Name
ID
Math 311
Final Exam
Spring 2002
Section 503
P. Yasskin
1 EC
/10 3
/30
3
/40 4
/30
Solutions
x
1.
10 points Extra Credit
Determine if and where the line
1
=
y
z
x
intersects the plane
2
=
y
1
−8
+r
3
z
4
+t
2
3
5
6
3
+s
−1
.
6
4
2
Equate x, y and z to get the equations
1 + 4t = 2 + r + 3s
r + 3s − 4t = −1
2 + 5t = −8 − r + 6s
or
3 + 6t = 3 + 2r + 4s
−r + 6s − 5t = 10
2r + 4s − 6t = 0
Solve for r, s and t:
3 −4
−1
−1 6 −5
2 4 −6
10
R2 + R1
0
R 3 − 2R 1
1



1 3 −4
−1
0 1 −1
1
0 1 −1
−1
0 = −2

−4
0 9 −9
0 −2 2
1

R3 − R2
Contradiction
3


−1
1
R
9 2
1
R
2 3
9
−
2
1 3 −4
−1
0 1 −1
1
0 0
−2
0
No solutions
The line does not intersect the plane.
1
2.
2 1
 
40 points
be the vector space of p × q matrices, and let P =
Let M p, q
4 2
.
6 3
Consider the linear function
   M3, 2 
and compute LX .
L : M 2, 2

a b
a. 5 Let X =
2 1
a b
L
=
c d

c d
2a + c
a b
4 2
=
c d
6 3

given by L X = PX
2b + d
4a + 2c 4b + 2d
6a + 3c 6b + 3d
b. 2 Identify the domain of L. What is its dimension?

 

Dom L = M 2, 2

dim Dom L = 4
c. 2 Identify the codomain of L. What is its dimension?

 

Codom L = M 3, 2

dim Codom L = 6
d. 8 Identify the kernel (null space) of L. Give a basis and the dimension.

LX =0

2a + c = 0
2b + d = 0
4a + 2c = 0
4b + 2d = 0
6a + 3c = 0
6b + 3d = 0


X=
a
Ker L =
b
1
1
−2 0
,
d = −2b
b
=
0
−2 0
0
c = −2a
−2a −2b
−2a −2b
= Span
Basis is
a

0
,
1
0 −2
1
a
0
−2 0
0
+b
0
1
0 −2
1
0 −2
.

dim Ker L = 2
2
Problem 2 continued:
e. 8 Identify the image (range) of L. Give a basis and the dimension.

2a + c

Im L =
2b + d
4a + 2c 4b + 2d
6a + 3c 6b + 3d
2 0
=
a
0 2
+b
4 0
6 0
0 4
+c
4 0
0 4
6 0
+d
3 0
0 2
,
0 1
2 0
0 6
2 0
= Span
1 0
0 3
1 0
,
2 0
0 6
0 2
0 1
,
3 0
0 2
0 3
(Not Linearly Independent!)
1 0
= Span
2 0
0 1
,
0 2
3 0
1 0
Basis is
2 0
0 3
0 1
,
3 0
.
0 2

dim Im L = 2
0 3

f. 3 Is L onto? Why?




L is NOT onto because dim Im L = 2 but dim Codom L = 6, so Im L ≠ Codom L .

g. 3 Is L one-to-one? Why?
 
L is NOT one-to-one because Ker L ≠ 0 .

h. 2 Check that the dimensions of the kernel and image are consistent with the dimensions of the domain and
codomain.



dim Ker L + dim Im L = 2 + 2 = 4 = dim Dom L
3
Problem 2 continued:
i. 7 Find the matrix of L relative to the bases

1 0
E1 =
E2 =
0 0
0 1
0 0
1 0
F1 =
0 0
F4 =
Recall:
L
 
L E1 = L
 
L E2 = L
 
L E3 = L
 
L E4 = L
=
1 0
0 0
0 1
0 0
0 0
1 0
0 0
0 1
1 0
F2 =
1 0
E4 =
F3 =
1 0
0 1
0 0
0 0
F5 =
0 1
0 0
0 1
0 0
0 0
0 0
0 0
0 0
0 0
2a + c
c d
0 0
0 0
0 0
a b
E3 =
F6 =
0 0
0 1
2b + d
4a + 2c 4b + 2d
6a + 3c 6b + 3d
2 0
=
4 0
= 2F 1 + 4F 2 + 6F 3
6 0
0 2
=
0 4
= 2F 4 + 4F 5 + 6F 6
0 6
1 0
=
2 0
= 1F 1 + 2F 2 + 3F 3
3 0
0 1
=
0 2
= 1F 4 + 2F 5 + 3F 6
0 3
2 0 1 0
4 0 2 0
A=
6 0 3 0
0 2 0 1
0 4 0 2
0 6 0 3
4
3.
30 points On the vector space P 2 =
polynomials given by
polynomials of degree less than 2
consider the function of two
⟨p, q ⟩ = ∫ ∞ pxqxe− dx
x
 
⟨p, q ⟩ is an inner product.
∞
∞
i. ⟨q, p ⟩ = ∫ qx px e − dx = ∫ px qx e − dx = ⟨p, q ⟩
∞
∞
∞
ii. ⟨p, q + r ⟩ = ∫ px q + r x e − dx = ∫ px qx e − dx + ∫ px rx e − dx = ⟨p, q ⟩ + ⟨p, r ⟩
∞
∞
iii. ⟨p, aq ⟩ = ∫ px aq x e − dx = a ∫ px qx e − dx = a⟨p, q ⟩
∞
iv. ⟨p, p ⟩ = ∫ px  e − dx ≥ 0 because px  e − is non-negative. Further
⟨p, p ⟩ = 0  ∫ ∞ px e− dx = 0
 px e− = 0 because px e− is non-negative and continuous
 px = 0
So ⟨p, q ⟩ is an inner product.
15  Find the angle θ between the polynomials px = 1 and qx = x.
0
a. 15 Show
x
x
0
0
x
x
0
0
x
0
x
x
0
0
2
2
x
x
0
2
x
0
2
b.
2
x
x
You may use these integrals without proof:
∞
∞
∫ e −x dx = 1, ∫ xe −x dx = 1,
0
0
∞
∫ x 2 e −x dx = 2,
0
∞
∫ x 3 e −x dx = 6
0
⟨p, q ⟩ = ∫ ∞ 1 ⋅ x ⋅ e− dx = 1 ⟨p, p ⟩ = ∫ ∞ 1 ⋅ 1 ⋅ e− dx = 1 ⟨q, q ⟩ = ∫ ∞ x ⋅ x ⋅ e− dx = 2
⟨p, q ⟩ = 1 = 1  θ = 45° = π
cos θ =
4
|p||q|
1⋅ 2
2
x
0
x
0
x
0
5
4.
30 points Gauss’ Theorem states that if V is a volume in R 3 and ∂V is its boundary surface oriented
⃗ is a nice vector field on V then
outward from V and F
⋅
⋅
⃗ dV = ∫∫ F
⃗ dS
∫∫∫ ⃗∇ F
∂V
V

⃗ = xz 2 , −yz 2 , x 2 z + y 2 z
Verify Gauss’ Theorem if F

and V is the volume above the paraboloid
z = x 2 + y 2 and below the plane z = 9.
Notice that ∂V consists of the paraboloid P
and a disk D.
Be sure to use the correct
orientations for P and D.
⋅

⃗ dV :
a. 6 Compute ∫∫∫ ⃗
∇ F
V
⋅
⋅
= ⃗∇
⃗=
i. ⃗
∇ F
⃗∇ F
⃗ = r2
⋅ xz , −yz , x z + y z  = z
2
2
2
2
2
− z2 + x2 + y2 = x2 + y2
(in cylindrical coordinates)
⋅
⃗ dV =
ii. ∫∫∫ ⃗
∇ F
V
2π
3
9
0
0
r
 
= ∫ ∫ ∫ 2 r 2 r dz dr dθ = 2π ∫


3
3
4
6
= 2π ∫ 9r 3 − r 5 dr = 2π 9r − r
0
4
6
6
= 3 π = 243π
6
2
= 36π 1 − 1
2
3
0
3
0
9
r3z
⋅4
dr
r2
4
6
= 2π 9 3 − 3
6
⋅
 
⃗ dS :
b. 10 For the paraboloid P compute ∫∫ F
 

⃗ r, θ =
i. R
= r cos θ, r sin θ, r 2
⃗r =
ii. R
=
⃗θ =
iii. R
= −r sin θ, r cos θ, 0
cos θ,


P
sin θ, 2r
 
 

 
⃗=
= î −2r 2 cos θ − ̂ 2r 2 sin θ + k̂ r cos 2 θ + r sin 2 θ = −2r 2 cos θ, −2r 2 sin θ, r
iv. N
This points up. We need it down. Reverse it.
⃗ = 2r 2 cos θ, 2r 2 sin θ, −r
N
v.
vi.
vii.

⃗ R
⃗ r, θ  =
F
= r cos θ, −r sin θ, r 
⃗ R
⃗ r, θ  ⋅ N
⃗=
= 2r cos θ − 2r sin θ − r
F
⃗ ⋅ dS =
= ∫ ∫ 2r cos θ − 2r sin θ − r  dr dθ = ∫
∫∫ F
5
5
7
P
2π
3
0
0
2
7
3
2π
0
0
= ∫ 2r 7 sin 2θ − r 5 θ
2
4
2
7
2

5
2
7
3
5
6
dr = −2π ∫ r 5 dr = −2π r
6
0
3
0
3
0
∫
2π
0
2r cos 2θ − r  dθ dr
7
5
= −3 5 π = −243π
6
Problem 4 continued:

⋅
 

⃗ = xz 2 , −yz 2 , x 2 z + y 2 z .
Recall F
⃗ dS :
c. 10 For the disk D compute ∫∫ F
D
 

⃗ r, θ =
i. R
= r cos θ, r sin θ, 9

⃗r =
ii. R
=
cos θ,
⃗θ =
iii. R
=
−r sin θ, r cos θ, 0

⃗=
iv. N
= 0, 0, r
  =
⃗ R
⃗ r, θ  ⋅ N
⃗=
F
⃗ R
⃗ r, θ
v. F
vi.
⋅
⃗ dS =
vii. ∫∫ F

sin θ, 0
This points up which is correct.

= 81r cos θ, 81r sin θ, 9r 2

= 9r 3
2π
3
0
0
 
= ∫ ∫ 9r 3 dr dθ = 2π 9r
4
D
4
3
0
6
= 3 π = 729π
2
2

d. 4 Verify the two sides of Gauss’ Theorem are equal.
⋅
⃗ dV = 243π
∇ F
i. ∫∫∫ ⃗
2
V
⋅
⋅
⃗ dS + ∫∫ F
⃗ dS = −243π + 729π = 729π − 486π = 243π
ii. ∫∫ F
2
2
2
P
D
iii. They are equal.
7