Name
ID
MATH 311
Final Exam
Fall 2001
Section 200
Solutions
P. Yasskin
1.
1
/10
4
/20
2
/20 5-7
/30
3
/20
10 points Let P 02 be the subset of P 2 consisting of those polynomials of degree 2 or less whose
constant term is zero. In particular
P 02 = p = ax + bx 2 
a. 8  Show P 02 is a subspace of P 2 .
Let p, q ∈ P 02 . Then p = ax + bx 2 and q = cx + dx 2 . So p + q = a + c x + b + d x 2 ∈ P 02 . Further
kp = ka x + kb x 2 ∈ P 02 . So P 02 is closed under addition and scalar multiplication and so is a
subspace.
b. 2  What is the dimension of P 02 ? Why?
dim P 02 = 2 because a basis is x, x 2  which has 2 vectors.
1
2.
20 points Again let P 02 be the subset of P 2 consisting of those polynomials of degree 2 or less
whose constant term is zero. Consider the function ⟨∗, ∗ ⟩ of two polynomials given by
1 4pq
dx.
⟨p, q ⟩ = ∫
2
0 x
a. 5  Show the function ⟨∗, ∗ ⟩ is an inner product on P 02 .
1 4pq
4qp
dx = ∫
dx = ⟨p, q ⟩
2
2
0 x
0 x
1 4paq + br 
1 4pq
1 4pr
ii. bilinear:
dx = a ∫
dx + b ∫
dx = a⟨p, q ⟩ + b⟨p, r ⟩
⟨p, aq + br ⟩ = ∫
2
2
2
0
x
0 x
0 x
1 4p 2
iii. positive definite:
dx ≥ 0 because the integral of a non-negative quantity is
⟨p, p ⟩ = ∫
2
0 x
non-negative.
1 4p 2
4p 2
dx = 0, then 2 = 0, or p = 0.
If ⟨p, p ⟩ = ∫
2
0 x
x
i. symmetric:
⟨q, p ⟩ = ∫
1
b. 10  Apply the Gram Schmidt procedure to the basis p 1
= x, p 2 = x 2 to produce an
orthogonal basis q 1 , q 2 and an orthonormal basis r 1 , r 2 .
q1 = p1 = x
⟨q 1 , q 1 ⟩ = ∫
1
4q 21
4x 2 dx = ∫ 1 4 dx = 4x|1 = 4
dx
=
∫
0
2
x2
0 x
0
1
0
|q 1 | = 2
q
r1 = 1 = x
2
|q 1 |
1 4p q
1
2 1
dx = ∫
⟨p 2 , q 1 ⟩ = ∫
2
0
x
0
⟨p 2 , q 1 ⟩
2
q1 = x −
q2 = p2 −
⟨q 1 , q 1 ⟩
4x 2 x dx = ∫ 1 4x dx = 2x 2 |1 = 2
0
x2
0
2 x = x2 − x
4
2
2
x
2
4 x− 1
1 4q 2
1 4 x −
1
2
2
2
2
1
dx = ∫
dx = ∫ 4 x −
dx =
⟨q 2 , q 2 ⟩ = ∫
2
2
3
0 x
0
x2
0
3
1
= 1 − −1 = 1
6
6
3
0
|q 2 | =
1
3
q2
=
|q 2 |
Summary:
r2 =
3 x2 − x
2
q 1 = x,
q2 = x2 − x
2
c. 5  Find the change of basis matrices
r1 =
r2 =
C

r p
=
r1 = x ,
2
C

= 1 p 1 + 0p 2
x
2
2
C

p r
−1
=
3
0
3
2
1
2

p r
=
C .

1 − 3
2
2
0
3
2
1
3
2
3 x2 − x
2
p r
C=
= − 3 p1 + 3 p2
3 x2 − x
2
and
r p
r2 =
2 1
0
1
3
2
3.
20 points Again let P 02 be the subset of P 2 consisting of those polynomials of degree 2 or less
whose constant term is zero. Consider the function L : P 02  P 2 given by
dp
L p  = p −
.
dx
a. 4  Show the function L is linear.
Lap + bq  = ap + bq  −
dap + bq 
= a p − dp
dx
dx
+ b q − dq
dx
= aLp  + bLq 
b. 6  Find the kernel of L. Give a basis.
Let p = ax + bx 2 . Then Lp  = 0 says
dax + bx 2 
= ax + bx 2  − a + 2bx  = −a  + a − 2b x + bx 2 = 0.
ax + bx 2  −
dx
This implies a = b = 0, and so p = 0.
Therefore KerL  = 0 .
There is no basis.
c. 6  Find the image of L. Give a basis.
dax + bx 2 
= ax + bx 2  − a + 2bx  = ax − 1  + bx 2 − 2x 
dx
Therefore ImL  = ax − 1  + bx 2 − 2x  = Spanx − 1, x 2 − 2x 
Basis is x − 1, x 2 − 2x .
Lp  = ax + bx 2  −
d. 2  Is L onto? Why?
L is not onto because CodomL  = P 2 while ImL  = Spanx − 1, x 2 − 2x  ≠ P 2
e. 2  Is L one-to-one? Why?
L is one-to-one because KerL  = 0 .
3
4.
20 points Again let P 02 be the subset of P 2 consisting of those polynomials of degree 2 or less
whose constant term is zero. Again consider the function L : P 02  P 2 given by
dp
L p  = p −
.
dx
a. 10  Find the matrix of L relative to the bases
and
e 1 = 1, e 2 = x, e 3 = x 2 for P 2 .
p 1 = x, p 2 = x 2 for P 02
Call it A.
Lp 1  = Lx  = x − dx = x − 1 = −e 1 + e 2
dx
2
Lp 2  = Lx 2  = x 2 − dx = x 2 − 2x = −2e 2 + e 3
dx
−1 0
1 −2
A=

e p
0
1
b. 5  Find the matrix of L relative to the bases
and
e 1 = 1, e 2 = x, e 3 = x 2 for P 2
r 1 , r 2 for P 02
is the orthonormal basis you found in problem 2. Call it B.
where r 1 , r 2
B

e r
=
A
C
e p
p r


−1 0
1 −2
=
0
1 − 3
2
2
0
3
1
3
2
−1
2
=
1
2
0
−
5 3
2
3
c. 5  Recompute B by another method.
L r 1  = L x
2
L r 2  = L
= x −
2
3 x2 − x
2
= 3 x2 − x
2
dx
2 = x − 1 = −1e + 1e
2
2
2 1 2 2
dx
= 3 x2 − x
2
− 3 2x − 1
2
=
d 3 x2 − x
2
dx
3
5 3
e −
e + 3 e3
2 1
2 2
−
3
2
−1
2
B=

e r
1
2
0
−
5 3
2
3
4
5.
30 points
Do this problem, if you did the Volume of Desserts or Planet X Project.
Find the z-component of the center of mass of the apple
whose surface is given in spherical coordinates by
ρ = 1 − cos ϕ
and whose density is 1.
HINT: The ϕ-integrals can be done using the substitution
u = 1 − cos ϕ.
x = ρ sin ϕ cos θ
u = 1 − cos ϕ
dV = ρ 2 sin ϕ dρ dϕ dθ
y = ρ sin ϕ sin θ
du = sin ϕ dϕ
z = ρ cos ϕ
M = ∫∫∫ 1 dV = ∫
2π
0
π
1−cos ϕ
0
0
∫ ∫
ρ 2 sin ϕ dρ dϕ dθ = 2π ∫
π
1 − cos ϕ  4
= 2π ∫ 1 − cos ϕ  3 sin ϕ dϕ = 2π 
3 0
3
4
z − mom = ∫∫∫ z dV = ∫
2π
0
π
1−cos ϕ
0
0
∫ ∫
π
ρ3
0
3
π
0
1−cos ϕ
2 4
= 2π   = 8π
3
5
3
15
4
ρ cos ϕ ρ 2 sin ϕ dρ dϕ dθ = 2π ∫
2
π
= π ∫ 1 − cos ϕ  4 cos ϕ sin ϕ dϕ = π ∫ u 4 1 − u  du = π
2 0
2 0
2
5
2
π
1
1
32
π
4 3−5
=
−
=2 π
=−
2
sin ϕ dϕ
0
3
π
ρ4
0
4
u5 − u6
5
6
1−cos ϕ
0
2
0
cos ϕ sin ϕ dϕ
= π
2
25 − 26
5
6
15
z̄ = z − mom = − 32π 3 = − 4
M
5
15 8π
5
6.
30 points
Do this problem, if you did the Interpretation of Div and Curl Project.
⃗ = xz 2 , yz 2 , 0  at the point x, y, z  = 0, 0, c .
Find the divergence of the vector field F
a. by using the derivative defintion:
⃗⋅F
⃗ = z 2 + z 2 = 2z 2
∇
⃗⋅F
⃗
∇
0,0,c
= 2c 2
b. by using the integral definition:
HINTS: For a sphere of radius ρ centered at a, b, c , if you use standard spherical coordinates,
the normal vector is
⃗=
N
ρ 2 sin 2 ϕ cos θ, ρ 2 sin 2 ϕ sin θ, ρ 2 cos ϕ sin ϕ
The ϕ-integral can be done using the substitution u = cos ϕ.
You can ignore terms in the integral proportional to ρ n with n > 3 since they drop out of the limit.
⃗⋅F
⃗
∇
0,0,c
= lim
3
 4πρ 3
ρ 0
⃗ ⋅ dS⃗
∫∫ F
where the integral is over the sphere of radius ρ centered at 0, 0, c .
⃗ ϕ, θ  = ρ sin ϕ cos θ, ρ sin ϕ sin θ, c + ρ cos ϕ 
R
⃗ = ρ 2 sin 2 ϕ cos θ, ρ 2 sin 2 ϕ sin θ, ρ 2 cos ϕ sin ϕ
N
(Given)
⃗ = xz 2 , yz 2 , 0  = ρ sin ϕ cos θc + ρ cos ϕ  2 , ρ sin ϕ sin θc + ρ cos ϕ  2 , 0
F
⃗⋅N
⃗ = ρ 2 sin 2 ϕ cos θρ sin ϕ cos θc + ρ cos ϕ  2 + ρ 2 sin 2 ϕ sin θρ sin ϕ sin θc + ρ cos ϕ  2
F
= ρ 3 sin 3 ϕc + ρ cos ϕ  2
π
2π
0
0
⃗ ⋅ dS⃗ = ∫ ∫
∫∫ F
π
2π
0
0
⃗⋅N
⃗ dθ dϕ = ∫ ∫
F
π
ρ 3 sin 3 ϕc + ρ cos ϕ  2 dθ dϕ = 2πρ 3 ∫ sin 3 ϕc + ρ cos ϕ  2 dϕ
Drop terms of order greater than ρ 3 :
0
π
π
0
0
⃗ ⋅ dS⃗ ≈ 2πρ 3 ∫ c 2 sin 3 ϕ dϕ = 2πρ 3 c 2 ∫ 1 − cos 2 ϕ  sin ϕ dϕ
∫∫ F
⃗ ⋅ dS⃗ ≈ −2πρ 3 c 2 ∫ −1 1 − u 2  du = −2πρ 3 c 2 u − u 3
∫∫ F
3
1
⃗⋅F
⃗
∇
0,0,c
= lim
3
 4πρ 3
ρ 0
8πρ 3 c 2
3
−1
1
u = cos ϕ
= −2πρ 3 c 2 − 2
3
du = − sin ϕ dϕ
+ 2πρ 3 c 2 2
3
=
8πρ 3 c 2
3
= 2c 2
6
7.
30 points
Do this problem, if you did the Gauss’ and Ampere’s Laws Project.
Find the total charge in the cylinder
x2 + y2 ≤ a2,
⃗ = r̂ = ⃗r =
E
r
r2
where ⃗r = x, y, 0  and r =
0≤z≤1
if the electric field is
y
x ,
,0
2
2
2
x + y x + y2
x2 + y2 .
a. using the derivative form of Gauss’ Law.
y
∂
x
⃗= 1
ρ = 1 ∇⃗ ⋅ E
+ ∂
4π
4π ∂ x x 2 + y 2
∂y x 2 + y 2
x 2 + y 2  − x2x 
x 2 + y 2  − y2y 
= 1
+
2
2
4π
x 2 + y 2 
x 2 + y 2 
2x 2 + y 2  − 2x 2 − 2y 2
= 1
=0
2
4π
x 2 + y 2 
Q = ∫∫∫ ρ dV = 0
C
b. using the integral form of Gauss’ Law.
⃗ ⋅ dS⃗ = ∫∫ E
⃗ ⋅ dS⃗ + ∫∫ E
⃗ ⋅ dS⃗ + ∫∫ E
⃗ ⋅ dS⃗
4πQ = ∫∫ E
∂C
bottom
top
sides
⃗ = ±k̂ while E
⃗ is horizontal. So E
⃗⋅N
⃗ = 0 and
On the ends of the cylinder, the normal is N
⃗ ⋅ dS⃗ = ∫∫ E
⃗ ⋅ dS⃗ = 0
∫∫ E
bottom
top
The sides are parametrized by Rθ, z  = a cos θ, a sin θ, z .
⃗θ =
R
−a sin θ, a cos θ, 0
⃗ z =  0,
R
⃗=
E
0,
x
2
,
y
1
,0
x + y x + y2
⃗⋅N
⃗ = cos 2 θ + sin 2 θ = 1
E
2
2
⃗=R
⃗θ × R
⃗ z = a cos θ, a sin θ, 0 
N
=
a cos θ , a sin θ , 0
a2
a2
=
⃗ ⋅ dS⃗ = ∫ ∫ E
⃗⋅N
⃗ dθ dz = ∫ 1 ∫ 2π 1 dθ dz = 2π
4πQ = ∫∫ E
sides
0
0
cos θ , sin θ , 0
a
a
Q= 1
2
c. What do these results tell you about the location of the electric charge? Why?
⃗⋅E
⃗ and ∇
⃗ are defined which is everywhere
Part (a) says ρ = 0. So there is no charge wherever E
1
but r = 0 which is the z-axis. However, part (b) says Q = . So there must be charge along the
2
z-axis.
7