hw6

Homework 6 Solutions

Math 53 – Summer 2010

August 8, 2010

Exercise 16.1 #36. (a) Sketch the vector field F(x, y) = i + xj and then sketch some flowlines. What shape do these flow lines appear to have? (b) If parametric equations of the flowlines are x = x(t), y = y(t), what differential equations do these functions satisfy? Deducethat dy

dx= x. (c) If a particle starts at the origin in the velocity field given by F, find an

equation of the path it follows.

Solution. (a) See Figure 1. The flow lines appear to be parabolas.

(b) Flow lines satisfy

x′(t) i + y′(t) j = F(x(t), y(t)) = i + x(t) j,

and matching components we see that x′(t) = 1 and y′(t) = x(t).

(c) The fundamental theorem of calculus gives x(t)− x(0) =∫ t

0x′(s) ds, so x(0) = 0 and

x′(s) = 1 give x(t) = t. Similarly,

y(t) − y(0) =

∫ t

0

y′(s) ds =

∫ t

0

x(s) ds =

∫ t

0

s ds =1

2t2,

so y(0) = 0 implies y(t) = 1

2t2. So the flow line passing through the origin has parametric

equations x(t) = t, y(t) = 1

2t2. Eliminating t gives the Cartesian equation y = 1

2x2.

Exercise 16.2 #8. Evaluate∫

Csin x dx + cos y dy, C consists of the top half of the circle

x2 + y2 = 1 from (1, 0) to (−1, 0) and the line segment from (−1, 0) to (−2, 3).

Solution. Let C1 be the circular portion of C and C2 the line segment. Then

∫

C

sin x dx + cos y dy =

∫

C1

sin x dx + cos y dy +

∫

C2

sin x dx + cos y dy.

4

-4

y

12

-2

x

2 40

8

0

Figure 1: The field F together with several flow lines.

We parametrize

C1 : r1(t) = cos t i + sin t j, 0 ≤ t ≤ π

C2 : r2(t) = (−1 − t) i + 3t j, 0 ≤ t ≤ 1.

Then

r′1(t) = − sin t i + cos t j

r′2(t) = −i + 3j.

2

We find∫

C1


∫

C1

sin x dx +

∫

C1

cos y dy

=

∫ π

0

sin(cos t)(− sin t) dt +

∫ π

0

cos(sin t) cos t dt

= [− cos(cos t)]t=πt=0 + [sin(sin t)]t=π

t=0

= (− cos(−1) + cos 1) + (0 − 0)

= 0∫

C2


∫

C2

sin x dx +

∫

C2

cos y dy

=

∫

1

0

sin(−1 − t)(−1) dt +

∫

1

0

cos 3t(3) dt

=

∫

−2

−1

sin u du + [sin 3t]t=1

t=0

= [− cos u]u=−2

u=−1 + sin 3

= cos 1 − cos 2 + sin 3.

Adding, we have∫

C

sin x dx + cos y dy = cos 1 − cos 2 + sin 3.


Cxyz2 ds, where C is the line segment from (−1, 5, 0) to

(1, 6, 4).

Solution. Parametrize the line segment:

r(t) = (1 − t)〈−1, 5, 0〉 + t〈1, 6, 4〉 = 〈−1 + 2t, 5 + t, 4t〉, 0 ≤ t ≤ 1.

Then r′(t) = 〈2, 1, 4〉 and |r′(t)| =√

22 + 12 + 42 =√

21, and∫

C

xyz2 ds =

∫

1

0

(−1 + 2t)(5 + t)(4t)2(√

21) dt

= 16√

21

∫

1

0

(−5 + 9t + 2t2)t2 dt

= 16√

21

∫

1

0

2t4 + 9t3 − 5t2 dt

= 16√

21

[

2

5t5 +

9

4t4 − 5

3t3

]t=1

t=0

= 16√

21

(

2

5+

9

4− 5

3

)

=236

√21

15.

3


CF · dr, where F(x, y, z) = (x + y) i + (y − z) j + z2 k and

C is given by r(t) = t2 i + t3 j + t2 k, 0 ≤ t ≤ 1.

Solution. We find r′(t) = 2t i + 3t2 j + 2tk, so

∫

C

F · dr =

∫

1

0

〈t2 + t3, t3 − t2, t4〉 · 〈2t, 3t2, 2t〉 dt

=

∫

1

0

2t3 + 2t4 + 3t5 − 3t4 + 2t5 dt

=

∫

1

0

2t3 − t4 + 5t5 dt

=

[

1

2t4 − 1

5t5 +

5

6t6

]t=1

t=0

=1

2− 1

5+

5

6

=17

15.

Exercise 16.3 #12. Find a function f such that F = ∇f and use it to evaluate∫

CF · dr,

where F(x, y) = x2 i + y2 j and C is the arc of the parabola y = 2x2 from (−1, 2) to (2, 8).

Solution. If F = ∇f , then

{

fx = x2

fy = y2.

}

⇔

f =1

3x3 + g1(y)

f =1

3y3 + g2(x)

.

We must find a function f which is compatible with both of the above; one choice is f(x, y) =1

3(x3 + y3). Then, by the Fundamental Theorem of Calculus for Line Integrals,

∫

C

F · dr = f(2, 8) − f(−1, 2) =1

3(23 + 83 − (−1)3 − 23) = 171.

Exercise 16.3 #20. Show that the line integral is independent of path and evaluate theintegral:

∫

C

(1 − ye−x) dx + e−x dy,

where C is any path from (0, 1) to (1, 2).

4

Solution. The vector field 〈P, Q〉 = 〈1 − ye−x, e−x〉 is defined on all of R2 (which is simply

connected) so we could simply compute Qx − Py, find that it is zero, and conclude that〈P, Q〉 is conservative, hence independent of path, but since we need to evaluate the integralanyway it is easier to find f(x, y) with ∇f = 〈P, Q〉. We therefore seek f(x, y) satisfying

{

fx = 1 − ye−x

fy = e−x

}

.

The first equation implies f(x, y) = x + ye−x + g(y) for some function g(y); differentiatingwe find fy = e−x + g′(y). Comparing with the second equation we learn g′(y) = 0 and gmust be constant. Since adding a constant will not change ∇f , we set g(y) = 0. Thusf(x, y) = x + ye−x, and the fundamental theorem of calculus for line integrals gives

∫

C

(1 − ye−x) dx + e−x dy = f(1, 2) − f(0, 1) = (1 + 2e−1) − (1) = 2e−1.

Exercise 16.3 #26. Let F = ∇f , where f(x, y) = sin(x−2y). Find curves C1 and C2 thatare not closed and satisfy

∫

C1

F · dr = 0 and

∫

C2

F · dr = 1.

Solution. Since F is conservative with potential function f , we have for any curve C∫

C

F · dr = f(final point of C) − f(initial point of C).

So we may recast the problem as finding C1 and C2 so that

f(final point of C1) − f(initial point of C1) = 0

f(final point of C2) − f(initial point of C2) = 1.

Noting that f(0, 0) = f(π, 0) = 0 and f(π/2, 0) = 1, we can let C1 (respectively C2) be anycurve (say the line segment, just to be concrete) from (0, 0) to (π, 0) (respectively from (0, 0)to (π/2, 0)).

Exercise 16.3 #30. Let F(x, y) = −y i+x j

x2+y2 . (a) Show Qx − Py = 0. (b) Show that∫

CF · dr

is not independent of path. Does this contradict Theorem 6?

Solution. (a) We simply calculate the partial derivatives:

Qx =

(

x

x2 + y2

)

x

=(1)(x2 + y2) − x(2x)

(x2 + y2)2=

y2 − x2

(x2 + y2)2

Py =

( −y

x2 + y2

)

y

=(−1)(x2 + y2) + y(2y)

(x2 + y2)2=

y2 − x2

(x2 + y2)2,

5

and note that Qx − Py = 0.

(b) To show that the line integral is not independent of path we show an example ofa closed curve C along which the line integral of F is nonzero. Notice that F is definedeverywhere except the origin (0, 0). If C is a curve which does not enclose the origin, weare doomed to fail in this endeavor: for a curve C which does not enclose the origin wecan surround C with a simply connected region D on which F is defined everywhere, andour theorems tell us that Qx − Py = 0 implies F is conservative on D. So we try a curveC which does not enclose the origin; the simplest example that comes to mind is the unitcircle, oriented counterclockwise, which we parametrize as usual with r(t) = cos t i + sin t jfor 0 ≤ t ≤ 2π. Then

∫

C

F · dr =

∫

2π

0

− sin t i + cos t j

cos2 t + sin2 t· (− sin t i + cos t j) dt

=

∫

2π

0

sin2 t + cos2 t dt

= 2π 6= 0.

This does not contradict Theorem 6 because the vector field in this example is defined on aregion (R2 minus the origin) which is not simply connected, and we cannot use Qx −Py = 0to conclude that F is conservative.

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