Math 212 Spring 2008: Solutions: HW #10
Instructor: S. Cautis
1. section 7.1 #2a
p
√
cos2 t + sin2 t + 1 = 2.
c′ (t) = (cos t, − sin t, 1). Hence ||c′ (t)|| =
Hence the path integral is
Z 2π
√
(sin t + cos t + t) 2dt = [− cos t + sin t + t2 /2]2π
0
0
= (−1 + 0 + (2π)2 /2) − (−1 + 0) = 2π 2
2. section 7.1, #6
a) In cartesian coordinates the path is c(θ) = (r(θ) cos θ, r(θ) sin θ). Hence
c′ (θ) = (r′ (θ) cos θ − r(θ) sin θ, r′ (θ) sin θ + r(θ) cos θ)
so
||c′ (θ)||
p
(r′ cos θ − r sin θ)2 + (r′ sin θ + r cos θ)2
p
r′2 cos2 θ − 2rr′ cos θ sin θ + r2 sin2 θ + r′2 sin2 θ + 2rr′ sin θ cos θ + r2 cos2 θ
=
p
r′2 + r2
=
=
Hence the path integral is
r
Z θ2
dr
f (r cos θ, r sin θ) r2 + ( )2 dθ
dθ
θ1
b) If r = 1 + cos θ then
Z
0
2π
dr
dθ
= − sin θ and we get the integral
Z
q
(1 + cos θ)2 + sin2 θdθ =
2π
√
2 + 2 cos θdθ
0
3. section 7.1, #12
a) The parametrized curve in this case is c(t) = (t, f (t)) so c′ (t) = (1, f ′ (t))
and
p
||c′ (t)|| = 1 + f ′ (t)2
Thus the length of the graph is
Z b
Z
′
||c (t)||dt =
a
a
b
p
1 + (f ′ (t))2
b) f (x) = log(x) = ln(x)/ ln(10) in this case. So f ′ (x) =
the length is
Z 2r
1
1 + 2 2 dx
x ln 10
1
1
x ln 10 .
Hence
4. section 7.2, #2
a) dx = − sin tdt while dy = cos tdt so the line integral is
Z
2π
(cos t)(cos tdt) − (sin t)(− sin tdt) =
0
Z
2π
dt = 2π
0
c) There are two integrals we need to compute here. The first path is
parametrized by c1 (t) = (1 − t, t, 0) where 0 ≤ t ≤ 1. Then dx = −dt,
dy = dt and dz = 0 and the integral is
Z 1
(0)dt + (0)dt + (1 − t)(t)(0dt) = 0
0
The second path is c2 (t) = (0, 1 − t, t) where 0 ≤ t ≤ 1. Then dx = 0, dy =
−dt and dz = dt. So the integral is
Z 1
(1 − t)(t)(0dt) + (0dt) + (0dt) = 0
0
So both integrals are zero and hence their sum, which is the integral over
the whole path, is also zero.
5. section 7.2, #4
a) F is perpendicular to c′ (t) means that F (c(t)) · c′ (t) = 0. Then
Z
Z
Z
F · ds = F (c(t)) · c′ (t)dt = 0dt = 0
c
c
c
b) F is parallel to c′ (t) means that F (c(t)) = λ(t)c′ (t) for some λ(t) > 0.
Then
Z
Z
Z
′
′
F · ds = λ(t)c (t) · c (t)dt = λ(t)||c′ (t)||2 dt
c
c
c
while
Z
c
so that
R
c
||F ||ds =
F · ds =
6. section 7.2, #10
R
c
Z
c
′
′
||λ(t)c (t)||||c (t)||dt =
Z
λ(t)||c′ (t)||2 dt
c
||F ||ds.
R
R
is the curve obtained
We have that − c2 F · ds = copp F · ds where copp
2
2
from c2 by moving in the opposite direction. Then
Z
Z
F · ds = 0
F · ds −
c1
c2
is the same as
Z
c1
F · ds +
Z
copp
2
F · ds = 0
The left side is by definition
Z
c
F · ds
where c isR the closed curve obtained byR first going along
c1 and then along
R
F
·
ds = 0 (which is
F
·
ds
−
F
·
ds
=
0
precisely
when
.
So
copp
2
c2
c1
c
what we needed to show).
7. section 7.2, #16
We know that if c(t) is any path from p1 = (0, 0, 0) to p2 = (1, 1, 2) then
Z
∇f · ds = f (p2 ) − f (p1 )
c
In our case we know f (p1 ) = 5 and we want to find f (p2 ). To do this
we calculate the left side of the equality above by choosing the simplest
possible path – namely a straight line. This is parametrized by c(t) =
t(1, 1, 2) where 0 ≤ t ≤ 1.
Then the left side is
Z
Z 1
2
2
2
(4t3 et , 2tet , tet ) · (1, 1, 2)dt =
0
1
2
et (4t3 + 2t + 2t)dt
0
Hence
f (1, 1, 2) = f (0, 0, 0) +
Z
0
1
t2
3
e (4t + 4t)dt = 5 + 4
Z
1
2
et (t3 + t)dt
0
8. section 7.2 #17
The point is that if the force field is a gradient vector field then the work
done only depends on the end-points of the paths. So we want to find a
function f such that ∇f = F .
A little trial and error leads us to the function f = √ 2 1 2 2 . It’s easy
x +y +z
to check that ∇f = F . Thus the
p work done only depends on the points
√ 2 1 2 2 = R1 and on f rac1 x22 + y22 + z22 = R1 which only depend
1
2
x1 +y1 +z1
on the radii R1 and R2 .
9. section 7.2 #18
√
The rider starts at position ( 2π, 0, 0) and finishes at (0, 0, 2π). At this
point one could try to find the path. It might look like (r(t) cos t, r(t) sin t, αt)
where constant α and function r(t) are yet to be determined (here 0 ≤
t ≤ 2π). Since the z coordinate at time t = 2π is 2π the constant α = 1.
Since the path lies on the surface x2 + y 2 + z = 2π we must have
r(t)2 cos2 t + r(t)2 sin2 t + t = 2π
√
so that r(t)2 = 2π − t and r(t) = 2π − t. Thus the path she takes is
√
√
c(t) = ( 2π − t cos t, 2π − t sin t, t)
where
0 ≤ t ≤ 2π. Then one could go ahead and compute the line integral
R
c (y, x, 1)·ds. However, there are two points worth making. First, it is not
clear to me that the path described above is the unique path satisfying the
description in the problem. Second, point one does not matter because
our field is actually a gradient field.
We know how to test for gradient fields (∇ × F = 0) and doing so tells
us that F should be a gradient field. By trial and error we find that
∇(xy + z) = (y, x, 1) = F . Thus the integral we want is
Z
Z
F ·ds = ∇(xy+z)·ds = (xy+z)(0,0,2π) −(xy+z)(√2π,0,0) = 2π−0 = 2π
c
c
which is the work the cyclists does.
What is unrealistic about the model is that the velocity of the cyclist as
t → 2π tends to infinity (as the expression of c′ (t) obtained from that of
c(t) above illustrates).