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.