Solutions to Sample Questions for Final Exam

Solutions to Sample Questions for Final Exam
1 Find the points on the surface xy 2 z 3 = 2 that are closest to the origin. We use the
method of Lagrange Multipliers, with f (x, y, z) = x2 + y 2 + z 2 for the square of the distance
from the origin, and the constraint is given by g = 2 for g(x, y, z) = xy 2 z 3 . We get
∇f = h2x, 2y, 2zi and ∇g = hy 2z 3 , 2xyz 3 , 3xy 2z 2 i.
Setting ∇f = λ∇g gives us the system of equations
2x = λy 2 z 3
2y = 2λxyz 3
2z = 3λxy 2 z 2
xy 2 z 3 = 2.
Note that the last equation implies that x, y, z 6= 0. Dividing both sides of the second
equation by 2y gives 1 = λxz 3 , so λ = xz1 3 . Plug that into the first equation and clear
2
fractions to obtain 2x2 = y 2 . Plug this value of λ into the third equation to get 2z = 3yz , so
√
√
2z 2 = 3y 2. Hence z 2 = 3x2 , since 2x2 = y 2. Consequently, either z = 3x or z = − 3x.
√
Case 1: z = 3x. This and y 2 = 2x2 allow us to write the constraint equation entirely in
terms of x:
√
6 3x6 = 2 =⇒ x6 = 3−3/2 =⇒ x = ±3−1/4 .
√
Case 2: z = − 3x. That implies x and z have opposite signs. Therefore xy 2 z 3 is negative,
since y 2 is positive. This is not a possibility since our constraint equation requires xy 2 z 3 = 2.
Therefore Case 2 does not occur.
We also have y 2 = 2x2 = 2(3−1/4 )2 =
√2 ,
3
√
so y = ± 31/42 .
Therefore, the solutions of the system of equations above are
(3−1/4 , 2(3−1/4 ), 31/4 ),
(−3−1/4 , 2(3−1/4 ), −31/4 )
(3−1/4 , −2(3−1/4 ), 31/4 ),
and
(−3−1/4 , −2(3−1/4 ), −31/4 ).
Every one of these points has the same distance from the origin:
r
√
q
5
8
−1/4
2
−1/4
2
1/4
2
1/2
(±3
) + (±2(3
) + (±3 ) =
+ 3 = 1/4 .
1/2
3
3
2 Let f (x, y) = 4xy 2 − x2 y 2 − xy 3 . Let D be the closed triangular region in the xy-plane
with vertices (0, 0), (0, 6) and (6, 0). Find the absolute maximum and minimum values of f
on D.
First we find the critical points of f . The gradient is ∇f = h4y 2 −2xy 2 −y 3 , 8xy−2x2 y−3xy 2i.
This gives us the system of equations
4y 2 − 2xy 2 − y 3 = 0
8xy − 2x2 y − 3xy 2 = 0.
We can rewrite this after some factoring as
y 2(4 − 2x − y) = 0
xy(8 − 2x − 3y) = 0.
When one of the factors y 2 or xy is zero, the critical point will not be inside the domain –
it will be on the boundary, or maybe even outside the domain entirely, so we do not need
to worry about those here. (We will check the boundary later, anyway.) So we really only
need to consider the system of equations
4 − 2x − y = 0
8 − 2x − 3y = 0.
Subtracting the second equation from the first yields 4 − 2y = 0, so y = 2. Plug that back
into either equation to obtain x = 1. So the only critical point inside the domain is (1, 2).
Notice that f(1,2)=4 .
Next we check the boundary. There are three line segments that form the boundary of D:
x = 0, y = 0 and x = 6 − y. On the first two pieces, f(x,0)=0 and f(0,y)=0 . On the last
piece, we have
f (6 − y, y) = 4(6 − y)y 2 − (6 − y)2 y 2 − (6 − y)y 3
= 2y 3 − 12y 2.
This segment corresponds to 0 ≤ y ≤ 6, so we use the methods of 1-variable calculus to
find the absolute maximum and minimum there. We take an ordinary derivative of thie
expression with respect to y and set it equal to zero to obtain the equation
6y 2 − 24y = 0,
so y = 4 (or y = 0, but we’ve already checked that). In that case, we have x = 2 on the line
segment, so
f(2,4) = -64 .
Consequently, we see that the absolute maximum of f on the domain D is 4, and it occurs
at the point (1, 2), and the absolute minimum is −64 which occurs at the point (2, 4).
3 Find the critical points of the function f (x, y) = 3xy − x2 y − xy 2 and classify them as
local maxima, local minima or saddle points.
The gradient of f is ∇f = h3y − 2xy − y 2, 3x − x2 − 2xyi. Setting this equal to the zero
vector gives us the system of equations
3y − 2xy − y 2 = 0
3x − x2 − 2xy = 0.
Factoring yields
y(3 − 2x − y) = 0
x(3 − x − 2y) = 0.
Since either factor of either equation can be zero, this really gives us 4 systems of equations:
(I) y = 0 and x = 0
(II) y = 0 and 3 − x − 2y = 0
(III) 3 − 2x − y = 0 and x = 0
and
(IV) 3 − 2x − y = 0 and 3 − x − 2y = 0.
Solving each of these systems gives us the following four solutions:
(0, 0),
(3, 0),
(0, 3)
and
(1, 1).
Now we classify these critical points. The second derivatives of f are
fxx = −2y,
fx,y = 3 − 2x − 2y
and
fyy = −2x.
Thus the discriminator is
D(x, y) = fxx fyy − (fxy )2 = 4xy − (3 − 2x − 2y)2.
Therefore
D(0, 0) = −9 < 0,
so (0, 0) is a saddle point;
D(3, 0) = −9 < 0,
so (3, 0) is a saddle point;
D(0, 3) = −9 < 0,
so (0, 3) is a saddle point; and
D(1, 1) = 3 > 0 and fxx (1, 1) = −2 < 0,
so (1, 1) is a local maximum.
4 Find the direction in which f (x, y, z) = zexy increases most rapidly at the point (0, 1, 2).
What is the maximum rate of increase?
The gradient is ∇f = hzyexy , zxexy , exy i. Hence
∇f (0, 1, 2) = h2, 0, 1i.
This is the direction in which f increases most rapidly (equivalently, it increases fastest in
the direction of the unit vector
h2, 0, 1i
∇f (0, 1, 2)
= √ .
|∇f (0, 1, 2)|
5
The rate of change is
|∇f (0, 1, 2)| = |h2, 0, 4i| =
√
5.
∂z
∂z
5 If z = f (x2 − y 2), where f is differentiable, show that y ∂x
+ x ∂y
= 0.
We use the chain rule to calculate
∂z
∂
= f ′ (x2 − y 2) [x2 − y 2 ] = f ′ (x2 − y 2)2x
∂x
∂x
and
∂z
∂
= f ′ (x2 − y 2) [x2 − y 2] = −f ′ (x2 − y 2)2y.
∂y
∂y
Consequently,
y
6 Calculate
RRR
B
∂z
∂z
+x
= y(f ′(x2 − y 2)2x) + x(−f ′ (x2 − y 2)2y)
∂x
∂y
= 2xyf ′(x2 − y 2) − 2xyf ′ (x2 − y 2 )
= 0.
x2 +y 2 dV , where B is the unit ball in R3 : B = {(x, y, z); x2 +y 2 +z 2 ≤ 1}.
We use spherical coordinates to calculate the triple integral:
Z Z Z
Z π Z 1 Z 2π
2
2
x + y dV =
(ρ sin(φ) cos(θ))2 + (ρ sin(φ) sin(θ))2 ρ2 sin(φ) dθ dρ dφ
B
Z0 π Z0 1 Z0 2π
=
ρ4 sin3 (φ)dθ dρ dφ
0
0
0
Z πZ 1
= 2π
ρ4 sin3 (φ) dρ dφ
Z0 0
2π π 3
sin (φ) dφ
=
5 0
Z
2π π
=
(1 − cos2 (φ)) sin(φ) dφ
5 0
Z
2π −1
=−
(1 − s2 ) ds
(s = cos(φ), ds = − sin(φ) dφ)
5 1
−1
2π
s3 =−
s−
= 8π
.
15
5
3 1
7 Calculate
RR
S
x2 +y 2 dS, where S is the unit sphere in R3 : S = {(x, y, z); x2 +y 2 +z 2 = 1}.
We parametrize the sphere using spherical coordinates, keeping in mind that ρ = 1 everywhere on the sphere:
~r(θ, φ) = hsin(φ) cos(θ), sin(φ) sin(θ), cos(φ)i.
The partial derivatives are
~rθ = h− sin(φ) sin(θ), sin(φ) cos(θ), 0i
and
~rφ = hcos(φ) cos(θ), cos(φ) sin(θ), − sin(φ)i.
Therefore
~rθ × ~rφ = h− sin2 φ cos θ, − sin2 φ sin θ, − sin φ cos φi.
Using the pythagorean identity to simplify yields
q
|~rθ × ~rφ | = sin4 φ cos2 θ + sin4 φ sin2 θ + sin2 φ cos2 φ
q
= sin4 φ + sin2 φ cos2 φ
q
= sin2 φ
= sin φ.
In the last line, we used the fact that 0 ≤ φ ≤ π implies sin φ ≥ 0, so tat we didn’t have to
be concerned with absolute values. Now we can compute
Z Z
Z π Z 2π
2
2
x + y dS =
(sin φ cos θ)2 + (sin φ sin θ)2 sin φ dθ dφ
0
S
0
Z π Z 2π
=
sin3 φ dθ dφ
0
Z 0π
= 2π
sin3 φ dφ
0
Z π
= 2π
(1 − cos2 φ) sin phi dφ
0
Z −1
= −2π
(1 − s2 ) ds
(s = cos φ, ds = − sin φ dφ)
1
−1
s3 = −2φ s −
3 1
8π
=
.
3
2
2
2
2
R8 RLet
R E2 be the region bounded by the paraboloids z = 4−x −y and z = x +y . Compute
x dV .
E
The paraboloid z = 4 − x2 − y 2 opens downward, while z = x2 + y 2 opens upward. The
paraboloids intersect in a circle:4 − x2 − y 2 = x2 + y 2 implies 4 = 2x2 + 2y 2, so 2 = x2 + y 2 ;
hence z = 4−2 = 2. That is to say, the intersection of the paraboloids is the circle x2 +y 2 = 2
in the plane z = 2. If we “stand on top of the z-axis and look down”
we see that the domain
√
in the xy-plane isthe circle centered at the origin with radius 2. Now if we use cylindrical
coordinates for the integral, we get
√
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π and r 2 ≤ z ≤ 4 − r 2 .
Therefore
Z Z Z
2π
x dV =
Z
2π
=
Z
2π
=
Z
Z
2π
=
Z
Z
Z
2π
2
E
0
0
0
0
Z
Z
√
2
0
√
2
Z
4−r 2
(r cos θ)2 r dz dr dθ
r2
4−r 2
Z
r 3 cos2 θ dz dr dθ
r2
0
√
2
√
2
0
0
4−r2
zr 3 cos2 θr2 dθ dr
(4r 3 − 2r 5 ) cos2 θ drdθ
√2
6
r
cos2 θ dθ
=
r4 −
3
0
0
Z 2π 8
cos2 θ dθ
=
4−
3
0
Z 2π
4
=
cos2 θ dθ
3 0
Z
4 2π 1 + cos(2θ)
dθ
=
3 0
2
2π
2
sin(2θ) θ+
=
3
2
0
4π
=
.
3
9 Let F~ (x, y, z) = (2xy 3 + z 2 )~i + (3x2 y 2 + 2yz)~j + (y 2 + 2xz)~k. (a) Find a potential function
R
f for F~ . (b) Compute C F~ · d~r, where C is the curve ~r(t) = ht, t2 , t3 i, with 0 ≤ t ≤ 1.
(a) The potential function needs to satisfy
∂f
∂x
= 2xy 3 + z 2 , so we have
f = x2 y 3 + xz 2 + C(y, z).
Differentiating with respect to y yields
∂f
∂C
= 3x2 y 2 +
,
∂y
∂y
but this must equal 3x2 y 2 + 2yz, so we have
∂C
∂y
= 2yz; hence C(y, z) = y 2z + D(z). Thus
f = x2 y 3 + xz 2 + y 2 z + D(z),
and differentiating with respect to z yields
∂f
= 2xz + y 2 + D ′ (z).
∂z
We need to have
be
∂f
∂z
= 2xz + y 2 , so D ′ (z) = 0. Thus we may take the potential function to
f (x, y, z) = x2 y 3 + xz 2 + y 2z.
(b) The given path C begins at (0, 0, 0) and ends at (1, 1, 1). Therefore, using the Fundamental Theorem for Line Integrals gives us
Z
F~ · d~r = f (1, 1, 1) − f (0, 0, 0) = 1 + 1 + 1 − 0 − 0 − 0 = 3.
C
H
10 Compute C (y 3 + tan x) dx − (x3 + sin y) dy, where C is the positively oriented boundary
of the region between the circles x2 + y 2 = 1 and x2 + y 2 = 4.
We use Green’s Theorem with P (x, y) = y 3 + tan x and Q(x, y) = −(x3 + sin y) and D the
region between these circles to get
I
ZZ ∂Q ∂P
−
dA
P dx + Q dy =
∂x
∂y
C
D
ZZ
=
(−3x2 − 3y 2) dA
D
Z 2 Z 2π
=−
3r 2 r dr dθ
1
0
Z 2
= −6π
r 3 dr
1
4 2
−3πr =
.
= − 45π
2
2 1
~ such that curl G
~ = h2x, 3yz, −xz 2 i.
11 Show that there is no vector field G
~ Then div F~ = div curl G
~ = ~0,
Write F~ = h2x, 3yz, −xz 2 i. Suppose that F~ = curl G.
because the divergence of the curl of any vector field is the zero vector field. However, for
this vector field F~ , we have
∂
∂
∂
div F~ = ∇ · h2x, 3yz, −xz 2 i =
[2x] +
[3yz] + [−xz 2 ] = 2 + 3z − 2xz.
∂x
∂y
∂z
~
This is not the zero vector field, so F~ cannot be the curl of G.
12 Let S be the surface of the bottomless box with corners (±1, ±1, ±1) and with outward
orientation (that is to say, choose the outward orientation as if the box was closed). Let
RR
RR
~ (b) Find
~
F~ = hx + y 2 , x + z 3 , zi. (a) Find
F~ · dS.
(curl F~ ) · dS.
S
S
(a) Let S2 be the bottom of the box – that is to say, S2 is the square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1
in the plane z = −1. Let S2 have the downward orientation. Then together, S and S2 are the
outward-oriented boundary of the box E given by −1 ≤ x ≤ 1, −1 ≤ y ≤ 1 and −1 ≤ z ≤ 1.
According to the Divergence Theorem,
ZZ
ZZ
ZZZ
~+
~=
F~ · dS
F~ · dS
div F dV
S
S2
E
Z 1Z 1Z 1
=
∇ · hx + y 2 , x + z 3 , zi dV
−1 −1 −1
Z 1Z 1Z 1
=
2dV
−1
−1
−1
= 2(V olume of E)
= 16.
To finish solving for the integral over S, we need to know the value of the integral over S2 .
We can compute that directly with the parametrization
~r(x, y) = hx, y, −1i.
We calculate
~rx (x, y) = h1, 0, 0i
and
~ry (x, y) = h0, 1, 0i.
So
~rx (x, y) × ~ry (x, y) = h0, 0, 1i.
This has upward orientation, but the normal vector we need has downward orientation:
~ry (x, y) × ~rx (x, y) = h0, 0, −1i.
Therefore
ZZ
S2
~=
F~ · dS
Z
=
Z
1
1
Z
−1 −1
1 Z 1
−1
hx + y 2 , x + 1, −1i · h0, 0, −1i dV
1 dS
−1
= (Area of [−1, 1] × [−1, 1])
= 4.
Hence
ZZ
S
~ + 4 = 16,
F~ · dS
so
ZZ
S
~ = 12.
F~ · dS
(b) Let C be the positively-oriented boundary of C: it is the square curve from (−1, −1, −1)
to (1, −1, −1) to (1, 1, −1) to (−1, 1, −1) and back to the starting corner. The according to
Stoke’s Theorem, we get
ZZ
I
~
~
curl F · dS =
F~ · d~r.
S
C
RR
~ where S2 is the square
But notice that this is the same value that we get for S2 curl F~ · dS,
above but with upward orientation:
ZZ
I
~
~
curl F · dS =
F~ · d~r.
S2
Thus
ZZ
S
C
~=
curl F~ · dS
ZZ
S2
~
curl F~ · dS.
The point is that the surface integral over S2 is much easier to calculate. First we need the
curl:
curl F~ = ∇ × F~ = h−3z 2 , 0, 1 − 2yi.
Therefore
ZZ
S2
~=
curl F~ · dS
=
=
1
Z
−1
Z 1
−1
Z 1
−1
1
=
Z
1
−1
Z 1
−1
h−3, 0, 1 − 2yi · h0, 0, 1i dx dy
1 − 2y dx dy
x − 2xy|1−1 dy
Z
2 − 4y dy
1
= 2y − 2y 2−1
−1
= 4.
That is to say,
ZZ
S
~ = 4.
curl F~ · dS