Math 2263 Multivariable Calculus
Homework 25: 16.3 #8, 16
July 20, 2011
16.3#8
Determine whether or not F is a conservative vector field. If it is, find a function f such that
F = ∇f , where F (x, y) = (xy cos xy + sin xy)i + (x2 cos xy)j.
∂P
= xy(− sin xy)(x) + cos xy(x) + cos xy(x)
∂y
= −x2 y sin xy + 2x cos xy
∂Q
= x2 (− sin xy)(y) + 2x cos xy
∂x
Since ∂P
= ∂Q
, the field is conservative.
∂y
∂x
Looking at the first term, we have
∂f
= xy cos xy + sin xy
∂x
and from the second term, we have
∂f
= x2 cos xy
∂y
From this second equation, we have
f (x, y) = x sin xy + . . . .
Check the partial with respect to x.
∂
(x sin xy) = x cos xy(y) + sin xy
∂x
This shows that f (x, y) = x sin xy satisfies ∇f = F .
16.3#16
R
(a) Find a function f such that F = ∇f and (b) use part (a) to evaluate C F · dr along the
given curve C.
F (x, y, z) = (2xz + y 2 )i + 2xyj + (x2 + 3z 2 )k, C : x = t2 , y = t + 1, z = 2t − 1, 0 ≤ t ≤ 1
(a)
fx = 2xz + y 2 ⇒ f (x, y, z) = x2 z + xy 2 + . . .
fy = 2xy ⇒ f (x, y, z) = xy 2 + . . .
fz = x2 + 3z 2 ⇒ f (x, y, z) = x2 z + z 3 + . . .
Together, f (x, y, z) = xy 2 + x2 z + z 3 . We may check this:
∇f (x, y, z) = hy 2 + 2xz, 2xy, x2 + 3z 2 i
(b)
Z
Z
∇f · dr = f (r(1)) − f (r(0))
F · dr =
C
C
= f (1, 2, 1) − f (0, 1, −1)
2
1
3
3
= 1(2) + 1 (1) + (1) − 0 + 0 + (−1)
= 6 − (−1) = 7