Math 53, Sample Final - Solutions

9 December 2013
Christopher A. Wong
Math 53, Sample Final - Solutions
1. Find and classify all critical points of the function f (x, y) = y 3 + 3x2 y − 12y as local maxima,
local minima or saddle points.
Solution. We calculate ∇f = (6xy, 3x2 +3y 2 −12). Then ∇f = ~0 only at (0, ±2), (±2, 0); thus
there are four critical points. Calculating the second derivatives, fxx = 6y, fxy = 6x, fyy =
6y. Then, applying the second derivative test with discriminant D = fxx fyy − (fxy )2 , we
have D = 144 for the points (0, ±2) and D = −144 for (±2, 0). Using the sign of fxx
to distinguish between maxima and minima, we ultimately find that (0, 2) is a minimum,
(0, −2) is a maximum, and (±2, 0) are saddle points.
2. Find the maximum and minimum values of the function x+y+z on the surface of the ellipsoid
x2 y 2 z 2
+
+
= 12.
2
4
6
Solution. We will use the method of Lagrange multipliers to find the constrained extrema.
If f = x + y + z is the function to be optimized, and g(x, y, z) = x2 /2 + y 2 /4 + z 2 /6 is the
function whose level set g = 12 is the ellipsoid, then we set
∇f = λ∇g,
which yields
(1, 1, 1) = (λx, λy/2, λz/3),
from which we obtain x = 1/λ, y = 2/λ, z = 3/λ. Substituting this into our original constraint, we have
(1/λ)2 (2/λ)2 (3/λ)2
+
+
= 12,
2
4
6
and hence λ = ±1/2, and hence f has extreme points on the ellipsoid at the points (2, 4, 6)
and (−2, −4, −6), and so its maximum and minimum values are ±12.
3. Change the order of integration in the triple integral
Z 1Z xZ y
Z ?Z ?Z
f (x, y, z) dz dy dx =
0
0
0
?
?
?
f (x, y, z) dy dx dz.
?
Solution. By drawing the picture of the region of integration, we find that the region is a
tetrahedron bounded by the four planes z = y, x = y, z = 0, and x = 1. Therefore, when
rearranging the order of integration, we obtain
Z 1Z 1Z x
f (x, y, z) dy dx dz.
0
z
z
4. A homogeneous ball of radius R and mass density µ rotates with the angular velocity ω about
a line passing through its center. Compute the kinetic energy accumulated in the ball due to
this rotation.
Solution. The rotational kinetic energy is given by K = 12 Iω 2 , where I is the moment of
inertia of the rotating body. This moment of inertia is given by
Z
I=
µr2 dV,
B
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where r = r(x, y, z) is the distance function from the point (x, y, z) to thep
axis of rotation,
which we can take to be the z-axis, and B is the ball. Explicitly, r(x, y, z) = x2 + y 2 . Then,
in spherical coordinates, r = ρ sin φ, so this integral becomes
Z R Z 2π Z π
I=µ
(ρ2 sin2 φ)ρ2 sin φ dφ dθ dρ
0
0
0
Z
2πµR5 π 3
=
sin φ dφ
5
0
Z
2πµR5 π
=
sin φ(1 − cos2 φ) dφ
5
0
π
2πµR5 1
3
=
cos φ − cos φ
5
3
0
=
8πµR5
.
15
Therefore the kinetic energy is K =
4
5 2
15 πµR ω .
5. An ink spot of initially round shape (x − 1)2 + y 2 ≤ a2 is carried by the flow of a twodimensional fluid with the velocity vector field
~ = (x + e−x2 cos y)~i + (2y + 2xe−x2 sin y)~j.
V
Find the area of the ink spot at the moment t.
Solution. Let C(t) be the boundary of the ink spot, which changes in t, let E(t) be its interior,
~ means that
and let A(t) be its area. Then having the ink spot be carried by the flow of V
I
dA(t)
~ · ds.
=
V
dt
C(t)
By the two-dimensional divergence theorem, we have that
I
ZZ
ZZ
~ · ds =
~ dA =
V
div V
3 dA = 3A(t),
C(t)
E(t)
E(t)
d
A(t) = 3A(t), which has the solution A(t) = A(0)e3t . Since the initial
and hence we have dt
2
area A(0) = πa , then the area of the ink spot is A(t) = πa2 e3t at any time t.
6. Calculate the area of the surface
2x2 2y 2
z =a 1− 2 − 2 ,
a
a
z ≥ 0.
Solution. Appealing to polar coordinates we can parametrize this surface by
√
2u2
, 0 ≤ u ≤ a/ 2, 0 ≤ v ≤ 2π.
r(u, v) = u cos v, u sin v, a 1 − 2
a
Calculating the tangent vectors in this surface:
ru = (cos v, sin v, −4u/a)
rv = (−u sin v, u cos v, 0),
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then we can calculate the differential surface element
p
p
dS = kru × rv k du dv = (4u2 /a)2 + u2 du dv = u 16u2 /a2 + 1 du dv.
Therefore the surface area of the surface in question is given by
Z
1 dS
SA =
S
2π
Z
Z
√
a/ 2
=
=
0
a2
Z
0
2π
32 0
13
= πa2 .
12
Z
p
u 16u2 /a2 + 1 du dv
9√
k dk du,
k = 16u2 /a2 + 1
1
7. Compute the flux of the vector field F~ = ~r/|~r|3 , where ~r = x~i + y~j + z~k across the surface
(x + 1)2 = y 2 + z 2 , x ≤ 0 equipped with an orientation of your choice.
Solution. This surface is a part of an infinite double cone rotationally symmetric about the
x-axis, with the point at (−1, 0, 0). Let us first notice that the divergence of F~ is zero (we
calculated this in a homework exercise). Therefore, if S1 is the upper (finite) part of the cone
and S2 is the lower (infinite) part of the cone, then S2 is a closed surface, and by applying
the divergence theorem, the flux of F~ through S2 is zero.
Now let us compute the flux of F through S1 directly; give S1 the orientation pointing in the
negative x-axis. Then we can parametrize the surface by σ(r, θ) = (r − 1, r cos θ, r sin θ). We
find that
σr = (1, cos θ, sin θ), σθ = (0, −r sin θ, r cos θ),
and
σr × σθ = (r, −r cos θ, −r sin θ).
Notice that this vector points to the positive x-axis, so we need σθ × σr . Then the flux of F~
through S1 is given by
Z
Z 2π Z 1
1
~
F · n dS =
(r − 1, r cos θ, r sin θ) · (σθ × σr ) dr dθ
2
2 3/2
S
0
0 ((r − 1) + r )
Z 2π
r
=
dr dθ,
2
(2r − 2r + 1)3/2
0
which can be simplified, but not easily.
8. Give an example of a divergence-free vector field which is not the curl of any vector field.
Justify your answer.
Solution. Consider the vector field F (~r) = ~r/k~rk3 as before. Then the divergence of F is zero
whenever F is not the origin. Now, let S be the outward-oriented unit sphere. Then
ZZ
ZZ
F · n dS =
1 dS = 4π.
S
S
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However, if F = curl G for some vector field G, then since F is differentiable on an open
neighborhood of S, then applying Stokes’ theorem would yield
I
ZZ
ZZ
G · dr,
curl G · n dS =
F · n dS =
∂S
S
S
where ∂S is empty, and hence the above equals zero. But the calculation previously showed
that the integral was 4π, and hence this is a contradiction, so it is impossible that F = curl G.
9. Find the domain D on the plane for which the line integral
Z
(x2 y − 2y) dx + (2x − y 2 x) dy
∂D
takes on its maximal possible value, and compute this value.
Solution. Applying Green’s theorem, the integral in question is equal to
ZZ
4 − x2 − y 2 dA,
D
which is maximized when D is the region on which the integrand is non-negative, which is
the disc x2 + y 2 ≤ 4. We can compute this value by using the parametrization σ(r, θ) =
(r cos θ, r sin θ) with 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. The Jacobian for this change of variables is
cos θ
sin θ
⇒ | det J| = r,
J=
−r sin θ r cos θ
so then the integral becomes
Z
0
2 Z 2π
4r − r3 dθ dr = 8π.
0
10. On grid paper, two ants crawl with velocities (2, 5) and (−2, 3), starting at the initial positions
(0, 0) and (15, 0) respectively. Find the shortest distance between the ants.
Solution. Let t be the time parameter. Then the trajectories of the two ants can be simultaneously parametrized by
r1 (t) = (2t, 5t),
r2 (t) = (15 − 2t, 3t).
Then their squared distance is given by
kr1 − r2 k2 = (15 − 4t)2 + (−2t)2 = 20t2 − 120t + 225,
√
which is minimized at t = 3, which yields a distance 45.
11. A comet follows an elliptical orbit with the major semiaxis 13 au and the minor semiaxis 5
au. Find how close the comet comes to the Sun.
Solution. It is implied in this problem that the Sun is at one focus of the ellipse. The equation
for the curve traced by the comet’s orbit is the ellipse
x2
y2
+
= 1,
132
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and we can calculate its foci (±p, 0) by recall that the sum of the distances of any point on
the ellipse to the two foci is constant. In particular, at the point (13, 0), the sum of these
distances is 26, and at the point (0, 5), the distance to each focus is 13 and forms a right
triangle with hypotenuse 13 and leg 5. Therefore the remaining leg has length 12, so the
focus is at the point (12, 0), and the ellipse is closest to this point at (13, 0), so the comet
comes as close as one astronomical unit from the Sun.
12. Find out if the cross section of the quadratic surface 2x2 = 3y 2 +5z 2 −1 by the plane z = x−y
is a hyperbola.
Solution. Calculate the cross-section by substituting z = x − y, yielding
2x2 = 3y 2 + 5x2 − 10xy + 5y 2 − 1 =⇒ 1 = 3x2 − 10xy + 8y 2 .
This is a level set f (x, y) = 1 of the quadratic surface f (x, y) = 3x2 − 10xy + 8y 2 . We can
classify this quadratic surface by using the discriminant D = (3)(8) − 52 = −1, so the surface
is a hyperbolic paraboloid, and its level sets are hyperbolas, so indeed this cross-section is a
hyperbola.
13. For a square-shaped thin board of mass density σ and area a2 , compute the moment of inertia
about the axis perpendicular to the plane of the square and passing through one of its vertices.
Solution. Let D be the square region [0, a] × [0, a]. Then the moment of inertia of D with
mass density σ through the vertex at (0, 0) is given by
ZZ
Z aZ a
2σa4
2
2
I=σ
(x + y ) dA = σ
x2 + y 2 dx dy =
.
3
D
0
0
14. Compute the area of the part inside the cylinder x2 + y 2 = 12a2 of the surface obtained by
rotating the parabola z = x2 /a lying in the plane y = 0 about the z-axis.
Solution. The surface in question satisfies the equation z = x2 /a + y 2 /a, with the restriction
that x2 + y 2 ≤ 12a2 in order to be contained inside the cylinder. We can parametrize this
surface by
√
σ(r, θ) = (r cos θ, r sin θ, r2 /a), 0 ≤ r ≤ a 12, 0 ≤ θ ≤ 2π.
Its non-normalized normal vector is given by
2
2
σr × σθ = − 2ra cos θ, − 2ra sin θ, r ,
and hence kσr × σθ k =
p
4r4 /a2 + r2 . Then the surface area of this surface is given by
ZZ
Z
√
a 12 Z 2π
kσr × σθ k dS =
S
0
Z
0
√
a 12
= 2π
r
p
p
4r4 /a2 + r2 dθ dr
4r2 /a2 + 1 dr
0
Z
πa2 49 √
u du,
4 1
= 57πa2 .
=
u = 4r2 /a2 + 1
15. Give an example of a curl-free vector field which is not the gradient of any function.
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Solution. Use the example from class F (x, y, z) = (−y/(x2 + y 2 ), x/(x2 + y 2 ), 0). We can
easily calculate that the curl of F is ~0. Now, let C be the unit circle x2 + y 2 = 1, z = 0. Then,
using the standard polar parametrization of the curve,
Z 2π
I
(− sin θ, cos θ) · (− sin θ, cos θ) dθ = 2π.
F · dr =
0
C
Now, suppose that F = ∇f for some scalar function f . Then, since C is a closed curve, the
fundamental theorem of calculus implies that
I
I
∇f · dr = f (1, 0, 0) − f (1, 0, 0) = 0,
F · dr =
C
C
a contradiction to our previous calculation. Therefore it is impossible that F = ∇f .
16. A smooth vector field F~ has zero flux across any closed surface. Prove that div F~ = 0.
Solution. Let B(p, r) be the ball of radius r about a center p. Then, by Gauss’s theorem,
ZZ
ZZZ
F~ · dS =
div F~ dV = 0,
∂B(p,r)
B(p,r)
which holds for every ball B(p, r). By the mean value theorem, there exists a point p0 in
B(p, r) such that div F~ (p0 ) = 0. But since this holds for every r, then by fixing p constant
but taking r → 0, we demonstrate that div F~ (p) = 0 for every p, which implies that F~ has
zero divergence everywhere.
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