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MATH 20 (Spring 2014) Final Exam - Sketch of Solutions
1. (a) Let f (x, y) be a differentiable function of x and y. Suppose y is a function of x such that f (x, y) = 0, show
that:
dy
f
=− x
dx
fy
/5
wherever fy , 0. [Hint: Use the multivariable chain rule.]
Solution: See Textbook P.779-780
(b) Given that:
Using part (a), find
y 2 − x2 − sin(xy) = 0.
/5
dy
wherever it is defined. Your answer can be in terms of both x and y.
dx
Solution: Straight-forward.
2. Consider the function
f (x, y) = x2 − 4x + y 2 + 9.
(a) Find the maximum and minimum values of f (x, y) on the ellipse
/6
4x2 + 9y 2 = 36.
Solution: Use Lagrange’s Multiplier. Let g(x, y) = 4x2 + 9y 2 . Solve the system ∇f = λ∇g under the
constraint g(x, y) = 36:
2x − 4 = 8λx
2y = 18λy
2
4x + 9y 2 = 36
Note that you can’t simply cancel out y on both sides of the second equation, because y may be 0.
You should divide into two cases: (i) y = 0; and (2) y , 0.
Case (i) gives you two candidate points by substituting y = 0 into the three equation: (x, y) = (3, 0)
or (x, y) = (−3, 0).
Case (ii) gives λ = 1/9. After substituting this into the first equation you can find the value of x.
However, by substituting this x into the three equation you will find there is no real solution for y.
Therefore, there is no candidate points in this case.
Evaluate the two candidate points: f (3, 0) = 6 (the min) and f (−3, 0) = 30 (the max).
(b) Find the absolute maximum and minimum values of f (x, y) on the region:
4x2 + 9y 2 ≤ 36.
Solution: Lagrange’s Multiplier deals with the boundary ellipse (which was done in (a)). In this
case, we set ∇f = 0 to solve for interior critical point(s). One should get only (x, y) = (2, 0) which is
in the interior. f (2, 0) = 5 which is less than the min on the boundary. Therefore, the min and max
over the region are: 5 and 30.
/4
3. Let C be the path of a particle travelling from (x, y, z) = (0, 0, 0) to (x, y, z) = −π2 , 0, 54 π5/2 . The cylindrical
coordinates of the particle at time t are given by
r = t2,
θ = t,
z=
4 5/2
t .
5
(a) Parametrize C in xyz-coordinates, i.e. express x, y and z in terms of t so that (x, y, z) are the rectangular
coordinates of the particle at time t. Express your final answer in vector form
r(t) =? i + ? j + ? k,
/3
? ≤ t ≤?
Solution: Recall that the cylindrical to (x, y, z) conversion is given by:
x = r cos θ,
y = r sin θ,
z = z.
Therefore, the parametrization of the curve C is:
4
r(t) = t 2 cos t i + t 2 sin t j + t 5/2 k,
5
0 ≤ t ≤ π.
(b) Compute, and simplify as much as possible, the speed function |r0 (t)| of the path, where r(t) is the
parametrization found in the previous part.
/5
Solution: Straight-forward computation. Group all sin2 t and cos2 t pairs together. Answer:
q
q
|r0 (t)| =
...
= t 2 (t 2 + 4t + 4) = t 2 (t + 2)2 = t(t + 2).
|{z}
computations
(c) Find the arc length of the curve C.
/2
Solution: Apply the arc length formula:
Z
π
0
Z
π
|r (t)|dt =
0
Z
π
t(t + 2)dt =
0
t 2 + 2tdt =
0
π3
+ π2 .
3
4. (a) Find the equation for the plane containing the lines
r1 (t) = h1, 2, 3i + th1, −1, −1i,
r2 (s) = h2 + s, 3 − 3s, −2 + 3si.
Solution: Parametrization of a straight-line in vector form is given by: x0 +tv where x0 is one of the
point on the line, and v is the direction of the line. Therefore, the direction of line r1 is h1, −1, −1i
and that of line r2 is h1, −3, 3i. These two directions are parallel to the plane and therefore their
cross product is a normal vector to the plane. Therefore,
n = h1, −1, −1i × h1, −3, 3i = h−6, −4, −2i.
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Pick one of the point on the plane, say r1 (0) = h1, 2, 3i. The equation of the plane is:
−6x − 4y − 2z = (−6)(1) + (−4)(2) + (−2)(3) = −20.
After simplication: 3x + 2y + z = 10.
(b) There is one (and only one) sphere x2 + y 2 + z2 = a2 centered at the origin which intersects the above
plane at exactly one point. Find the radius a of this sphere.
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Solution: The plane above must be a tangent plane to the sphere. Denote the contact point by Q,
then a = |OQ| and the vector OQ is normal to the plane. Since we have already found the normal
vector of the plane, which is h3, 2, 1i, we know
OQ
h3, 2, 1i
.
= ±n̂ = ± √
|OQ|
14
Pick one of the point on the plane, say P (1, 2, 3). Then, the projection of OP onto the normal vector
of the plane is given by OQ (sketch a diagram to make sense of this!). Therefore,
√
h3, 2, 1i 5 14
a = |OQ| = |OP | cos ∠P OQ = |OP · n̂| = h1, 2, 3i · √
=
.
7
14
5. (a) Evaluate the double integral:
/6
Z 4Z
0
2
√
x3
e dx dy
y
3
Solution: Not possible to integrate ex by x, so you need to switch the order of integration to dy dx.
Sketch the region using the lower/upper limits given. Rewrite the integral using the dy dx order.
2
(b) Evaluate the surface integral:
/6
1
0
"
-1
-2
(x2 + y 2 ) dσ ,
2.0
S
1.5
1.0
p
where S is the portion of the cone z = x2 + y 2
with 1 ≤ z ≤ 2, as shown in the figure.
-2
-1
0
1
2
Solution: The term x2 + y 2 suggests that it’s best to use cylindrical coordinates. The cone is given
by the equation z = r. Parametrize the cone by:
r(r, θ) = hr cos θ, r sin θ, ri,
1 ≤ r ≤ 2,
The rest is straight-forward computation of a surface integral.
Page 3
0 ≤ θ ≤ 2π.
Z
6. (a) Show that
1
r
dr = −
+ constant.
2
2
(4 + r )
2(4 + r 2 )
/2
Solution: Single-variable calculus substitution: Let t = 4 + r 2 .
Z
∞
(b) Show that
−∞
Z
∞
1
du dv = 2π
2
2 2
−∞ (4 + u + v )
Z
∞
0
r
dr.
(4 + r 2 )2
/4
Solution: Use polar coordinates: Let u = r cos θ and v = r sin θ.
(c) Evaluate:
Z
∞
−∞
Z
/6
∞
−∞
1
dx dy.
4 + (x − y)2 + (x + 2y)2 2
Solution: Do a change of variable: Let u = x − y and v = x + 2y, then use the results in (b) and (a) to
complete the problem.
I
7. (a) Calculate the circulation
F · dr of the vector field:
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T
F(x, y) = −y + sinA x i + 2x − ln(y B + 1) j
around the clockwise oriented triangle T with vertices (0, 0), (2, 0), and (1, 5). Here A is the sum of
David Wigyul’s and Edward Newkirk’s ages, and B is the last six digits of your Banner’s ID.
Solution: It is almost impossible to calculate this circulation directly (not possible even for Mathematica). Since T is a 2D closed path, and F is defined everywhere in T (here we assume that A
and B are positive, of course), the only way out (within the course syllabus) is to use the Green’s
Theorem.
By applying the Green’s Theorem, you should get a double integral constant 3 as the integrand, so
the answer is simply 3 times the area of the triangle T . Note also that T is oriented clockwise so
you should negate the final answer.
Z
(b) Find the line integral
from t = − π2 to
π
2,
H · dr, where H(x, y) = y 2 + x i + (2xy − 3) j, C is the path r(t) = ti + (cosN t)j
C
and N is the 4-digit pin of Frederick Fong’s ATM card.
Solution: Again, some prior experience from single-variable calculus should tell you that integrating a higher power of trig functions is usually not possible by hand. Even with Mathematica, it will
take a very long time.
Now C is NOT a closed path so you can’t use the Green’s Theorem, but a simple calculation of ∇×H
should tell you that H is conservative. Therefore, the line integral is path-independence, and so
you can choose your favorite path, with the same end-points, to evaluate this line integral.
Needless to say, the easiest path should be the straight-line connecting − π2 , 0 and π2 , 0 , i.e.
r(t) = ti + 0j,
−
π
π
≤t≤ .
2
2
The line integral of H along this path can be calculated easily.
Alternatively, you may also find the potential function of H, then you may simply substract values
the potential function at the end-points to find the line integral.
Page 4
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8. (a) Find the volume of the solid D in R3 bounded on
top by part of the spherepx2 +y 2 +z2 = 4 and on the
bottom by the cone z = x2 + y 2 , as shown in the
figure.
1
0
-1
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2.0
1.5
1.0
0.5
1
0.0
0
-1
Solution: Use spherical coordinates to set-up a triple integral. The lower/upper limits for each
coordinates are:
π
0 ≤ ρ ≤ 2, 0 ≤ φ ≤ , 0 ≤ θ ≤ 2π.
4
(b) Find the surface flux of the vector field F(x, y, z) = 2xi + 2yj + 2zk through the boundary surface S of the
above solid D.
/4
Solution: It would take a very long time to compute the surface flux since by doing so you would
have to parametrize both the cone and the sphere. Prior experience of computing flux of these two
kind of surfaces should tell you that it’s not possible by hand within a short time.
The boundary surface is closed. It suggests that you should use the Divergence Theorem!
$
F · n̂ dσ =
$
∇ · F dV =
S
6 dV = 6 · answer in (a).
D
D
9. Let C be a simple closed curve on the xy-plane, R be the region enclosed by C in the xy-plane, and S be a
surface above the xy-plane in the xyz-space with boundary curve C. Consider the two vector fields:
F(x, y, z) = yz2 i + xzj + (x2 + y 2 )k
G(x, y, z) = (2y − x)i − (2x − 2yz)j + (z − z2 )k.
(a) Verify that ∇ × F = G.
/4
Solution: Straight-forward.
"
(b) Show that
G · n̂dσ = 0.
/6
S
Solution: Note that S is not!closed (since it has a boundary curve C), you cannot apply the Divergence Theorem directly for S G · n̂ dσ . However, you may glue the planar region R to S to form a
closed surface S + R. Denote D be the solid enclosed by S + R. Apply the Divergence Theorem on
the closed surface S + R, you can get:
$
$
$
G · n̂ dσ =
∇ · G dV =
∇ · (∇ × F) dV =
0 dV = 0.
S+R
D
D
D
Recall that ∇ · (∇ × F) = 0 for any vector field F. By additivity of integrals:
"
"
G · n̂ dσ +
G · n̂dσ = 0.
We next argue that
!
S
R
G · n̂ dσ = 0, then it will imply
R
Page 5
!
S
G · n̂ dσ = 0 as well.
Note that R is a region on the xy-plane, so:
"
"
"
G · n̂ dσ =
G · kdA =
(z − z2 )dA.
Since z = 0 on the xy-plane,
!
R
R
R
R
(z − z2 )dA = 0, as desired.
Alternatively, you may use a combination of the Stokes’ and Green’s Theorems:
"
"
G · n̂ dσ =
(∇ × F) · n̂ dσ
(from (a))
S
I S
=
F · dr
(Stokes’ Theorem)
"C
=
(∇ × F) · k dA
(Green’s Theorem. Note that C is on the xy-plane)
R
"
=
G · k dA
(using (a) again)
R
"
=
(z − z2 ) dA
(definition of G)
R
"
=
0 dA
(z = 0 on the xy-plane)
R
= 0.
y
x
10. (a) Let F(x, y) = − x2 +y 2 i + x2 +y
2 j. It can be verified that ∇ × F = 0 for any point (x, y) , (0, 0). It seems like
the Green’s Theorem asserts that for any simple closed curve C enclosing region R, we have:
I
"
F · dr =
(∇ × F) · kdA = 0.
C
/2
R
What is wrong with this claim?
Solution: F is not defined at (0, 0). Green’s Theorem doesn’t hold if the curve C encloses the origin.
(b) The Faraday’s Law of Induction in differential form is stated as:
∇×E = −
/2
∂B
,
∂t
where E is the electric field and B is the magnetic field. Assume E and B are defined everywhere in
R3 . Using the differential form of the Faraday’s Law and one of the Green’s, Stokes’ or Divergence
Theorems, derive the integral form of the Faraday’s Law:
I
"
∂
E · dr = −
B · n̂dσ ,
∂t S
C
where S is a surface in the three dimensional space with simple closed boundary curve C.
Solution:
"
I
"
E · dr =
C
(∇ × E) · n̂ dσ = −
S
S
Page 6
∂
∂
B · n̂ dσ = −
∂t
∂t
"
B · n̂ dσ .
S
Note that the last step relies on the fact that both C and S are stationary and so the normal vector
n̂ is unchanged. In your future E & M class you may deal with a wire C which is moving. In such
case, the result in this problem does not hold.
(c) Let E be the electric field defined everywhere in R3 . Assume B = 0 everywhere, i.e. no magnetic field.
Explain briefly, using the Faraday’s law ∇×E = − ∂B
, why there is a potential function ϕ(x, y, z) such that
∂t
E = ∇ϕ.
/2
Solution: B = 0 so ∇ × E = 0 according to the Faraday’s Law. Therefore, E is conservative by the
curl-test. There exists a potential function ϕ such that E = ∇ϕ.
(d) One of the four Maxwell’s Equations asserts that ∇ · B = 0 where B is the magnetic field. Explain briefly
why this model rules out the possibility of magnetic monopole, i.e. a magnet with only one pole (either
south or north).
/2
Solution: If a magnetic monopole exists, then the magnetic field will be radial from the magnet
(like the electric field of a charge). Should it happen, the magnetic flux through a closed surface S
(such as a sphere) enclosing the magnet must be non-zero. It will violate the law ∇ · B = 0 since the
Divergence Theorem asserts that:
$
B · n̂ dσ =
∇ · B dV = 0.
S
D
Therefore, the magnetic flux through S must be zero.
(e) Let f be a scalar function and F be a vector field. Determine whether each of the following quantities
is a vector or a scalar. Circle the correct answer.
i.
ii.
iii.
iv.
/2
∇f is a vector.
∇ · F is a scalar.
∇ × F is a vector.
∇ × ∇f is a vector.
(f) Explain why the Green’s Theorem is regarded as the two-dimensional case of the Stokes’ Theorem.
Solution:
"
I
Green’s Theorem :
F · dr =
IC
Stokes’ Theorem :
(∇ × F) · k dA
"R
F · dr =
C
(∇ × F) · n̂ dσ .
S
Green’s Theorem is the two-dimensional case of the Stokes’ Theorem because when applying the
Stokes’ Theorem on a closed path C on the xy-plane, the unit normal vector is given by k. The
surface S is a region R on the xy-plane and so dσ = dA.
Page 7
/2
11. Suppose f (x, y, z) is a function such that
fxx + fyy + fzz = 0
where fxx , fyy and fzz are the second partial derivatives with respect to x, y and z respectively.
(a) The gradient ∇f is a vector field. Multiplying this vector field by the scalar f (x, y, z) gives another vector
field f ∇f . Show that this new vector field’s satisfies the identity
∇ · (f ∇f ) = |∇f |2 .
[Hint: f ∇f = f
∂f
∂x
i+ f
∂f
∂y
j+ f
∂f
∂z
/4
(*)
k. Be careful that some of the ∇’s mean the gradient of a scalar
function, while some of them mean the divergence of a vector field.]
Solution:
∇ · (f ∇f ) =
!
!
!
∂f
∂f
∂f
∂
∂
∂
f
+
f
+
f
∂x ∂x
∂y ∂y
∂z ∂z
(by the definition of divergence)
= (fx fx + f · fxx ) + (fy fy + f · fyy ) + (fz fz + f · fzz )
= fx2 + fy2 + fz2 + (fxx
= fx2 + fy2 + fz2 + 0
+ fyy + fzz )
(product rule)
(by rearranging the terms)
(given)
On the other hand, |∇f |2 = |fx i + fy j + fz k|2 = fx2 + fy2 + fz2 . Therefore, ∇ · (f ∇f ) = |∇f |2 as desired.
(b) Let D be a solid region bounded by a closed surface S with outward unit normal n̂. Using the identity
(*) from the previous part, show that
"
$
f ∇f · n̂ dσ =
|∇f |2 dV .
S
/3
D
Solution: Integrate both sides of (*) on D, we get:
$
$
∇ · (f ∇f )dV =
|∇f |2 dV .
D
D
Then apply Divergence Theorem on the left-hand side:
$
∇ · (f ∇f )dV =
(f ∇f ) · n̂dσ .
D
S
These prove the desired result.
(c) Using the result of the previous part, argue that if f = 0 everywhere on the boundary surface S, then in
fact f = 0 everywhere in D.
#
|∇f |2 dV = 0.
(f
∇f
)
·
n̂dσ
=
0dσ
=
0.
Using
(b),
we
also
have
S
S
D
!
Since |∇f |2 is non-negative, the integral D |∇f |2 dV = 0 happens only if |∇f | = 0 on D. Therefore,
∇f = 0 which implies f must be constant on D. Since f = 0 on S which is the boundary of D, by
continuity this constant must be 0.
Therefore f = 0 on D.
Solution: If f = 0 on S, then
Page 8
/3
(d) Recall from one of the homework problems that the temperature distribution u satisfies the following
heat equation:
∂u
uxx + uyy + uzz =
,
∂t
where t is the time variable. Suppose the temperature at every point is unchanged over time (i.e. steady
state), and that the temperature is zero on the shell of a sphere. Base on the results of previous part(s),
what do you know about the temperature inside the sphere?
Solution: When u is unchanged over time, we have ∂u
= 0, and so uxx + uyy + uzz = 0, which is
∂t
exactly the condition on f at the beginning of the problem.
Given further that u = 0 on the shell of a sphere, using (c) we know that u = 0 in the interior of the
sphere as well. Therefore, the temperature inside the sphere must be 0.
Page 9
/2