Math 113 - Review Worksheet Solutions
1. We have |c~v | = |c||~v |. A unit vector is a vector with length 1 and so we want |c||~v | = 1 and
so |c| = 1/|~v |. We have
|~v | =
p
√
√
42 + (−1)2 + 82 = 16 + 1 + 64 = 81 = 9.
Thus, the vector h4/9, −1/9, 8/9i is a unit vector parallel to ~v .
2. Find the angle between the vectors ~v1 = h1, 3, 1i and ~v2 = h3, 2, −9i.
We have that ~v1 · ~v2 = |~v1 ||~v2 | cos(θ), where θ is the angle between ~v1 and ~v2 . Also, ~v1 · ~v2 =
(1)(3) + (3)(2) + (1)(−9) = 3 + 6 − 9 = 0. Thus, cos(θ) = 0 and so θ = π/2 is the angle between
them.
3. Let ~v = h7, 8i and w
~ = h−5, 4i. Find a constant c and a vector ~x so that
~v = cw
~ + ~x
and ~x is perpendicular to w.
~
We have
cw
~ = projw~ ~v
~v · w
~
w
~
=
w
~ ·w
~
3
−35 + 32
w
~ =− w
~
= 2
2
41
5 + 4 15 12
=
,−
.
41 41
3
Hence c = − 41
. Then,
~x = ~v − cw
~
15 12
= h7, 8i −
,−
41 41
287 328
15 12
=
,
−
,−
41 41
41 41
272 340
=
,
.
41 41
One can double check that this vector is perpendicular to w.
~
1
2
4. (a) If ~v = h1, 3, 7i and w
~ = h4, −1, 2i, we have
~i ~j ~k ~v × w
~ = 1 3 7
4 −1 2
= h3 · 2 − (−1) · 7, −(1 · 2 − 4 · 7), 1 · (−1) − 4 · 3i
= h13, 26, −13i.
(b) The vector above is a normal vector for the plane containing P , Q and R (note that ~v = P~Q
and w
~ = P~R). To get an equation for the plane we use the coordinates of the normal vector,
as well as the coordinates of one of the points. We get
13(x − 2) + 26(y + 5) − 13(z − 3) = 0.
~ We have ~v × w
~ = 13h1, 2, −1i, and so the area
(c) The area of the triangle 4P QR is 12 |~v × w|.
is
√
1 p 2
13
6
13 1 + 22 + (−1)2 =
.
2
2
5. Find the distance between the point (5, 1, −5) and the plane x + 2y + 3z = 6.
~ is
Let P = (5, 1, −5) and Q be the point on the plane closest to P . Then the vector QP
perpendicular to the plane, and so it is parallel to the normal vector ~v = h1, 2, 3i.
~ can be written as a
So, if we take any point R in the plane (say R = (6, 0, 0)), the vector RP
~
~
~
sum of something in the plane, plus QP . It follows that QP = proj~n RP . We get
!
~ · ~n
RP
~ =
QP
~n
~n · ~n
h−1, 1, −5i · h1, 2, 3i
~n
=
h1, 2, 3i · h1, 2, 3i
−1 + 2 − 15
= 2
~n
1 + 2 2 + 32
−14
=
~n = −~n.
14
√
√
~ | = | − ~n| = 12 + 22 + 32 = 14. [ This
Thus, the distance from the point to the plane is |QP
is the same flavor of problem as Example 8 in Section 12.5. ]
6. For each of the three surfaces below, sketch traces of the surfaces in the xy-plane, xz-plane
and yz-plane. Use these to match the equation of a surface with its picture. Give reasons for
your choices.
3
For (a), the trace in the xy-plane (z = 0) is the parabola y = x2 + 1, the trace in the xz-plane
(y = 0) is z 2 − x2 = 1 is a hyperbola that opens in the z-direction, and the trace in the yz-plane
(x = 0) is the parabola y = −z 2 + 1 that opens down. These things tell us that (a) corresponds
to the middle picture - which is a hyperbolic paraboloid.
For (b), the trace in the xy-plane (z = 0) is y 2 = x2 + 1 which is a hyperbola that opens in
the y-direction. The trace in the xz-plane (y = 0) is z 2 − x2 = 1, a parabola that opens in
the z-direction. The trace in the yz-plane (x = 0) is y 2 + z 2 = 1, which is a circle. All of
these traces have points on them and this means that the surface is connected. Therefore, it
corresponds to the left picture - a hyperboloid of one sheet.
For (c), the trace in the xy-plane (z = 0) is y 2 = x2 − 1, or x2 − y 2 = 1, a hyperbola that
opens in the x direction. The trace in the xz-plane (y = 0) is x2 − z 2 = 1, another hyperbola
that opens in the x-direction. Finally, the trace in the yz-plane (x = 0) is y 2 + z 2 = −1, which
doesn’t have any points on it. Therefore, the surface is disconnected (it doesn’t intersect the
plane x = 0) and it corresponds to the right picture - a hyperboloid of two sheets.
7. Curve (a) involves only trig functions and this will mean that it is closed - it doesn’t have
starting or ending points. This means that it is the bottom right picture.
Curve (c) involves trig functions but also functions of t. If you think about this in polar
coordinates, the “radius” is increasing as the height does as well. This gives the “tornado”
shape which is the bottom left picture.
3
0
2
For curve
√ (b), the y-component is y(t) = t − t. We have y (t) = 3t − 1 which is zero if
t = 1/ 3. This means that y(t) has a local minimum here, and this is reflected in the top left
picture. Hence curve (b) matches this picture.
By the process of elimination the last picture must correspond to curve (d). [The x-component
of this curve will have local minima and maxima, but the range of t-values isn’t large enough
to show this, I believe.]
8. The arc length of a curve is given by the formula
Z
b
|~r 0 (t)| dt.
a
p
In this case, ~r 0 (t) = h−a sin(t), b cos(t)i and so |~r 0 (t)| = a2 sin2 (t) + b2 cos2 (t). Hence, the
arc length is
Z 2π q
a2 sin2 (t) + b2 cos2 (t) dt.
0
p
√
9. (a) We have ~r 0 (t) = h2t, 2t2 , 1i and so |~r 0 (t)| = (2t)2 + (2t2 )2 + 12 = 1 + 4t2 + 4t4 =
0
1
2
~
(1 + 2t2 ). Thus, T~ (t) = |~~rr 0 (t)
= 1+2t
2 h2t, 2t , 1i. We differentiate T (t) using the product rule
(t)|
4
and get
1
4t
h2t, 2t2 , 1i +
−
h2, 4t, 0i
2
2
(1 + 2t )
1 + 2t2
h−8t2 , −8t3 , −4ti + h2 + 4t2 , 4t + 8t3 , 0i
=
(1 + 2t2 )2
h2 − 4t2 , 4t, −4ti
=
.
(1 + 2t2 )2
T~ 0 (t) =
The curvature is
~
| ddsT |
dT~ dt = |~r 0 (t)| . Thus, we get
p
√
2 )2 + (4t)2 + (−4t)2
(2
−
4t
4 − 16t2 + 16t4 + 16t2 + 16t2
0
|T~ (t)| =
=
(1 + 2t2 )2
(1 + 2t2 )2
√
4 + 16t2 + 16t4
=
(1 + 2t2 )3
2
2 + 4t2
=
.
=
2
3
(1 + 2t )
(1 + 2t2 )2
Hence κ(t) =
|T~ 0 (t)|
|~
r 0 (t)|
=
2
.
(1+2t2 )3
(b) We find that T~ (1) = 31 h2, 2, 1i = h2/3, 2/3, 1/3i. We have T~ 0 (1) = h−2, 4, −4i/9 =
~ is the unit vector parallel to this. We have
h−2/9,
4/9, −4/9i. The vector N
p
p
~ (1) = (3/2)h−2/9, 4/9, −4/9i =
(−2/9)2 + (4/9)2 + (4/9)2 =
36/81 = 2/3. Thus, N
h−1/3, 2/3, −2/3i. Finally,
~i
~k ~
j
~
~ (1) = 2/3 2/3 1/3 B(1)
= T~ (1) × N
−1/3 2/3 −2/3
= h−2/3, 1/3, 2/3i.
10. The plot on the left in the Problem 10-part 1 picture is f (x, y) = sin(x2 + y 2 ) - the value
of the function is oscillating and this corresponds to a trigonometric function.
The plot on the right in the Problem 10-part 1 picture is f (x, y) = x2 +y 2 . This surface appears
to be an elliptic paraboloid, and this corresponds to the equation z = x2 + y 2 .
The plot on the right in the Problem 10-part 2 picture is f (x, y) = x2 +yx2 +1 . Notice that as
|hx, yi| → ∞, f (x, y) → 0, and this corresponds to the value of the function getting small near
the edges of the picture (and with the value of the function being positive on one half and
negative on the other half).
The plot on the left in the Problem 10-part 2 picture is f (x, y) = x4 − 2x2 y 2 + y 4 . You can see
this by the process of elimination. Also, notice that there are two perpendicular lines in the
5
picture along which the function is constant. These corresponds to y = x and y = −x where
f (x, x) = f (x, −x) = x4 − 2x4 + x4 = 0.
11. (a) If we switch to polar, and get x = r cos(θ), y = r sin(θ), we get 7r4 sin2 (θ) cos2 (θ) in the
numerator. In the denominator, we get (x2 + y 2 )3/2 = (r2 cos2 (θ) + r2 sin2 (θ))3/2 = (r2 )3/2 = r3 .
The condition (x, y) → (0, 0) becomes r → 0. We get
7r4 sin2 (θ) cos2 (θ)
= lim r(7 sin2 (θ) cos2 (θ)).
r→0
r→0
r3
lim
Now, recall that θ is independent of r, but regardless of θ, we have that |7 sin2 (θ) cos2 (θ)| ≤ 7,
so the biggest the thing inside the limit can be is 7r. Since 7r → 0 as r → 0, the limit exists
and is zero.
(You can also do this with the squeeze theorem. A squeeze theorem argument is below.)
The square root in the denominator makes this a bit tricky to work with. Instead of working
with that, we’ll square the function inside the limit. Let
7x2 y 2
, and
(x2 + y 2 )3/2
49x4 y 4
g(x, y) = f (x, y)2 = 2
.
(x + y 2 )3
f (x, y) =
Notice that the combined power of x and y in the numerator is 8 and the combined power
of x and y in the denominator is 6. This suggests that the limit will tend to zero, since the
numerator will go to zero faster than the denominator.
To justify this rigorously, we can use the squeeze theorem. Notice that
(x2 + y 2 )4 = x8 + 4x6 y 2 + 6x4 y 4 + 4x2 y 6 + y 8 .
This expression is greater than or equal to 6x4 y 4 since all of the terms on the right hand side
are ≥ 0. Thus, we have
6x4 y 4 ≤ (x2 + y 2 )4
and so
6x4 y 4
≤ 1.
(x2 + y 2 )4
Multplying by 49/6 gives us
49x4 y 4
49
≤
.
(x2 + y 2 )4
6
This is the same as g(x, y), except for the power of (x2 + y 2 ) in the denominator. We have that
g(x, y) = (x2 + y 2 ) ·
49x4 y 4
49
≤ (x2 + y 2 ).
2
2
4
(x + y )
6
6
Taking the square root gives
7 p 2
0 ≤ f (x, y) ≤ √
x + y2.
6
We have
7 p 2
√
x + y2 = 0
(x,y)→(0,0)
(x,y)→(0,0)
6
and so by the squeeze theorem lim(x,y)→(0,0) f (x, y) = 0 as well.
lim
0=
lim
(b) Along the approach y = 0, the numerator is zero and so the limit is zero. Along the
approach y = x, we get
2
sin2 (x)
sin(x)
1
1
lim
lim
= .
=
2
x→0
2x
2 x→0 x
2
Therefore the limit does not exist.
12. The cross section of the graph with x = −1.5, y varying and z = f (x, y) looks like a
parabola facing downwards. This would imply that g(y) = f (−1.5, y) would be concave down,
and so g 00 (y) < 0. This implies that fyy (−1.5, 0) < 0.
13. We have
∇f = hfx , fy , fz i
= hsin(xy) + xy cos(xy) − 2y 2 z sin(2xy 2 z), x2 cos(xy) − 4xyz sin(2xy 2 z), −2xy 2 sin(2x2 yz)i.
= 3x2 y − 2xy 2 , ∂f
= x3 − 2x2 y. Therefore fx (1, 1) = 3 − 2 = 1 and
14. We have ∂f
∂x
∂y
fy (1, 1, ) = 1 − 2 = −1. The equation for the tangent plane is
z = f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1)
= 0 + 1 · (x − 1) + (−1)(y − 1)
= x − 1 − y + 1 = x − y.
15. It is natural to take x0 = y0 = 1/2 and compute the linear approximation to f (x, y) at
(x0 , y0 ). We have
∂f
−(−y)
y
=
=
2
∂x
(1 − xy)
(1 − xy)2
∂f
−(−x)
x
=
=
.
2
∂y
(1 − xy)
(1 − xy)2
Thus,
fx (1/2, 1/2) = fy (1/2, 1/2) =
1/2
1/2
1/2
8
=
=
= .
2
2
(1 − 1/4)
(3/4)
9/16
9
Then, the linear approximation is
L(x, y) = f (1/2, 1/2) + fx (1/2, 1/2)(x − 1/2) + fy (1/2, 1/2)(y − 1/2).
7
We have f (1/2, 1/2) = 1/(1 − 1/4) = 4/3 and so
L(x, y) =
4 8
8
+ (x − 1/2) + (y − 1/2).
3 9
9
Then
f (0.5001, 0.4997) ≈ L(0.5001, 0.4997) =
4 8
8
+ (0.001) + (−0.003)
3 9
9
4 8
+ (−0.002) = 1.3333 . . . + (0.8888888 . . .)(−0.002)
3 9
= 1.3333 . . . − 0.00177777 = 1.3315555 . . . .
=
∂z
∂z dy
= ∂x
· dx
+ ∂y
· dt . At t = 0, we have x = 0 and y = 1.
16. We use the chain rule and get dz
dt
dt
We have
dx
dx
= 2 cos(2t), so
|t=0 = 2
dt
dt
dy
dy
= et cos(t) − et sin(t), so
|t=0 = 1
dt
dt
∂z
= 3x2 y 2 + 2y 3
∂x
∂z
= 2x3 y + 6xy 2 + 1.
∂y
At (0, 1) we have
∂z
∂x
= 2. At (0, 1) we have
∂z
∂y
= 1. Thus, we have
dz
∂z dx ∂z dy
=
·
+
·
dt
∂x dt
∂y dt
= 2 · 2 + 1 · 1 = 5.
√
√
17. This is a directional derivative problem. Let ~u = h1/ 2, 1/ 2i be a unit vector pointing
in the direction where x = y. We have
D~u (f )(0, 0) = ∇f (0, 0) · ~u.
We have ∇f = hfx , fy i = hex−2y , −2ex−2y i and so ∇f (0, 0) = h1, −2i. Thus, we have
√
√
√
D~u (f )(0, 0) = h1, −2i · h1/ 2, 1/ 2i = −1/ 2.
18. Let f (x, y) be the magnetic field strength and say the point we start at is the origin.
Let ~i = h1, 0i be the east direction and −~j = h0, −1i be the south direction. The problem
statement is saying that
D~i (f )(0, 0) = −12 and D−~j (f )(0, 0) = −5.
8
Since D~u (f )(0, 0) = ∇f · ~u, it follows that if ∇f (0, 0) = ha, bi, then
ha, bi · ~i = a = −12
ha, bi · −~j = −b = −5.
Thus, a = −12 and b = 5 and so ∇f (0, 0) = h−12, 5i. The direction of the gradient is the
direction of greatest increase, and so the optimal direction
to go is √
a unit vector√parallel to
p
∇f (0,0)
2
2
∇f (0, 0), namely ~u = |∇f (0,0)| . We have |∇f (0, 0)| = (−12) + 5 = 144 + 25 = 169 = 13.
Thus
~u = h−12/13, 5/13i
which could be described as west-northwest.
19. The critical points occur when
fx (x, y) = 4x3 + 4y 2 − 4x = 0
fy (x, y) = 4y 3 + 8xy − 8y = 0.
In the second equation every term has y as a factor. This gives y(4y 2 + 8x − 8) = 0. Thus,
either y = 0 or 4y 2 + 8x − 8 = 0.
If y = 0, then the first equation becomes 4x3 − 4x = 0. This factors as 4x(x − 1)(x + 1) and
so we get the three critical points (−1, 0), (0, 0) and (1, 0).
If y 6= 0, then 4y 2 = 8 − 8x. The first equation has a y 2 term in it, and plugging this expression
into the first equation gives us
4x3 + (8 − 8x) − 4x = 0.
Dividing by 4 gives x3 − 3x + 2 = 0. Notice that x = 1 is a root of x3 − 3x + 2 and so if we can
factor out x − 1 and get (x − 1)(x2 + x − 2). We can factor x2 + x − 2 as (x − 1)(x + 2) = 0.
Hence, if y 6= 0 then x = 1 or x = −2. If x = 1, we get 4y 2 = 8 − 8 = 0 and this gives us the
critical point (1, 0) we already have. If√x = −2, we get 4y 2 = 8 − 8(−2) = 24 and so y 2 = 6.
This gives us the critical points (−2, ± 6). Thus, there are a total of 5 critical points.
We have fxx = 12x2 − 4, fxy = 8y and fyy = 12y 2 + 8x − 8. Recall that if (a, b) is a critical
point, the second derivative test says that if D(a, b) = fxx (a, b)fyy (a, b) − fxy (a, b)2 < 0, then
(a, b) is a saddle point. If D(a, b) > 0 and fxx (a, b) > 0, then (a, b) is a local minimum and if
D(a, b) > 0 and fxx (a, b) < 0, we have a local maximum. Here’s a table:
a
b
D(a, b) fxx (a, b)
Type
−1
0
−128
8
Saddle point
0
0
32
−4
Local maximum
1
0
0
8
Saddle point
√
−2 −√ 6 1728
44
Local minimum
−2
6
1728
44
Local minimum.
9
The point (1, 0) is actually a saddle point, but the 2nd derivative test is not capable of determining this. It requires a much more careful and much more detailed analysis.
20. We apply the method of Lagrange multipliers. We have f (x, y, z) = xy 2 z 3 and g(x, y, z) =
x + y + z. We get
∇f (x, y, z) = λ∇g(x, y, z)
2 3
hy z , 2xyz 3 , 3xy 2 z 2 i = hλ, λ, λi.
Thus, y 2 z 3 = 2xyz 3 and y 2 z 3 = 3xy 2 z 2 . Since we are given that x, y, and z are positive, we
can divide by anything we like. Dividing by yz 3 in the first equation gives y = 2x. Dividing by
y 2 z 2 in the second equation gives z = 3x. Plugging into the constraint gives x + 2x + 3x = 1
and so x = 1/6, y = 1/3 and z = 1/2. Evaluating the function here gives
f (1/6, 1/3, 1/2) =
1 1 1
1
1
1
· 2· 3 =
=
=
.
6 3 2
6·9·8
54 · 8
432
21. The solid region described is sitting above the region in the plane D = {(x, y) ∈ R2 : 0 ≤
x ≤ 1, 1 − x ≤ y ≤ 1}. The bottom
RR of the solid region is z = 0 and the top is z = f (x, y) =
2
2
4 − x − y . Hence the volume is D f (x, y) dA. This is
Z
1
Z
0
1
1−x
1
(4 − x2 − y 2 ) dy dx
1
y3
dx
=
4y − x y −
3 1−x
0
Z 1
1
(1 − x)3
2
2
=
4−x −
− 4(1 − x) − x (1 − x) −
dx
3
3
0
Z 1
4 3
2
=
− x + x + 3x dx
3
0
1
1 4 1 3 3 2
= − x + x + x
3
3
2
0
3
1 1 3
=− + + = .
3 3 2
2
Z
2
p
22. There is no nice antiderivative for sin(y). Therefore, we need to switch the order of
integration. The region is 0 ≤ x ≤ 1, sin−1 (x2 ) ≤ y ≤ π2 . The curve y = sin−1 (x2 ) becomes
p
sin(y) = x2 or x = sin(y) (this doesn’t sound familiar does it?). Since
psin(x) is increasing on
−1 2
2
[0, π/2], the inequality sin (x ) ≤ y becomes x ≤ sin(y) and so x ≤ sin(y).
10
p
Therefore, the curve x = sin(y) is the right boundary, the curve x = 0 is the left boundary,
the top boundary is y = π/2 and the bottom boundary is y = 0. This gives us
Z π/2 Z √sin(y) p
sin(y) dx dy
0
0
Z π/2 h p
ix=√sin(y)
dy
x sin(y)
=
x=0
0
Z
π/2
p
p
( sin(y)) · ( sin(y)) dy
=
0
Z
π/2
sin(y) dy
=
0
= [− cos(y)]π/2
= − cos(π/2) + cos(0) = −0 + 1 = 1.
0
23. We switch to polar coordinates. The original region of integration is the disk of radius
1 centered at the origin. Thus, the limits on r and θ are 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 1. Our
integrand becomes 43 π(1 − r2 )3/2 and dy dx = r dθ dr. This gives us
Z 1 Z 2π
Z 1
Z 1
4π
4
4
2 3/2
2 3/2 2π
πr(1 − r ) dθ , dr = π
θr(1 − r )
dr =
· 2π ·
r(1 − r2 )3/2 dr.
0
3
3
3
0
0
0
0
Set u = 1 − r2 . Then, du = −2r dr. The new limits become 1 to 0 and r dr = − 21 du. This
gives us
1
Z
Z
1
4π 2 2 5/2
4π 2 2
8π 2 0 3/2
4π 2 1 3/2
8π 2
u
− du =
u du =
=
u
· =
.
3 1
2
3 0
3 5
3 5
15
0
24. The rod is in the shape of a cylinder, and cutting away the sphere will sepearte the rod
into two pieces. The smallest value of z in the top piece will occur when x2 + y 2 = 3 and this
means that
z 2 = (x2 + y 2 + z 2 ) − (x2 + y 2 ) = 25 − 3 = 22.
In particular, the top piece will be described by x2 + y 2 ≤ 3 and 25 − x2 − y 2 ≤ z 2 ≤ √
25. We
set up this integral in polar coordinates. Saying that x2 + y 2 ≤ 3 means thatpr ≤ 3 and
2
2
0 ≤ θ ≤ 2π. The highest
√ z-coordiante minus the smallest z-coordinate is 5 − 25 − x − y
and this becomes 5 − 25 − r2 . Hence, the total volume (which is twice the volume of the top
piece is)
√ 3
Z 2π Z √3
Z 2π √
1
2
(5 − 25 − r2 )r dr dθ = 2
(5/2)r2 + (25 − r2 )3/2
dθ
3
0
0
0
0
√
Z 2π 5
1 3/2 1 3/2
(88 22 − 410)π
=2
· 3 + 22 − 25
dθ =
.
2
3
3
3
0
11
25. We have
∂ 2
∂ x
∂
(x ) +
(e sin(y)) + (−zex cos(y))
∂x
∂y
∂z
x
x
= 2x + e cos(y) − e cos(y) = 2x.
∇ · F~ =
We have
~i
~k
~j
∂
∂
∂
∇ × F~ = ∂x
∂y
∂z
x2 ex sin(y) −zex cos(y)
= hzex sin(y), zex cos(y), ex sin(y)i.
26. The vector field on the left is the one that has a curl that always points out of the picture,
and the one on the right is the one where the divergence is always zero.
The curl in the left hand picture is perhaps the easiest to spot. If a line integral is done around
circle centered at the origin then Green’s theorem says that
Z
F~ · d~s =
ZZ
C
(∇ × F~ ) · ~k dA
D
and since the vector field would always be parallel to the circle, the line integral would be
positive (and hence the curl would be as well). Another way to see this is to look along the
positive x-axis. The arrows point in the y-direction and get longer as one moves farther from
the origin. This means that a leaf would spin in the counterclockwise direction if it were put
into the flow of the fluid there, as the force on the right side of the leaf would overcome the
force on the left side.
The flow of the fluid in the right hand picture doesn’t “bunch up” anywhere, and this is a
reflection of the fact that the divergence is always zero. This means that the divergence zero
vector field is the one on the right.
27. We parametrize C by ~r(t) = hcos(t), sin(t)i, 0 ≤ t ≤ 2π.
h− sin(t), cos(t)i. We get
Z
C
F~ · d~s =
Z
2π
F~ (~r(t)) · ~r 0 (t) dt =
Z
We then have ~r 0 (t) =
2π
hsin(t), − cos(t)i · h− sin(t), cos(t)i dt
Z 2π
Z 2π
2
2
=
− sin (t) − cos (t) dt =
−1 dt = −2π.
0
0
0
0
12
28. We have
~k
~i
~j
∂
∂
∂
∇ × F~ = ∂x
∂y
∂z
2xyz + y 2 + z 2 + 1 x2 z + 2xy + z x2 y + 2xz + y = h(x2 + 1) − (x2 + 1), −(2xy + 2z) + (2xy + 2z), (2xz + 2y) − (2xz + 2y)i = ~0.
Thus, the curl is zero. Since F~ is defined on R3 , which is open and simply-connected, it follows
that F~ is conservative (this is a variant of Theorem 6 from Section 16.3 for the R3 case). Now,
we will actually find a potential function f for it.
We need to have
∂f
= 2xyz + y 2 + z 2 + 1, and so
∂x
f = x2 yz + xy 2 + xz 2 + x + g(y, z).
We take the formula above and compute the partial derivative with respect to y. We get
∂g
∂f
= x2 z + 2xy +
(y, z).
∂y
∂y
This should equal x2 z + 2xy + z and so we should have
g(y, z) = yz + h(z) and so
∂g
(y, z)
∂y
= z so we should have
f (x, y, z) = x2 yz + xy 2 + xz 2 + x + yz + h(z).
Now, when we compute
∂f
∂z
we get
∂f
= x2 y + 2xz + y + h0 (z).
∂z
We should get x2 y + 2xz + y and so h0 (z) = 0. We may take h(z) = 0 and the function f (x, y, z)
as given above.
29. We don’t want to do this using the definition. Notice that
~r(0) = h0 + 1 − cos(sin(0)), sin(0)i = h1 − 1, 0i = h0, 0i
~r(2π) = h2π + 1 − cos(sin(2π), sin(2πesin(2π) )i = h2π + 1 − cos(0), sin(2π)i = h2π, 0i.
R
We will find a function f so that ∇f = F~ . Then, C F~ · d~r = f (2π, 0) − f (0, 0). We want
∂f
= x3 − 3xy 2
∂x
1
3
f = x4 − x2 y 2 + q(y)
4
2
for some function q. Then
∂f
= −3xy 2 + q 0 (y) = −3xy 2 y + y 3 .
∂y
13
Thus, q 0 (y) = y 3 and so q(y) = 14 y 4 and so
1
3
1
f (x, y) = x4 − x2 y 2 + y 4 .
4
2
4
1
4
4
Thus, f (2π, 0) − f (0, 0) = 4 (2π) = 4π , and this is equal to the line integral above.
30. (a) This is true. If F~ = ∇f , then
∂Q
∂P
= fyx = fxy =
.
∂x
∂y
(b) This is false. Theorem 6 in Section 16.3 says this is only true if the domain of F~ is
simply-connected. In fact, the vector field
y
x
F~ (x, y) =
,−
x2 + y 2 x2 + y 2
is not conservative even though Qx = Py because the domain is {(x, y) ∈ R2 : (x, y) 6= (0, 0)}
which is NOT simply-connected.
(c) This is true. It is the statement of Theorem 4 from Section 16.3.
0
0
31. We have ~r(0) = h0 + ee , 2i and ~r(2π) = h0 + ee , 2i, and so ~r(t) is a closed curve. Let
F~ (x, y) = hP (x, y), Q(x, y)i. Then,
∂Q
= −2y
∂x
∂P
= −2y.
∂y
It follows that ∂Q = ∂P . Since F~ (x, y) is defined on all of R2 (note that esin(x) ≥ 0 for all x),
∂x
∂y
which is simply-connected, Theorem 6 of Section 16.3 implies sthat F~ is conservative. This
means that the line integral around the closed curve is equal to zero.
32. The definition would be quite unpleasant again here. If we let P (x, y) = 6xy 3 + y and
Q(x, y) = 9x2 y 2 − 2x, then
∂Q ∂P
−
= (18xy 2 − 2) − (18xy 2 + 1) = −3.
∂x
∂y
The orientation of the octogon matches that in Green’s theorem, and so we have
Z
ZZ ∂Q ∂P
~
F · d~r =
−
dA = (−3) · (the area of D).
∂x
∂y
C
D
√
√
One triangle that makes
up
D
has
vertices
(0,
0),
(1,
0),
and
(
2/2,
2/2). Thus, it has base
√
√
of length 1, height 2/2, and so its area is 2/4. The octogon is made
8 triangles that
√ up of √
are each the same size, and
so
the
area
of
the
entire
octogon
is
8(
2/4)
=
2
2. Hence, the
√
√
line integral equals (−3)(2 2) = −6 2.
14
33. The equation ~r(t) = ht2 − 1, t3 − ti parametrizes a closed curve for −1 ≤ t ≤ 1 with a
counterclockwise orientation. Find the area of the region enclosed by this curve.
It’s tricky to actually get a formula for what this curve is in the form y = f (x), so the natural
way to approach this is to use Green’s theorem. Green’s theorem says that
Z
ZZ ∂Q ∂P
P dx + Q dy =
−
dA
∂x
∂y
C
D
where C is the boundary of D and we have an orientation on C so that as we
RR move around
it, C is on the left. Since we want to find the area of D we want to compute D 1 dA and so
we pick a vector field F~ (x, y) = hP (x, y), Q(x, y)i so that Qx − Py = 1. I will take Q = x and
P = 0.
Now, we evaluate the line integral on the left hand side using the definition. We have x = t2 − 1
and dy = (3t2 − 1) dt. Thus,
Z
Z 1
Z 1
2
2
x dy =
(t − 1) · (3t − 1) dt =
3t4 − 4t2 + 1 dt
−1
C
−1
1
3 4
3 5 4 3
3 4
= t − t +t
=
− +1 − − + −1
5
3
5 3
5 3
−1
6 8
18 40 30
8
= − +2=
−
+
= .
5 3
15 15 15
15
By Green’s theorem, this line integral is equal to the area enclosed by the curve.