MATH 260 – Homework assignment 6 – February 19, 2013

1. Sketch level curves for

(a) fpx, yq � xy

(b) fpx, yq � ex�y

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(c) fpx, yq � x2 � y � 4

(d) fpx, y, zq � cospx2 � y2 � z2q

(e) fpx, y, zq � lnp1� x2 � 2y2q.

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2. Describe the interior, exterior and boundary of each of the following sets. Which are open/closed/neither?

(a) S � tpx, yq P R2 | 2x2 � y   8u

This is an open set, the set of points below the parabola y � 8 � 2x2 (being open, S � ³S.

The boundary of the set is the parabola itself. The exterior of the set is the set of points above the

parabola.

(b) S � tpx, y, zq P R3 | 1 ¤ x ¤ 3, 0 ¤ y ¤ 2, 1� z2 ¤ 0u

This set is closed. It consists of two parts: the points both inside an on the surface of the square

“column” where 1 ¤ x ¤ 3 and 0 ¤ y ¤ 2 and z ¥ 1 (which extends infinitely in the positive z

direction) and the points inside and on the square “column” with the same xs and ys but z ¤ �1

(which extends infinitely in the negative z direction). The points shown inside and on the following

(where the two pieces extend infinitely in the z direction:

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The interior of S consists of the points inside but not on the surface of the columns, the boundary

is the points on the surface of the columns, and the exterior is all the points outside the columns

(note that the exterior is “connected” i.e., consists of one piece only).

(c) S � tpx, y, zq P R3 | |x� y � z|   1u

This open set lies between the two parallel planes x � y � z � 1 and x � y � z � �1. is the

interior of the octahedron whose surface consists of the eight triangles, one in each “octant” of R3,

as shown here:

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The boundary consists of the two planes, and the exterior is the set of points above the top plane

and below the bottom one.

(d) S � tpx, yq P R2 |x2 � y2 � 0u

This set consists of the two lines y � x and y � �x in the plane. It has no interior (intS � H),

it is its own boundary (BS � S) so it is closed. The exterior is the set of points not on the two lines

(so the exterior consists of four pieces).

3. Show that intS X BS � H. Also, show that a set S is closed if and only if S � intS Y BS.

Suppose x P intS. Then there is an R ¡ 0 such that for all r with 0   r   R, the ball Brpxq is

contained entirely in S. But then there are no small balls containing points both inside and outside

S, therefore x R BS. Likewise, if x P BS then every ball Brpxq contains at least one point not in S,

so x R intS. Thus intS X BS � H.

First, suppose S is closed. Writing SC for the complement of S, we have that SC is open.

Therefore, if x P SC then there is a ball Brpxq such that Brpxq X S � H. Therefore x is neither

in the interior nor on the boundary of S, so x P pintS Y BSqC , and so SC P pintS Y BSqC . Taking

complements, we have intSYBS � S. On the other hand, since every point of Rn is either in intS,

BS or extS, and no point in extS is in S by definition, we also have S � intS Y BS. Therefore

S � intS Y BS.

Next, suppose S � intSYBS, and assume x P extS. Since there is a ball Brpxq with BrpxqXS �H, we have that extS is open. Therefore intS Y BS is closed, so S is closed.

4. Suppose f : Rn Ñ Rn and we write the function f in component form as

fpxq � pf1pxq, f2pxq, . . . , fmpxqq.

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Show that f is continuous (at a) if and only if all of the components of f are continuous (at a).

First, suppose that f is continuous at a. Since fkpxq � xfpxq , eky where ek is the (constant) unit

vector in the direction of the kth coordinate axis in Rm, and since constant functions are continuous

and inner products of continuous functions are continuous, we get fkpxq is also continuous (at a).

On the other hand, if all the component functions fkpxq are continuous at x � a, we know that

limxÑa

fkpxq � fkpaq for all k � 1, . . . ,m. Therefore

limxÑa

fpxq � limxÑa

rf1pxqe1 � f2pxq � � � � � fmpxqems� r lim

xÑaf1pxqse1 � r lim

xÑaf2pxqse2 � � � � � r lim

xÑafmpxqsem

� f1paqe1 � f2paqe2 � � � � � fmpaqem� fpaq.

Therefore f is continuous at a.

5. Explain why an alternative definition of limit would be

limxÑa

fpxq � b if and only if lim}h}Ñ0

}fpa� hq � b} � 0

and an alternative definition of continuity would be that f is continuous at a if and only if

lim}h}Ñ0

}fpa� hq � fpaq} � 0.

Show that linear transformations from Rn to Rm are continuous.

If we set x � a�h, then h � x� a and clearly xÑ a is equivalent to hÑ 0, which shows the

equivalence of the limit definitions and the continuity definitions.

Now let A : Rn Ñ Rm be a linear transformation. Then Apa � hq � Apaq � Aphq, so if

fpxq � Apxq we’ll have fpa� hq � fpaq � Aphq. So we need to show that }Aphq} Ñ 0 as h Ñ 0.

By problem 4 above, this is equivalent to showing that each component of Aphq goes to 0 as hÑ 0.

But the kth component of Aphq is

Akphq � ak1h1 � � � � � aknhn

and we have

|Akphq| ¤ |ak1||h1| � � � � |akn||hn|.If }h}   δ then this expression is less than Mδ, where M is max

i,j|aij |. Therefore given ε ¡ 0 we

choose δ   nε{M and we’ll have |Akphq|   ε. Therefore Akpxq is continuous at x � a for all

1 ¤ k ¤ m and so f is continuous.

6. Give an ε-δ definition of continuity of f : Rn Ñ Rm at the point a P Rn.

For every ε ¡ 0 there is a δ ¡ 0 such that if }x�a}   δ then }pfpxq�fpaq}   ε (this implicitly

assumes that fpaq exists).

Alternatively, for every ε ¡ 0 there is a δ ¡ 0 such that if }h}   δ then }fpa� hq � fpxq}   ε.

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7. For what a P Rn are the following functions continuous?

(a) fpxq � }x}

Given ε ¡ 0, choose δ such that 0   δ   ε. If }h}   δ then

|}x� h} � }x}| ¤ }h}   ε

for all x. Therefore fpxq is continuous for all x P Rn.

(b) fpxq � 1� }x}21� }x}2

Since we know that }x} is continuous from part (a), f will be continuous wherever the de-

nominator is non-zero, namely for all x except where }x} � 1, i.e., except on the unit sphere in

Rn.

(c) fpxq � 1

}x}2 � 3 xx , by � }b}2x for a fixed vector b.

The denominator can be written

Bx� 3�?

5

2b , x� 3�?

5

2b

F

so the function will be discontinuous for those points x for which the vector x � 3�?52 b is perpen-

dicular to x� 3�?52 b. This is the equation of a sphere in Rn with center at the point 3

2b and radius?52 }b}. (To see this, recall that an angle inscribed in a circle is a right angle if and only if its sides

intersect the circle at antipodal points, i.e., at the endpoints of a diameter).

(d) fpx1, x2, x3q � arcsin

�x1 � x2x2 � x3

The domain of this function not all of R3, rather it is the set of points where

|x1 � x2| ¤ |x2 � x3|

The set of points where these two quantities are equal is the union of the two planes x1�2x2�x3 � 0

and x1 � x3 � 0, which are the red and blue planes in the following figure:

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The green plane is where x2 � x3, i.e., where the denominator of the fraction is zero.

The domain of f is thus the union of two sets of points: the set of points above the red and

blue planes together with the points on the planes except for the line where they cross, which has

x2 � x3, and the set of points below both planes together with the points on the planes (except for

the intersection line). The function is continuous at every point in the domain.

8. For which of the following functions, defined for px, yq � p0, 0q can fp0, 0q be defined so that f is

continuous at p0, 0q?

(a) fpx, yq � x� y

|x| � |y|

Certainly not this one – if y � 0, this function has limit 1 as xÑ 0� and limit �1 as xÑ 0�.

(b) fpx, yq � x3 � y4

x2 � y2

For this function, we have that the limit is zero as px, yq Ñ p0, 0q, so if we define fp0, 0q � 0 the

function will be continuous. To see that the limit is zero, rewrite the function and the limit using

polar coordinates:

limpx,yqÑp0,0q

x3 � y4

x2 � y2� lim

rÑ0

r3 cos3 θ � r4 sin4 θ

r2� lim

rÑ0rpcos3 θ � r sin4 θq � 0,

since the expression in parentheses is bounded as r Ñ 0 no matter what θ is.

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(c) fpx, yq �

$''''''''''''&''''''''''''%

x3

x2 � xy � y2if x P Q, y P Q

y3

x2 � xy � y2if x P Q, y R Q

2y3

x2 � xy � y2if x R Q, y P Q

sinx if x R Q, y R Q

The function will have a limit at p0, 0q if the limits of all four parts are the same. Clearly

the last part (sinx) approaches 0 at the origin, so we have to check that the other three parts

do. Since the degrees of the numerators are 3 while the degrees of the denominators are 2, the

polar coordinate trick of part (b) will show that the limits are zero provided the functions are

defined (if the denominators are zero for points other than the origin there might be trouble). But

x2 � xy � y2 � r2p1 � 12 sin 2θq ¥ 1

2r2, so the denominators are zero only at the origin and so if

fp0, 0q � 0 then f will be continuous at the origin.

(d) fpx, yq �

$''''''''''''&''''''''''''%

x3

x2 � xy � y2if x P Q, y P Q

y3

x2 � 2xy � y2if x P Q, y R Q

2y3

x2 � 2xy � y2if x R Q, y P Q

sinx if x R Q, y R Q

For this function, there will be trouble near the line y � x (for the third line) and y � �x for the

second line. The first and fourth approach zero as px, yq Ñ p0, 0q (whether they are rational or not).

Let’s think about the third line: we need y rational and x irrational, and we want to find a sequence

tpxn, ynqu of such points that approach the origin but for which fpxn, ynq does not approach zero.

So let

yn � 1

nand xn � 1

n�?

2

n4

Then

fpxn, ynq � 2y3npxn � ynq2 �

2n3

2n8

� n5

which goes to infinity as nÑ8. This shows that f cannot be defined at p0, 0q to be continuous.

9. Gotta practice! Find the first and second partial derivatives of the following functions. Also find

the derivative of the functions along the indicated vectors at the indicated points:

(a) fpx, yq �ax2 � y2, At px, yq � p1, 2q, v � i� 2j, v � 2i� j

We have BfBx � xa

x2 � y2BfBy � ya

x2 � y2

B2fBx2 � y2

px2 � y2q3{2B2fBxBy �

B2fByBx � � xy

px2 � y2q3{2B2fBy2 � x2

px2 � y2q3{2 .

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Since ∇fp1, 2q � p 1?5, 2?

5q, we have

Di�2jfp1, 2q � 1?5xp1, 2q , p1, 2qy �

?5 and D2i�jfp1, 2q � 1?

5xp1, 2q , p2,�1qy � 0.

(b) fpx, yq � arctan�yx

, At px, yq � p1, 2q, v � i� 2j, v � 2i� j

For this function BfBx � �y

x2 � y2BfBy � x

x2 � y2

B2fBx2 � 2xy

px2 � y2q2B2fBxBy �

B2fByBx � y2 � x2

px2 � y2q2B2fBy2 � �2xy

px2 � y2q2 .

Since ∇fp1, 2q � p� 25 ,

15 q, we have

Di�2jfp1, 2q � 1

5xp�2, 1q , p1, 2qy � 0 and D2i�jfp1, 2q � 1

5xp�2, 1q , p2,�1qy � �1.

(c) fpx, y, zq � e�px2�2y2�3z2q, At px, y, zq � p2, 2, 1q, v � �2i� 2j� k, v � i� j� 2k

We have

BfBx � �2xe�px

2�2y2�3z2q BfBy � �4ye�px

2�2y2�3z2q BfBx � �6ze�px

2�2y2�3z2q

B2fBx2 � p4x2�2qe�px2�2y2�3z2q B2f

By2 � p16y2�4qe�px2�2y2�3z2q B2fBz2 � p36z2�6qe�px2�2y2�3z2q

B2fBxBy �

B2fByBx � 8xye�px

2�2y2�3z2q B2fBxBz �

B2fBzBx � 12xze�px

2�2y2�3z2q B2fByBz �

B2fBzBy � 24yze�px

2�2y2�3z2q.

Since ∇fp2, 2, 1q � p�4e�15 , �8e�15 , �6e�15q, we have

D�2i�2j�kfp2, 2, 1q � e�15 xp�4,�8,�6q , p�2,�2,�1qy � 30e�15

and

Di�j�2kfp2, 2, 1q � e�15 xp�4,�8,�6q , p1, 1, 2qy � �24e�15.

(d) f : Rn Ñ R, fpxq � xx , xyp, all x, all v.

First off, since xx , xy � x21 � x22 � � � � � x2n, we have Dkpxx , xyq � 2xk. Therefore

BfBxk � 2pxk xx , xyp�1

,

and so

Dvfpxq �n

k�1

2pxk xx , xyp�1vk � 2p xx , xyp�1 xx , vy .

For the second derivatives, we can use the chain rule. If k � ` we have

B2fBxkBx` �

BBxk

�2px` xx , xyp�1

� 4ppp� 1qxkx` xx , xyp�2

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and if k � ` we have

B2fBx2k

� BBxk

�2pxk xx , xyp�1

� 2p xx , xyp�1 � 4ppp� 1qx2k xx , xyp�2

(e) f : Rn Ñ R, fpxq � xTAx for a constant matrix A, all x, all v.

If we let ek be the vector with 1 in the kth place and zeros in all the other places, then

BfBxk � eTkAx� xTAek �

n

`�1

pAk` �A`kqx`.

Therefore

Dvfpxq �n

k�1

n

`�1

pAk` �A`kqx`vk � vTAx� xTAv.

For the second derivatives, we have

B2fBxkBx` � Ak` �A`k.

10. Prove the following properties of the gradient operator (assume f and g are differentiable on

some open set):

(a) ∇f � 0 if f is a constant (challenge: is this if and only if?)

Since f is constant, we haveBfBxi � 0 for all i and so ∇f � 0.

Because of the mean-value theorem, a function f with ∇f � 0 at every point will be constant on

connected open sets. But if the open set consists of several disconnected pieces, then f can assume

different constant values on each piece.

(b) ∇pf � gq � ∇f �∇g

Let hpxq � fpxq � gpxq. ThenBhBxi �

BfBxi �

BgBxi . These are the components of ∇h, and so

∇h � ∇f �∇g.

(c) ∇pcfq � c∇f if c is a constant

Likewise, if hpxq � cfpxq, thenBhBxi � c

BfBxi , so ∇h � c∇f .

(d) ∇pfgq � f∇g � g∇f .

Let hpxq � fpxqgpxq. ThenBhBxi �

BfBxi g � f

BgBxi . Putting these all together in a vector gives

∇h � p∇fqg � fp∇gq � f∇g � g∇f .

(e) ∇�f

g

� g∇f � f∇g

g2at points where g � 0.

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Let hpxq � fpxqgpxq . Then

BhBxi �

gDif � fDig

g2. Putting these all together in a vector gives

∇h � g∇f � f∇gg2

11. It is common in physics to let rpxq � }x}.

(a) Describe ∇rpxq geometrically.

Since ∇r � x

}x} �x

rwe have that ∇r is a unit vector that points directly away from the origin.

(b) Calculate ∇rp.

∇rp � prp�1∇r � prp�1x

r� prp�2x.

(c) For which p can you find a scalar field fpxq such that ∇f � rpx? What is f?

From part (b), for p � �2 we’ll have

∇�rp�2

p� 2

� 1

p� 2pp� 2qrpx � rpx.

For p � �2 we need to find a function f such that ∇f � x

r2. From our experience with single-variable

calculus (where the offending exponent is �1), it makes sense to try logarithms:

∇ logprq � 1

r∇r � 1

r

x

r� x

r2.

So we can do this for all values of p.