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 = [tex]\int\limits^b_a {f(r(t)).r(t)} \, dt[/tex]
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
