MAT2615 ASSIGNMENT 4 2023

MAT2615
ASSIGNMENT 4
2023

, QUESTION 1




Solution:



a).



The integral is independent of path if there exist a function f(x, y) such that
∇f = F



Let: F = (ex sin(y) , ex cos(y))



∇f = F
∇f = (ex sin(y) , ex cos(y))
∂f ∂f
( , ) = (ex sin(y) , ex cos(y))
∂x ∂y


∂f
= ex sin(y)
∂x

f(x, y) = ∫ ex sin(y) dx

f(x, y) = ex sin(y) + g(y)
∂f ∂ x
= [e sin(y) + g(y)]
∂y ∂y
∂f ∂f
= ex cos(y) + g ′ (y) ∴ Remember = ex cos(y)
∂y ∂y

, ex cos(y) = ex cos(y) + g ′ (y)
ex cos(y) − ex cos(y) = g ′ (y)
0 = g ′ (y)
g ′ (y) = 0
g(y) = c



f(x, y) = ex sin(y) + g(y)
f(x, y) = ex sin(y) + c



Since the function f(x, y) exist such that ∇f = F then the integral is independent of path.



b).



i).


π
r(t) = (1 − t)r0 + tr1 ∴ r0 = (0,0) , r1 = (1, )
2
π
r(t) = (1 − t) ∙ (0,0) + t (1, )
2
π
r(t) = (t, t)
2


π
Let: x = t and y = t
2
dx
=1 ⇒ dx = dt
dt
dy π π
= ⇒ dy = dt
dt 2 2


π
(1, ) 1
2 πt πt π
∫ ex sin(y) dx + ex cos(y) dy = ∫ et sin ( ) dt + et cos ( ) ∙ ( dt)
(0,0) 0 2 2 2
1
πt π πt
= ∫ et sin ( ) dt + et cos ( ) dt
0 2 2 2
1
πt π πt
= ∫ et [sin ( ) + cos ( )] dt
0 2 2 2