W. Liu - Training to all new Teaching Assistants
Linear Algebra
Wanmin Liu
wanminliu@gmail.com
Department of Mathematics
Hong Kong University of Science and Technology
2012 August
This is a tutorial demo training to all new Teaching Assistants. It contains ten
typical problems in Linear Algebra. The new TA would be randomly assigned a
problem, and present the problem on blackboard to other new TAs. Pros and
Cons of each presentation would be highlighted and discussed after the presen-
tation. Typical teaching techniques/tricks/mistakes would also be emphasised.
Problem 1. Solve the following system of linear equations:
2x
1
+7x
2
+3x
3
+x
4
= 6
3x
1
+5x
2
+2x
3
+2x
4
= 4
9x
1
+4x
2
+x
3
+7x
4
= 2
Keywords. Elementary row operations, rank, solution structures of linear system of equa-
tions.
Suggested Solution. Let A be the coefficient matrix. We do elementary row operations to the
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augmented matrix
¯
A:
¯
A =
2 7 3 1 6
3 5 2 2 4
9 4 1 7 2
→
−1 2 1 −1 2
0 11 5 −1 10
0 −11 −5 1 −10
→
1 −2 −1 1 −2
0 1
5
11
−
1
11
10
11
0 0 0 0 0
→
1 0 −
1
11
9
11
−
2
11
0 1
5
11
−
1
11
10
11
0 0 0 0 0
.
Since rank(A) = rank(
¯
A) = 2 < n = 4, the system has infinite many solutions. Let x
3
= x
4
= 0,
we find x
1
=−
2
11
, x
2
=
10
11
, and a special solution
η =
−
2
11
10
11
0
0
.
Let x
3
=1, x
4
=0, we obtain x
1
=
1
11
, x
2
=−
5
11
. Let x
3
=0, x
4
=1, we obtain x
1
=−
9
11
, x
2
=
1
11
.
Let
ξ
1
=
1
11
−
5
11
1
0
, ξ
2
=
−
9
11
1
11
0
1
.
The general solutions of the system are x =η+k
1
ξ
1
+k
2
ξ
2
for arbitrary numbers k
1
and k
2
.
Problem 2. For various cases of the two numbers a and b, find out the solution(s) of the
following system of linear equations:
x
1
+x
2
−x
3
= 1
2x
1
+(a +2)x
2
−(b +2)x
3
= 3
−3ax
2
+(a +2b)x
3
= −3
Keywords. Determinant, Cramer’s rule, elementary row operations.
Suggested Solution. The number of equations and the number of unknowns are the same.
Let A be the coefficient matrix of the system. Then
|A|=
¯
¯
¯
¯
¯
¯
1 1 −1
2 a +2 −(b +2)
0 −3a a +2b
¯
¯
¯
¯
¯
¯
=a(a −b).
We discuss the following three cases.
1). |A| 6= 0, i.e., a 6=0 and a 6=b. In this case, the system has a unique solution. We can use
Cramer’s rule to find it out.
x
2
=
¯
¯
¯
¯
¯
¯
1 1 −1
2 3 −(b +2)
0 −3 a +2b
¯
¯
¯
¯
¯
¯
a(a −b)
=
1
a
, x
3
=
¯
¯
¯
¯
¯
¯
1 1 1
2 a +2 3
0 −3a −3
¯
¯
¯
¯
¯
¯
a(a −b)
=0, x
1
=1 −x
2
+x
3
=1 −
1
a
.
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2). a =0. In this case we do elementary row operations to the augmented matrix:
1 1 −1 1
2 2 −(b +2) 3
0 0 2b −3
→
1 1 −1 1
0 0 −b 1
0 0 2b −3
→
1 1 −1 1
0 0 −b 1
0 0 0 −1
.
So the system has no solution.
3). a =b 6=0. We do elementary row operations to the augmented matrix:
1 1 −1 1
2 a +2 −(a +2) 3
0 −3a 3a −3
→
1 1 −1 1
0 a −a 1
0 −3a 3a −3
→
1 1 −1 1
0 1 −1
1
a
0 0 0 0
.
So
x
1
x
2
x
3
=
1 −
1
a
1
a
+k
k
=
1 −
1
a
1
a
0
+k
0
1
1
,
where k is an arbitrary number.
Problem 3. Let
α
1
=
1
0
2
3
, α
2
=
1
1
3
5
, α
3
=
1
−1
a +2
1
, α
4
=
1
2
4
a +8
, β =
1
1
b +3
5
.
1). For which values of a and b will β not be represented as linear combination of α
1
, α
2
,
α
3
, α
4
.
2). For which values of a and b will β be uniquely represented as linear combination of α
1
,
α
2
, α
3
, α
4
.
Keywords. Linear combination of vectors, solution structure of linear system of equations,
augmented matrix, elementary row operations, rank.
Suggested Solution. By the definition, whether β can be represented as linear combination
of α
1
, α
2
, α
3
, α
4
or not is equivalent to whether the system
x
1
α
1
+x
2
α
2
+x
3
α
3
+x
4
α
4
=β
has solution or not. Let us solve the system:
x
1
+x
2
+x
3
+x
4
= 1
x
2
−x
3
+2x
4
= 1
2x
1
+3x
2
+(a +2)x
3
+4x
4
= b +3
3x
1
+5x
2
+x
3
+(a +8)x
4
= 5
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Let the coefficient matrix be A. We do elementary row operations on the augmented matrix
¯
A:
¯
A =
1 1 1 1 1
0 1 −1 2 1
2 3 a +2 4 b +3
3 5 1 a +8 5
→
1 1 1 1 1
0 1 −1 2 1
0 1 a 2 b +1
0 2 −2 a +5 2
→
1 1 1 1 1
0 1 −1 2 1
0 0 a +1 0 b
0 0 0 a +1 0
.
1). When a =−1, b 6=0, rank(A) =2 <3 =rank(
¯
A), the system has no solution. In this case,
β cannot be represented as linear combination of α
1
, α
2
, α
3
, α
4
.
2). When a 6= −1, for arbitrary b we have rank(A) = rank(
¯
A) = 4, the system has a unique
solution:
x
1
= −
2b
a+1
x
2
= 1 +
b
a+1
x
3
=
b
a+1
x
4
= 0
In this case,
β =−
2b
a +1
α
1
+(1 +
b
a +1
)α
2
+
b
a +1
α
3
.
Problem 4. Let A =
µ
1 −sin(θ)cos(θ) cos
2
(θ)
−sin
2
(θ) 1 +sin(θ)cos(θ)
.
1). Find eigenvalues of A.
2). Find the corresponding eigenvectors.
3). Find an invertible 2 ×2 matrix P such that
˜
A =P
−1
AP is in Jordan normal form.
Keywords. Eigenvalue, eigenvector, Jordan normal form.
Suggested Solution.
1).
|A −λI|=
¯
¯
¯
¯
1 −sin(θ)cos(θ)−λ cos
2
(θ)
−sin
2
(θ) 1 +sin(θ)cos(θ)−λ
¯
¯
¯
¯
=(λ −1)
2
.
The eigenvalue of A is 1, with mutiplicity 2.
2). Let us solve the equation (A −λI )x =0 for λ =1.
µ
−sin(θ)cos(θ) cos
2
(θ)
−sin
2
(θ) sin(θ) cos(θ)
µ
x
1
x
2
=
µ
0
0
=⇒p
1
=
µ
x
1
x
2
=
µ
cos(θ)
sin(θ)
.
So the eigenvector corresponding the the eigenvalue 1 is p
1
(or nonzero scalar multi-
plication of p
1
).
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3). We would like to find a vector p
2
satisfying Ap
2
= p
1
+λp
2
, i.e., (A −λI)p
2
= p
1
, where
λ =1. Solving this equation, we obtain p
2
=
µ
−sin(θ)
cos(θ)
.
Let P =(p
1
, p
2
) =
µ
cos(θ) −sin(θ)
sin(θ) cos(θ)
. Then
AP =(Ap
1
, Ap
2
) =(λp
1
, p
1
+λp
2
) =(p
1
, p
2
)
µ
λ 1
0 λ
=P
µ
λ 1
0 λ
.
So
˜
A =P
−1
AP =
µ
1 1
0 1
.
Problem 5. Let A =
µ
3 4
5 2
.
1). Diagonalize A.
2). Let t be a formal variable, compute (t A)
2
, (t A)
3
.
3). Define exp(t A) =
P
∞
n=0
1
n!
(t A)
n
as the formal sum (regardless the convergence). Com-
pute exp(t A).
Keywords. Matrix diagonalization, eigenvalue, eigenvector, exponential of matrix, applica-
tion of linear algebra.
Suggested Solution.
1). Let us compute the eigenvalues and corresponding eigenvectors.
¯
¯
¯
¯
3 −λ 4
5 2 −λ
¯
¯
¯
¯
=λ
2
−5λ −14 =(λ +2)(λ −7),=⇒λ
1
=−2, λ
2
=7.
For λ
1
=−2, we have
µ
5 4
5 4
µ
x
1
x
2
=
µ
0
0
,=⇒
µ
x
1
x
2
=
µ
4
−5
.
For λ
1
=7, we have
µ
−4 4
5 −5
µ
x
1
x
2
=
µ
0
0
,=⇒
µ
x
1
x
2
=
µ
1
1
.
Let P =
µ
4 1
−5 1
, we have
P
−1
AP =
µ
−2 0
0 7
.
Here P
−1
=
µ
1
9
−
1
9
5
9
4
9
.
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2).
t A =P
µ
−2t 0
0 7t
P
−1
=⇒
(t A)
2
=P
µ
−2t 0
0 7t
P
−1
P
µ
−2t 0
0 7t
P
−1
=P
µ
(−2t )
2
0
0 (7t )
2
P
−1
=
µ
29t
2
20t
2
25t
2
24t
2
.
Similarly, (t A)
3
=P
µ
(−2t )
3
0
0 (7t )
3
P
−1
=
µ
187t
3
156t
3
195t
3
148t
3
.
3). We find that
(t A)
n
=P
µ
(−2t )
n
0
0 (7t )
n
P
−1
.
So we obtain
exp(t A) =
∞
X
n=0
1
n!
P
µ
(−2t )
n
0
0 (7t )
n
P
−1
=P
µ
P
∞
n=0
1
n!
(−2t )
n
0
0
P
∞
n=0
1
n!
(7t )
n
P
−1
= P
µ
exp(−2t ) 0
0 exp(7t )
P
−1
=
1
9
µ
4exp(−2t) +5exp(7t ) −4exp(−2t) +4exp(7t )
−5exp(−2t) +5exp(7t ) 5exp(−2t) +4exp(7t )
.
Remark. When you learn some knowledge on linear differential equations, you will see above
are typical processes on solving the equation x
0
(t ) = Ax(t), where x(t) is a unknown vector
function of t. The general solution is x(t) =exp(t A)x
0
, where x
0
is given as the initial vector.
This is an application of using linear algebra to solving ODE, and you already see the power
of matrix diagonalization.
Problem 6. Let u =(6,a +1,3)
T
, v =(a,2,−2)
T
, w =(a,1,0)
T
.
1). For which value of a will u and v be linear dependent or linear independent? When
they are linear dependent, write down their linear relation.
2). For which value of a will u, v and w be linear dependent or linear independent? When
they are linear dependent, write down their linear relation.
Keywords. Linear dependence/independence, elementary row operations, determinant.
Suggested Solution.
1). Let x
1
u +x
2
v =0, we obtain a system of linear equations:
6x
1
+ax
2
= 0
(a +1)x
1
+2x
2
= 0
3x
1
−2x
2
= 0
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We do elementary row operations to the augmented matrix:
6 a 0
a +1 2 0
3 −2 0
→
6 a 0
0 −
a
2
+a−12
6
0
0 −
a
2
−2 0
.
When
(
−
a
2
+a−12
6
= 0
−
a
2
−2 = 0
i.e., a =−4, the system has nonzero solutions hence u and v are linear dependent. And
we can take x
1
= 2, x
2
= 3, so 2u +3v = 0. When a 6= −4, the system only has zero
solution hence u and v are linear independent.
2). Let us compute the determinant of matrix generated by u, v, w:
¯
¯
¯
¯
¯
¯
6 a a
a +1 2 1
3 −2 0
¯
¯
¯
¯
¯
¯
=−2a
2
−5a +12 =−(a +4)(2a −3).
When a =−4 or a =
3
2
, u, v, w are linear dependent.
i). The case a =
3
2
. Let us write x
1
u+x
2
v +x
3
w =0. We do elementary row operations
for the augmented matrix:
6
3
2
3
2
0
5
2
2 1 0
3 −2 0 0
→
1 −
2
3
0 0
0 1
3
11
0
0 0 0 0
.
We may take x
3
=11, x
2
=−3, x
1
=−6, i.e., −6u −3v +11w =0.
ii). The case a =−4. We still use the relation as step 1): 2u +3v +0w =0.
Problem 7. Suppose that A is a n ×n matrix with eigenvalues λ
1
,λ
2
,...,λ
n
, and the corre-
sponding eigenvectors ξ
1
,ξ
2
,...,ξ
n
.
1). What are the eigenvalues and the corresponding eigenvectors of k A (k is a nonzero
constant)?
2). What are the eigenvalues and the corresponding eigenvectors of A
m
(m is a positive
integer)?
3). Assume that A is invertible. What are the eigenvalues and the corresponding eigenvec-
tors of A
−1
?
4). Assume that A is invertible. Let adj(A) be the adjugate matrix of A. So we have the
Laplace’s formula A adj(A) =adj(A) A =|A|I
n
. What are the eigenvalues and the corre-
sponding eigenvectors of adj(A) ?
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5). Let P be a n ×n invertible matrix. What are the eigenvalues and the corresponding
eigenvectors of P
−1
AP?
6). Let f (x) = x
m
+c
1
x
m−1
+···+c
m−1
x +c
m
(m is a positive integer). What are the eigen-
values and the corresponding eigenvectors of f (A)?
7). What are the eigenvalues of A
T
?
Keywords. Eigenvalue, eigenvector, adjugate matrix, matrix polynomial, determinant.
Suggested Solution. By definition we have Aξ
i
=λ
i
ξ
i
.
1). Since k Aξ
i
= kλ
i
ξ
i
, k A has eigenvalues kλ
1
, kλ
2
, ..., kλ
n
, and the corresponding
eigenvectors ξ
1
,ξ
2
,...,ξ
n
.
2). Since A
m
ξ
i
= A
m−1
Aξ
i
= λ
i
A
m−1
ξ
i
= ··· = λ
i
m
ξ
i
, A
m
has eigenvalues λ
1
m
, λ
2
m
, ... ,
λ
n
m
, and the corresponding eigenvectors ξ
1
,ξ
2
,...,ξ
n
.
3). Since A is invertible, |A|=λ
1
λ
2
···λ
n
6=0. Each λ
i
is nonzero. Left multiplying λ
i
−1
A
−1
to the equation Aξ
i
= λ
i
ξ
i
, we obtain A
−1
ξ
i
= λ
i
−1
ξ
i
. So A
−1
has eigenvalues λ
1
−1
,
λ
2
−1
, ..., λ
n
−1
, and the corresponding eigenvectors ξ
1
,ξ
2
,...,ξ
n
. Therefore, when A is
invertible, we can only assume that m is a nonzero integer in step 2) and still obtain the
same results.
4). By the Laplace formula, we obtain adjA = |A|A
−1
. By step 1) and 3), A has eigenvalues
|A|
λ
1
,
|A|
λ
2
, ...,
|A|
λ
n
and the corresponding eigenvectors ξ
1
,ξ
2
,...,ξ
n
.
5). We have APP
−1
ξ
i
= λ
i
ξ
i
. Left multiplying P
−1
, we get (P
−1
AP)(P
−1
ξ
i
) = λ
i
(P
−1
ξ
i
).
So P
−1
AP has eigenvalues λ
1
, λ
2
, ..., λ
n
, and the corresponding eigenvectors P
−1
ξ
1
,
P
−1
ξ
2
, ..., P
−1
ξ
n
.
6). By step 1) and 2), we obtain that f (A) has eigenvalues f (λ
1
), f (λ
2
), .. ., f (λ
n
), and the
corresponding eigenvectors ξ
1
,ξ
2
,...,ξ
n
.
7). Since |(A
T
−λI )|=|(A −λI)
T
|=|(A −λI )|, so A
T
and A have the same eigenvalues.
Problem 8. Let A be the real symmetric matrix
4 2 2
2 4 2
2 2 4
. Find an orthogonal matrix T to
diagonalize A.
Keywords. Real symmetric matrix, orthogonal matrix, inner product, Gram-Schmidt process.
Suggested Solution. We first find the eigenvalues and the corresponding eigenvectors of A.
|A −λI| =
¯
¯
¯
¯
¯
¯
4 −λ 2 2
2 4 −λ 2
2 2 4 −λ
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
¯
¯
8 −λ 2 2
8 −λ 4 −λ 2
8 −λ 2 4−λ
¯
¯
¯
¯
¯
¯
=(8 −λ)
¯
¯
¯
¯
¯
¯
1 2 2
1 4 −λ 2
1 2 4 −λ
¯
¯
¯
¯
¯
¯
= (8 −λ)
¯
¯
¯
¯
¯
¯
1 0 0
1 2 −λ 0
1 0 2 −λ
¯
¯
¯
¯
¯
¯
=(8 −λ)(2 −λ)
2
.
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• For λ
1
=8, we have
−4 2 2
2 −4 2
2 2 −4
x
1
x
2
x
3
=
0
0
0
.
We obtain an eigenvector ξ
1
=(1,1, 1)
T
.
• For λ
2
=λ
3
=2(multiplicity 2), we have
2 2 2
2 2 2
2 2 2
x
1
x
2
x
3
=
0
0
0
.
We obtain two linear independent eigenvectors ξ
2
=(−1,1, 0)
T
, ξ
3
=(−1,0, 1)
T
.
ξ
1
is already orthogonal to ξ
2
and ξ
3
(this is not a coincidence, why?). Next we apply Gram-
Schmidt process for ξ
2
and ξ
3
. Take
η
2
=ξ
2
,
η
3
=ξ
3
−
(ξ
3
,η
2
)
(η
2
,η
2
)
η
2
=
−1
0
1
−
1
2
−1
1
0
=
−
1
2
−
1
2
1
.
We normalize ξ
1
, η
2
, η
3
and obtain
α
1
=
1
p
3
1
p
3
1
p
3
, α
2
=
−
1
p
2
1
p
2
0
, α
3
=
1
p
6
1
p
6
−
2
p
6
.
Let
T =(α
1
,α
2
,α
3
) =
1
p
3
−
1
p
2
1
p
6
1
p
3
1
p
2
1
p
6
1
p
3
0 −
2
p
6
.
Then
T
−1
AT =
8 0 0
0 2 0
0 0 2
.
Remark.
1). We have many ways to choose ξ
2
and ξ
3
so the matrix T is not unique.
2). For a real symmetry matrix A, if Aξ = λ
1
ξ, Aη = λ
2
η, and λ
1
6= λ
2
, then ξ and η are
orthogonal. The proof is as following: λ
1
(ξ,η) = (Aξ, η) =(ξ, Aη) =λ
2
(ξ,η), and λ
1
6=λ
2
,
hence (ξ,η) =0.
Problem 9. Let C[0, 1] be the set of all real valued continuous functions on the interval [0,1].
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1). For any f , g ,∈C[0,1] and a scalar a ∈R, define
(f +g )(t) = f (t)+g (t),
(a f )(t) = a f (t ).
Check that the C[0,1] with above addition and scalar multiplication is a vector space
over R. This is an example of infinite-dimensional vector space.
2). For any f , g ∈C[0,1], define 〈f , g 〉=
R
1
0
f (t)g (t)d t . Check that 〈·,·〉 is an inner product
on the vector space C[0,1].
3). Let V be the subspaces of functions generated by the two functions f (t) = t , g (t) = t
2
.
Find an orthonormal basis for V .
Keywords. Example of vector space, inner product, orthonormal basis, Gram-Schmidt pro-
cess, application of linear algebra.
Suggested Solution.
1). For any f , g , h ∈C[0,1], and a,b ∈R, we check the following axioms.
– Associativity of addition: f +(g +h) =(f +g ) +h.
– Commutativity of addition: f +g = g + f
– Identity element of addition: there exists an element 0 ∈ C[0,1], called the zero
vector, such that f +0 = f . Here 0 is the function which maps every element in
[0,1] to the value 0.
– Inverse elements of addition: for every f ∈ C [0,1], there exists an element −f ∈
C[0,1], called the additive inverse of f , such that f +(−f ) =0.
– Distributivity of scalar multiplication with respect to vector addition: a(f +g) =
a f +ag .
– Distributivity of scalar multiplication with respect to field addition:(a+b) f = a f +
b f .
– Compatibility of scalar multiplication with field multiplication: a(b f ) =(ab)f .
– Identity element of scalar multiplication: 1 f = f , where 1 denotes the multiplica-
tive identity in R.
2). We can check easily that:
– 〈f , g 〉=
R
1
0
f (t)g (t)d t =〈g, f 〉.
– 〈a f , g 〉=
R
1
0
a f (t)g (t )d t = a〈f , g 〉.
– 〈f +g ,h〉=
R
1
0
(f (t) +g (t))h(t)dt =〈f ,h〉+〈g,h〉.
– 〈f , f 〉=
R
1
0
f (t)f (t )d t ≥0, with equality if and only if f =0.
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3). We apply Gram-Schmidt process to f (t) =t and g (t ) = t
2
. Let h =g −
〈g ,f 〉
〈f , f 〉
f .
kf k
2
=〈f , f 〉=
Z
1
0
t
2
d t =
1
3
, h(t) =t
2
−3(
Z
1
0
t
3
d t)t = t
2
−
3
4
t.
Let us normalize f and h: khk
2
=
R
1
0
(t
2
−
3
4
t)
2
d t =
1
80
. So
f
kf k
=
p
3t
3
and
h
khk
=
p
80(t
2
−
3
4
t) will be an orthonormal basis for V .
Remark. We can play exactly the same game for the space C[−π,π] with inner product 〈f , g 〉
=
1
π
R
π
−π
f (t)g (t)d t . Let n be integer. Then
g
n
=sin(nt), n >0 and h
n
=cos(nt), n ≥0
will form an orthonormal basis. For example, we can express t in term of above basis. This is
the linear algebra theory parts of the Fourier analysis, and it is easy. The difficulty parts are
how to handle the infinite sum, i.e., the convergent problem and analysis.
Problem 10.
1). Let V = R
3
and take the standard basis e
1
,e
2
,e
3
. Let P : V → V such that P(e
1
) = e
1
,
P(e
2
) = e
2
, P(e
3
) = 0. Write down the matrix A which represents the operator P in the
standard basis. Check that P
2
=P (here P
2
means P ◦P : V →V ) and A
2
= A. What are
the eigenvalues of P?
2). Let V be a vector space and P : V →V be a linear transform such that P
2
=P.
i). What are the possible eigenvalues of P?
ii). Let U =ker(P), W =im(P). Let Q = I −P : V →V .
*
Check that P is the zero operator on U and the identity operator on W .
*
Check that Q
2
=Q.
*
Check that Q is the zero operator on W and the identity operator on U .
*
Check that for any x ∈V , we can write x =u +w for some u ∈U and w ∈W .
*
Check that the above decomposition is unique. Therefore V =U ⊕W .
Keywords. Linear transform, projection operator, kernel, image, decomposition of space,
invariant subspace
Suggested Solution.
1). We would like to find a matrix A such that
P(x) = Ax,∀x =
x
1
x
2
x
3
.
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Since
x =
x
1
x
2
x
3
=x
1
1
0
0
+x
2
0
1
0
+x
3
0
0
1
=(e
1
,e
2
,e
3
)
x
1
x
2
x
3
,
We have
P(x) =P(e
1
,e
2
,e
3
)
x
1
x
2
x
3
=(Pe
1
,Pe
2
,Pe
3
)
x
1
x
2
x
3
.
So we take
A =(Pe
1
,Pe
2
,Pe
3
) =(e
1
,e
2
,0) =
1 0 0
0 1 0
0 0 0
.
It is easy to see that P
2
= P and A
2
= A. It is obvious that P has eigenvalue 1 with the
corresponding eigenvectors e
1
,e
2
and eigenvalue 0 with the corresponding eigenvector
e
3
.
2). i). Suppose that P(x) = λx for some nonzero vector x. Then λx = P(x) = P
2
(x) =
P(λx) = λ
2
x, so (λ
2
−λ)x = 0. Since x is nonzero, the possible eigenvalues are
λ =1 or λ = 0.
ii). We already know that both the kernel U and the image W are invariant subspaces
of V .
*
By definition of kernel, for any u ∈ U , P (u) = 0. Hence the restriction of P
to the subspace U is the zero operator. By the definition of image, for any
w ∈W , we can write w =P(v) for some vector v ∈V . So
P(w) =P(P(v)) =P
2
(v) =P (v) =w,
and the restriction of P to the subspace W is the identity operator.
*
Q
2
=(I −P)
2
= I −2P +P
2
= I −P =Q.
*
For any w ∈ W we have Q(w) = w −P(w) = w −w = 0. So the restriction of
Q on W is the zero operator. For any u ∈U we have Q(u) =(I −P)(u) =u. So
the restriction of Q on U is the identity operator.
*
For any x ∈V , we can write x =(x −P (x))+P(x) where u =x −P (x) =Q(x) ∈U
and w =P(x) ∈W .
*
If x =
˜
u +
˜
w for some
˜
u ∈U and
˜
w ∈W , we have u =Q(x) =Q(
˜
u +
˜
w) =
˜
u and
w =P(x) =P(
˜
u +
˜
w) =
˜
w. So the decomposition is unique.
Remark Geometrically, we have the projection operator which is a linear operator. You can
think above P as the projection of V onto W along the "direction" U ; and Q as the projection
of V onto U along the "direction" W . Projection twice is the same as projection once, so we
have P
2
=P and Q
2
=Q. If we have a nontrivial projection, which means none of P nor Q is
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the identity operator on V , then we can decompose V into direct sum of smaller spaces as
above. Algebraically, if a linear operator (or a matrix) P : V →V satisfies P
2
=P, we just call P
a projection. Such a matrix is called an idempotent matrix.
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Calculus I
Wanmin Liu
wanminliu@gmail.com
Department of Mathematics
Hong Kong University of Science and Technology
2012 August
This is a tutorial demo training to all new Teaching Assistants. It contains ten
typical problems in one variable Calculus. The new TA would be randomly as-
signed a problem, and present the problem on blackboard to other new TAs. Pros
and Cons of each presentation would be highlighted and discussed after the pre-
sentation. Typical teaching techniques/tricks/mistakes would also be empha-
sised.
Problem 1. Let
f (x) =
(
e
−
1
x
2
x 6=0;
0 x =0.
1). Show that f (x) is continuous at 0.
2). Show that f (x) is differentiable, and compute f
0
(x).
3). Show f
0
(x) is continuous at 0.
Keywords. Continuity, limit, differentiability, example of infinitely differentiable function.
Suggested Solution.
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1). By taking t =
1
x
2
, we have
lim
x→0
f (x) = lim
t→+∞
1
e
t
=0 = f (0).
So f (x) is continuous at x =0.
2). When x 6= 0, f (x) is the composition of some basic function and hence differentiable.
We need to check that lim
∆x→0
f (0+∆x)−f (0)
∆x−0
exists.
lim
∆x→0
f (0+∆x) − f (0)
∆x −0
= lim
∆x→0
e
−
1
∆x
2
∆x
= lim
t→∞
t
e
t
2
taking t =
1
∆x
= lim
t→∞
1
2t e
t
2
L’Hôpital’s rule
= 0,
so f
0
(0) =0. Hence
f
0
(x) =
(
2
x
3
e
−
1
x
2
x 6=0;
0 x =0.
3).
lim
x→0
f
0
(x) = lim
x→0
2
x
3
e
−
1
x
2
= lim
t→∞
2t
3
e
t
2
taking t =
1
x
= lim
t→∞
6t
2
2t e
t
2
= lim
t→∞
12t
(2t e
t
2
)
0
= lim
t→∞
12
(2t e
t
2
)
00
L’Hôpital’s rule three times
= 0.
So lim
x→0
f
0
(x) =0 = f
0
(0), and f
0
(x) is continuous.
Remark. By the mathematical induction, it can be shown that f (x) is infinitely differentiable
at x =0, and f
(n)
(0) =0 for any positive integer n.
Problem 2. Let α and β be two positive numbers, and
f (x) =
½
x
α
sin(
1
x
β
) x 6=0;
0 x =0.
1). Show that f (x) is continuous at 0.
2). For what values of α and β is f (x) differentiable?
3). For what values of α and β is f
0
(x) continuous?
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Keywords. Continuity, differentiability, example of differentiable function but its differential
is not continuous.
Suggested Solution.
1). Since sin(
1
x
β
) is bounded, it is clear that
lim
x→0
f (x) = lim
x→0
x
α
sin(
1
x
β
) =0 = f (0).
2). When x 6=0, we have
f
0
(x) = αx
α−1
sin(
1
x
β
) +x
α
cos(
1
x
β
)(
−β
x
β+1
)
= αx
α−1
sin(
1
x
β
) −βx
α−β−1
cos(
1
x
β
).
We need to compute lim
∆x→0
f (0+∆x)−f (0)
∆x−0
.
lim
∆x→0
f (0+∆x) − f (0)
∆x −0
= lim
∆x→0
∆x
α−1
sin(
1
∆x
β
).
When α > 1, the above limit exists and f
0
(0) = 0. When 0 < α ≤ 1, the limit does not
exist. Therefore, for α >1 and arbitrary positive β, f (x) is differentiable and
f
0
(x) =
½
αx
α−1
sin(
1
x
β
) −βx
α−β−1
cos(
1
x
β
) x 6=0;
0 x =0.
When 0 <α ≤ 1, f (x) is not differentiable.
3). We need to compute lim
x→0
f
0
(x). When α > 1 and α −β −1 > 0, the limit exists and
equals 0, then f
0
(x) is continuous. When α > 1, and α−β −1 ≤0, f
0
(x) exists but is not
continuous.
Remark. We can further ask similar questions. For what values of positive α and β does f
00
(x)
exist? For what values of positive α and β is f
00
(x) continuous?
Problem 3. Find the limit
lim
x→∞
a
1
x
1
+a
1
x
2
+···+a
1
x
n
n
nx
where a
1
, a
2
, ··· , a
n
are positive.
Keywords. Limit, L’Hôpital’s rule.
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Suggested Solution.
lim
x→∞
a
1
x
1
+a
1
x
2
+···+a
1
x
n
n
nx
= lim
x→∞
exp
nx ln
a
1
x
1
+a
1
x
2
+···+a
1
x
n
n
= exp
n lim
x→∞
ln(a
1
x
1
+a
1
x
2
+···+a
1
x
n
) −ln n
1
x
continuity of limit
= exp
n lim
x→∞
a
1
x
1
ln a
1
+a
1
x
2
ln a
2
+···+a
1
x
n
ln a
n
a
1
x
1
+a
1
x
2
+···+a
1
x
n
(−
1
x
2
)
−
1
x
2
L’Hôpital’s rule
Since
lim
x→∞
a
1
x
=1 for a > 0,
we have
lim
x→∞
a
1
x
1
ln a
1
+a
1
x
2
ln a
2
+···+a
1
x
n
ln a
n
a
1
x
1
+a
1
x
2
+···+a
1
x
n
=
ln(a
1
a
2
···a
n
)
n
.
So the original limit equals
exp
µ
n
ln(a
1
a
2
···a
n
)
n
=a
1
a
2
···a
n
.
Problem 4. Find the second derivative
d
2
y
dx
2
.
1). y = tan(x +y). 2). y =
¡
x
1+x
¢
x
.
3).
½
x = f
0
(t )
y = t f
0
(t ) − f (t)
where we assume that f
00
(t ) exists and does not equal zero.
Keywords. Implicit function, chain rule, parameter function, logarithmic differentiation.
Suggested Solution.
1). This is an implicit function of y. Taking derivative with respect to x on both sides, we
have
y
0
=sec
2
(x +y)(1 +y
0
).
So
y
0
=
−sec
2
(x +y)
sec
2
(x +y) −1
=
−sec
2
(x +y)
tan
2
(x +y)
=−
1 +y
2
y
2
=−1 −
1
y
2
.
Taking derivative w.r.t. x again,
y
00
=2y
−3
y
0
=2y
−3
µ
−
1 +y
2
y
2
=−
2(1 +y
2
)
y
5
.
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2). Since both the base and the exponential have the variable x, we need to take the natural
logarithm first.
ln y = x(ln x −ln(1 +x)).
Taking derivative with respect to x, we obtain
1
y
y
0
=(lnx −ln(1 +x)) +x(
1
x
−
1
1 +x
).
So
y
0
= y
µ
ln
x
1 +x
+
1
1 +x
.
Taking derivative with respect to x again, we have
y
00
= y
0
µ
ln
x
1 +x
+
1
1 +x
+y
µ
1
x
−
1
1 +x
−
1
(1 +x)
2
= y
µ
ln
x
1 +x
+
1
1 +x
2
+y
µ
1
x(1 +x)
2
=
³
x
1 +x
´
x
µ
³
ln
x
1 +x
´
2
+
2
1 +x
ln
x
1 +x
+
1
x(1 +x)
.
3). This is the parameter equation with parameter t . We have
dy
dx
=
y
0
t
x
0
t
=
f
0
(t ) +t f
00
(t ) − f
0
(t )
f
00
(t )
=t .
d
2
y
dx
2
=
³
dy
dx
´
0
t
x
0
t
=
t
0
f
00
(t )
=
1
f
00
(t )
.
Problem 5. Find the maximal value of the sequence {
n
p
n} for positive integer n.
Keywords. Maximal of a function, logarithmic differentiation.
Suggested Solution. Let f (x) =x
1
x
where (x >0). Then ln f (x) =
1
x
ln x. So
f
0
(x) = f (x)(
1
x
ln x)
0
= f (x)
1 −ln x
x
2
=x
1
x
−2
(1 −ln x).
Let f
0
(x) = 0. We have x = e. When 0 < x < e, f
0
(x) > 0. So f (x) is strictly increasing in the
interval (0, e], and hence f (1) < f (2). When e < x, f
0
(x) < 0. So f (x) is strictly decreasing in
the interval (e, ∞), and hence for any integer n ≥ 3, we have f (n) ≤ f (3). Since f (2)
6
= 2
3
<
3
2
= f (3)
6
, then f (2) < f (3). So max{
n
p
n} =max{f (n)} = f (3) =
3
p
3.
Problem 6. Compute the following indefinite integral:
1).
Z
x ln(x −1) dx. 2).
Z
e
−x
cos x dx.
Keywords. Integral by parts.
Suggested Solution.
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1).
Z
x ln(x −1) dx =
1
2
Z
ln(x −1) dx
2
=
1
2
µ
x
2
ln(x −1) −
Z
x
2
x −1
dx
=
1
2
µ
x
2
ln(x −1) −
Z
x
2
−1 +1
x −1
dx
=
1
2
µ
x
2
ln(x −1) −
Z
(x +1) dx −
Z
1
x −1
d(x −1)
=
1
2
x
2
ln(x −1) −
1
4
(x +1)
2
−
1
2
ln(x −1) +C.
2).
Z
e
−x
cos x dx =
Z
e
−x
d(sin x)
= e
−x
sin x +
Z
e
−x
sin x dx
= e
−x
sin x −
Z
e
−x
d(cos x)
= e
−x
sin x −e
−x
cos x −
Z
e
−x
cos x dx.
So
Z
e
−x
cos x dx =
e
−x
2
(sin x −cos x) +C.
Problem 7.
1). Calculate
R
secθ dθ. (Hint: let u =secθ +tan θ.)
2). Calculate
R
dx
p
x
2
−1
, x >1 by substituting x =secθ.
3). Recall sinh(t) =
e
t
−e
−t
2
, cosh(t) =
e
t
+e
−t
2
. Let x = cosh(t), it is obvious that x is an even
function of t, so the usual inverse function does not exist. However, we can restrict to
t ≥0. Find the inverse function of x =cosh(t ) for t ≥0.
4). Calculate
R
dx
p
x
2
−1
, x >1 by substituting x =cosht.
Keywords. Using the substitution to find integral, hyper-trigonometric functions.
Suggested Solution.
1) 1st method. By the hint, setting u =secθ +tanθ, we find
du = (sec θ tan θ +sec
2
θ)dθ =u sec θ dθ.
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So
Z
secθ dθ =
Z
u secθ dθ
u
=
Z
du
u
= ln|u|+C
= ln|secθ +tan θ|+C.
2nd method.
Z
secθ dθ =
Z
dθ
cosθ
=
Z
cosθ dθ
cos
2
θ
=
Z
dsinθ
1 −sin
2
θ
=
Z
dt
(1 −t)(1 +t)
let t =sinθ
=
1
2
Z
1
1 −t
+
1
1 +t
dt =
1
2
(−ln|1 −t|+ln|1 +t|)+C
= ln
s
|1 +t|
|1 −t|
+C =ln
s
|1 +t|
2
|(1 −t)(1 +t)|
+C
= ln
¯
¯
¯
¯
1 +sin θ
cosθ
¯
¯
¯
¯
+C =ln|secθ +tan θ|+C.
2). Let x =secθ for 0 <θ <
π
2
. Then
dx =secθ tanθ dθ.
Z
dx
p
x
2
−1
=
Z
secθ tanθ dθ
tanθ
=
Z
secθ dθ
= ln|secθ +tan θ|+C
= ln|x +
p
x
2
−1|+C.
Since (x +
p
x
2
−1)(x −
p
x
2
−1) = 1 and x +
p
x
2
−1 ≥ x −
p
x
2
−1 > 0, we have x +
p
x
2
−1 >1. So
Z
dx
p
x
2
−1
=ln(x +
p
x
2
−1) +C for x >1.
3). Let x =cosht =
e
t
+e
−t
2
where t ≥0. Then
(e
t
)
2
−2x(e
t
) +1 =0.
Since t ≥0, we have e
t
≥1. So e
t
=x +
p
x
2
−1, where the other root is dropped because
it is smaller than 1. Hence t =cosh
−1
(x) =ln(x +
p
x
2
−1).
4). Let x =cosht >1. So dx =sinh t dt. Since
cosh
2
t −sinh
2
t =1,
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we find
Z
dx
p
x
2
−1
=
Z
sinh t dt
sinh t
=
Z
dt = t +C
= ln(x +
p
x
2
−1) +C.
Problem 8. Consider the improper integral
R
+∞
2
dx
x(ln x)
p
.
1). For what values of p is the improper integral divergent?
2). For what values of p is the improper integral convergent?
3). In the case of convergence, for what value of p is the improper integral minimal? Find
the minimal value of the improper integral.
4). (Logarithmic p-series) For what values of p is the series
∞
X
n=2
1
n(lnn)
p
convergent?
Keywords. Improper integral, convergent, divergent, minimal value.
Suggested Solution. Let A >2.
Z
A
2
dx
x(lnx)
p
=
Z
A
2
d(ln x)
(ln x)
p
=
(
lnln A −lnln2 p =1;
1
1−p
¡
(ln A)
1−p
−(ln2)
1−p
¢
p 6=0.
1). When p =1,
Z
+∞
2
dx
x(lnx)
p
= lim
A→+∞
(lnln A −lnln2) =+∞.
When p <1,
Z
+∞
2
dx
x(lnx)
p
= lim
A→+∞
(
1
1 −p
¡
(ln A)
1−p
−(ln2)
1−p
¢
) =+∞.
So the improper integral is divergent if p ≤1.
2). When p >1,
Z
+∞
2
dx
x(lnx)
p
= lim
A→+∞
(
1
1 −p
¡
(ln A)
1−p
−(ln2)
1−p
¢
) =
(ln2)
1−p
p −1
.
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3). For p >1, let f (p) =
(ln2)
1−p
p−1
. So
f
0
(p) =
(ln2)
1−p
ln(ln2)(−1)(p −1) −(ln2)
1−p
(p −1)
2
=
(ln2)
1−p
(−ln(ln2))
¡
p −(1 −
1
lnln2
)
¢
(p −1)
2
.
Since 2 <e, we know ln2 <lne =1. So lnln 2 <ln1 = 0. Let p
0
=1 −
1
lnln2
. Then p
0
>1.
When 1 < p < p
0
, we obtain f
0
(p) <0. When p >p
0
, we find f
0
(p) >0. So when p = p
0
,
the improper integral is minimal. The minimal value is
f (p
0
) =
(ln2)
1
lnln 2
−
1
lnln2
=−(lnln 2)(ln2)
1
lnln 2
.
4). By the integral test for the convergence of series, when p > 1, the logarithmic p-series
converges.
Remark. Let us recall the integral test. Let {a
n
} be a sequence of positive terms. Suppose
that a
n
= g (n), where g is a continuous, positive, decreasing function of x for all x ≥ N (N a
positive integer). Then the series
P
∞
n=N
a
n
and the integral
R
∞
N
g (x)dx both converge or both
diverge.
Problem 9. Consider the alternating series
P
∞
n=1
(−1)
n
1
n
p
.
1). For what values of p is the series divergent?
2). For what values of p is the series absolutely convergent?
3). For what values of p is the series conditionally convergent?
Keywords. Alternating series, divergence, absolutely convergence, conditional convergence,
Leibniz’s theorem for the alternating series test.
Suggested Solution. Let a
n
=(−1)
n
1
n
p
.
1). When p ≤0, we find lim
n→∞
a
n
6=0. So the alternating series diverges.
2). We have
∞
X
n=1
|a
n
|=
∞
X
n=1
1
n
p
.
The p-series is convergent for p >1 (by the integral test). So when p >1, the alternating
series is absolutely convergent.
3). When 0 <p ≤1, the alternating series satisfies the assumption of the Leibniz’s theorem
and conditionally converges.
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Remark. Let us recall the Leibniz’s theorem. The series
∞
X
n=1
(−1)
n+1
a
n
=a
1
−a
2
+a
3
−a
4
+···
converges if all three of the following conditions are satisfied.
1). The a
n
’s are all positive.
2). a
n
≥a
n+1
for n ≥ N , for some integer N.
3). a
n
→0.
The theorem is proved by the trick of pairing: let b
1
= (a
1
−a
2
), b
2
= (a
3
−a
4
), ··· Then the
alternating sum of a
n
will become the positive sum of b
n
.
Problem 10. Let h ∈(0, π). Consider the function
f (x) =
½
1 0 ≤x ≤h;
0 h < x ≤π.
1). Find the Fourier sine series for f (x). What is the value of the Fourier sine at the discon-
tinuity point x =h?
2). Find the Fourier cosine series for f (x). What is the value of the Fourier cosine at the
discontinuity point x =h?
Keywords. Fourier cosine series, Fourier sine series.
Suggested Solution.
1). Regardless the discontinuity points, we define the odd extension of f by
F (x) =
1 0 <x <h;
0 h < x <π;
−1 −h < x <0;
0 −π <x <−h.
Then we apply a periodic extension with period 2π to F (x), and still denote it by F (x).
The Fourier coefficients are
a
n
= 0, n = 0, 1,2,···
b
n
=
2
π
Z
π
0
F (x)sinnx dx =
2
π
Z
h
0
sinnx dx =
2(1 −cos nh)
nπ
n = 1, 2,··· .
So the Fourier sine of F (x) is
g (x) =
2
π
∞
X
n=1
1 −cos nh
n
sinnx.
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Applying the convergence theorem of Fourier series, we have
f (x) =g (x) for x ∈(0,h) ∪(h, π).
At the jump discontinuity point x =h, the Fourier series converges to the average
g (h) =
f (h
+
) + f (h
−
)
2
=
1
2
.
2). Regardless the discontinuity points, we define the even extension of f by
F (x) =
1 0 <x <h;
0 h < x <π;
1 −h < x <0;
0 −π <x <−h.
Then we apply a periodic extension with period 2π to F (x), and still denote it by F (x).
The Fourier coefficients are
a
0
=
2
π
Z
π
0
F (x)dx =
2
π
Z
h
0
dx =
2h
π
,
a
n
=
2
π
Z
π
0
F (x)cosnx dx =
2
π
Z
h
0
cosnx dx =
2sinnh
nπ
n = 1, 2,··· ;
b
n
= 0, n = 1, 2,··· .
So the Fourier cosine of F (x) is
g (x) =
h
π
+
2
π
∞
X
n=1
sinnh
n
cosnx.
Applying the convergence theorem of Fourier series, we have
f (x) =g (x) for x ∈(0,h) ∪(h, π).
At the jump discontinuity point x =h, the Fourier series converges to the average
g (h) =
f (h
+
) + f (h
−
)
2
=
1
2
.
Remark The Fourier series of a function f (x) defined on the interval −L < x <L is
g (x) =
a
0
2
+
∞
X
n=1
³
a
n
cos
nπx
L
+b
n
sin
nπx
L
´
,
where a
0
=
1
L
R
L
−L
f (x)dx, a
n
=
1
L
R
L
−L
f (x)cos
nπx
L
dx, b
n
=
1
L
R
L
−L
f (x)sin
nπx
L
dx.
The convergence theorem of Fourier series states that:
If the function f and its derivative f
0
are piecewise continuous over the interval −L < x < L,
then f (x) equals its Fourier series g (x) at all points of continuity. At a point h where a jump
discontinuity happens in f , the Fourier series converges to the average
g (h) =
f (h
+
) + f (h
−
)
2
,
where f (h
+
) and f (h
−
) means the right and the left limits of f at h, respectively.
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Calculus II
Wanmin Liu
wanminliu@gmail.com
Department of Mathematics
Hong Kong University of Science and Technology
2012 August
This is a tutorial demo training to all new Teaching Assistants. It contains ten
typical problems in multi-variable Calculus. The new TA would be randomly as-
signed a problem, and present the problem on blackboard to other new TAs. Pros
and Cons of each presentation would be highlighted and discussed after the pre-
sentation. Typical teaching techniques/tricks/mistakes would also be empha-
sised.
Problem 1. Determine whether the following limits exist. If they do exist then find their value.
1). lim
(x,y)→(0,0)
sin x y
y
. 2). lim
(x,y)→(0,0)
1 −cos(x
2
+y
2
)
(x
2
+y
2
)x
2
y
2
.
3). lim
(x,y)→(0,0)
x
2
y
2
x
2
y
2
+(x −y)
2
. 4). lim
(x,y)→(0,0)
x y
p
x
2
+y
2
.
Keywords. Multi-variable limit.
Suggested Solution.
1).
lim
(x,y)→(0,0)
sin x y
y
= lim
(x,y)→(0,0)
sin x y
x y
x =1·0 =0.
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2).
lim
(x,y)→(0,0)
1 −cos(x
2
+y
2
)
(x
2
+y
2
)x
2
y
2
= lim
(x,y)→(0,0)
2sin
2
x
2
+y
2
2
(
x
2
+y
2
2
)
2
x
2
+y
2
4x
2
y
2
(1 −cos θ =2sin
2
θ
2
)
=
1
2
·1
2
lim
(x,y)→(0,0)
µ
1
x
2
+
1
y
2
=+∞.
The limit does not exist.
3). We take two ways to compute the limit. The first way is taking y = x. The second way is
taking y =2x.
lim
(x,y)→(0,0)
y=x
x
2
y
2
x
2
y
2
+(x −y)
2
= lim
x→0
x
4
x
4
=1.
lim
(x,y)→(0,0)
y=2x
x
2
y
2
x
2
y
2
+(x −y)
2
= lim
x→0
4x
4
4x
4
+x
2
= lim
x→0
4x
2
4x
2
+1
=0.
These two ways of computing are not consistent so the original two-variable limit does
not exist.
4). We apply the basic inequality |x y|≤
x
2
+y
2
2
. So
¯
¯
¯
¯
¯
x y
p
x
2
+y
2
¯
¯
¯
¯
¯
≤
x
2
+y
2
2
p
x
2
+y
2
=
p
x
2
+y
2
2
.
For any ε >0, taking δ =2ε, when 0 <
p
x
2
+y
2
<δ, we have
¯
¯
¯
¯
¯
x y
p
x
2
+y
2
−0
¯
¯
¯
¯
¯
≤
p
x
2
+y
2
2
<
δ
2
=ε.
By the definition of limit, lim
(x,y)→(0,0)
x y
p
x
2
+y
2
=0.
Problem 2. Let z(x, y) =
y
x
2
+y
2
.
1). Find ∆z at the point (x, y) =(3,4) with ∆x =0.1, ∆y =0.1.
2). Compute dz.
3). Compute dz at the point (x, y) = (3,4) with ∆x = 0.1, ∆y = 0.1. What is the difference
between ∆z and dz at this time?
Keywords. total differential, change of function, partial derivative.
Suggested Solution.
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1). By definition ∆z = z(x +∆x, y +∆y)−z(x, y). Putting the given data inside, we obtain
∆z|
(x,y)=(3,4)
(∆x,∆y)=(0.1,0.1)
=z(3.1,4.1) −z(3,4) =
4.1
26.42
−
4
25
≈−0.00481.
2).
dz =
∂z
∂x
dx +
∂z
∂y
dy =
−y ·2x
(x
2
+y
2
)
2
dx +
(x
2
+y
2
) −y ·2y
(x
2
+y
2
)
2
dy
=
−2x y
(x
2
+y
2
)
2
dx +
x
2
−y
2
(x
2
+y
2
)
2
dy.
3). By the definition, we know dx =∆x, dy =∆y. So
dz|
(x,y)=(3,4)
(∆x,∆y)=(0.1,0.1)
=
−24
25
2
·0.1 +
−7
(25)
2
·0.1 =−
3.1
25
2
=−0.00496.
We find that
∆z − dz|
(x,y)=(3,4)
(∆x,∆y)=(0.1,0.1)
≈−0.00481 −(−0.00496) =0.00015,
which is higher order smaller than the size of the change of variable 0.1.
Remark. In general we have an estimation on the error in the standard linear approximation.
If z = z(x, y) has continuous first and second partial derivatives throughout an open set con-
taining a rectangle R centered at (x
0
, y
0
) and if M is any upper bound for the values of |z
xx
|,
|z
x y
|, and |z
y y
| on R, then the error E(x, y) = z(x, y) −L(x, y) of z(x, y) and its standard linear
approximation
L(x, y) =z(x
0
, y
0
) +z
0
x
(x −x
0
) +z
0
y
(y −y
0
)
satisfies the inequality
|E(x, y)|≤
1
2
M(|x −x
0
|+|y −y
0
|)
2
.
So
∆z − dz =E(x, y), |∆ z − dz|≤
1
2
M(|∆x|+|∆y|)
2
.
∆z − dz is higher order smaller than ∆x or ∆y.
Problem 3. Define the 3-dimensional Laplace operator as 4=
∂
2
∂x
2
+
∂
2
∂y
2
+
∂
2
∂z
2
, where (x, y, z)
are variables in the Cartesian coordinates in R
3
.
1). Express the Laplace operator in cylinder coordinates.
2). Express the Laplace operator in spherical coordinates. (This part is optional.)
Keywords. The Laplace operator, cylinder coordinates, spherical coordinates, change of vari-
able, chain rule.
Suggested Solution.
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Let us recall that the change of variables
form the Cartesian coordinate to the cylin-
der coordinates and to the spherical coordi-
nates. We have the relations
x =r cos θ, y =r sinθ, z = z,
r =
q
x
2
+y
2
, tanθ =
y
x
,
where θ ∈[0,2π), and the relations
x =ρ sinφcosθ, y =ρ sinφ sinθ , z =ρ cosφ,
ρ =
q
x
2
+y
2
+z
2
, r =ρ sinφ,
where θ ∈[0,2π) and φ ∈[0, π).
1). Let us compute the first order partial derivatives.
∂r
∂x
=
1
2
2x
p
x
2
+y
2
=
x
r
,
∂r
∂y
=
y
r
,
∂θ
∂x
=
1
1 +
¡
y
x
¢
2
−y
x
2
=−
y
r
2
,
∂θ
∂y
=
1
1 +
¡
y
x
¢
2
1
x
=
x
r
2
.
Caution. We notice that
∂r
∂x
=
x
r
=cosθ. We also know x =r cosθ. So
∂x
∂r
=cosθ. There-
fore
∂r
∂x
6=
1
∂x
∂r
.
Let us compute the second order partial derivatives.
∂
2
r
∂x
2
=
1
r
+x
∂
1
r
∂x
=
1
r
−
x
r
2
∂r
∂x
=
1
r
−
x
2
r
3
,
∂
2
r
∂y
2
=
1
r
−
y
2
r
3
.
∂
2
θ
∂x
2
=−y
∂
1
r
2
∂x
=
2y
r
3
x
r
=
2x y
r
4
,
∂
2
θ
∂y
2
=x
∂
1
r
2
∂x
=−
2x
r
3
y
r
=−
2x y
r
4
.
We find that
∂
∂x
=
∂
∂r
∂r
∂x
+
∂
∂θ
∂θ
∂x
,
∂
∂y
=
∂
∂r
∂r
∂y
+
∂
∂θ
∂θ
∂y
,
∂
2
∂x
2
=
∂
2
∂r
2
µ
∂r
∂x
2
+
∂
∂r
∂
2
r
∂x
2
+
∂
2
∂θ
2
µ
∂θ
∂x
2
+
∂
∂θ
∂
2
θ
∂x
2
∂
2
∂y
2
=
∂
2
∂r
2
µ
∂r
∂y
2
+
∂
∂r
∂
2
r
∂y
2
+
∂
2
∂θ
2
µ
∂θ
∂y
2
+
∂
∂θ
∂
2
θ
∂y
2
.
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Hence
4 =
∂
2
∂x
2
+
∂
2
∂y
2
+
∂
2
∂z
2
=
∂
2
∂r
2
x
2
+y
2
r
2
+
∂
∂r
µ
2
r
−
x
2
+y
2
r
3
+
∂
2
∂θ
2
x
2
+y
2
r
4
+
∂
∂θ
2x y −2x y
r
4
+
∂
2
∂z
2
=
∂
2
∂r
2
+
1
r
∂
∂r
+
1
r
2
∂
2
∂θ
2
+
∂
2
∂z
2
(also written as below)
=
1
r
∂
∂r
µ
r
∂
∂r
+
1
r
2
∂
2
∂θ
2
+
∂
2
∂z
2
.
2). We change the cylinder coordinates to spherical coordinates. The relations are
r =ρ sinφ, θ =θ, z =ρ cosφ;
ρ =
p
r
2
+z
2
, θ =θ, φ =arctan
r
z
.
Let us compute the first order partial derivatives.
∂ρ
∂r
=
r
ρ
=sinφ,
∂ρ
∂z
=cosφ,
∂φ
∂r
=
1
1 +
¡
r
z
¢
2
1
z
=
cosφ
ρ
,
∂φ
∂z
=−
sinφ
ρ
.
Let us compute the second order partial derivatives.
∂
2
ρ
∂r
2
=cosφ
∂φ
∂r
=
cos
2
φ
ρ
,
∂
2
ρ
∂z
2
=−sin φ
∂φ
∂z
=
sin
2
φ
ρ
;
∂
2
φ
∂r
2
=−
sinφ
ρ
∂φ
∂r
−
cosφ
ρ
2
∂ρ
∂r
=−
2sinφcosφ
ρ
2
,
∂
2
φ
∂z
2
=
2sinφcosφ
ρ
2
.
We have
∂
∂r
=
∂
∂ρ
∂ρ
∂r
+
∂
∂φ
∂φ
∂r
=sinφ
∂
∂ρ
+
cosφ
ρ
∂
∂φ
,
∂
∂z
=
∂
∂ρ
∂ρ
∂z
+
∂
∂φ
∂φ
∂z
=cosφ
∂
∂ρ
−
sinφ
ρ
∂
∂φ
.
So
∂
2
∂r
2
=sin
2
φ
∂
2
∂ρ
2
+
cos
2
φ
ρ
∂
∂ρ
+
cos
2
φ
ρ
2
∂
2
∂φ
2
−
2sinφcosφ
ρ
2
,
∂
2
∂z
2
=cos
2
φ
∂
2
∂ρ
2
+
sin
2
φ
ρ
∂
∂ρ
+
sin
2
φ
ρ
2
∂
2
∂φ
2
+
2sinφcosφ
ρ
2
.
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Therefore
4 =
∂
2
∂r
2
+
∂
2
∂z
2
+
1
r
∂
∂r
+
1
r
2
∂
2
∂θ
2
(cylinder coordinate)
=
∂
2
∂ρ
2
+
1
ρ
∂
∂ρ
+
1
ρ
2
∂
2
∂φ
2
+
1
ρ sinφ
µ
sinφ
∂
∂ρ
+
cosφ
ρ
∂
∂φ
+
1
ρ
2
sin
2
φ
∂
2
∂θ
2
=
∂
2
∂ρ
2
+
2
ρ
∂
∂ρ
+
1
ρ
2
∂
2
∂φ
2
+
1
ρ sinφ
µ
cosφ
ρ
∂
∂φ
+
1
ρ
2
sin
2
φ
∂
2
∂θ
2
(also written as below)
=
1
ρ
2
∂
∂ρ
µ
ρ
2
∂
∂ρ
+
1
ρ
2
sinφ
∂
∂φ
µ
sinφ
∂
∂φ
+
1
ρ
2
sin
2
φ
∂
2
∂θ
2
.
Problem 4. Fix the ellipsoid x
2
+2y
2
+z
2
=1.
1). Find the tangent planes on the ellipsoid which are parallel to the plane x −y +2z =
p
22.
2). Find the distance between the two surfaces. The distance dist between two surfaces α,
β is defined as: dist(α, β) =min{dist(P,Q)| P ∈α, Q ∈β}.
Keywords. Tangent plane, normal vector, distance.
Suggested Solution.
1). Let F(x, y, z) =x
2
+2y
2
+z
2
−1. Then
n =(F
x
,F
y
,F
z
) =(2x,4y,2z).
The normal vector of the plane is a =(1,−1,2). Since this plane is parallel to the tangent
planes, we obtain n ∥a, i.e.,
2x
1
=
4y
−1
=
2z
2
.
So we find x =
1
2
z, y =−
1
4
z. Substituting them into ellipsoid, we have
³
z
2
´
2
+2
³
−
z
4
´
2
+z
2
=1.
Then z =±2
q
2
11
, x =±
q
2
11
, y =∓
1
2
q
2
11
. The tangent points are
P
1
=
Ã
r
2
11
,−
1
2
r
2
11
,2
r
2
11
!
, P
2
=
Ã
−
r
2
11
,
1
2
r
2
11
,−2
r
2
11
!
.
The tangent planes are
Ã
x ±
r
2
11
!
−
Ã
y ∓
1
2
r
2
11
!
+2
Ã
z ±2
r
2
11
!
=0,
i.e., x −y +2z =±
p
22
2
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2). The distance of the two surfaces is the distance of P
1
to the original plane. Let P =
(x
0
, y
0
, z
0
) be a point outside the plane β : ax +by +cz = d. Let Q = (x, y,z) be any
point on the plane. Let R be the projection of P to β. We see that n = (a,b,c) is a
normal vector of the plane.
dist(P,β) = |PR|=|PQ||cosθ|=
¯
¯
¯
¯
−−→
PQ ·
n
|n|
¯
¯
¯
¯
=
¯
¯
(x −x
0
, y −y
0
, z −z
0
) ·(a,b, c)
¯
¯
p
a
2
+b
2
+c
2
=
¯
¯
a(x −x
0
) +b(y −y
0
) +c(z −z
0
)
¯
¯
p
a
2
+b
2
+c
2
=
¯
¯
ax
0
+by
0
+cz
0
−d)
¯
¯
p
a
2
+b
2
+c
2
,
where in the last equation, we use the condition Q ∈β. Now
dist(α,β) =dist(P
1
,β) =
|
p
22
2
−
p
22|
p
1
2
+(−1)
2
+2
2
=
p
33
6
.
Problem 5. Let S be the sphere x
2
+y
2
+z
2
=a
2
and C be the cylinder x
2
+y
2
=ax (a>0).
1). Compute the area of the part of the sphere inside the cylinder.
2). Write down the volume of the region Ω bounded by the cylinder in terms of triple inte-
gral. Explicitly express the triple integral in three repeated single integrations (no need
to compute the final integration).
Keywords. Double integration, triple integration.
Suggested Solution.
1). By the symmetry, we only need to computer in the first octant. In this case the sphere
is given by the function z =
p
a
2
−x
2
−y
2
. Then
∂z
∂x
=
−x
p
a
2
−x
2
−y
2
,
∂z
∂y
=
−y
p
a
2
−x
2
−y
2
. Let D
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be the projection of the inside parts of the sphere onto the x y plane. We find
Area = 4
Ï
D
s
1 +
µ
∂z
∂x
2
+
µ
∂z
∂y
2
dx dy =4
Ï
D
a
p
a
2
−x
2
−y
2
dx dy
= 4a
Ï
D
1
a
2
−r
2
r dr dθ =4a
Z
π
2
0
dθ
Z
aθ
0
1
p
a
2
−r
2
r dr
= 2a
Z
π
2
0
dθ
−1
p
a
2
−r
2
d(a
2
−r
2
) =2a
Z
π
2
0
h
−2(a
2
−r
2
)
1
2
i
a cosθ
0
dθ
= 4a
Z
π
2
0
a(1 −sinθ) dθ =2a
2
(π −2).
2). We sketch the region Ω. Let P =(0,0,z). The intersection of the horizontal plane pass-
ing through P and the region Ω is sketching as above picture, which is bounded by
the circle from P to T , and circle from T to Q. The point Q is given by (t ,0,0). So the
coordinate of T can be computed as the following equations.
x
2
+y
2
= t
2
x
2
+y
2
= ax
z
2
+t
2
= a
2
=⇒(x, y, z) =(
t
2
a
,
t
p
a
2
−t
2
a
,
p
a
2
−t
2
).
So the volume of Ω:
Vol(Ω) = 4
Ñ
Ω
dx dy dz =4
Z
a
0
dz
Z
t
p
a
2
−t
2
a
0
dy
Z
p
t
2
−y
2
a
2
−
q
¡
a
2
¢
2
−y
2
dx.
Problem 6. Let Ω be the region bounded by the cylinder x
2
+y
2
=1, and planes z = 1, z = 0,
x =1, y =0 in the first octant.
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1). Compute
Ð
Ω
x y dV by using the cylinder coordinate.
2). Compute
Ð
Ω
x y dV by using the Cartesian coordinate.
3). Compute
Ð
Ω
x y dV by using Gauss’ divergence theorem.
Keywords. Triple integration, the cylinder coordinate, Gauss’ divergence theorem.
Suggested Solution.
1).
Ñ
Ω
x y dV =
Z
π
2
0
dθ
Z
1
0
r dr
Z
1
0
r
2
sinθcosθ dz
=
1
4
Z
π
2
0
sin2θd(2θ)
Z
1
0
r
3
dr =
1
16
(−cos2θ)
¯
¯
¯
π
2
0
=
1
8
.
2).
Ñ
Ω
x y dV =
Z
1
0
x dx
Z
p
1−x
2
0
y dy
Z
1
0
dz =
Z
1
0
x dx
Z
p
1−x
2
0
y dy
=
1
2
Z
1
0
x(1 −x
2
)dx =
1
4
Z
1
0
(1 −t)dt =
1
8
.
3). Gauss’ divergence theorem states that
Ñ
Ω
div F dV =
Ó
S
F ·n dσ,
where F is a smooth vector field on Ω, div F =∇·F is the divergence of F, S is the bound-
ary of Ω, n is the outward normal vector of S. To use this theorem for the problem, we
need to construct a vector field on V . Let us define F = 0i +0j +x y zk = x yzk. Then
div F =
∂x yz
∂z
= x y. We split the surface into S
1
,···, S
5
. Denote n
i
the outward normal
vector of S
i
. Then
F ·n
i
=0, i =3,4,5.
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Notice that F =(0,0,0) on S
1
. We have
Ñ
Ω
x y dV =
Ñ
Ω
div F dV =Σ
5
i=1
Ó
S
i
F ·n
i
dσ
=
Ó
S
2
(0,0,x y) ·(0,0,1) dσ =
Ï
x
2
+y
2
≤1
x≥0,y≥0
x y dx dy
=
Z
1
0
Z
π
2
0
r
3
sinθcosθ dr dθ =
1
8
.
Problem 7. Let Γ be the intersection of the plane y = z and the sphere x
2
+y
2
+z
2
=1, in the
direction counterclockwise with respect to the z-axis.
1). Compute the integral
H
Γ
x y z dz by parametrization the curve Γ.
2). Compute the integral
H
Γ
x y z dz by Stocks’ theorem.
Keywords. Line integral, orientation, Stocks’ theorem.
Suggested Solution.
1). The intersection of plane and the unit sphere is the following.
½
x
2
+y
2
+z
2
=1
y = z
=⇒
½
x
2
+2z
2
=1
y = z
We take the parameter equations,
x = cos t
y =
p
2
2
sin t
z =
p
2
2
sin t
t : 0 →2π.
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I
Γ
x y z dz =
Z
2π
0
cos t
Ã
p
2
2
sin t
!Ã
p
2
2
sin t
!Ã
p
2
2
cos t
!
dt
=
p
2
4
Z
2π
0
sin
2
t cos
2
t dt =
p
2
16
Z
2π
0
sin
2
(2t )dt
=
p
2
16
Z
2π
0
1 −cos 4t
2
dt =
p
2
16
π.
2). The circulation of a vector field F = 0i +0j +xyzk around the boundary Γ of the ori-
ented disc S in the direction counterclockwise with respect to its normal vector n =
(0,−
p
2
2
,
p
2
2
) equals the integral of ∇×F ·n over S.
I
Γ
F · dr =
Ï
S
∇×F ·ndσ.
I
Γ
F · dr =
I
Γ
x y zk ·( dx, dy, dz) =
I
Γ
x y z dz.
curl F = ∇×F
=
¯
¯
¯
¯
¯
¯
¯
i j k
∂
∂x
∂
∂y
∂
∂z
0 0 xyz
¯
¯
¯
¯
¯
¯
¯
=
∂x yz
∂y
i −
∂x yz
∂x
j =xzi −y zj =(xz,−y z,0).
Ï
S
∇×F ·ndσ =
Ï
S
(xz, −y z,0) ·(0, −
p
2
2
,
p
2
2
)dσ =
p
2
2
Ï
S
yz dσ
Now S is in the plane z = y, so dσ =
r
1 +
³
∂z
∂x
´
2
+
³
∂z
∂y
´
2
dx dy =
p
1 +0
2
+1
2
dx dy =
p
2dx dy. Then we find that
p
2
2
Ï
S
yz dσ =
Ï
x
2
+2y
2
=1
y
2
dx dy.
Let x =r cosθ, y =
p
2
2
r sinθ. We obtain dx dy =
p
2
2
r dr dθ, and hence
Ï
x
2
+2y
2
=1
y
2
dx dy =
Z
1
0
Z
2π
0
1
2
r
2
sin
2
θ
Ã
p
2
2
!
r dr dθ
=
p
2
4
Z
1
0
r
3
dr
Z
2π
0
1 −cos 2θ
2
dθ =
p
2
16
π.
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Problem 8. Computer the surface integral
Î
Σ
x y z dx dy, where Σ is the parts of unit sphere in
the region x ≥0, y ≥0, with normal vector outward.
Keywords. Surface integral, oriented surface.
Suggested Solution. Let us sketch the graph.
We divide Σ in two parts. Σ
1
is given by the equation z =
p
1 −x
2
−y
2
, and Σ
2
is given by the
equation z =−
p
1 −x
2
−y
2
. The orientation of Σ
1
is the same as the xy-plane, and is upward.
(The really meaning is that the dot product of the two normal vector for the two surfaces are
positive.) The orientation of Σ
2
is the opposite to the x y-plane. Denote the projection of Σ
1
(or Σ
2
) to the x y-plane by R
x y
. Then we obtain
Ï
Σ
x y z dx dy =
Ï
Σ
1
x y z dx dy +
Ï
Σ
2
x y z dx dy
=
Ï
R
x y
x y
q
1 −x
2
−y
2
dx dy −
Ï
R
x y
x y(−
q
1 −x
2
−y
2
)dx dy
= 2
Ï
R
x y
x y
q
1 −x
2
−y
2
dx dy
= 2
Ï
R
x y
r
2
sinθcosθ
p
1 −r
2
r dr dθ x =r cosθ, y =r sinθ
=
Z
π/2
0
sin2θdθ
Z
1
0
r
3
p
1 −r
2
dr
=
Z
1
0
(1 −s
2
)s
2
ds s =
p
1 −r
2
=
2
15
.
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Problem 9. Using Green’s theorem to compute the integral
H
C
(2x y −x
2
+x) dx +(x+y
4
+y
2
)dy,
where C is the boundary of the region R cutting out by two parabolas y = x
2
and x = y
2
, with
counterclockwise orientation.
Keywords. Green’s theorem in the plane.
Suggested Solution. Let us recall Green’s theorem in flux-divergence form. The outward flux
of a field F = Mi +Nj across a simple closed curve C equals the double integral of div F over
the region R enclosed by C.
I
C
F ·n ds =
I
C
M dy −N dx =
Ï
R
µ
∂M
∂x
+
∂N
∂y
dx dy.
Now let us take M =x +y
4
+y
2
, N =x
2
−x −2x y. We have
I
C
(2x y −x
2
+x)dx +(x +y
4
+y
2
)dy =
Ï
R
(
1 −2x
)
dx dy =
Z
1
0
dy
Z
p
y
y
2
(1 −2x)dx
=
Z
1
0
£
x −x
2
¤
p
y
y
2
dy =
Z
1
0
(
p
y −y −y
2
+y
4
)dy
=
2
3
−
1
2
−
1
3
+
1
5
=
1
30
.
Problem 10. Finding the work done by F = yi+zy+xz over the curve C, which is the intersec-
tion of the plane x + y +z = 1 with three coordinate planes, and clockwise orientation from
z-axis.
Keywords. Work done by a variable force over a space curve, Stocks’ theorem.
Suggested Solution.
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Method 1). Let Σ be the region of the plane x +y +z = 1 in the first octant, with norm vector
n =−
1
p
3
(1,1,1). Then the work
W =
I
C
F · dr =
Ï
Σ
∇×F ·ndσ
=
Ï
Σ
¯
¯
¯
¯
¯
¯
¯
−
1
p
3
−
1
p
3
−
1
p
3
∂
∂x
∂
∂y
∂
∂z
y z x
¯
¯
¯
¯
¯
¯
¯
dσ
=
p
3
Ï
Σ
dσ =
p
3 Area(Equilateral triangle)
=
p
3 ×
1
2
(
p
2) ×sin
π
3
=
3
2
.
Method 2). Let us divide C in two three line segments C
1
, C
2
, C
3
.
W =
I
C
F · dr =
I
C
y dx +z dy +x dz
=
Z
C
1
y dx +z dy +x dz
+
Z
C
2
y dx +z dy +x dz
+
Z
C
3
y dx +z dy +x dz
= 3
Z
C
1
y dx +z dy +x dz by the symmetry of the functions and regions
= 3
Z
x+y=1,z=0
0≤x≤1
y dx +z dy +x dz =3
Z
1
0
(1 −x)dx =
3
2
.
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