrac{\partial f}{\partial x}=-y$
$\dfrac{\partial f}{\partial y}=x$

Integrating the first PDE with respect to $x$ gives

$f(x,y)=-xy+g(y)$

Differentiating with respect to $y$ gives

$\dfrac{\partial f}{\partial y}=-x+\dfrac{\mathrm dg}{\mathrm y}=x$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=2x$

But we assumed that $g(y)$ is a function of $y$ alone, which means there is no solution for $g$, and therefore no solution for $f$. Hence $f$ is not conservative.


Back to part (c). $\mathbf f$ is not conservative, so we have to compute the line integral the "long" way. Replacing $x=\cos t$ and $y=\sin t$, we have

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}(-\sin t\,\mathbf i+\cos t\,\mathbf j)\cdot(-\sin t\,\mathbf i+\cos t\,\mathbf j)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(\sin^2t+\cos^2t)\,\mathrm dt=\int_0^{2\pi}\mathrm dt=2\pi$

Answer 2

Answer:

(a) ∂Θ∂x = 1/(x² + y²) - 2x²/(x² + y²)²

(b) ∂Θ∂x = -1/(x² + y²) + 2y²/(x² + y²)²

(c) ∫cf⋅dr =2π

(d).y; conservative

Step-by-step explanation:

to begin, let us define the parameters from the question.

given that f(x,y) = -yī + xĴ / x² + y² .............(1)

where c is r(t)=(cost)i+(sint)j

from equation (1), f(x,y) = -y/x²+y² ī  +  x/ x²+y²Ĵ

let M be given as -y/x²+y² ī and N be given as x/ x²+y²Ĵ

(a). from the question, ∂Θ/∂x we have;

∂Θ/∂x = 1/x²+y² - x(x²+y²)⁻¹⁻¹(2x)

this becomes 1/(x² + y²) - 2x²/(x² + y²)²

(b). ∂p/∂y

we have ∂p/∂y = -1/(x² + y²) + y/(x² + y²)² (2y)

which becomes -1/(x² + y²) + 2y²/(x² + y²)²

(c). Given  ∫cf⋅dr

from this we have  ∫cf⋅dr = $\int\limits^b_a {f(r(t)).r(t)} \, dt$

where r(t) = cost ī + sint Ĵ

which is same as r(t) = -sint ī + cost Ĵ

f(r(t)) = -sint ī + cost Ĵ / cos²t + sin²t = -sint ī + cost Ĵ

this gives  f(r(t)). r(t) = sin²t + cos²t = 1

recall, ∫cf⋅dr = ∫2π to 0 dt= [t] 2π to 0 = 2π - 0 = 2π

∴ ∫cf⋅dr =2π

(d). is f conservative?

in this question we shall solve for a matrix of the function i.e.

2 × 2 matrix [∂/∂x            ∂/∂y

                   -y/x²+y²     x/x²+y²]

solving this 2 × 2 matrix = ∂/∂x(x/x²+y²) - ∂/∂y(-y/x²+y²)

this will give 2/x²+y² - (2x²+2y²)/(x²+y²)² = 2/x²+y² - 2(x²+y²)/(x²+y²)²

= 2/x²+y² - 2(x²+y²)/(x²+y²)² = 2/x²+y - 2/x²+y = 0

from this we can say f is conservative