Multivariable Calculus MT234P Problems/Homework/Notes
Recommended Reading:M. Spivac Calculus on Manifolds,C. H. Edwards Advanced Calculus of Several Variables,C. Goffman Calculus of Several Variables,S. Lang Calculus of Several Variables,W. Fleming Functions of Several Variables
1 Homework
1. For t, x ∈ R, t > 0, let
f(t, x) =1√te−x2
4t .
For t0, x0 ∈ R, s > 0, compute(∂
∂tf
)(t0, x0) =
∂
∂tf(t0, x0) :=
d
dt
∣∣∣∣t=t0
f(t, x0) and∂2
∂x2f(t0, x0) :=
d
dx1
∣∣∣∣x1=x0
d
dx
∣∣∣∣x=x1
f(t0, x) .
Solution:∂
∂tf(t0, x0) =
−1
2t−32 e
−x2
4t + t−12x2
4t2e−x2
4t
∂2
∂x2f(t0, x0) =
∂
∂x
∣∣∣∣x=x0
1√t
−2x
4te−x2
4t =−1
2t−32 e
−x2
4t +1√t
−2x
4t
−2x
4te−x2
4t
hence∂
∂tf(t0, x0) =
∂2
∂x2f(t0, x0) ,
f solves the heat equation on R.
2. All vector spaces in this course will be subspaces of Rn, i.e. subsets V ⊂ Rn so that
(a) 0 ∈ V ,
(b) V + V ⊂ V (i.e. ∀x, y ∈ V : x+ y ∈ V ),
(c) RV ⊂ V (i.e. ∀α ∈ R, x ∈ V : αx ∈ V ).
A map f : V → W between vector spaces V , W is linear if
∀α ∈ R, x, y ∈ V : f(αx+ y) = αf(x) + f(y) .
A norm on a vector space V is a function n : V → R+0 = [0,∞) so that for all v, w ∈ V and α ∈ R
we have
(a) n(v) = 0 if and only if v = 0,
(b) n(αv) = |α|n(v),
(c) n(v + w) ≤ n(v) + n(w).
Which of the following functions ni : R2 → R+0 , i = 21, . . . , 24, are norms?
(1) n1(x, y) = |x|+ |y|(2) n2(x, y) = x+ y
(3) n3(x, y) = max
|x| ,
√x2+y2
2
(4) n4(x, y) =√|x|+
√|y|
For each of these functions ni, draw the level sets
n−1i (1) = (x, y) ni(x, y) = 1
If n is a norm, you need not hand up a proof for this. If n is not a norm, show (by an explicitexample) that n violates one of the properties above.
Examples: Let m,n, l : R2 → R+0 be given by
m(x, y) = max|y| ,
√|x2 − y2|
, n(x, y) = 4
√x4 + x2y2 + y4 , l(x, y) = max 3 |x| , 3 |y| , 2 |x+ y| .
The 1-level sets of these are
x
y
m(x, y) = max|y| ,
√|x2 − y2|
= 1
n(x, y) = 4√x4 + x2y2 + y4 = 1
l(x, y) = max 3 |x| , 3 |y| , 2 |x+ y| = 1
m violates the triangle inequality:
m((2, 1) + (2,−1)) = m(4, 0) = 4 6≤ m(2, 1) +m(2,−1) =√
3 +√
3 ,
so m is not a norm.
n is a norm.
l is a norm. In the homework a proof for this is not required. To see that l is a norm, we check 2a,2b, 2c: For 2a, clearly l(0, 0) = 0 and for x, y ∈ R we have l(x, y) = 0 only if x = y = 0. For 2b,compute
l(α(x, y)) = l(αx, αy) = max 3 |αx| , 3 |αy| , 2 |αx+ αy|= |α|max 3 |x| , 3 |y| , 2 |x+ y| = |α| l(x, y) .
For the triangle inequality 2c, let x, x′, y, y′ ∈ R be arbitrary. We estimate
l((x, y) + (x′, y′)) = l(x+ x′, y + y′) = max 3 |x+ x′| , 3 |y + y′| , 2 |x+ x′ + y + y′|
≤ max 3 |x|+ 3 |x′| , 3 |y|+ 3 |y′| , 2 |x+ y|+ 2 |x′ + y′|≤ max (3 |x| , 3 |y| , 2 |x+ y|+ 3 |x′| , 3 |y′| , 2 |x′ + y′|)≤ max 3 |x| , 3 |y| , 2 |x+ y|+ max 3 |x′| , 3 |y′| , 2 |x′ + y′|
l(x, y) + l(x′, y′) .
Solution: The level sets are
x
y
|x|+ |y|
x+ y
max
|x| ,
√x2+y2
2
3. Sketch the region in R2 between the graps of the functions f(x) = e−|x| and g(x) = −f(x), i.e. theset
A =
(x, y) x, y ∈ R,−e−|x| ≤ y ≤ e−|x|.
Compute the area of A.
Hint: A rectangle in R2 is a set R ⊂ R2 of the form R = (a, b)× (u, v), a, b, u, v ∈ R, a ≤ b, u ≤ v.A simple set S is a finite disjoint union of rectangles, i.e. S =
⋃ki=1Ri, where Ri are rectangles and
Ri ∩Rj = ∅ if i 6= j. The area of a rectangle is (by definition)
vol (a, b)× (u, v) := (b− a)(v − u) .
The area of a simple set S =⋃ki=1 Ri is
volk⋃i=1
Ri =k∑i=1
vol (Ri) .
The area of an arbitrary subset of A ⊂ R2 is
volA = sup vol (S) S a simple set with S ⊂ A .
Solution:
x
y
The area is
volA = 4
∫ ∞0
e−t dt = 4
Please hand up your solutions to problems 1-3 in the class on Monday, 5/2.
4. Recall the following definition/theorem: For a sequence x ∈ (Rk)N, x(n) = (x1(n), . . . , xk(n)) andL = (L1, . . . , Lk) ∈ Rk the follwing are equivalent:
(a) ∀j = 1, . . . , k : limn→∞
xj(n) = Lj
(b) limn→∞
‖x(n)− L‖2 = limn→∞
√√√√ k∑j=1
(xj(n)− Lj)2 = 0
(c) limn→∞
‖x(n)− L‖ = 0 for any norm ‖·‖ on Rk.
By definition, the sequence x converges to L, limn→∞ x(n) = L, if and only if one (hence all) of theseconditions holds.
Which of the sequences in R2 given below converge? Prove your answer.
(a) xa(n) =(
sin(n)n, 15n3−16
3n3+n−1
)Solution: lim
n→∞xa(n) = (0, 5)
(b) xb(n) =
(n
1n ,
1
n
n∑j=1
1√j
)Hint: Recall that n
1n = exp
(1n
ln(n)). You might estimate
∑nj=1
1√j
by an integral.
Solution: limn→∞
xb(n) =
(limn→∞
exp
(1
nln(n)
), limn→∞
1
n
n∑j=1
1√j
)= (1, 0). For the first compo-
nent, for each n ∈ N choosing k(n) so that 2k(n) ≤ n < 2k(n)+1 we estimate
1
nln(n) ≤ 1
2k(n)ln(2k(n)+1
)=k(n) + 1
2k(n)ln(2)
n→∞−→ 0
because ln increases and 2k ≥ k2 for k ≥ 4. For the right hand component, we estimate
n∑j=1
1√j≤ 1 +
n∑j=2
1√j≤ 1 +
n∑j=2
∫ j
j−1
1√xdx = 1 +
∫ n
1
1√xdx = 1 + 2
√n− 2
Hence for the limit we get
0 ≤ limn→∞
1
n
n∑j=1
1√j≤ lim
n→∞
1 + 2√n− 2
n= 0 .
(c) xc(n) =
(0 −1
212
0
)n(11
)Hint: Recall matrix multiplication,(
a bc d
)(xy
)=
(ax+ bycx+ dy
)and the notation (
a bc d
)n(xy
)=
(a bc d
)· · ·(a bc d
)︸ ︷︷ ︸
n times
(xy
)
Solution: For any (a, b) ∈ R2 the components of(0 −1
212
0
)(ab
)
are±a
2, ±b
2
, hence ∥∥∥∥(0 −1
212
0
)(ab
)∥∥∥∥∞
=1
2
∥∥∥∥(ab)∥∥∥∥∞.
It follows that
‖xc(n)‖∞ =1
2‖xc(n− 1)‖∞ = · · · = 1
2n−1‖xc(1)‖∞ =
1
2n−1
n→∞−→ 0 .
5. Recall the definition of the limit of a function f : U → Rk at U ⊂ Rm 3 a. We say f(x) converges toL ∈ Rk as x tends to a,
limx→a
f(x) = L or to stress U , limx→ax∈U
f(x) = L ,
if for any norms ‖·‖ on Rm resp. Rk,
∀ε > 0∃δε > 0∀x ∈ U, 0 < ‖x− a‖ < δε : ‖f(x)− L‖ < ε .
There also is a sequence criterion for this limit: For f , a, L as above,
limx→ax∈U
f(x) = L ⇐⇒ ∀x ∈ (U \ a)N limn→∞
x(n) = a : limn→∞
f(x(n)) = L
Because of this, in order to disprove that limx→a f(x) exists, you can also supply two sequencesu, v ∈ (Rm \ a)N so that limn→∞ f(u(n)) and limn→∞ f(v(n)) exist but are different.
Determine the limits below or prove that they do not exist.
(a) lim(x,y)→0
x2y2
x4 + y4
Solution: Consider the sequences u =((
1n, 0))n∈N and v =
((1n, 1n
))n∈N. Clearly
limn→∞
u(n) = 0 = limn→∞
v(n)
but, with f(x, y) := x2y2
x4+y4,
limn→∞
f(u(n)) = limn→∞
0(1n
)4+ 0
= 0
6= limn→∞
f(v(n)) = limn→∞
(1n
)4(1n
)4+(
1n
)4 = limn→∞
1
2=
1
2.
(b) lim(x,y)→0
x sin(xy)
x2 + xy + y2
Hint: Recall that |sin(u)| ≤ |u| for all u ∈ R.
Solution: For all x, y ∈ R, we have |xy| ≤ x2+y2
2=‖(x,y)‖22
2(“AGM”). Since ‖·‖1 and ‖·‖2 are
equivalent, there is c ∈ R+ so that for the numerator we estimate
|x sin(xy)| ≤ (|x|+ |y|) |xy| = ‖(x, y)‖1 |xy| ≤ c ‖x, y‖2
‖(x, y)‖22
2=c
2‖(x, y)‖3
2 .
The denominator is bounded below,
x2 + xy + y2 ≥ x2 − |xy|+ y2 = ‖(x, y‖22 − |xy| ≥
1
2‖(x, y‖2
2 .
For the quotient this gives∣∣∣∣ x sin(xy)
x2 + xy + y2
∣∣∣∣ ≤ c2‖(x, y)‖3
2
12‖(x, y‖2
2
= c ‖(x, y‖2
(x,y)→0−→ 0 .
Hence the limit exists and is 0.
(c) lim(x,y)→0
sin(xy)
x2 + xy + y2
Solution: This limit does not exist. To see this, consider the sequences((
1n, 0))n∈N and((
1n, 1n
))n∈N:
limn→∞
sin(( 1n× 0)
)(1n
)+ 0 + 0
= limn→∞
0(1n
)+ 0 + 0
= 0
but
limn→∞
sin(
1n× 1
n
)(1n
)2+(
1n
) (1n
)+(
1n
)2 = limu→0
sin(u)
3u=
1
3.
6. Recall the Bolzano-Weierstrass Theorem.
For sets A,B we denote by BA the sets of functions f : A→ B. A sequence is a function x : N→ Xfrom the set N of natural numbers to any set X.
If X is a set and x, y ∈ XN are sequences in X, then y is a subsequence of x if there is m ∈ NN
(strictly) increasing (i.e. for all n ∈ N, m(n+ 1) > m(n)) so that y = x m, i.e. for all k ∈ N,
y(k) = x(n(k)) .
Theorem 1 (Bolzano-Weierstrass) Every bounded sequence in Rd has a convergent subsequence.
Thus, if d ∈ N, c ∈ R and x ∈([−c, c]d
)Nthen there is m ∈ NN, m(n + 1) > m(n) for all n ∈ N,
such that x m = (x(m(j)))j∈N converges.
Let x, y, z ∈ RN be sequences so that for all n ∈ N,
x(n+1) = 3√|x(n)y(n)z(n)| , y(n+1) =
x(n) + y(n) + z(n)
3, z(n+1) = sin(x(n)+y(n)+z(n)) .
Prove that the sequence (x(n), y(n), z(n))n∈N has a convergent subsequence.
Hint: Estimate ‖(x(n), y(n), z(n))‖∞ = max |x(n)| , |y(n)| , |z(n)|Solution: If M = max |x(n)| , |y(n)| , |z(n)|, then
|x(n+ 1)| = 3√|x(n)y(n)z(n)| ≤ 3
√M3 = M ,
|y(n+ 1)| = |x(n) + y(n) + z(n)|3
≤ |x(n)|+ |y(n)|+ |z(n)|3
≤M
|z(n+ 1)| ≤ 1 .
Hence, for all X(n) = (x(n), y(n), z(n)), n ∈ N, we have
‖X(n)‖∞ ≤ max ‖X(n− 1)‖∞ , 1 ≤ max ‖X(n− 2)‖∞ , 1 ≤ · · · ≤ max ‖X(1)‖∞ , 1 .
It follows that the sequence is bounded. By the Bolzano-Weierstrass Theorem the sequence has aconvergent subsequence.
7. Let (xn)n∈N ∈ (R2)N
be defined by
x1 = (1, 1) and for n ∈ N, n > 1, xn+1 := f(xn) ,
where f : R2 → R2 is the map with
f(u, v) := (1− e−|v|, u/2) .
Does (xn)n∈N converge? Prove your answer.
Hint:∣∣1− e−|v|∣∣ ≤ |v|. Compare xn and xn+2.
Solution: For v ∈ R, ev ≥ 1 + v because ev > 0, ln is increasing and for v > −1,
ln(ev) = v ≥ ln(1 + v) =
∫ 1+v
1
1
sds .
We compute
‖f(f(u, v))‖1 =∣∣1− e−|u/2|∣∣+
∣∣∣∣1− e−|v|2
∣∣∣∣ ≤ |u/2|+ |v/2| ≤ ‖(u, v)‖1
hence
‖xn+2‖1 ≤1
2‖xn‖1 .
Since ‖x1‖1 = 2 and ‖x2‖1 = 1− e−1 + 12≤ 3
2this gives
‖x2k+1‖1 ≤2
2k=
1
2k−1and ‖x2k+2‖1 ≤
3/2
2k=
3
2k
which both converge to 0. The sequence therefore converges to (0, 0).
Please hand up your solutions to problems 4-7 in the tutor’s box before Friday, 16/02/2018
8. Let U ⊂ Rn, a ∈ U . A function f : U → Rk is continuous at a if (for any norms ‖·‖ on Rn, Rk)
∀ε > 0∃δε > 0∀x ∈ U, ‖x− a‖ ≤ δε : ‖f(x)− f(a)‖ ≤ ε . (2)
This can be phrased in terms of sequences,
Theorem 3 A function f : U → Rk, U ⊂ Rn, is continuous at a ∈ U if
∀x ∈ UN, limx = a : lim f x = f(a) .
The following functions fi : R2 → R are not continuous at the point a = (0, 0). In each case findε > 0 so that for no δε > 0 we have that ‖x− a‖ ≤ δε implies that |f(x)− f(a)|. Also in each casefind a sequence x ∈ (R2)N converging to a but so that fi x does not converge to fi(a). Say whichnorm you use on R2.
(a) fa(x, y) =
x2y2
x5−y5 if x 6= y
0 if x = y,
(b) fb(x, y) =
xy2
x2+y4if (x, y) 6= (0, 0)
0 if x = 0 = y,
(c) fc(x, y) =
sin(yx
)if x 6= 0
0 if x = 0,
(d) fd(x, y) =
e−1
x2+y2 if (x, y) 6= (0, 0)1 if x = 0 = y
.
Prove your statements.
Example: The function f : R2 → R given by f(x, y) =
√|xy|
|x|+|y| if (x, y) 6= (0, 0)
0 if x = 0 = yis not continuous
at (0, 0). Set ε := 110
. Then for any δ > 0 let x = y = δ/2. Then
‖(x, y)− a‖∞ = ‖(x, y)‖∞ = δ/2 < δ
but
|f(x, y)− f(a)| = 1
26≤ ε .
If x ∈ (R2)N is the sequence with
x(n) =
(1
n,
1
n
)then
limx = limn→∞
(1
n,
1
n
)= (0, 0) ,
but
lim f x = limn→∞
f
(1
n,
1
n
)= lim
n→∞
1
2=
1
2.
9. Show that the function h : Rn → Bn1 (0) := x ∈ Rn ‖x‖ < 1 with
h(x) =x
1 + ‖x‖
is continuous, bijective and that its inverse map h−1 : Bn1 (0)→ Rn is also continuous.
10. For which α, β ∈ R is the function f : R2 → R2 with
f(x, y) =
(x3y3
x2+y2, 2x2−x2y2+2y2
x2+y2
)if (x, y) 6= (0, 0)
(α, β) if x = 0 = y
continuous in (0, 0)?
11. A function f : U → Rk, U ⊂ Rn open, is differentiable at a ∈ U if there is a linear map daf : Rn → Rk
and a function R : U × U so that
∀x ∈ U : f(x) = f(a) + daf(x− a) +R(a, x) (4)
and
limx→a
R(a, x)
‖x− a‖= lim
h→0
R(a, a+ h)
‖h‖= 0 .
The linear map daf is uniquely determined by these conditions and is called the derivative of f at a.
Compute the derivative of the following functions fi : R2 → R, i.e. find a formula for d(a,b)fi(x, y).
(1) f1(x, y) = ‖(x, y)‖42 =
(x2 + y2
)2
Hint: Calculate with the scalar product.
Solution: f1(v) = ‖v‖42 = 〈v v〉2, hence for h ∈ R2,
f1(v + h) = 〈v + h v + h〉2 = (〈v v〉+ 2 〈v h〉+ 〈h h〉)2
=(‖v‖2
2 + 2 〈v h〉+ ‖h‖22
)2
= ‖v‖42 + 4 ‖v‖2
2 〈v h〉︸ ︷︷ ︸dvf1h
+ 2 ‖v‖22 ‖h‖
22 + 4 〈v h〉 ‖h‖2
2 + 4 〈v h〉2 + ‖h‖42︸ ︷︷ ︸
R(v,v+h)
In order to show that
limh→0
R(v, v + h)
‖h‖2
= 0
we estimate ∣∣∣∣R(v, v + h)
‖h‖2
∣∣∣∣ =
∣∣∣∣∣2 ‖v‖22 ‖h‖
22 + 4 〈v h〉 ‖h‖2
2 + 4 〈v h〉2 + ‖h‖42
‖h‖2
∣∣∣∣∣=
∣∣∣∣∣2 ‖v‖22 ‖h‖2 + 4 〈v h〉 ‖h‖2 + ‖h‖3
2 +4 〈v h〉2
‖h‖2
∣∣∣∣∣≤∣∣2 ‖v‖2
2 ‖h‖2 + 4 〈v h〉 ‖h‖2 + ‖h‖32
∣∣+
∣∣∣∣∣4 〈v h〉2
‖h‖2
∣∣∣∣∣ by the triangle inequality
≤∣∣2 ‖v‖2
2 ‖h‖2 + 4 〈v h〉 ‖h‖2 + ‖h‖32
∣∣+
∣∣∣∣∣4 ‖v‖22 ‖h‖
22
‖h‖2
∣∣∣∣∣ by Cauchy-Schwarz
= ‖h‖2
(∣∣2 ‖v‖22 + 4 〈v h〉+ ‖h‖2
2
∣∣+ 4 ‖v‖22
) h→0−→ 0 .
(2) f2(x, y) = x3 + xy
Solution:
f2(a+ x, b+ y) = (a+ x)3 + (a+ x)(b+ y) = a3 + ab+ 3a2x+ bx+ ay︸ ︷︷ ︸d(a,b)f2(x,y)
+ 3ax2 + xy + x3︸ ︷︷ ︸R((a,b)(a+x,b+y)
In order to show that
lim(x,y)→0
R((a, b)(a+ x, b+ y)
‖(x, y)‖= 0
we use the max-norm to estimate∣∣∣∣R((a, b)(a+ x, b+ y)
‖(x, y)‖∞
∣∣∣∣ =|3ax2 + xy + x3|
max |x| , |y|
≤ |3ax2|max |x| , |y|
+|xy|
max |x| , |y|+
|x3|max |x| , |y|
by the triangle inequality
≤ |3a|max |x| , |y|+ max |x| , |y|+ max |x| , |y|2 because |x| , |y| ≤ max |x| , |y|
As max |x| , |y| → 0 this clearly tends to 0.
(3) f3(x, y) = sin xy
Solution: The sine function is differentiable with derivative the cosine function. Hence wehave
sin(u+ δ) = sin(u) + δ cos(u) + δc(u, δ)
with a function c so that limδ→0 c(u, δ) = 0. Hence
sin((a+x)(b+y)) = sin(ab+ay+bx+xy) = sin(ab)+(ay+bx+xy) cos(ab)+(ay+bx+xy)c(ab, ay+bx+xy)
= sin(ab) + (ay + bx) cos(ab)︸ ︷︷ ︸d(a,b)f3(x,y)
+xy cos(ab) + (ay + bx+ xy)c(ab, ay + bx+ xy)︸ ︷︷ ︸R((a,b),(a+x,b+y))
We estimate the remainder wrt the maximum norm,∣∣∣∣R((a, b), (a+ x, b+ y))
‖x, y‖∞
∣∣∣∣ =
∣∣∣∣xy cos(ab) + (ay + bx+ xy)c(ab, ay + bx+ xy)
‖x, y‖∞
∣∣∣∣≤ ‖x, y‖∞ + (|a|+ |b|+ ‖x, y‖∞)c(ab, ay + bx+ xy︸ ︷︷ ︸
h→0−→0
)h→0−→ 0
Example: To compute the derivative of the function f : R2 → R with f(x, y) = ex2+y2 = e‖(x,y)‖22 , we
expand
f(p+ h) = e‖p+h‖22 = e‖p‖
2+2〈p h〉+‖h‖22 (5)
We now use that the exponential function is differentiable. Thus
ex+δ = ex + δex + r(x, δ) = ex + δex + δc(x, δ) (6)
with a functions r so that limδ→0
r(x, δ)
δ= 0, or a function c so that lim
δ→0c(x, δ) = 0. Thus we can
continue (5),
f(p+ h) = e‖p‖22 +
(2 〈p h〉+ ‖h‖2
2
)e‖p‖
22 + r
(‖p‖2
2 , 2 〈p h〉+ ‖h‖22
)= f(p) + 2 〈p h〉 e‖p‖
22︸ ︷︷ ︸
dpfh
+ ‖h‖22 e‖p‖22 + r
(‖p‖2
2 , 2 〈p h〉+ ‖h‖22
)︸ ︷︷ ︸R(p,p+h)
.
We need to show that
limh→0
R(p, p+ h)
‖h‖= 0 .
To this end we estimate∣∣∣∣R(p, p+ h)
‖h‖2
∣∣∣∣ =
∣∣∣∣∣‖h‖22 e‖p‖22 + r
(‖p‖2
2 , 2 〈p h〉+ ‖h‖22
)‖h‖2
∣∣∣∣∣≤‖h‖2
2 e‖p‖22 +
∣∣r (‖p‖22 , 2 〈p h〉+ ‖h‖2
2
)∣∣‖h‖2
by the triangle inequality
≤ ‖h‖2 e‖p‖22 +
∣∣2 〈p h〉+ ‖h‖22
∣∣ c (‖p‖22 , 2 〈p h〉+ ‖h‖2
2
)‖h‖2
≤ ‖h‖2 e‖p‖22 +
(2 ‖p‖2 ‖h‖2 + ‖h‖2
2
)c(‖p‖2
2 , 2 〈p h〉+ ‖h‖22
)‖h‖2
by Cauchy-Schwarz
= ‖h‖2 e‖p‖22 + (2 ‖p‖2 + ‖h‖2) c(‖p‖2
2 , 2 〈p h〉+ ‖h‖22︸ ︷︷ ︸
h→0−→0
)h→0−→ 0 by(6) .
Please hand up your solutions to problems 8-11 in the tutor’s box before Friday, 09/03/2018
12. Recall matrix notation for linear maps. For i,m ∈ N, 1 ≤ i ≤ m, we denote by emi ∈ Rm the vector
emi =
0 ← 1
...01 ← i
0...0 ← m
For n, k ∈ N and ai,j ∈ R, 1 ≤ i ≤ k, 1 ≤ j ≤ n we denote bya11 · · · a1,n
......
ak,1 · · · ak,n
the linear map A : Rn → Rk so that
Aenj =k∑i=1
ai,jei .
for all j, 1 ≤ j ≤ n.
Compute the (matrix of the) derivative d(1,2,1)f of the function f : R3 → R4 given by
f(x, y, z) =
x2
x2 + y2
ex + y + zxyz
.
Solution: The ith column of the matrix of d(1,2,1)f is
d(1,2,1)fe3i =
d
dt
∣∣∣∣t=0
f
121
+ te3i
i.e.
d
dt
∣∣∣∣t=0
f
1 + t21
,d
dt
∣∣∣∣t=0
f
1 + t2 + t
1
,d
dt
∣∣∣∣t=0
f
12
1 + t
.
Thus
d(1,2,1)f =
2x 0 02x 2y 0ex 1 1yz xz xy
∣∣∣∣∣∣∣∣x=1,y=2,z=1
=
2 0 02 4 0e 1 12 1 2
13. Recall the chain rule: If U ⊂ Rm, V ⊂ Rn are open, a ∈ U , g : U → V , f : V → Rk, g differentiable
at a, f differentiable at f(a), then the composition f g is differentiable at a and
da(f g) = dg(a)fdag .
For the proof, expand g and f according to (4),
g(a+ h) = g(a) + dagh+Rg(a, h)
f(g(a) + h′) = f(g(a)) + dg(a)fh′ +Rf (g(a), h′)
with
limh→0
Rg(a, h)
‖h‖= 0 , lim
h′→0
Rf (g(a), h′)
‖h′‖= 0 .
Thenf(g(a+ h)) = f(g(a) + dagh+Rg(a, h)︸ ︷︷ ︸
h′
)
= f(g(a)) + dg(a)f(dagh+Rg(a, h)) +Rf (g(a), dagh+Rg(a, h))
= f(g(a)) + dg(a)fdagh︸ ︷︷ ︸da(fg)h
+ dg(a)fRg(a, h) +Rf (g(a), dagh+Rg(a, h))︸ ︷︷ ︸
Rfg(a,h)
and
limh→0
dg(a)fRg(a, h) +Rf (g(a), dagh+Rg(a, h))
‖h‖= 0 .
Compute the derivatives of the following functions fi : R2 → R2 at 0.
(a) fa(x, y) =
(det
(1 + xy y + y2
x 1 + 3x
), trace
(1 + 4x+ x2 5y6y + xy + x3 1 + 7x
))Hint: The trace of a matrix is the sum of its diagonal entries. Also recall that the trace is thederivative of the determinant at the identity, did det = trace.
Solution:
d(0,0)fa(x, y) =
(trace
(0 yx 3x
), trace
(1 + 4x 5y
6y 1 + 7x
))= (3x, 2 + 11x)
(b) fb(x, y) = trace
(1 + x yy 1 + x
)7
Hint: Do not compute
(1 + x yy 1 + x
)7
. What is the derivative of the function p7 : A 7→ A7
on 2× 2-matrices? You only need this at the identity, expand (id +h)7.
Solution: Since
(id +h)7 = id +7h+
(72
)h2 + . . .+ h7︸ ︷︷ ︸R(id,h)
and R satisfies limh→0
R(id, h)
‖h‖= 0 we have
didp7h = 7h .
The function in question is the trace of the composition of this with
g : R2 → R4 , (x, y) 7→(
1 + x yy 1 + x
)= id +
(x yy x
)hence
d(0,0)g(x, y) =
(x yy x
).
Since the trace is linear, the chain rule gives
d(0,0)fb(x, y) = trace dg(0,0)p7d(0,0)g(x, y) = trace
(7
(x yy x
))= 14x
(c) fc = g3 = g g g where g : R2 → R2 is the function with g(x, y) = (y2, 1 + x).
Solution: The derivative of g at (a, b) ∈ R2 is
da,bg(x, y) = (2by, x) =
(0 2b1 0
).
By the chain rule, since g(0, 0) = (0, 1), g(g(0, 0)) = (1, 1)
d(0,0)f = dg(g(0,0))gdg(0,0)gd(0,0)g
=
(0 21 0
)(0 21 0
)(0 01 0
)=
(0 21 0
)(2 00 0
)=
(0 02 0
)i.e.
d0,0f(x, y) = (0, 2x) .
(d) fd(x, y) =
⟨ ex+y
xy 3
√1+x2
1+x4
x
× sin(y)
2x+ cos(y)x
yy1
⟩2
Hint: Recall that the vector product of a, b ∈ R3 is the vector a × b so that for every z ∈ R3
the identity〈a× b z〉 = det(a, b, c)
holds.
Solution: The function fd is the composition g h of the functions
h : R2 →M(3× 3,R) , h(x, y) =
ex+y sin(y) y
xy 3
√1+x2
1+x42x+ cos(y) y
x x 1
g : M(3× 3,R)→ R2 , g(A, y) =
(detA
2
).
By the chain rule, since h(0) = id, did det = trace,
d0f = d0(g h) = dh(0)gd0h =
(trace d0h
0
)=
(d0(trace h)
0
)=
(3 10 0
).
14. For a norm n on Rn we denote by Bnr (p) the set
Bnr (p) = x ∈ Rn n(x− p) < r .
By definition, a subset A ⊂ Rn is open if
∀a ∈ A∃r ∈ R+ : Bnr (p) ⊂ A . (7)
(a) Prove that all norms on Rn define the same open subsets, i.e. (7) wrt a norm n is equivalent to(7) for any other norm.
(b) Sketch the following subsets of R2. Which are open?
i. (x, y) ∈ R2 x2 = y2 ,ii. (x, y) ∈ R2 x+ y < 3 ,iii.
(x, y) ∈ R2 1 < exy2< 3,
Prove your answer.
15. Let α ∈ Hom(Rn,R) and f : Rn → R be differentiable and so that f ≥ α (i.e. f(x) ≥ α(x) for allx ∈ Rn). Show that for p ∈ Rn we have dpf = α if f(p) = α(p). Does the converse hold?
Solution: Since f is differentiable at p we have
f(p+ h) = f(p) + dpfh+R(p, h) ≥ α(p+ h) = α(p) + α(h)
for all h ∈ Rn. If f(p) = α(p) this gives
dpfh− α(h) +R(p, h) ≥ 0 ,
hence for any t ∈ R
dpfth− α(th) +R(p, th) = t (dpfh− α(h)) +R(p, th) ≥ 0
and therefore
dpfh− α(h) = limt→0
t (dpfh− α(h))
t= lim
t→0
t (dpfh− α(h)) +R(p, th)
t≥ 0 .
Thus dpf − α : h 7→ dpfh − α(h) is a linear map Rn → R with image a subbset of R+0 , which is
impossible for a nontrivial linear map.
The converse does not hold. For an example let f : R→ R, f(x) = x3, α = 0, p = 0.
Please hand up your solutions to problems 12-15 in the tutor’s box before Tuesday, 3/4/2018
16. Let U ⊂ Rn be a closed bounded subset and f : U → R be a continuous function. Prove that fassumes its maximum on U , i.e. there is x ∈ U so that
f(x) = max f(U) = sup f(U) .
Hint: Bolzano-Weierstrass. You might first prove that f(U) is bounded. By completeness, sup f(U)then exists.
Solution: To see that f(U) is bounded above, assume the contrary. Then there is a sequence u ∈ UN
so that∀n ∈ N : f(un) > n .
By the Bolzano-Weierstrass Theorem, this sequence contains a convergent subsequence. Since U isclosed the limit of this subsequence lies in U . Renaming if necessary, we thus may assume limu =x ∈ U . By continuity, n < f(un)
n→∞−→ f(x), a contradiction. Thus f(U) is bounded above and bythe completeness of R, S = sup f(U) exists. Thus there is u ∈ UN with
∀n ∈ N : S ≥ f(un) ≥ S − 1
n. (8)
As before, by the Bolzano-Weierstrass Theorem, u contains a convergent subsequence with limit inU . Renaming the un if necessary, we can assume limu = x ∈ U . By continuity and because of (8),
S ≥ f(x) = limn→∞
f(un) ≥ limn→∞
S − 1
n= S .
Hence f(x) = sup f(U).
17. Let f : U → R, U ⊂ Rn and a ∈ U . Then f has a (loal) maximum at a if (there is a an open subsetV ⊂ U such that) f(a) ≥ f(x) for all x ∈ U (for all x ∈ V ), i.e. f(a) = max f(U) (f(a) = max f(V )).If f is differentiable, then a is a critical point of f if daf = 0.
Find all critical points of the function f : R3 → R,
f(x, y, z) = xyz − x4 − y4 − z4
and decide which are (local) maxima/minima.
Hint: The function is symmetric, i.e. invariant under permutation of the coordinates. For deter-mining which of the critical points are (local) maxima/minima, problem 16 might be useful.
Solution: The derivative of f is
d(x,y,z)f = (yz − 4x3, xz − 4y3, xy − 4z3) .
If (x, y, z) is critical, thenxyz = 4x4 = 4y4 = 4z4
hence
|x| = |y| = |z| = 0 or =1
4and either none or two of the coordinates are negative. Thus, up to permutation of the x, y, z, thecritical points are
(0, 0, 0) ,
(1
4,1
4,1
4
),
(−1
4,−1
4,1
4
)and f takes the values
f(0, 0, 0) = 0 , f
(1
4,1
4,1
4
)=
1
44, f
(−1
4,−1
4,1
4
)=
1
44.
If |x| , |y| , |z| > 1 then f(x, y, z) < 0. By problem 16 f has a maximum in (x, y, z) |x| , |y| , |z| ≤ 1.It follows that the non-zero critical points are all global maxima of f . The remaining critical pointn = (0, 0, 0) is no local minimum because f(x, 0, 0) = −x4 < 0 for all x. Thus arbitrarily close to nthere is a point p with f(p) < f(n). Nor is n a local maximum, because f(x, x, x) = x3 − 4x4 > 0 =f(n) for x ∈ (0, 1/4).
18. Find all critical points of the function f : R2 → R with
f(x, y) = ey cos(x) + ex cos(y) .
Solution: The derivative of f is
d(x,y)f = (−ey sin(x) + ex cos(y), ey cos(x)− ex sin(y)) = (ex, ey)
(cos(y) − sin(y)− sin(x) cos(x)
)︸ ︷︷ ︸
Q
.
SincedetQ = cos(x) cos(y)− sin(x) sin(y) = cos(x+ y) = 0
the point (x, y) can only be critical if x+ y is a zero of cos, i.e. an odd multiple of π/2, say
x+ y = (2k + 1)π
2.
If k = 2l is even, then x+ y = 2lπ + π2
and cos(y) = cos(π2− x) = sin(x) as well as cos(x) = sin(y).
In this case it follows that ex = ey, hence
x = y =(4l + 1)π
4.
If k = 2l − 1 is odd, then x + y = 2lπ − π2
and cos(y) = cos(−π2− x) = − sin(x) as well as
cos(x) = − sin(y). This would imply that ex = −ey, which is impossible.
Please hand up your solutions to problems 16-18 in the tutor’s box before Friday, 13/04/2018
19. Find all local/global maxima/minima of the function f : R4 → R with
f(x, y, z, w) = x3w3 + z2 + x2 + yzw .
Solution: In a critical point (x, y, z, w),
d(x,y,z,w)f = (3x2w3 + 2x, zw, 2z + yw, 3x3w2 + yz) = 0 ,
we must have zw = 0, hence z = 0 or w = 0. If w = 0 then x = 0 = z, and all points (0, y, 0, 0) arecritical, f(0, y, 0, 0) = 0.
If z = 0 and w 6= 0, then y = 0 = x, and all points (0, 0, 0, w) are critical, f(0, 0, 0, w) = 0. Thesecond derivative at these points is
d2(0,y,0,w)f =
0 0 0 00 0 w 00 w 2 y0 0 y 0
.
This is indefinite if w 6= 0 or y 6= 0. In case w 6= 0 this is because
d2((0,y,0,w))f
0ab0
,
0ab0
= 2b2 + 2wab
which is positive for a = 0, b 6= 0 and negative if a = −2b/w. Thus the critical points (0, 0, 0, w)with w 6= 0 and the points (0, y, 0, 0) with y 6= 0 can not be local extrema.
At 0 = (0, 0, 0, 0), the second derivative is positve semidefinite, not zero. Thus arbitrarily close to0 there are points q with f(q) > f(0) = 0, hence 0 is not a local maximum. To see that 0 is not alocal minimum, for w ∈ R+ consider points(
w3,−w,w3, w)
withf(w3,−w,w3, w) = w12 + 2w6 − w5 = w5(w7 + 2w − 1)
which is negative for w sufficiently small.
20. Find the best constants c, k (i.e. k as small as possible, c as large as possible) so that
c ‖x‖4 ≤ ‖x‖2 ≤ k ‖x‖4 for all x ∈ Rn . (9)
Solution: For k, we maximize
k = max
‖x‖2
‖x‖4
x ∈ Rn \ 0
= min ‖x‖2 ‖x‖4 = 1
k2 = maxx2
1 + . . .+ x2n x4
1 + . . .+ x4n = 1, x1, . . . , xn ≥ 0
If f , f(x) = ‖x‖2
2 has a maximum at x given g(x) = x41 + . . .+ x4
n = 1, then there is λ ∈ R so that
2xi = λ× 4x3i for all i = 1 . . . n
hence xi = 0 or x2i = 1/(2λ). Thus some, possibly none, of the xi are 0 and the others are equal up
to sign. Up to permutation of the coordinates, we may assume
x1 = ±x2 = · · · = ±xj = ± 14√j
, xj+1 = · · · = xn = 0 for some j .
Thus the critical points of f given g = 1 arepj :=
± 14√j, . . . ,± 1
4√j︸ ︷︷ ︸
j
, 0, . . . , 0︸ ︷︷ ︸n−j
j = 0 . . . n
.
The value of f does not depend on the sign, so
f(pj) =√j = 1,
√2, . . . ,
√n
of which√n is the largest. Thus k = 4
√n.
For c, we need to minimize f . Thus the largest c satisfying (9) is 1.
21. Let f ∈ C2(Rn,R) = f : Rn → R f is twice differentiable and the second derivative is continuousbe convex, i.e.
∀x, y ∈ Rn : f
(x+ y
2
)≤ f(x) + f(y)
2. (10)
(a) Prove that the second derivative of f is everywhere positive semidefinite.
Solution: Let p, h ∈ Rn, t ∈ R. Then since f is twice differentiable, we have
f(p± th) = f(p)± dpfh+ 2t2d2pf(h, h) +R(p,±th) , lim
t→0
R(p,±th)
t2= 0
Adding these two equations gives
f(p+ th) + f(p− th) = 2f(p) + 4t2d2pf(h, h) +R(p, th) +R(p,−th)
hence
d2pf(h, h) =
f(p+ th) + f(p− th)− 2f(p)−R(p, th)−R(p,−th)
4t2
= limt→0
f(p+ th) + f(p− th)− 2f(p)−R(p, th)−R(p,−th)
4t2
= limt→0
f(p+ th) + f(p− th)− 2f(p)
4t2≥ 0
by (10)
(b) Show that every critial point of f is a minimum.
Solution: Let x, y ∈ Rn, y = x + h, and f(x) > f(y). Without loss of generality, we mayassume x = 0, f(x) = 0. Then by (10),
f(h/2) ≤ f(0) + f(h)
2=f(h)
2.
Iterating this argument shows that
f(h/2n) ≤ f(0) + f(h)
2=f(h)
2n
hence
d0fh = limt→0
f(th)
t= lim
n→0
f(2−nh)
2−n≤ f(h) < 0 .
and x = 0 is not a critical point.
22. Sketch the domain
Ω =
(u cos(t)
t,u sin(t)
t
)t ∈ (2π, 4π), u ∈ (1, 2)
and compute its area.
Solution:
Ω = Φ(Q) where
Φ: R+ × R→ R2 , Φ(t, u) =
(u cos(t)
t,u sin(t)
t
),
Q = (2π, 4π)× (1, 2) .
Φ is a diffeomorphism of Q with Φ(Q). Its derivative is
d(t,u)Φ =
(−u sin(t)
t− u cos(t)
t2cos(t)t
u cos(t)t− u sin(t)
t2sin(t)t
)
whose determinant is
det d(t,u)Φ = det
(−u sin(t)
t− u cos(t)
t2cos(t)t
u cos(t)t− u sin(t)
t2sin(t)t
)= det
(−u sin(t)
tcos(t)t
u cos(t)t
sin(t)t
)
=u
t2det
(− sin(t) cos(t)cos(t) sin(t)
)=−ut2
.
Thus
vol Ω =
∫Q
∣∣det d(t,u)Φ∣∣ =
∫ 4π
2π
∫ 2
1
u
t2du dt =
3
2
(1
2π− 1
4π
)=
3
8π.
Please hand up your solutions to problems 19-22 in the tutor’s box before Friday, 27/04/2018
2 Some problems for the study period
1. Recall the definitions: What does it mean for a function f : U → Rk, U ⊂ Rn open, to be continuous,differentiable, twice differentiable at a point p ∈ U?
2. Recall the main theorems: (Fundamental Theorem of Calculus), Bolzano-Weierstrass, Chain Rule,Lagrange Multipliers, Transformation formula
3. Let v, w ∈ R3. Prove that the function f : R3 → R3 with
f(x) = ‖x‖2 v × x+ 〈v x〉x
is differentiable and compute its derivative.
Solution: The function is polynomial,
f(x+ h) = ‖x+ h‖2 v × (x+ h) + 〈v x+ h〉 (x+ h)
= ‖x‖2 v × x+ 〈v x〉x︸ ︷︷ ︸=f(x)
+2 〈x h〉 v × x+ ‖x‖2 v × h+ 〈v x〉h+ 〈v h〉x︸ ︷︷ ︸=:lx(h)
+ ‖h‖2 v × x+ 2 〈x h〉 v × h+ 〈v h〉h+ ‖h‖2 v × h︸ ︷︷ ︸=:R(x,h)
.
Clearly, lx(h) is linear in h. Since‖v × x‖ ≤ ‖v‖ ‖x‖
and by Cauchy-Schwarz, we can estimate R(x, h),
‖R(x, h)‖ ≤ 3 ‖h‖2 ‖v‖ ‖x‖+ ‖v‖ ‖h‖2 + ‖v‖ ‖h‖3 ,
hence ∥∥∥∥R(x, h)
‖h‖
∥∥∥∥ ≤ 3 ‖h‖ ‖v‖ ‖x‖+ ‖v‖ ‖h‖+ ‖v‖ ‖h‖2 h→0−→ 0 .
Thus f is differentiable at x and dxf = lx.
4. Let f : R2 → R be twice continuously differentiable with
d2(x,y)f =
(x 11 y
)(11)
for all x, y ∈ R and d0f = 0, f(0) = 1. Compute f(1, 1). Find all local/global extrema of thisfunction.
Hint: For a given point p = (x, y) ∈ R2 look at the function g : R→ R, g(t) = f(tp) = f(tx, ty) andcompute g′, g′′.
Solution: Equation (11) means that
d2(x,y)f(u, v) =
⟨u
(x 11 y
)v
⟩.
We first compute the differential at p = (x, y) ∈ R2. By the Fundamental Theorem of Calculus,
dpf = d0f +
∫ 1
0
d
dt
∣∣∣∣t=τ
dtpf dτ =
∫ 1
0
d2τpf · p dτ
=
∫ 1
0
((τx 11 τy
)(xy
))tdτ =
∫ 1
0
((τx2 + yx+ τy2
))tdτ =
(1
2x2 + y, x+
1
2y2
).
Again by the Fundamental Theorem of Calculus, we can now compute f(p),
f(x, y) = f(p) = f(0) +
∫ 1
0
d
dt
∣∣∣∣t=τ
f(tp) dτ = 1 +
∫ 1
0
dτpf · p dτ
= 1 +
∫ 1
0
(1
2(τx)2 + τy, τx+
1
2(τy)2
)·(xy
)dτ
= 1 +
∫ 1
0
1
2τ 2x3 + τxy + τxy +
1
2τ 2y3 dτ
= 1 +x3
6+ xy +
y3
6
hence f(1, 1) = 7/3.
The critical points of the function are the points (x, y) with d(x,y)f =(
12x2 + y, x+ 1
2y2)
= 0, i.e.with
x2 = −2y , y2 = −2x .
Thus the critical points are(0, 0) and (−2,−2) .
At (0, 0) the second derivative is
(0 11 0
), which is indefinite, hence (0, 0) is a saddle, no local
extremum. At (−2,−2) the second derivative is
(−2 11 −2
), which is negative definite, hence a local
maximum. Since f(x, x)x→∞−→ +∞, the function has no global maximum.
5. Find the extrema of f : (x, y, z) ∈ R3 x2 + y2 + z2 ≤ 1 → R, given by
f(x, y, z) = x3 − z2 .
Solution: By problem 16, since the domain of f is bounded and closed, f must have a minimumand a maximum.
If p = (x, y, z) is an extremum in the interior of the domain of f , then dpf = 0,
d(x,y,z)f = (3x2, 0,−2z) = 0 hence x = 0 = z .
The second derivative of f at a point (0, y, 0),
d2(0,y,0) =
6x 0 00 0 00 0 −2
∣∣∣∣∣∣x=0,z=0
=
0 0 00 0 00 0 −2
is negative semidefinite, so we can only have a maximum at such a point. However, for any x positivewe have f(x, y, 0) = x3 > f(0, y, 0), hence (0, y, 0) can not be a local maximum. Thus there are noextrema in the interior of the domain of f .
Thus the extrema of f must lie on the boundary of its domain. The critical points of f on(x, y, z) ∈ R3 x2 + y2 + z2 = 1 are the points (x, y, z) with
3x2 = λx
0 = λy
−2z = λz
x2 + y2 + z2 = 1
henceλ = 0 , x = z = 0 , y = ±1 or
y = 0 6= λ ,
In the second case, if z = 0, then x = ±1. If z 6= 0 then λ = −2 = 3x or x = 0. Thus the criticalpoints can only be
(0,±1, 0) , (±1, 0, 0) , (0, 0,±1) ,
(−2
3, 0,±
√5
3
)with
f(0,±1, 0) = 0 , f(±1, 0, 0) = ±1 , f(0, 0,±1) = −1 , f
(−2
3, 0,±
√5
3
)=−8
27−5
9=−23
27.
Hence the maximum of f is 1 at (1, 0, 0) and minimum is −1 assumed at (−1, 0, 0) and (0, 0,±1).
6. Maximize x+ y + z + u3 given x2 + y2 + z2 + u2 = 4/3.
Solution: Let f, g be the functions on R4 with g(x, y, z, u) = x2 + y2 + z2 + u2 and f(x, y, z, u) =x + y + z + u + u3. Since dpg 6= 0 for all p with g(p) = 4/3, we can use Lagrange multipliers. If(x, y, z, u) is a critical point of f restricted to g−1(4/3), then there is λ ∈ R so that
d(x,y,z,u)f = (1, 1, 1, 3u2) = λ(x, y, z, u) =λ
2d(x,y,z,u)g
hence x = y = z and u = 0 or
λ =1
x=
1
y=
1
z= 3u
x = y = z =1
3uhence
1
3u2+ u2 =
4
3, u2 =
2
3±√
4
9− 1
3=
2± 1
3= 1 or
1
3.
Thus the restricted critical points are
±(
2
3,2
3,2
3, 0
)or ±
(1
3,1
3,1
3, 1
)or ±
(1√3,
1√3,
1√3,
1√3
).
At these points, the function f takes the values
±8
3, ± 2 , ±
(√3 +
1
3√
3
)= ±10
9
√3
respectively. Since 83> 2 > 10
9
√3, the maximum is 8/3 at
(23, 2
3, 2
3, 0).
7. Let f : Rn → Rk be a twice continuously differentiable function with f(0) = 0 and d2pf = 0 for all
p ∈ Rn. Show that f is a linear map.
Hint: Show that f(x) = d0f · x for all x ∈ Rn. For x ∈ Rn consider the function t 7→ f(tx), t ∈ Rn.
Solution: Since the second derivative of f , i.e. the derivative of the function
df : Rn → Hom(Rn,Rk) , p 7→ dpf
vanishes, the function df is constant, i.e. dpf = d0f for all p ∈ Rn. For x ∈ Rn and let gx : R→ R bethe function with gx(t) = f(tx). Let c : R→ Rn be the function with c(t) = tx. Clearly, dtc · h = hxfor h ∈ R, t ∈ R. Thus gx = f c. By the Chain Rule
g′x(t) = dtgx · 1 = dc(t)f dtc · 1 = dtxf · x
Then by the Fundamental Theorem of Calculus,
f(x) = gx(1) = gx(0) +
∫ 1
0
g′x(t) dt = 0 +
∫ 1
0
dtxf · x dt =
∫ 1
0
d0f · x dt = d0f · x .
8. Compute the volume of(x− ty, y − tx, xy + t) ∈ R3 0 < x < 1, 0 < y < x, 0 < t < 1
Solution: We parametrize this domain by
Φ: (x, y, t) 0 < x < 1, 0 < y < x, 0 < t < 1 →
(x− ty, y − tx, xy + t) ∈ R3 0 < x < 1, 0 < y < x, 0 < t < 1
withΦ(x, y, t) = (x− ty, y − tx, xy + t) .
The differential of Φ is
d(x,y,t)Φ =
1 −t −y−t 1 −xy x 1
with determinant
det d(x,y,t)Φ = 1− t2 − x(−x− yt) + y(tx+ y) = 1− t2 + x2 + 2xyt+ y2 > 0 .
Hence the volume of the given domain is∫(x,y,t) 0<x<1,0<y<x,0<t<1)
det d(x,y,t)Φ dxdydt =
∫ 1
0
∫ 1
0
∫ x
0
1− t2 + x2 + 2xyt+ y2 dydxdt
=
∫ 1
0
∫ 1
0
x− xt2 + x3 + x2t+1
3x3 dxdt
=
∫ 1
0
1
2− 1
2t2 +
1
4+
1
3t+
1
12dt
=1
2− 1
6+
1
4+
1
6+
1
12=
10
12.
9. Sketch the domain (uv + v2, v − u) ∈ R2 u, v ∈ (0, 2), u < v
and compute its area.
Solution:
The domain, Ω, is parametrized by
Φ: (u, v) 0 < u < v < 2 → Ω , Φ(u, v) = (uv + v2, v − u) .
d(u,v)Φ =
(v u+ 2v−1 1
)det d(u,v)Φ = u+ 3v
vol Ω =
∫Ω
1 =
∫ 2
0
∫ 2
u
det d(u,v)Φ dvdu
=
∫ 2
0
∫ 2
u
u+ 3v dvdu =
∫ 2
0
u(2− u) + 3
(2− u2
2
)du
=
∫ 2
0
2u− u2 + 6− 3u2
2du = 4− 8
3+ 12− 4 = 9
1
3
10. Find all local extrema of the function f : R2 → R, f(x, y) = x3 − x2y + y + 1.
Solution: The derivative of the function,
d(x,y)f = (3x2 − 2xy,−x2 + 1)
vanishes if and only if
x = ±1 , y = ±3
2i.e. at the points (
1,3
2
)and
(−1,−3
2
)The second derivative,
d2(x,y)f =
(6x− 2y −2x−2x 0
)is indefinite if x 6= 0, hence none of the critical points is an extremum. The function has no localextrema.
11. Find all local extrema of the function f : R2 → R with
f(u, v) = u4 + v4 + uv .
Solution: The derivative of the function vanishes,
d(u,v)f = (4u3 + v, 4v3 + u) = 0
if and only ifv = −4u3 = −4(−4v3)3 = 44v9
i.e. if v = 0 = u or v = ±12
= −u. Thus the critical points are
(0, 0) ,
(−1
2,1
2
),
(1
2,−1
2
).
The second derivative of f ,
d(u,v)f =
(12u2 1
1 12v2
)at these points is (
0 11 0
),
(3 11 3
),
(3 11 3
)is indefinite, postive definite and positive definite respectively. Thus 0 = (0, 0) is no local extremumand p+ =
(−1
2, 1
2
), p−
(12,−1
2
)are local minima. Since the function is positive for sufficiently large
‖(u, v)‖, the function must have a global minimum and since f(p+) = f(p−), both p+ and p− areglobal minima.
12.