Section 8.3 - Conservative Fields
Problem 1. (Exercise 7.3.2) Determine which of the following vector fields F in the plane
is the gradient of a scalar function f . If such an f exists, find it.
(a) F (x, y) = (cos(xy) − xy sin(xy)) î − (x2 sin(xy)) ĵ
p
p
(b) F (x, y) = x x2 y 2 + 1 î + y x2 y 2 + 1 ĵ
(c) F (x, y) = (2x cos(y) + cos(y)) î − (x2 sin(y) + x sin(y)) ĵ
Solution.
(a) To check if F is a gradient field (i.e. is conservative) we want to take its curl. In this
case, we mean the scalar curl since we are in R2 . If this value is equal to 0, then we
know that F is in fact a gradient field.
∂
∂
(cos(xy) − xy sin(xy))
−x2 sin(xy) −
∂x
∂y
= −x2 · cos(xy) · y + sin(xy) · −2x − (− sin(xy) · x − (xy · cos(xy) · x + sin(xy) · x))
= −x2 y cos(xy) − 2x sin(xy) − −x sin(xy) − x2 y cos(xy) − x sin(xy)
= −x2 y cos(xy) − 2x sin(xy) + 2x sin(xy) + x2 y cos(xy)
= 0.
curl F =
Thus, F is in fact a gradient field. Now to find f (x, y, z) such that ∇f (x, y, z) = F , we
notice that
∂f
∂f
= cos(xy) − xy sin(xy)
= −x2 sin(xy).
∂x
∂y
So we want to integrate,
Z
−x2 sin(xy) dy
f (x, y) =
1
+C
x
= x cos(xy) + C
= x2 cos(xy) ·
Let us choose C = 0, since it doesn’t matter what our constant is as when we differentiate
it will disappear. Now, let us check to make sure that when we take the derivative with
respect to x that we get what we’re supposed to,
∂
(x cos(xy)) = x · − sin(xy) · y + cos(xy) · 1
∂x
= cos(xy) − xy sin(xy).
Thus, f (x, y) = x cos(xy).
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(b) The method here remains the same.
∂ p 2 2
∂ p 2 2
y x y +1 −
x x y +1
curl F =
∂x
∂y
− 12 2
− 12 2
1 2 2
1 2 2
= y·
x y +1
· y · 2x − x ·
x y +1
· x · 2y
2
2
− 1
− 1
= xy 3 x2 y 2 + 1 2 − x3 y x2 y 2 + 1 2 6= 0
Thus, F is not a gradient field.
(c) Once again, we look at the scalar curl
∂
∂
−x2 sin(y) + x sin(y) −
(2x cos(y) + cos(y))
∂x
∂y
= (−2x sin(y) + sin(y)) − (2x · − sin(y) + sin(y))
= −2x sin(y) + sin(y) + 2x sin(y) − sin(y)
= 0.
curl F =
Thus, F is in fact a gradient field. Now to find f (x, y, z) such that ∇f (x, y, z) = F . As
we did previously, we can integrate and we find that f (x, y) = x2 cos(y) + x cos(y).
Problem 2. (Exercise 7.3.19(b)) Show that the vector field
F =
2x
2y(x2 + 1)
î
−
ĵ
y2 + 1
(y 2 + 1)2
R
is conservative. Calculate C F · ds for the curve C where C is parametrized by x = t3 − 1,
y = t6 − t, where 0 ≤ t ≤ 1.
Solution. First we check that the vector field is in fact conservative:
2y(x2 + 1)
∂
2x
∂
− 2
−
curl F =
∂x
(y + 1)2
∂y y 2 + 1
−2y
−1
=
· 2x − 2x · 2
· 2y
(y 2 + 1)2
(y + 1)2
4xy
4xy
=− 2
+ 2
=0
2
(y + 1)
(y + 1)2
and thus, the vector field is conservative. Now we find a function f (x, y, z) such that
∇f (x, y, z) = F . Just like we did in Problem 1, we integrate and we find that f (x, y, z) =
(x2 + 1)(y 2 + 1)−1 . The path integral can be found by simply taking the difference between
the values of f at the endpoints of the curve in each case. So then, we see that
Z
F · ds = f (0, 0) − f (−1, 0) = 1 − 2 = −1.
C
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Note: These notes and problems are meant to follow along with Vector Calculus by Jerrold
Marsden and Anthony Tromba, Sixth Edition. The pictures were generated using Wolfram
Alpha.
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