Download Solutions to some problems (Lectures 15-20)

Solutions to some problems (Lectures 15-20)
Problem Set 17, problem 3(c)
Decide whether the vector field could be a gradient field. Justify your answer. Hint. Find potential functions for the vector fields.
~i + √ y ~j + √ x ~k
F~ (x, y, z) = √x−z
2 +z 2
x2 +z 2
x2 +z 2
Solution. Let us try to find the potential function f (x, y, z) for the vector
field F~ . We have to solve the system of equations
∂f
∂x
∂f
∂y
∂f
∂z
=
=
=
−z
x2 + z 2
y
√
2
x + z2
x
√
.
2
x + z2
√
Consider the second equation and find the antidervative,
∂f
y
y2
=√
⇒ f (x, y, z) = √
+ C(x, z).
∂y
x2 + z 2
2 x2 + z 2
If we differentiate the obtained function f (x, y, z) with respect to x, we get
xy 2
∂f
∂C(x, z)
=−
+
.
3/2
2
2
∂x
∂x
2(x + y )
Obviuosly,
−
xy 2
∂C(x, z)
−z
6= √
+
3/2
2
2
∂x
2(x + y )
x2 + z 2
because the left hand side of the equation depends on y. Thus, the antiderivative obtained from the second equation does not satisfy the first
equation of the system of equations. Consequently, there is no potential
function for this vector field.
Problem set 20, problem 6
Let
~
~
~ = y i − xj ;
F~ = y~i + x~j; G
x2 + y 2
~ =
H
x~i + y~j
.
(x2 + y 2 )1/2
(a) By finding the potential functions, show that each of the vector fields
~ H
~ is a gradient vector field.
F~ , G,
~ H
~ around the unit circle in the xy-plane,
(b) Find the line integrals of F~ , G,
centered at the origin, and traversed counterclockwise.
(c) For which of the three vector fields can Green’s Theorem be used to
calculate the line integral in part (b)? Why?
Solution.
(a) Find the potential function f for the vector field F~ . We have
∂f
∂x
∂f
∂y
= y ⇒ f (x, y) = xy + C(y)
= x ⇒ f (x, y) = xy + C(x)
Combining these two equations, we get f (x, y) = xy.
~ We have
Find the potential function g for the vector field G.
∂g
∂x
∂g
∂y
y
+ y2
x
= − 2
x + y2
=
x2
Consider the first equation and find the antiderivative.
Z
g(x, y) =
y
dx =
2
x + y2
Z
1
d(x/y) = arctan(x/y),
(x/y)2 + 1
since (1 + z 2 )−1 dz = arctan z. (Substitute z = x/y in the integral.)
Similarly, for the second equation we have
R
g(x, y) = −
Z
x
dy = −
x2 + y 2
Z
1
d(x/y) = arctan(x/y),
(x/y)2 + 1
(Here we make subsitution z = x/y, dz = −x/y 2 dy.)
Thus, we obtain g(x, y) = arctan(x/y).
~ We have
Find the potential function h for the vector field H.
∂h
∂x
∂h
∂y
x
(x2 + y 2 )1/2
y
2
(x + y 2 )1/2
=
=
Consider the first equation and find the antiderivative.
Z
h(x, y) =
x
1
dx =
2
(x2 + y 2 )1/2
Z
1
d(x2 ) = (x2 +y 2 )1/2 +C(y)
(x2 + y 2 )1/2
Similarly, for the second equation we have
Z
h(x, y) =
y
1
dy =
1/2
2
2
2
(x + y )
Z
(x2
1
d(y 2 ) = (x2 +y 2 )1/2 +C(x)
+ y 2 )1/2
Combining these equations we obtain h(x, y) = (x2 + y 2 )1/2 .
(b) The integral C F~ · d~r = 0 because F~ is a gradient field, and f is a continuous function, so we can apply the fundamental theorem of calculus
and show thatR the circulation of F~ is equal to zero.
~ ·d~r, this integral is equal to zero, since H is a gradient
The same for C H
field and h is a continuous function. We can apply the fundamental
theorem of calculus and show that
R
Z
~ · d~r =
H
C
Z
gradh · d~r = 0,
C
sinceR C is a closed curve.
~ · d~r we cannot apply the fundamental theorem of calculus to
For C G
find the integral. Indeed, the function g is discontinuous at the points
with coordinates y = 0. Thus, for the unit circle C the potential function
g is undefined at the points (1, 0) and (−1, 0). We can find the integral
by direct calculation.
Parameterize C by ~r(t) = cos t~i + sin t~j, 0 ≤ t ≤ 2π. We have ~r0 (t) =
− sin t~i + cos t~j. The vector field F~ along the curve in terms of the
parameter t is given by F~ = sin t~i − cos t~j. Thus,
Z
F~ · d~r(t) =
Z 2π
(sin t~i − cos t~j) · (− sin t~i + cos t~j) dt
0
C
Z 2π
=
0
− sin2 t − cos2 t dt = −
Z 2π
dt = −2π.
0
(c) Green’s Theorem does not apply to the computation of the line integrals
~ and H
~ because their domains do not include the origin, which is
for G
in the interior of the circle. Green’s Theorem does apply to F~ .