Section 1.1

1. (1.1.1). Let x = (3,−1,−1, 1) and y = (−2, 2, 1, 0). Compute the norms of x and y and the anglebetween them.

2. (1.1.2). Given x, y ∈ Rn, show that

a. |x + y|2 = |x|2 + 2x · y + |y|2.

b. |x + y|2 + |x− y|2 = 2 (|x|2 + |y|2).

3. (1.1.4). Under what conditions on a and b is Cauchy’s inequality an equality? (Examine theproof.)

Solution. When is |a · b| = ‖a‖ ‖b‖? We can use the formula a · b = ‖a‖ ‖b‖ cos θ. Therefore,

|a · b| = ‖a‖ ‖b‖ ⇔ | cos θ| = 1

⇔ θ = kπ, k ∈ Z,

that is, vectors are colinear (or linearly dependent). Also, if a and/or b are/is zero, then a · b = 0 =‖a‖ ‖b‖, and Cauchy’s inequality is also an equality.

4. (1.1.5). Under what conditions on a and b is the triangle inequality an equality?

Solution. When is ‖a + b‖ = ‖a‖+ ‖b‖? We have

‖a + b‖ = ‖a‖+ ‖b‖ ⇔ ‖a + b‖2 = (‖a‖+ ‖b‖)2

⇔ ‖a‖2 + 2a · b + ‖b‖2 = ‖a‖2 + 2‖a‖ ‖b‖+ ‖b‖2

⇔ a · b = ‖a‖ ‖b‖

and we are back to Exercise 1.1.4, so vectors a and b must be colinear or one or both must be zero.

5. (1.1.6). Show that∣∣∣‖a‖ − ‖b‖∣∣∣ ≤ ‖a− b‖ for every a, b ∈ Rn.

6. (1.1.7). Suppose a, b ∈ R3.

a. Show that if a · b = 0 and a× b = 0, then either a = 0 or b = 0.

b. Show that if a · c = b · c and a× c = b× c for some nonzero c ∈ R3, then a = b.

c. Show that (a× a)× b = a× (a× b) if and only if a and b are proportional.

Section 1.2

1. (1.2.1). For each of the following sets S in the plane R2, do the following: (i) Draw a sketch ofS. (ii) Tell whether S is open, closed, or neither. (iii) Describe Sint, S and ∂S. (These descriptionsshould be in the same set-theoretic language as the description of S itself given here.)

a. S = {(x, y) : 0 < x2 + y2 ≤ 4}.

b. S = {(x, y) : x2 − x ≤ y ≤ 0}.

1

c. S = {(x, y) : x > 0, y > 0, and x + y > 1}.

d. S = {(x, y) : y = x3}.

e. S = {(x, y) : x > 0 and y = sin(1/x)}.

2. (1.2.3). Show that if S1 and S2 are open, so are S1 ∪ S2 and S1 ∩ S2.

Solution. Let x ∈ S1 ∪ S2. Then x ∈ S1 or x ∈ S2. Assume x ∈ S1. Since S1 is open, there is aball Br(x) ⊂ S1. But this ball is also contained in S1 ∪ S2, since the latter set is larger than S1.Therefore x is an interior point of S1 ∪ S2, and reasoning similarly for all points in S2, we see thatall points of S1 ∪ S2 are interior points, and thus S1 ∪ S2 is open. Let x ∈ S1 ∩ S2. Since x ∈ S1 andS1 open, there exists r1 > 0 such that Br1(x) ⊂ S1. Since x ∈ S2 open, there exists r2 > 0 such thatBr2(x) ⊂ S2. Let r = min(r1, r2). Then Br(x) ⊂ S1 and Br(x) ⊂ S2, and thus Br(x) ⊂ S1 ∩ S2, andthus S1 ∩ S2 is open.

3. (1.2.4). Show that if S1 and S2 are closed, so are S1 ∪ S2 and S1 ∩ S2.

Solution. If S1 and S2 are closed, then Sc1 and Sc

2 are open, and, from the previous exercise, so areSc

1 ∩ Sc2 and Sc

1 ∪ Sc2. Since Sc

1 ∪ Sc2 is open,

(Sc1 ∪ Sc

2)c

is closed. From De Morgan’s law,

(Sc1 ∪ Sc

2)c = (Sc

1)c ∩ (Sc

2)c = S1 ∩ S2,

and thus S1 ∩ S2 is closed. Similarly, since Sc1 ∩ Sc

2 is open,

(Sc1 ∩ Sc

2)c

is closed and from De Morgan’s law,

(Sc1 ∩ Sc

2)c = (Sc

1)c ∪ (Sc

2)c = S1 ∪ S2

is closed.

4. (1.2.8). Give an example of a set S such that the interior of S is unequal to the interior of theclosure of S.

Solution. Take the set S = {(x, y) : 0 < x2 + y2 < 4} from Exercise 1.2.1, for example. We haveSint = S. On the other hand, the boundary ∂S of S is {(0, 0)} ∪ {x2 + y2 = 4}, and thus the closureS of S is

S = {(x, y) : x2 + y2 ≤ 4}.

Thus, the interior of S is (S)int

= {(x, y) : x2 + y2 < 4} 6= Sint.

Section 1.3

1. (1.3.1). For the following functions f , show that lim(x,y)→(0,0) f(x, y) does not exist.

2

a. f(x, y) =x2 + y√x2 + y2

b. f(x, y) =x

x4 + y4

c. f(x, y) =x4y4

(x2 + y4)3

Solution. Use the fact that the limit does not exist if limx→a f(x) differs when a is approached alongtwo different paths. Try the usual suspects: (x, cx), (cy, y), (x, cx2), (cy2, y), (x,

√cx), etc., for c ∈ R.

2. (1.3.2). For the following functions f , show that lim(x,y)→(0,0) f(x, y) = 0.

a. f(x, y) =x2y2

x2 + y2

b. f(x, y) =3x5 − xy4

x4 + y4

3. (1.3.5). Let f(x, y) = y(y − x2)/x4 if 0 < y < x2, f(x, y) = 0 otherwise. At which point(s) if fdiscontinuous?

Solution. First, represent the domain of f , as in Figure 1.

−5 −4 −3 −2 −1 0 1 2 3 4 5

0

5

10

15

20

25

y≥ x2

0<y<x2 0<y<x2

y≤ 0

Figure 1: Domain of f in Exercise 1.3.3.

If there are any discontinuities, they will occur at the boundary between the domains of f , soalong the curves y = x2 or y = 0.

4. (1.3.8). Suppose f : Rn → Rk has the following property: For any open set U ⊂ Rk, {x : f(x) ∈U} is an open set in Rn. Show that f is continuous on Rn. Show also that the same result holds if“open” is replaced by “closed”.

3

5. Let f : Rn → R and g : Rn → R be continuous functions. Show that the functions

f 2g :Rn → Rx 7→ (f(x))2g(x)

and

f 2 + g :Rn → Rx 7→ (f(x))2 + g(x)

are continuous.

6. Use the (ε, δ) formulation to show that x2 → 4 as x → 2.

7. Let f : A ⊂ Rn → Rm. Suppose that there are positive constants K, α such that

‖f(x)− f(y)‖ ≤ K‖x− y‖α

for all x, y ∈ A. Show that f is continuous. (We say f is Holder-continuous, or, when α = 1,Lipschitz-continuous.)

Solution. We want to show that for all a ∈ A and all ε > 0, there exists δ > 0 such that ‖x− a‖ <δ ⇒ ‖f(x)− f(a)‖ < ε.

Given ε > 0, suppose we take

δ =( ε

K

)1/α

.

Then

‖x− a‖ < δ ⇔ ‖x− a‖ <( ε

K

)1/α

⇔ ‖x− a‖α <ε

K⇔ K‖x− a‖α < ε

But we know that for all x, y ∈ A,

‖f(x)− f(y)‖ ≤ K‖x− y‖α,

so, taking y = a,‖f(x)− f(a)‖ ≤ K‖x− a‖α < ε

for all x ∈ A, given ε > 0, provided ‖x− a‖ < δ with δ =(

εK

)1/α. Therefore f is continuous.

Section 2.1

1. (2.1.1). Suppose that f is differentiable on the interval I and that f ′(x) > 0 for all x ∈ I exceptfor finitely many points at which f ′(x) = 0. Show that f is strictly increasing on I.

2. (2.1.8). Suppose f and g are differentiable functions on R with values in Rn.

a. Show that (f · g)′ = f ′ · g + f · g′.

b. Suppose also that n = 3, and show that (f × g)′ = f ′ × g + f × g′.

Solution. Just compute the dot and cross products of f and g, and differentiate the results. Witha bit of reordering, the result is evident.

4

Section 2.2

1. (2.2.1). For each of the following functions f , (i) compute ∇f , (ii) find the directional derivativeof f at the point (1,−2) in the direction (3

5, 4

5).

a. f(x, y) = x2y + sin πxy.

b. f(x, y) = e4x−y2.

c. f(x, y) = (x + 2y + 4)/(7x + 3y).

Solution. a) ∇f = (2xy + πy cos πxy, x2 + πx cos πxy), and thus

D( 35, 45)

f(1,−2) = ∇f(1,−2) ·(

3

5,4

5

)= −8

5− 2

5π.

b) ∇f =(4e4x−y2

,−2ye4x−y2)

= 2e4x−y2(2,−y), so

D( 35, 45)

f(1,−2) = ∇f(1,−2) ·(

3

5,4

5

)= (4, 4) ·

(3

5,4

5

)=

28

5.

c) We have

∇f =

((7x + 3y)− (x + 2y + 4)7

(7x + 3y)2,(7x + 3y)2− (x + 2y + 4)3

(7x + 3y)2

)=

(−11y − 21

(7x + 3y)2,

11x− 12

(7x + 3y)2

),

and thus ∇f(1,−2) = (1,−1), and

D( 35, 45)

f(1,−2) =3

5− 4

5= −1

5.

2. (2.2.4). Show that u = xe2x+y−1e5x satisfies the partial differential equation x∂u∂x

u+2y ∂u∂y

u+ ∂u∂z

u =3u.

3. Compute the directional derivative of

f(x, y, z) = ex2+y2+z2

in the direction u = (1, 1,−1) at the point (x0, y0, z0) = (1, 10, 100).

Solution. We have

∇f =(2xex2+y2+z2

, 2yex2+y2+z2

, 2zex2+y2+z2)

= 2ex2+y2+z2

(x, y, z)

5

So

Duf(1, 10, 100) = ∇f(1, 10, 100) · (1, 1,−1)

= 2× e1+100+10000(1 + 10− 100)

= −188e10101

4. Find ∇f forf(x, y, z) = exy − x cos(yz2).

Solution.

∇f =

(∂f

∂x,∂f

∂y,∂f

∂z

)=(yexy − cos(yz2), xexy + xz2 sin(yz2), 2xyz sin(yz2)

)Section 2.3

1. Suppose that a duck is swimming in the circle x = cos t, y = sin t, and that the water temperatureis given by T = x2ey − xy3. Find dT/dt, i) using the chain rule and ii) by expressing T in terms of tand differentiating.

Solution. i) T : R2 → R, D : R → R2, thus T ◦D : R → R. By the mean value theorem, we have

dT

dt= ∇T ·D′,

with ∇T evaluated at D(t). NowD′(t) = (− sin t, cos t)

and∇T = (2xey − y3, x2ey − 3xy2).

Evaluating ∇T at D(t) gives

∇T (D(t)) =(2 cos tesin t − sin3 t, cos2 tesin t − 3 cos t sin2 t

)and thus

dT

dt= ∇T ·D′

= sin4 t− 3 cos2 t sin2 t + cos tesin t(cos2 t− 2 sin t).

ii) Evaluating T at x = cos t, y = sin t, we get

T = cos2 tesin t − cos t sin3 t.

To compute the derivative, first note that

(sin3 t)′ = (sin t sin2 t)′ = cos t sin2 t + sin t(2 cos t sin t) = 3 cos t sin2 t.

Therefore,

dT

dt= cos2 t cos tesin t − 2 cos t sin tesin t + sin t sin3 t− cos t(3 cos t sin2 t)

= sin4 t− 3 cos2 t sin2 t + cos tesin t(cos2 t− 2 sin t).

6

2. Use the chain rule and differentiation under the integral sign to show that

d

dx

∫ x

0

f(x, y)dy = f(x, x) +

∫ x

0

∂f

∂xdy.

3. Let (x(t), y(t)) be a path in the plane, 0 ≤ t ≤ 1, and let f(x, y) be C1. Assume that

dx

dt

∂f

∂x+

dy

dt

∂f

∂y≤ 0.

Show that f(x(1), y(1)) ≤ f(x(0), y(0)).

4. (2.3.4). Let u = f(r) and r = ‖x‖ =√

x21 + · · ·+ x2

n. Show that

n∑k=1

(∂u

∂xk

)2

= (f ′(r))2.

Solution. From the chain rule, we have

∂u

∂xk

=df

dr

∂r

∂xk

.

Here,∂r

∂xk

= xk

(x2

1 + · · ·+ x2n

)−1/2,

and therefore, (∂u

∂xk

)2

=

(df

dr

)2

x2k

(x2

1 + · · ·+ x2n

)−1.

It follows thatn∑

k=1

(∂u

∂xk

)2

=n∑

k=1

((df

dr

)2

x2k

(x2

1 + · · ·+ x2n

)−1

)

=

(df

dr

)2 (x2

1 + · · ·+ x2n

)−1n∑

k=1

x2k

=

(df

dr

)2 (x2

1 + · · ·+ x2n

)−1 (x2

1 + · · ·+ x2n

)=

(df

dr

)2

.

5. (2.3.5). Consider the surface defined in R3 by z = f(x), for x ∈ R2. Then the tangent plane tothis surface at the point f(a) (i.e., for x = a) has equation

z = f(a) +∇f(a) · (x− a),

for x ∈ R2. Show that this expression is the same as the one for the tangent plane to

S = {(x, y, z) ∈ R3; F (x, y, z) = 0}

given as a result in this section, if F (x, y, z) = f(x, y)− z.

6. (2.3.6). Find the plane tangent to the surface in R3 described by the given equation at the givenpoint a ∈ R3.

7

a. z = x2 − y3, a = (2,−1, 5).

b. x2 + 2y2 + 3z2 = 6, a = (1, 1,−1).

c. z =√

x + arctan y, a = (9, 0, 3).

d. xyz2 − ln(z − 1) = 8, a = (−2,−1, 2).

Solution. a) We can use for example the formula z = f(a)+∇f(a)·(x−a). We have∇f = (2x,−3y2).Here, a = (2,−1) and f(a) = 5. So ∇f(a) = (4,−3). Therefore,

z = 5 + (4,−3) · ((x, y)− (2,−1)) = 4x− 3y − 10.

c) Here again, we can use the result for surfaces z = f(x, y), with a = (9, 0) and f(a) = 3. We have

∇f =

(1

2x−1/2,

1

1 + y2

),

and thus ∇f(a) = (1/6, 1). Therefore,

z = 3 +

(1

6, 1

)· ((x, y)− (9, 0)) =

x

6+ y +

3

2.

7. Compute the matrix of partial derivatives for the following functions.

a. f(x, y) = (x, y).

b. f(x, y) = (xey + cos y, x, x + ey).

c. f(x, y, z) = (x + ez + y, yx2).

d. f(x, y) = (xyexy, x sin y, 5xy2).

Solution. a)

(1 00 1

).

b)

ey xey − sin y1 01 ey

.

c)

(1 1 ez

2xy x2 0

).

d)

xexy + xy2exy xexy + x2exy

sin y −x cos y5y2 10xy

.

8. Find the equation of the plane tangent to the surface z = x2 + y3 at (3, 1, 10).

Solution. We use the formz = f(a) +∇f(a) · (x− a)

with a = (3, 1) and x = (x, y). We have ∇f = (2x, 3y2), and thus ∇f(a) = (6, 3). Of course,f(a) = 10 (we are working at the point (3, 1, 10)), and thus

z = 10 + (6, 3) · ((x, y)− (3, 1)) = 6x + 3y − 11.

8

Section 2.4

1. (2.4.1). State and prove two analogues of Rolle’s theorem for functions of several variables, whosehypotheses are, respectively, the following:

a. f is differentiable on a set containing the line segment from a to b, and f(a) = f(b).

b. f is differentiable on a bounded open set S, continuous on the closure of S, and constant onthe boundary of S.

Solution. a) We want to prove something along the lines of “there exists c on the line segment (a, b)

such that Dhf(c) = 0, with h = ~ab”. See Figure 2 for an example.

Figure 2: An example for Exercise 2.4.1.a. At f(c), one directional derivative is zero.

b) This is more general. In a), we can only be sure that the directional derivative is going to be zerosomewhere on the line segment between a and b. Here, there must be a point c ∈ S such that alldirectional derivatives are zero. If we consider the case in R3 (for example with a surface defined byz = f(x, y)), then there must exist a c such that the tangent plane to the surface at f(c) is horizontal.See Figure 3 for an example.

Section 2.5

1. (2.5.1). Compute ∂z/∂x and ∂z/∂y when z is determined as a function of y and x by the followingequations:

a. x + y2 + z3 = 3xyz.

9

Figure 3: An example for Exercise 2.4.1.b.

b. 2x2 + 3y2 + z2 = e−z.

Solution. a) We write the equation as

F (x, y, z) = x + y2 + z3 − 3xyz,

and thus define implicitly z as a function of x and y, i.e., z(x, y), by considering the level setF (x, y, z) = 0. We have

∂xF + ∂zF∂xz = 0,

and thus

∂xz = −∂xF

∂zF

= − 1− 3yz

3z2 − 3xy

=3yz − 1

3z2 − 3xy.

Similarly,∂yF + ∂zF∂yz = 0,

and thus

∂xz = −∂yF

∂zF

= − 2y − 3xz

3z2 − 3xy

=3xz − 2y

3z2 − 3xy

10

b) We proceed as in a), with

F (x, y, z) = 2x2 + 3y2 + z2 − e−z.

2. (2.5.2). Suppose y and z are determined as functions of x by the equations z = x2 − y2 andz = 2x + 4y. Find dy/dx and dz/dx i) by solving the equations explicitly for y and z and, ii) byimplicit differentiation.

Section 2.6

In the exercises from the Folland book, it is assumed that all functions involved are C2.

1. Verify by explicit calculation that ∂x∂yf = ∂y∂xf :

a. f(x, y) = x2y + sin πxy.

b. f(x, y) = e4x−y2.

c. f(x, y) = (x + 2y + 4)/(7x + 3y).

Solution. a) ∂yf = x2 + πx cos πxy, so ∂x∂yf = 2x + π cos πxy − π2xy sin πxy. On the other hand,∂xf = 2xy + πy cos πxy, so ∂y∂xf = 2x + π cos πxy − π2xy sin πxy. We have equality.

2. (2.6.4). Let u = F (x + g(y)). Show that uxuxy = uyuxx.

3. (2.6.7). Suppose u = f(x− ct) + g(x− ct), where c ∈ R is a constant. Show that

∂2xu = c−2∂2

t u.

(This equation is known as the heat equation.)

4. (2.6.10). Derive the following version of the product rule for partial derivatives:

∂α(fg) =∑

β+γ=α

α!

β!γ!∂βf∂γg.

(These are multi-indexes.)

Section 2.7

1. (2.7.1). Let f(x) = x2(x− sin x) and g(x) = (ex − 1)(cos 2x− 1)2.

a. Compute the Taylor polynomials of order 5 at a = 0 of f and g, without computing derivatives(i.e., using the known forms).

b. Use (a) to find limx→0 f(x)/g(x) without using L’Hospital’s rule.

11

Section 2.8

1. (2.8.1). Find all the critical points of the following functions. Tell whether each nondegeneratecritical point is a local maximum, local minimum, or saddle point. If possible, tell whether thedegenerate critical points are local extrema too.

a. f(x, y) = x2 + 3y4 + 4y3 − 12y2.

b. f(x, y) = x4 − 2x2 + y3 − 6y.

c. f(x, y) = (x− 1)(x2 − y2).

Section 2.9

1. (2.9.1). Find the extreme values of f(x, y) = 2x2 + y2 + 2x on the set {(x, y) : x2 + y2 ≤ 1}.

2. (2.9.2). Find the extreme values of f(x, y) = 3x2 − 2y2 + 2y on the set {(x, y) : x2 + y2 ≤ 1}.

For the following exercises, I will post the texts later; for now, I am only giving numbers, for exercisesin Folland.

3. (2.9.10).

4. (2.9.13).

Section 2.10. Vector-valued functions and their derivatives

I will post the texts later, for now, I am only giving numbers, for exercises in Folland.

1. (2.10.1).

2. (2.10.2).

3. (2.10.3).

4. (2.10.4).

5. (2.10.5).

6. (2.10.7).

7. (2.10.9).

Section 3.1. The implicit function theorem

1. (3.1.1). Investigate the possibility of solving the equation x2 − 4x + 2y2 − yz = 1 for each of itsvariables in terms of the two others near the point (2,−1, 3). Do this both by checking the hypothesesof the IFT and by explicitly computing the solutions.

12

2. (3.1.2). Show that the equation x2 +2xy+3y2 = c can be solved either for y as a C1 function of xor for x as a C1 function of y (but perhaps not both) near any point (a, b) such that a2+2ab+3b2 = c,provided c > 0. What happens if c = 0 or if c < 0?

3. (3.1.3). Can the equation (x2 + y2 + 2z2)1/2 = cos z be solved uniquely for y in terms of x and znear (0, 1, 0)? For z in terms of x and y?

Solution. Define F (x, y, z) = (x2 + y2 + 2z2)1/2 − cos z. This function is C1 on R3: the polynomialx2 + y2 + 2z2 is C1, of course (actually, it is C∞), but also nonnegative, which means that thecomposition with the square root gives a C1 function on R3 \ {(0, 0, 0)}. Lastly, the sum of two C1

functions is C1, and therefore F ∈ C1 since cos z ∈ C1. Also, F (0, 1, 0) = 0. So we can try to applythe implicit function theorem.

First, to solve for y in terms of x and z, there must hold that

∂yF (0, 1, 0) 6= 0.

We have

∂yF =2y

2√

x2 + y2 + 2z2=

y√x2 + y2 + 2z2

,

and so ∂yF (0, 1, 0) = 1 6= 0. Therefore, we can use the implicit function theorem, so there existsunique functions f(y) and g(y), defined on some neighborhood of (0, 1, 0), such that F (f(y), y, g(z)) =0 on a neighborhood of (0, 1, 0).

Then, to solve for z in terms of x and y, there must hold that

∂zF (0, 1, 0) 6= 0.

We have

∂zF =4z

2√

x2 + y2 + 2z2=

2z√x2 + y2 + 2z2

,

and so ∂yF (0, 1, 0) = 0. Therefore, we cannot use the implicit function theorem, and it is not possibleto solve for z in terms of x and y, at (0, 1, 0).

4. (3.1.4). Sketch the graph of the equation x − e1−x − y3 = 0. Show graphically that for each xthere is a unique y satisfying this equation, and vice versa.

5. (3.1.5). Suppose F (x, y) is a C1 function such that F (0, 0) = 0. What conditions on F willguarantee that the equation F (F (x, y), y) = 0 can be solved for y as a C1 function of x near (0, 0)?

6. (3.1.6). Investigate the possibility of solving the equations xy + 2yz − 3xy = 0, xyz + z − y = 1for two of the variables as functions of the third near the point (x, y, z) = (1, 1, 1).

7. (3.1.7). Investigate the possibility of solving the equations u3 + xv − y = 0, v3 + yu− x = 0 fortwo of the variables as functions of the other two near the point (x, y, u, v) = (0, 1, 1,−1).

8. (3.1.8). Investigate the possibility of solving the equations xy2 +xzu+ yv2 = 3 and u3yz +2xv−u2v2 = 2 for u and v as functions of x, y and z near x = y = z = u = v = 1.

9. (3.1.9). Can the equations x2 + y2 + z2 = 6, xy + tz = 2 and xz + ty + et = 0 be solved for x, yand z as C1 functions of t near (x, y, z, t) = (−1,−2, 1, 0)?

13

10. Show that for all (u, v) close to (1 + e, 0), the system

x + y + ez = u

xyez = v

has a solution.

3.2

1. (3.2.1). For each of the following functions F , determine whether the set S = {(x, y) ∈ R2 :F (x, y) = 0} is a smooth curve. Draw a sketch of S. Examine the nature of S near any points where∇F = 0. Near which points of S is S the graph of a function y = f(x)? x = f(y)?

a. F (x, y) = x2 + 3y2 − 3.

b. F (x, y) = x2 − 3y2 − 3.

c. F (x, y) = x−√

3(y2 + 1).

d. F (x, y) = xy(x + y − 1).

e. F (x, y) = (x2 + y2)(y − x2 − 1).

f. F (x, y) = (x2 + y2)(y − x2).

g. F (x, y) = (ex − 1)2 + (sin y − 1)2.

Solution. a) F is a C1 function on R2. We have

∇F (x, y) = (2x, 6y),

and thus ∇F = 0 iff (x, y) = (0, 0). But (0, 0) 6∈ S, since F (0, 0) = −3 6= 0, so at all points in S,∇F 6= 0. Therefore, S is a smooth curve. At any point on S, S can be expressed as a function of xor as a function of y.b) F is a C1 function on R2. We have

∇F (x, y) = (2x,−6y),

and thus ∇F = 0 iff (x, y) = (0, 0). But (0, 0) 6∈ S, since F (0, 0) = −3 6= 0, so at all points in S,∇F 6= 0. Therefore, S is a smooth curve. At any point on S, S can be expressed as a function of xor as a function of y. [So note that the solution page 444 is wrong.]c) F is C1 on R2. We have

∇F (x, y) =

(1,− 3y√

3y2 + 3

),

and thus ∇F 6= 0 for all (x, y) ∈ R2; S is thus smooth.d) F is C1 on R2, and

∇F (x, y) = (y(2x + y − 1), x(x + 2y − 1)).

Thus ∇F = 0 iff

y(2x + y − 1) = 0 (1a)

x(x + 2y − 1) = 0 (1b)

14

From (1a), y = 0 or 2x + y− 1 = 0, i.e., y = 1− 2x. Substituting y = 0 in (1b) gives x = 0 or x = 1.Now suppose y = 1− 2x and substitute in (1b), giving

x(x + 2(1− 2x)− 1) = 0 ⇔ x(1− 3x) = 0 ⇔ x = 0 and x =1

3.

Therefore, ∇F = 0 iff (x, y) = (0, 1),

(1

3,1

3

), (0, 0) or (1, 0).

We must check that these points do indeed belong to S. We have

F (0, 0) = 0, F (1, 0) = 0, F (0, 1) = 0, F

(1

3,1

3

)=

1

9

(2

3− 1

)= − 1

27,

so (0, 0), (0, 1), (1, 0) ∈ S, (13, 1

3) 6∈ S.

S consists of the union of three curves, x = 0, y = 0 and x+y−1 = 0, and the points (0, 0), (0, 1)and (1, 0) lie at the intersections of these curves. At these points, S is not smooth.e) F is a C1 function, and

∇F (x, y) = (−2x(2x2 + y2 − y + 1), x2 + 3y2 − 2x2y − 2y).

Therefore, ∇F = 0 iff

2x(2x2 + y2 − y + 1) = 0 (2a)

x2 + 3y2 − 2x2y − 2y = 0 (2b)

From (2a), x = 0 or 2x2 + y2 − y + 1 = 0. Substituting x = 0 in (2b), it follows that 3y2 − 2y = 0,that is, y = 0 or y = 2/3.

S is the union of two curves: x2 + y2 = 0 and y − x2 − 1 = 0. The first is the circle of radius 0,that is, the point (0, 0), the second the parabola y = x2 + 1.

These two curves do not intersect, and so S is not smooth, because it is not connected.f) F is a C1 function, and

∇F (x, y) = (−2x(2x2 + y2 − y), x2(1− 2y) + 3y2).

Therefore, ∇F = 0 iff

2x(2x2 + y2 − y) = 0 (3a)

x2(1− 2y) + 3y2 = 0 (3b)

From (3a), x = 0 or 2x2 + y(y − 1) = 0. Substituting x = 0 in (3b), we get y = 0.S is the union of two curves: x2 + y2 = 0 and y − x2 = 0. The first is the circle of radius 0, that

is, the point (0, 0), the second the parabola y = x2.These two curves intersect (and in fact, are equal) at (0, 0). Therefore S is smooth.

g) F is C1, and∇F = (2(ex − 1)ex, 2(sin y − 1) cos y).

∇F = 0 iff

2(ex − 1)ex = 0 (4a)

2(sin y − 1) cos y = 0 (4b)

15

From (4a), ex = 0 or ex − 1 = 0, i.e., for x ∈ R, x = 0. Now, considering (4b), we have sin y = 1 orcos y = 0; this gives y = π/2 + 2kπ or y = π/2 + kπ, for k ∈ Z, that is, y = π/2 + kπ, for k ∈ Z.Thus points of the form

(x, y) =(0,

π

2+ kπ

), k ∈ Z,

are such that ∇F (x, y) = 0. Do they belong to S? We have, for k ∈ Z,

F(0,

π

2+ kπ

)=(sin(π

2+ kπ

)− 1)2

,

and soF(0,

π

2+ kπ

)= 0

when k is even, so (0, π/2 + 2kπ) ∈ S, for k ∈ Z, and are points such that ∇F = 0.Points on S must satisfy

(ex − 1)2 = −(sin y − 1)2.

Because both sides of the equality are nonnegative, the only possibility is for both sides to equalzero. But

(ex − 1)2 = (sin y − 1)2

is equivalent toex − 1 = sin y − 1,

that is,ex = sin y.

So, finally,x = ln sin y.

Therefore, the curve is not smooth, since it consists of infinitely many paraboloids, defined fory ∈ (2k, (2k + 1)π), k ∈ Z; see Figure 4. Since S is not connected, it is not a smooth curve.

−6 −5 −4 −3 −2 −1 0 1 2−15

−10

−5

0

5

10

15

Figure 4: The curve level set 0 of F (x, y) = (ex − 1)2 + (sin y − 1)2 in Exercise 3.2.1.g.

16

2. (3.2.2).

3. (3.2.3).

4. (3.2.4).

5. (3.2.6).

3.3

1. (3.3.1). For each of the following maps f : R2 → R3, describe the surfaces S = f(R2) and find adescription of S as the locus of an equation F (x, y, z) = 0. Find the points where ∂uf and ∂vf arelinearly dependent, and describe the singularities of S (if any) at these points.

a. f(u, v) = (2u + v, u− v, 3u).

b. f(u, v) = (au cos v, bu sin v, u), a, b > 0.

c. f(u, v) = (cos u cosh v, sin u cosh v, sin hv).

d. f(u, v) = (u cos v, u sin v, u2).

Solution. a) S is a plane. We have

∂uf = (2, 1, 3), ∂vf = (1,−1, 0),

and thus

∂uf × ∂vf =

∣∣∣∣∣∣~i ~j ~k2 1 31 −1 0

∣∣∣∣∣∣ = (3, 3,−3) 6= 0

so ∂uf and ∂vf are linearly independent for all (x, y, z) ∈ R3. Therefore, S is a smooth surface.b) See Figure 5. We have

∂uf = (a cos v, b sin v, 1), ∂vf = (−au sin v, bu cos v, 0)

and thus

∂uf × ∂vf =

∣∣∣∣∣∣~i ~j ~k

a cos v b sin v 1−au sin v bu cos v 0

∣∣∣∣∣∣= (−bu cos v,−au sin v, abu cos2 v + abu sin2 v)

= (−bu cos v,−au sin v, abu).

So (∂uf × ∂vf)(0, v) = 0.

2. (3.3.2).

3. (3.3.3).

4. (3.3.5).

5. (3.3.6).

17

20

10

0

−10

−20

2010

0−10−20

−20

−10

0

10

20

Figure 5: The range of f(u, v) = (u cos v, u sin v, u) in Exercise 3.3.1.b.

18

3.4

1. (3.4.1).

2. (3.4.2). Let (u, v) = f(x, y) = (x− 2y, 2x− y).

a. Compute the inverse transformation (x, y) = f−1(u, v).

b. Find the image in the uv-plane of the triangle bounded by the lines y = x, y = −x andy = 1− 2x.

c. Find the region in the xy-plane that is mapped to the triangle with vertices (0, 0), (−1, 2) and(2, 1) in the uv-plane.

Solution. a) First, note that we can apply the inverse mapping theorem, since f is C1 and that theFrechet derivative

Df =

(1 −22 −1

)has determinant 3. Therefore, f is invertible, and f−1 is C1.

To find the inverse, we need to find (u, v) as a function of (x, y). We have

u = x− 2y

v = 2x− y

From the first equation, x = u + 2y, which, when substituted into the second equation gives v =2u + 4y − y = 2u + 3y, that is,

y =v − 2u

3.

Then,

x = u + 2y = u + 2v − 2u

3=

3u + 2v − 4u

3=

2v − u

3.

Therefore,

f−1(u, v) =

(2v − u

3,v − 2u

3

).

b) The triangle is as shown in Figure 6. The first intuition is to compute the image of the vertices ofthe triangle. Here, it will work because the transformation is linear (and thus transforms lines intolines. But this is not always that case, so it is better to proceed as follows.

The vertices of the triangle are (0, 0) and the intersection of y = x and y = 1− 2x, and y = −xand y = 1− 2x, that is, (1/3, 1/3) and (1,−1), respectively. Knowing the vertices and looking at thefigure, we write the different line segments in parametric form.

Firstly, the segment y = x runs from (0, 0) to (1/3, 1/3). Thus, it is written in parametric formas (x, x), for 0 ≤ x ≤ 1/3. The image in the uv-plane is thus

(u, v) = f(x, x) = (x− 2x, 2x− x) = (−x, x),

for 0 ≤ x ≤ 1/3, that is, the line segment from (−1/3, 1/3) to (0, 0).Secondly, the segment y = −x runs from (0, 0) to (1,−1). In parametric form, it takes the form

(x,−x), for 0 ≤ x ≤ 1. The image of the segment in the uv-plane is

(u, v) = f(x,−x) = (x− 2(−x), 2x− (−x)) = (3x, 3x),

19

−0.25

0.0

0.0 0.75

1.5

0.25

−0.5

1.0

0.5

−1.0

−0.5 0.5 1.0

2.0

x

Figure 6: The triangle bounded by the lines y = x, y = −x and y = 1− 2x in Exercise 3.2.

for 0 ≤ x ≤ 1, i.e., the line segment between (0, 0) and (3, 3).Finally, the segment y = 1 − 2x runs from (1/3, 1/3) to (1,−1), so its parametric equation is

given for 1/3 ≤ x ≤ 1 by (x, 1− 2x). Therefore, the image of the segment is given in the uv-plane by

(u, v) = f(x, 1− 2x) = (x− 2(1− 2x), 2x− (1− 2x)) = (5x− 2, 4x− 1)

for 1/3 ≤ x ≤ 1. So, in the uv-plane, the line segment from (−1/3, 1/3) to (3, 3).So, in the uv-plane, we obtain the triangle with vertices (0, 0), (−1/3, 1/3) and (3, 3).

c) Once again, the transformation is linear; so it has linear inverse (you can check that by lookingat the function f−1). So a triangle in the uv-plane is the image of a triangle in the xy-plane. Asremarked above, the standard procedure would be to compute the pre-image of the line segments(i.e., the image of the lines segments through the function f−1). Here, we take the “linear shortcut”:the vertices of the triangle in the xy-plane that has vertices (0, 0), (−1, 2) and (2, 1) in the uv-planeare obtained by computing

f−1(0, 0) = (0, 0), f−1(−1, 2) =

(5

3,4

3

), f−1(2, 1) = (0,−1) .

3. (3.4.4).

4. (3.4.6).

20