CONTENTS

How to Use This Book IV

Acknowledgments IV

CHAPTER 1 The Geometry of Euclidean Space 1

CHAPTER 2 Differentiation 21

CHAPTER 3 Higher-Order Derivatives; Maxima and Minima 43

CHAPTER 4 Vector-Valued Functions 63

CHAPTER 5 Double and Triple Integrals 77

CHAPTER 6 The Change of Variables Formula and Applications of Integration 97

CHAPTER 7 Integrals over Paths and Surfaces 17

CHAPTER 8 The Integral Theorems of Vector Analysis 143

CHAPTER 9 Sample Exams 161

APPENDIX Answers to Chapter Te ts and Sample Exams 167

1

THE GEOMETRY OF EUCLIDEAN SPACE

1.1: VECTORS IN TWO- A ND T H REE-D IM ENSIO N A L SP ACE

GOALS

1. Be able to perform the following operations on vectors: addition, subt raction , scalar multiplication.

2. Given a vector and a point , be able to write the equation of the line passing through the point in the direction of the vector.

3. Given two points, be able to write the equation of the line passing through them.

STUDY HINTS

1. Space notation. The symbol ]R or ]R I refers to all points on the real number line or a onedimensional space. ]R 2 refers to all ordered pairs (X , y) which lie in the plane, a two-dimensional space. ]R3 refers to all ordered triples (x, y, z) which lie in three-dimensional space. In general, the "exponent" in ]R71 tells you how many components there are in each vector.

2. Vectors and scalars. A vector has both length (magnitude) and direction. Scalars are just ~umbers . Sc~lars do not have direction. Two vectors are equal If and only If they both have the same length and the same ~~ direction . Pictorially, they do not need to originate from the same starting point. The vectors shown here are equal.

3. Vector notation. Vectors are often denoted by boldface let ters , underlined letters, arrows over letters, or by an n-tuple (Xl, X2 , ... , x 71 ) . Each Xi of the n-tuple is called the lth component. BEWARE that the n-tuple may represent either a point or a vector. T he vector (0,0, ... , 0) is denoted O. Your instructor or other textbooks may use other notations such as a squiggly line underneath a letter. A circumflex over a letter is sometimes used to represent a unit vector.

4. Vector addition. Vectors may be added componentwise , e.g., in ]R2

(XI , yr) + (X2, Y2) =(Xl +YI,X2+Y2). Pictorially, two vectors may be thought of as the sides of a parallelogram. Star ting from the vertex formed by the two vectors , we form a new vector which ends at the opposite corner

;;;?J v

of the parallelogram. This new vector is the sum of the other two. Alternatively, one could simply translate v so that the tail of v meets the head of u . The vector joining the tail of u

U I to the head of v is u + v . ?:jU + V I

______ __ 1

CHAPTER 1 2

5. Vector subtmction. Just as with addition, vectors may be subtracted componentwise. Think of this as adding a negative vector . Pictorially, the vectors a , b

a b and a - b form a triangle. To determine the correct direction , hb ~ you should Ibe able to add a b and b to get a. a- - Thus a b ~ go," f

3THE GEOM ETRY OF EUCLIDEAN SPACE

7. To sketch v, tart at the origin and move 2 units along the x axis , then move 3 units parallel to the y axis , and then move -6 units parallel to the z axis. The vector w is sketched analogously. The vector - v has the same length as v, but it points in the opposite direction . To sketch v + w , t ranslate the tail of w to the head of v and draw the vector from the origin to the head of the translated w . The vector v - w goes from the

y head of w to the head of v.

9. On t he y axis, points have the coordinat s (0 , y, 0) , so we must restrict x and z to be 0. On the z axis, points have z (~~~'51 _ . (O,y,z) the coordinates (0,0, z) , so we must restrict x and y to be (x,O,z)/ / y 0. In the xz plane, points have the coordinates (:I: ,O , z). so ( (o, y, o) we must restrict y to be 0. In the yz plane, points have the coordinates (0 , y , z ), so we must restrict x to be O.x

12. Every point on the plane spanned by the given ve tors can be writ ten as aV I + bV2, where a and b are real numbers; therefore, the plane is described by

a(3 , -1,1) + b(O , 3,4). 15. Given two points a and b , a line through them is l(t) = a + t(b - a). In this case, a =

(-1 , - 1, -1) and b = (1, - 1, 2), so we get l(t ) = (-1, -1, -1) + t(2, 0, 3) = (2t - 1, -1, 3t - 1).

19. Substitute v = (x, y , z) =(2 + i, - 2 + t, -1 + i) into the equation for x, yand z and get 2:1: - 3y + z - 2 2(2 + t) - 3(-2 + t) + (- 1 + t) - 2

4 + 2t + 6 - 3t -- 1 + t - 2 = 7.

Since 7 f. 0, there are no points (:I: , y , z) satisfying the equation and lying on v . 23. J ust as the parallelogram of example 17 was described by v = sa + tb for sand t in [0, 1], the

paralielpiped can be described by w = sa + tb + rc, fo r s, t and r in [0 , 1] . Let a , band c be the sides of the triangle as shown, and let Vij denote the vector from point i to poin j . We assume that each median is divided into a ratio of 2 : 1 by the point of intersection. Then we have

V I2 -a/ 2 = - (c - b) / 2; V23 = (1 / 3)(a / 2 + b ); V34 (-2/ 3)(a+b/ 2);

28 .

1

V45 (a + b)/ 2. The vector V I S should be the sum V1 2 + V23 + V34 + V45, or

c - b 1 (a ) 2 ( b ) a+ b c V1 5 =--2- + 3 2 + b - 3 a+ 2' + - 2- = b - 2'

which is the median of the vector that ends on c. The other two median ' are analyzed the same way.

CHAPTER 1 4

30. (a) Using x, the number of C atoms; y, the number of H atoms; and z, the number of 0 atoms, as coordinates, we get

p(3, 4, 3) + q(O, 0, 2) = r(l, 0, 2) + s(O, 2,1). {b) To find the smallest integer solution for p, q, rand s, we balance the equation componentwise:

3p= r (equating x)

2s =4p i. e., s = 2p (equating y)

2q + 3p = 2r + s ~ 6p+ 2p i.e., q = (5/2)p. (equating z)

Let p = 2, then the smallest integer solu t ion is p = 2, q = 5, r = 6, s = 4. (c) In the diagram, P is (6,8 ,6), Q is (0,0,10) , R is (6,0 ,1 2) and

S is (0,8,4). Both sides of the equation add up to the vector

(6,8,16).

R

p y

x

1.2: THE INNER PRODUCT, LENGTH AN D DISTANCE

GOA LS

l. Be able to compute a dot product.

2. Be able to explain the geometric significance of the dot product.

3. Be able to normalize a vector.

4. Be able to compute the projection of one vector onto another.

STUDY HINTS

l. Inne1' product. This is also commonly called the dot product, and it is denoted by a . b or (a, b). The dot product is the sum 2.:::7=1 ajbj, where aj and bj are the ~th components of a and b , respectively. For example, in lR 2, a . b =al b1 + a2b2. Note that the dot product is a scalar.

2. Length of a vector. The length or the norm of a vector x = (x, y, z) is Jx2 + y2 + z2. It is denoted by Ilxll and is equal to VX-:-X. This is derivable from the fact that x . y = X1YI + X2Y2 + X3Y3 with x = y.

3. Unit vector. These vectors have length l. You can make any non-zero vector a unit vector by normalizing it. To normalize a vector, divide the vector by its length, i.e., compute a/liali .

4. Cauchy-Schwarz inequality. Knowing that la hi ~ Ilallllbll is most important for doing proofs in the optional sections of this text and in more advanced courses. .

5. Important geometric properties. Know that a b = Ilallllbll cos e, where B is the angle between the two vectors. As a consequence, a . b = 0 implies that a and b are orthogonal. The zero vector is orthogonal to all vectors.

6. Projections. The orthogonal projection of b onto a is the "shadow" of b falling onto a . The projection of b onto a is a vector of length (a b)/llall , in the direction of a/liali. Thus, the projection of b onto a is (a ' b)) a (ab)(

-W W=~a.

.. .

THE GEOMETRY OF EUCLIDEAN SPACE 5

7. Problem solving. Since vectors have magnitude and direction, they can be represented pictorially. It is often useful to sketch a diagram to help you visualize a vector word problem.

SOLUTIONS TO SELECTED EXERCISES

3. From the defin it ion of t he dot product, we get

uv (0 7 19) (- 2 - 1 0) -7

cos (j = = ' , " = ~ -0 1546Ilullllvll V72 + 192V22 + 12 V410V5 . .

From a hand calculator, we find that (j ~ 99 .

7. If w = ai + bj + ck, then Ilwll = va2 + b2 + c2 , and so

Ilull = vT+4 = ..)5; Ilvll = v1+l = h .

Using the formula for the dot product, we get

u v = (-1 )(1) + (2)( - 1) = - 3.

10. Using the same fo rmulas as in exercise 7, we get

Ilull =VI + 0 + 9 = JiO; Il vl l = v O + 16 + 0 = 4.

Since u does not have a j component and v does not have any i or k component, the vectors are perpendicular; therefore n . v =O.

12. A vector w is normalized by constructing the vector w/llwll. For the vectors in exercise 7: u 1 ( . 2 ') v 1 (' ' ) W = V5 -) + J; IRI = V2 ]- J .

15. The projection of v onto u is

nv _(-1)(2)+(1)(1) + (1)( - 3)(_, , k)-_i(_, , k) Ilul12u - (V1 + 1 + 1)2 l + J + - 3 I+ J + .

16. For orthogonality, we want the dot product to be O. (a) The dot product is (2i + bj ) . (- 3i + 2j + k ) = -6 + 2b , so b must be 3. (b) The dot product is (2i + bj ) . k = 0, so b can be any real number.

21. (a) Looking at the x componen s, the pilot needs to get from 3 to 23 . His velocity in the x direction is the i component , 400 km/ hr . T hus,

Llt = ~d = 23 - 3 = ~. v 400 20

The pilot flies over the airport (1/ 20) hour or 3 minutes later. The sam e answer could have been obtained by analyzing the y components. (b) We look at the z components and use the formula

~d = vLlt , i.e., h - 5 = (-1)( 1/ 20) ,

so h = 99/20. Thus, the pilot is 4.95 km above the airport when he passes over.

24. (a) It is convenient to draw the diagram with A on the .r axis.

CHAPTER 16

y

A x

(b) From the diagram in part (a), we get A = 150i and B = (llO cos 60 0 )i + (llO sin 60 0 )j.

A + B = (150 + llO cos 60 0 )i + (llO sin 60 0 )j = 205i + 55V3j. The angle that A + B makes with A is

() =tan- 1 ( ; ) (5~:) ~ tan - 1 (0.4647) ~ 250.= tan- 1 Alt.ernat.ively, the definition of the dot product gives us

() = cos- 1 ( A (A + B ) ) = cos- 1 (1 50)(205) + (0)(55V3) IIAII IIA + BII (150)(J51l00) ~ cos-l(O.gO~g) ~ 25 .

27. (a) Geometrically, we see that the i component of F is

IIFII cos (). Similarly, the j component of F is IIFII sin B. T here y Ffore,

F = IIFII cos ()i + IIFII sin ()j,

where () is the angle from the x axis. Since the angle from t he

y axis is 7r/4, () is also 7r/4, so F =3J2(i + j ).

(b) We compute D = 4i+ 2j, so FD = (3)2)(4) +(3)2)(2) =

18"\/2. Also, IIFII = 6 and IIDII = J20. From the definition of

the dot product,

F D 18)2 3 cos() = IIFII IIDII = 6J20 = vT5 ~ 0.9487,

i. e., () ~ 18, or equivalently, () ~ 0.322 radians. (c) From part (b), we had computed F D = 18)2. Knowing that cos () =3/.JfO, we calculate IIFII liD II cos() = (6J20)(3/JIO) = 18)2, also .

1.3: M ATRICES, DETERMINANTS AND THE CROSS PRODUCT

GOALS

1. Be able to compute a cross product.

2. Be able to explain the geometric significance of the cross product.

3. Be able to write the equation of a plane from given information regarding points on the plane or normals to the plane.

x

7THE GEOMETRY OF EUCLIDEAN SPACE

STUDY HINT S 1. Matri ces and determi1wnts. A matrix is just a rectangular array of numbers. The array is

written between a set of brackets. The determinant of a matrix is a. number; a matrix has no numerical value. The determinant is defined only for square matrices and it is denoted by vertical bars.

2. Computing determinants. Know that I ~ ! I= ad - be. Also know that abc d e f 9 h

Note the minus sign in front of the second term on the right-hand side. The general method for computing determinants is described next .

3. Computing n x n determinants. Use the checkerboard pattern shown here which begins with a plus sign in the upper left corner. Choose any column or row - usually picking the one wit,h the most zeroes saves work.

+ + Draw vertical and horizontal lines through the first number of + t he row or column. The numbers remaining form an (n -1) x

+ + (n -1) determinant, which should be multiplied by the number (with the sign determined by the checkerboard) through which both lines are drawn . Repeat for the remaining numbers of the row or column . Finally, sum the results . This process , called expansion by minors , works for any row or column. The best way to remember the process is by practicing. Be sure to use the correct signs.

4. Simplifying determinants . Determinants are easiest to compute when zeroes are present. Adding a non-zero multiple of one row or one column to another row or column does not change the value of the determinant, and this can often simplify the computations. See exa.mpIe 3.

5. Computing a cross product. If a = (aI, a 2 , a3) and b = (b 1 , b2 , b3), then J k

a x b = a1 a2 a3 b1 b2 b3

The order matters: a x b =-(b x a). The cross product is not commutative. Also, note that a x b is a vector , not a scalar .

6. Properties of the cross product. T he vectors a , b and a X b form a right-handed system (see figure 1.3.2 of the text) . The cross product a x b is orthogonal to both a and b . The length of a x b is IIallllbill sinOI, where 0 is the angle between a and b. Note that the formula for the cross prod uct involves sin 0, whereas t he dot product involves cos O.

7. More properties. If the cross product is zero , then either: (i) t he length of one of the vectors must be zero, or (ii) sinO = 0, i. e. , 0 = 0, so the vectors must be parallel.

8. Geometry. The absolute value of the determinant I ~ ! I is the area of the parallelogram spanned by the vectors (a, b) and (c, d) originat ing from the same point. The absolute value of

a bc the determinant d e f is the volume of the parallelpiped spanned by the vectors (a, b, c),

g h i (d, e, J) and (g, h, i) originating from the same point. The length of the cross product 11a x hll is the area of the parallelogram spanned by the vectors a and b. The vector a x b gives a vector normal to the plane spanned by a and b .

CHAPTER 1 8

9. Equation of a plane. Recall that the equation of the plane is ax + by + cz + d = O. The vector (a ,b, c) is orthogona] to the plane . Knowing two vectors in the plane , we can determine an orthogonal vector by using the cross product. Compare methods 1 and 2 of example 11.

10. Distance from point to plane. You should understand the derivation of the equation in the box preceding example 12. If necessary, review the geometric properties of the dot product in section 1.2.

SOLUTIONS TO SELECTED EXERCISES

2. (b) We subtract 12 times the third row from the first row and subtract 15 times the third row from the second row. 'fhen we expand along the first col umn:

36 18 17 o -42 41 -42 41 1 . - . 45 24 20 o -51 50 =3 -51 50 =3(-2100 + 2091) = -27.

13 5 -23 5 -2

5. The area of the parallelogram is /l a x hll . We compute

j k a x b = 1 -2 1 =-3i + j + 5k

2 1 1

and so the area of the parallelogram is

lin x hll = v'9 + 1 + 25 =55.

8. The volume of the parallelpiped is the absolute value of the 3 x 3 determinant made up of the

vectors ' components. Expand along the first row:

1 0 0 _1 3 -10 3 -1 - 2 'I =-l.

-14 2 -1

Thus, the volume is 1.

11. We want to find the cross product and then normalize it . We compute

So

J k v = -5 9 -4 I = 113i + 17j 103k.

789

There are two orthogonal vectors in opposing directions; they are given by

v/llvll =: (113i + 17j - 103k)/v'23667. Since all vectors are orthogonal to 0, the inclusion of that vector into the problem does not affect the answer.

15. (a) The equation of a plane with normal vector (A ,B ,C) and passing through the point (xo, Yo, zo) is A(x - xo) + B(y - Yo) + C(z - zo) = O. In this case, the equation is

l(x - 1) + l(y - 0) + l(z - 0) = 0 or x + y + z = 1. (d) Here, the normal vector is parallel to the line, so it is (-1 , - 2,3). Hence, the equation of the desired plane is

-1(x - 2) - 2(y - 4) + 3(z + 1) = 0 or - x - 2y + 3z + 13 = O.

9THE GEOMETRY OF EUCLIDEAN SPACE

16. (b) Two vectors in the desired plane are v = (0 - 1, 1 - 2, - 2 - 0) = (-1,-1 , - 2) and w = (4 - 0, 0 - 1, 1 + 2) = (4, - 1,3). The cross pr duct v x w is orthogonal to both vectors, and hence normal to the desired plane. We compute

v x w = - 5i - 5j + 5k,

so the desired equation is

-5(x - 1) - 5(y - 2) + 5(z - 0) =0 or - x - y + z + 3 =O.

22. (a) Let D be the matrix with rows u, v, w . T hen

Ul U2 U3 det D = u (vxw)= VI V2 V3

WI Wz W3

Use the fo llowing property of determinants: v (w x u ) corresponds to two row exchanges of the matrix D, 80 we hav

v . (w x u ) = (-1)( -1) det D =det D and w (u x v ) = (-1)( -1) det D =det D. To prove the other three, recall that u x v = -(v xu).

26. The line perpendicular to the plane is parallel to the normal of the plane, so the equation of the line is

l (t) = (1, - 2, - 3) + t (3, -1, -2). 29. Let u be the vector normal to the plane. Then u is perpendicular to 3i + 2j + 4k since v

is on the plane. Alsu u is perpendicular to 2i + j - 3k, because the vector Ai + Bj + Ck is erpendicular to all vectors in the plane Ax + By + Cz = D . To find u, we take t he cross

produ t of 3i + 2j +4k and 2i +j - 3k: J k

u = 32 4 =-10i + 17j - k. 2 1 - 3

When t = 0, we find that a point on the plane is (- 1,1 , 2), so the equation of the plane is - 10(x+ l) +17(y -l)- (z-2}=0 or - 10x + 17y-z - 25 =0.

30. First, find the normal ofthe plane. The normal ofthe plane is perpendicular 0 the line passing through (3, 2, - 1) and (1, - 1, 2). The equation of the line is l(t) = (3,2, - 1) + t (2, 3, - 3). The normal of the plane is also perpendicular to v = (1 , - 1, 0) +t (3, 2, -2). Therefore, two vectors on the desired plane are 2i + 3j - 3k and 3i + 2j - 2k, an the normal is (2i + 3j - 3k) x (3i + 2j - 2k) = - 5j + 5k. Now, we need a point on the plane, say, (3,2, -1). Thus, the equat ion of the plane is

O(z - 3) - 5(y - 2) + 5(z + 1) = 0 or - y + z = l. 34. T he plane passing through the origin and perpendicular to i - 2j + k is x - 2y + z = O. By

the distance formu la with (A, B , C , D) = (1, - 2, 1, 0) and (Xl, Y1, zt) = (6 ,1, 0), d = IAx l + BYI + C Zl + DI = 6 - 2 _ --- _ 2V6

VA2 + B2 + C2 VI + 4 + 1 V6 3 ' 37. Since all vectors in this exercise are unit vectors, liN x all = sin 01 and II N x hll = sin O2 . From

Snell 's law, n l sin 01 = n2 sin O2 . Hence,

nlllN x all =n211N x hll. To establish that N x a and N x h have the same direction, we assume t hat N, a and hall lie in the same plane, and a and h are on the same side of N. Hence N x aand N x h both are perpendicular to this plane and parallel to ach other . Thus nl llN x al l and n 211N X hll are equal.

10 CHAPTER 1

38. First, 4 times the first row is subtracted from the second row. Next, 7 times the first row is subtracted from the third row . The next step is expansion by minors along the first column. Finally, the 2 x 2 determinant is computed.

1.4: CYLINDRICAL AND SPHERICAL COORDINATES

GOALS l. Be able to convert back and forth between the cylindrical, spherical and cartesian coordinate

systems.

2. Be able to describe geometric objects with cylindrical and spherical coordinates. 3. Be able to describe the geometric effects of changing a coordinate.

STUDY HINTS l. Review. You should review polar coordinates in your one-variable calculus text .

. .~

2. Cylindrical coordinates. Denoted (r, {}, z), this is just like polar coordinates except that a z coordinate has been added. Know the formulas x = rcos{}, y = rsin{}, r = Jx 2+ y2 and {} ::::; tan-1(y/x).

3. Spherical cooridinates. Denoted by (p, {}, , p is the distance from the origin, > is the angle from the positive z axis and {} is the same as in cylindrical coordinates. Know the formulas

x=pcos{}sin>, y=psin{}sin> and z=pcos>. Also know t.hat

p = Jx2 + y2 + Z2, {} = tan-1(y/x) and > = cos-1(z/ Jx 2 + y2 + z2) = cos- 1(z/p). 4. Graphs of r, p = constant. Note that r = constant in cylindrical coordinates describes a

cylinder and that p = constant in spherical coordinates describes a sphere. You may have suspected this from the name of the coordinate system.

5. Computing {} and >. Remember that, in this text, > takes values from 0 to rr and {} ranges from 0 to 2iT. In some instances, it is more convenient to define {} in the range -rr to iT . You should be very careful about computing {}. If x = y = -1, t.hen tan-1(y/x) = iT/4, but plotting the point (-1,-1) in the xy plane shows that, in reality, {} = 5rr/4. This is why the authors fuss with tan- 1(y/x) in the definition. Plotting the x and y coordinates is very helpful for determining {}.

6. Negative r, p. Note that we have defined r and p to be non-negative. If the distance is given as a negative number, we need to reflect the given point across the origin .

7. Unit vector's in spherical and cylindrical coordinates. The unit vectors in cylindrical coordinates are e r , eo and ez . The vector e r points along the direction of r, while eo goes in the direction in which {} is measured , and e z = k. As one might expect., those three unit vectors form an orthogonal basis, and e r x eo = e z . Those vectors, however, lire not fixed as is the case with i, j and k , that is, if you change the point (7', {}, z), the set of unit vectors rotates. For spherical coordinates, there is also a set of unit vectorsep , eo and e",. Those vectors , in terms of i, j, k and the cartesian coordinates of the point are worked out in exercise 7 (see below) of this section.

11THE GEOMETRY OF EUCLIDEAN SPACE

SOLUTIONS TO SELECTED EXERCISES

1. (a) To conver t to rectangular coordinates, use x = r cos & and y = 1'Sin 11:

x = 1cos 45 = v2/2 and y = 1 sin 45 = V2/2. Next , use p = J x 2 + y2 + z2 and rP = cos - 1(z / p) to get the spherical coordinates:

Ir. - 1 ( 1l'P :::;; VI 2" + 2"1+ 1 = V 2 and = cos )21) = "4 ' Hence, for the cylindrical coordinates (1,45,1) , the rectangular coordinates are given by (/2/ 2,)2/ 2,1) and hesph r ical coordinates are (V2,rr/ 4, 1l' / 4) . (b) To convert to cylindrical coordinates, we use l' = ..jX2 + y2 and (J = tan-1(y/x):

r = .)4+1 = v'5 and B= tan - 1(1/ 2). Next , use the same formulas as in part (a) to get th spherical coordinates:

p=V4+ 4+ 1 = 3 and = cos-1(-2/ 3). Hence, the rectangular coordinates (2, 1, - 2) convert to the corresponding cylindrical coord inat s (/5, tan- 1(1/2), - 2) and to the spherical coordinates (3, tan- 1(1/2),cos-1(-2/ 3)).

2. (b) T his mappin g takes a point and rotates it by 1l' radians about the z axis . This is followed by a reflection across the xy plane. The net effect is that the point is reflected through the origin.

3. (b) Recall that the angle is measured from the "Nor th pole." If is 7i radians , t hen the location is at the "Soutl pole." The effect. of changing to 1[' - is taking a point and reflecting it across the xy plane.

5. Let p ~ 0, then (p, 0, 0) is the positi ve z axis. Now, let vary from 0 to 1l'. Then (p, 0, ) is the half plane in the xz plane wi th z ~ 0. By allowing & to vary from 0 to 21l', we rotate the half plane described above. Therefore, p ~ 0, ~ (J < 21l' and 0 ~ ~ 1l' describes all points in JR3.

If p < 0 also , the coordinates are not unique . For example, (x, y, z) = (1, 1,0) has spherical coordinates (V2,1l'/4,1l'/2) and (- V2, 51l'/ 4,1l' /2).

7. (a) First , ep is the unit vector along the vector (x, y, z) ; therefore, the formula is

xi + yj + zk

e p = . J x 2 + y2 + z2

z

y -----k------~

ry

xx (x, y,O)

Next, eo is parallel to the xy plane and denotes the direction in which the angle B is measured . It is perpendicular to r = (x , y, 0), so (ai + bj) . (xi + yj ) = O. Since we want B measured counterclockwise, a = -y (instead of y) and b = x. Therefore,

- yi + zj eo = J x 2 + y2'

12 CHAPTER 1

To fi nd e ,p , we note that e p , eo and e,p is a set of orthogonal vectors; they form a right-handed coordinate system with ep x eo = - e,p. So .

(x, y , z ) x (- y, x, 0) xzi + yzj - (x 2 + y2 )k ~ =- = .

rp J x2 + y2Jx2 + y2 + Z2

9. (a) The length of xi + yj + zk is J x2 + y2 + z2 , which is the definition of p. (b) Note that Il v ll = Jx 2+ y2 + z2 = P and v k = z; therefore,

13 THE GEOMETRY OF EUCLIDEAN SPACE

4. Matrices. A matrix is a rect angular array of numbers . Unlike a determinant, a matrix has no numerical value. You should remember that an n x m matrix has n rows and m columns. The (i, j) entry is the number located in row i, column j.

5. Matrix multiplication. You should practice until matrix multiplication becomes second nature to you. Let the components of A be a i j and let those of B be bkl, where A is an m x p matrix and B is a p x n matrix. Then the components of AB are

p

(A B )mn = L amjbjn . j =l

We can only multiply an m x p matrix with a p x n matrix, i.e., [m x p][P x n]. Note that the number of columns of A and the number of rows of B must be equal (p in this case). The result is an m x n matrix ("cancelling" the p).

6. Non-commutativity of matrix multiplication. In general, AB # B A . In fact, AB may be defined when BA is undefined . However , matrix multiplication is associat ive; i.e., (AB)C = A(BC) if the product AB C is defined .

7. Matrices and mappings. An m x n matrix can represent a mapping from ]Rn to ]Rm. To see this, let A be the matrix and let x be a vector in ]R n, represented as an n x 1 matrix, and y be a vector in ]Rm, an m x 1 matrix. Then the matrix A takes a point in ]R n to a point in ]Rm by the equation Ax = y.

SOLUTIONS TO SELE CTED EXERCISES

2. (a) Use the properties of leng hs and dot products:

(x + y) . (x + y) + (x - y) . (x - y) x . x + 2x . y + y . y + x . x - 2x . y + y . y 2x x+ 2y y =211xl12 + 211Y11 2.

The figure at the left depicts the equation geometrically. By the law of cosines, we have12

14 CHAPTER 1

8. We compute 1 nAB~ r: -1 ~31 and A+B~[: 0 2 1 1 -1 1

Expanding by minors across the first row gives

det A 3:12 -1 I 11 ~ II = 6 - 2= 4,0 1 + 1 1 detE 11

0 1 ~ 1- 11 ~ ~ 1= -3,

det(AE) 1-1 1 1_ 115 and3 1 -1 1 ~1 1- 31 5 1 -1 1 1=-12

det(A + B) - 41 ~ ~ I = 8.

11. (a) For n = 2,

I 'xall

det('xA) = 'xa21

For n = 3,

'xall Aal2 'xa13

det( AA) Aa21 'xa22 'xa23

Aa31 'x a32 'xa33 'xall det('xAd - 'xa12 det('xA2 ) + 'xa13 det(AA3) ,X . ,X2(a11 det Al - al2 det A2 + a1 3det A3) ,X3 det A,

where AI , A2 and A3 are 2 x 2 matrices obtained by expanding across the first row. Assume that for n = k, det('xA) = Ak det A. The for n = k + 1, det(AA) can be found by

a process analogous to the 3 x 3 case:

det(AA) 'xall det('xAd - 'xa12 det('xA2 ) + ... + (_l)k 'xal,k+l det('xAk+d ,Xk+l det A,

where AI, A 2 , ... , Ak+l are k x k matrices obtained by expanding across the first row. By induction, det('xA) =An det A for an n x n matrix A.

14. Assume, as in the book, that det(AE) = (det A)(det E). Then det(AEC) = det[(AB)C] = det(AE) det C = (det A)(det E)(det C).

17. Multiply the two matrices to get the identity matrix:

[ a b] 1 [d -b] e d ad - be -e a

Similarly, we can show that

15 THE GEOMETRY OF EUCLIDEAN SPACE

SOLUTIONS TO SELECT ED R EVIEW EXERCISES FOR CHAPT ER 1

1. v + w = (3,4, 5) + (I , -I , 1) = (4, 3, 6) =4i + 3j + 6k; v + w is the diagonal of a parallelogram whose sides are v and w.

3v =3(3,4 ,5) = (9 , 12, 15) =9i + 12j + 15k; 3v has the same direction as v with length 3 times the length of v.

6v + 8w =6{3, 4, 5) + 8(1, -I , 1) = (26, 16,38) =26i + 16j + 38k; 6v + 8w is the diagonal of a parallelogram whose sides are 6v and 8w. 6v has the same direction as v with length 6 times that of v and simi larly for 8w.

- 2v = - 2(3 , 4, 5) = (-6, - 8, -10) = - 6i - 8j - 10k; - 2v is a vector in the opposite direct ion of v with length twice that of v.

y y

x

v w = (3)(1) + (4)(-1) + (5)(1) =4; v w is the number II v llllw ll cos B, where B is the angle between v and w.

v x w is perpendicular to both v and w . Its length is the area of the parallelogram spanned by v and w.

4. (a) Using the point-direction form of the line, we get l(t ) = (0 , 1,0) + t (3, 0,1). (b) Using the point-point form of the line, we get

l(t ) =a + t(b - a) = (0 , 1, 1) + t [(O, 1, 0 - 0, 1, 1)] = (0 , 1, 1) + t (O, 0, -1) .

(c) T he normal to the plane is (a, b, c) = (-1,1, -I) , so th equation of the plane is

a(x - xo) + b(y - Yo) + c(z - zo ) = 0,

i.e.,

- 1(x - 1) + l(y - 1) - 1(z - 1) = or x - y + z = 1.

5. (b) v w = (1)(3) + (2)(1) + (- l ){O) =5.

i j k 1 1 6. (b) v x w = ~ ; ~1 = 1 ~ ~ 1j -I ~ ~ 1j + 1 ~ ~ 1k = i - 3j - 5k. 7. (b) We compute IIvll = VI + 4 + 1 = V6 and Il w ll = V9 + 1 + 0= v'lO. T hen the definit ion

of the dot product and the result of xercise 5(b) gives v w 5 5 5

cosO = = =-- --Jlv llllw il Y6v'lO V60 2.Jl5

16 (HAPTER 1

8. (b) The area of the parallelogram is the length of v x w. Using the result of exercise 6(b), we get Ilv x wll =VI + 9 + 25 = v'35.,

12. We compute the following dot products using the fact that u, v and w are orthonormal:

a . u (au + f3v + , 'w) . u = a. a v (au + (3v + ,w) . v = ,B. aw (au + (3v + ,w) . w = ,.

v Geometrically, a . u is the projection of a on u; similarly for the others.

15. From the definition of the dot product and the fact that U u = lIull 2 , we compute

v . a ~ lIall,b . a + IIblillall2 _ b a + IIblilia ll cos ()

II v II II all 11vlliiall IIvll v . b lIallllbW + IIblla . b lIalillbl ,1+ b . a

cos () = IIvllllbll IIvllllbll Ilvll

Since the angle between a and v is the same as the angle between b and v, the vector v must bisect the angle between a and b.

18. (a) Given that a b = a' . b, this implies that a b - a' . b = or (a - a') . b = 0 for all b. Choose b = a - a' to get lIa - a'1I2 = 0, so a - a' = 0, which means that a = a'. (b) The answer is yes . Ifax b = a' x b for all b, we can conclude that (a - a') x b = 0 for all b. Choose b to be a unit vector orthogonal to a - a'. By definition of the cross product, this implies lI a - a'il = 0, so a = a'.

22. (e) Note that the x and y coordinates lie in the third quadrant

of the xy plane. The definitions from section 1.4 are used: For

cylindrical coordinates, we compute

r JX2 + y2 = v'12 + 4 = 4.

() 7r + tan- 1(y/x) = 7r + tan- 1(1/V3)

7r + 7r/6 = 77r/6.

Thus, the cylindrical coordinates are (4,77r/6,3). x

For spherical coordinates, we compute

p = J x 2 + y2 + z2 = V12 + 4 + 9 = 5. 4> =cos-1(z/p) =cos-1(4/5).

Thus, the spherical coordinates are (5, 77r/6, cos- 1(4/5)),

/ I

y

23 , (b) Using the formulas from section 1.4, we calculate z

x = r cos () = (3)( V3/2) y

and

y = rsin() = (3)(1/2).

Thus, the corresponding cartesian coordinates are

(3V3/2, 3/2, ~4).

For the spherical coordinates, we compute

THE GEOMETRY OF EUCLIDEAN SPACE 17

p= Jx2 + y2 + Z2 =/27 +~+ 16 = 5.V 4 4

cjJ = cos- 1 (z /p) = cos- 1(-4/ 5).

Hence, the corresponding spherical coordinates are (5,1r/ 6,cos- 1(- 4/5)).

24. (b) Using th formulas from section 1.4, we compute:

z x pcosBsincjJ = 2(0)(1/ 2) = 0, y psinB sin

18 CHAPTER 1

44 . A vector which is normal to t.he first plane is v = 8i +j + k. A vector which is normai to the second plane is w = i - j - k. The cross product v x w is orthogonal to both normal vectors, so it should be parallel to both planes. We compute v x w = 2i - 7j - 9k , so the unit vector in question is (2i - 7j - 9k)/v'f34.

47. We want v = ai + .Bj + ,k to be such that !lvll = 1. From the definition of the dot product, we know that

o v'3 v j v'3, /3.' kso v = 2 1+ J +, .cos 30 = 2 = Ilvllllill' Since v makes equal angles wi h j and k , we must have .B = , . Since Ilvll == 1, we determine that .B =1 = 1/20. T herefore,

TEST F OR CHAPTER 1 (The answers are at the back of this book.)

1. True or false . If false, explain why.

(a) Given vectors u= i + j in IR 2, v = i + j +k in IR3 and w = 2i -k in IR3, we have uv = 2 and v w = 1.

(b) Given two matrices A and B, det(AB) is defined if and only if det(BA ) is defined. (c) For non-zero vectors a and b, the set a , b , and a x b, in that order, always forms a

right-handed system. (d) Suppose a , b, and c are non-zero vectors in IR3 and a b =a c =0. Then either b is a

multiple of c or b and c are parallel vectors. (e) A cross product is 0 only if one of the vectors is O.

2. Find an equation for the plane passing through the point (0,0,1) and parallel to the line containing the points (1, 0, 2) and (-1,2, 1).

3. Let A = (2 , 1,3) , B = (-3,2, 4) and C = (1 , 2, 2) be the vertices of a triangle. Let a be the vector from A to B and let b be the vector from B to C.

(a) Calculate a and b . (b) Describe the triangle ~ABC using vectors of the form sa + tb . (Find restrictions on the

parameters sand t.) (c) Calculate the area of ~ABC .

4. Let u = 3i + OJ + k and v = i - 2j + k have a common tail . Let () be the acute angle between u and v. Calculate sin ().

5. A parallelpiped spanned by the vectors (3, - 2,1), (0 ,2 ,-1) and (-1, 0, 1) is filled with sand. The sand will fill another parallelpiped spanned by the vectors - i, 2j and zk. Compute all possible values of z.

6. (a) In ]R4, fi nd an equation for the line passing through A = (1,3, - 2, 0) and B == (0,1, -1,1). (b) Show that the line in part (a) is not orthogonal to the line l(r) = (1 , 3, -2, 0) +r(2, 0, 3, 1).

7. Let

(a) Compute AB.

,. "

THE GEOMETRY OF EUCLIDEAN SPACE

(b) Compute det(AB) and interpret geometrically. (c) If det(BA) exists, compute it and interpret it geometrically.

8. Convert the spherical coordinates (-2, -11"/4, 27r/3) to (a) Cylindrical coordinates. (b) Rectangular coordinates.

9. Let u = (2, -1, 0,1) and v = (-1, - 2, 1,2). (a) Verify the Cauchy-Schwarz inequality for the given vectors. (b) Calculate the projection of v onto u.

10. A unique house has an inclined floor where the owner keeps his dog. The dog 's play area is a fenced en z closure with vertices at A = (2,4,1), B = (1, 2, 0) , C = (0,0, -1), D = (1, -1, -1) and E = (3,0,0), as shown in the diagram.

(a) All of the vertices lie on a plane. Find the equation

of the plane.

Dx (b) Find the area of the dog's play area. (c) An object is located at (1, 1, 3). What is the min

imum distance the dog must leap off the floor to

reach the object?

19

A

y

---------------- - - ------ --- --

21

2 DIFFERENTIATION

2.1: THE GEOMETRY OF REAL-VALUED F UNCTIONS

GOALS

1. Be able to define a graph, a level curve, a level set, and a section.

2. Be able to graph a function f : ]R2 --+ ]R .

ST UDY HINTS

1. Notation. f : A C ]Rn --+ ]Rm describes a function f The domain is A, which is a subset of ]Rn. Points in A are mapped to points in the range, which is a subset of ]R m . The only restriction on nand m is that they have to be positive integers.

2. Real-valued functio n . In the notation, f : A C ]Rn --+ ]R m, we restrict m to be 1 in this section. A real-valued function assigns a unique real number to each point in A.

3. Graphs. T he graph of f : A C ]Rn --+ ]R is drawn in the space ]Rn+l. If x is a point of A, the the graph consists of all points in ]Rn+l wi th the form (x,f(x)), where f (x) is a real number.

4. Level set. This is the set of x such that f(x) is a constant. In ]R2, such a set is called a level curve and in ]R3, it is called a level surface. Level sets are important for graphing.

5. Sections. These are intersections of graphs with a vertical plane. Usually, the most helpful sections for graphs in ]R3 are he intersections with the planes x = constant and y =consta.nt.

6. Graphing in JR3. Often, the best way to draw a graph in JR3 is to draw level curves for z = constant. Then lift or drop the curves to the appropriate "height" for z = constant. Analyzing the sections helps complete the graph. It is a good idea to review how to sketch the graph of an ellipse, a hyperbola, a circle and a parabola from your calculus or precalculus text.

7. Sketching planes. Many of us are poor artists, and as a result , three-dimensional geometry may be frustrating due to this problem rather than a lack of mathematical understanding. Planes are sometimes easily sketched by plotting three non-collinear points (usually on the coordinate axes) and then passing a plane through them.

8. Spheres. The equation (x-a )2+(y-b)2+(z-c)2 = r2 represents a sphere ofradius reentered at (a, b, c) .

9. Cylinders . A surface in ]R3 is called a cylinder if x, yor z is missing from the equation . A cylinder can be sketched by drawing the level curve in the plane where the missing variable is zero. Then move the curve along the axis of the missing variable.

10. Graphs in ]R4. Example 5 gives a fu nction whose graph cannot be drawn on paper. To see the "graph," one can make a movie which shows the concentric spheres expanding.

22 CHAPTER 2

SOLUTIONS T O SELECTED EXERCISES

1. (a) To determine the level curves, we look at the equation c = x - y +2, where c is a constant . T he equation is the same as y = x + (2 - c), which is the equation of a line with slope 1 and y intercept 2 - c.

The graph of f is a plane with intercepts at (-2, 0, 0), (0 ,2, 0) and (0 , 0,2) .

22. (b) We look at c = 1 - x - y 2 for c =constant . This rearran ges to x2 + y 2 = 1 - c, which is the equation for a circle of radius v'1='C, centered at the origin if c < 1. If c = 1, then the level set is a point at the origin. Th re is no level curve if c > 1.

3. (3) Substitute x =rcos (} and y = rsinB to get

Since z = 1'2 does not depend on () , the shape of the graph does not depend on () .

x 2 c2 25. For constant c, the equation c = y'100 - - y2 is equivalent to = 100 - x - y2 or 2x 2 + y2 = 100 - c . T h I vel curves are circles with radii VI00 - c2 , centered at the origin.

So, for c = 0, 2,4,6 8, 10 the radii are 10, v'96, j84, 8, 6 and 0, respectively. Drawing the level curves and raising them to the appropriate z values, we obtain the following graph. T he graph of f (x , y) is a hemisphere. T he level curves and the graph are shown here.

y ,

x

10. The level curves have the equation c = x / y or y = x / c for onstant c. Th se level curves are lines through the origin with slope l /e. One restriction is that y f:. 0. Next we discuss the graph z = x/ yo Noti e that when x is held constant, the sect ion are the hyperbolas c = yz , and when y i held constant , we get the lines c = x / z . Putting all this information together, we get t. he "twisted plane" as the graph.

23DIFFERENTIATIOI\l

y

12. The level surfaces have the equation c == 4:z;2 + y2 + 9z 2. T here is no level surface if c < O. If c = 0, then the level surface is the origin. If c > 0, then we should look at the level curves for constant k, i.e., analyze c - 9k2 = 4x2 + y2 . We recognize that if 9k 2 < c, then the level curves are ellipses which get smaller as Ikl approaches v'C/3. Similarly, we see that level sections parallel to the yz plane have the equation c - 4k 2 = y2 + 9z2 , which are ellipses with decreasing "radii" as Ikl approaches .;c/2. Also , the level sect ions parallel to the xz plane have

k2the equation c - = 4x2 + 9z2 , which are again ellipses which get smaller as Ikl approaches .;c. The level surfaces are ellipsoids if c is positive.

17. If c < 0, the level curve is empty. If c = 0, t he level curve is the x-axis. If c > 0, it is the pair of parallel lines Iyl = c; that is, the lines y = c and y = -c. In the yz plane, we sketch the graph of z = Iyl. Since x does not appear in the equation , we get a "cy,)inder" and this sketch is shifted along the x axis to obtain the graph in 1R3 .

y

x

22. The equation can be written as (x 2 - 2x + 1) + y2 = 1 or (x - 1)2 + y'2 = 1. In the xy plane, this is a circle with radius 1, centered at (1, 0). Since z is not in the equation, z may take on any value, so the circle is sh ifted up and down the z axis.

24 CHAPTER 2

y

-.---:;,.-

25 DIF FERENTIATION

x

2.2: LIMITS AND CONTINUITY

GOALS

1. Be able to defi ne the following: open disk, open set , neighborhood, boundary point , limit , continuous.

2. Be able to determine where a function is continuous.

3. Given a function, be able to compute a limi t or show that a limit does not exist .

STUDY HINTS

1. Theoretical section. This section is not essential for computational purposes. Your instructor may choose not to emphasize the material in this sect ion . Use your lectures to determine how important the material is for your course.

2. Definitions , (a) An open disk around Xo is the se of points x such that Il x - xoll < 1' . It is denoted Dr (xo) . Note the strict inequality. (b) An open set U is a set such that every xo has an open disk entirely within U. You will need to find an r when proving that a set is open. (c) A neighborhood of Xo is an open set containing Xo. (d) A boundary point of a set A has no neighborhood en irely inside or entirely outside of A.

3. Review. You should review the concepts of limit and continuity from your one-variable calculus textbook before continuing.

4. Limits. In the definition, be aware that Xo doesn't have to be in A; Xo may be on the boundary. Also, f (xo) does not have to be defined. We are only interested in the points x n ar Xo . For proofs, we need to choose U, which is dependent upon N.

5. Properties of limits . Most of th se are what you would intuitively xpect . Note hat for multiplication and division, the mapping is into }R I.

6. Continuity. Analogous to the one-variable definition, a function is continuous at Xo if

lim f(x ) = f( xo), X~Xo

I.e. , the limit equals the function value. T he limi t on the left-hand side is concerned about points near Xo. The right-hand side, f(xo), is concerned about the point Xo itself.

26 CHAPTER 2

7. Non-existent limits . Showing that the limit of f(x, y) does not exist is sometimes simple. To show a limit does not exist , one can, for example, look at the limit of f by first holding x constant, then holding y constant. If the two values differ, the limit does not exist.

SOLUTIONS TO SELECTED EXERCISES

2. Take < r ~ y, th n for all points (x , y), the open disk Dr (x,y) C B. Thus, B is open. 5. (c) Recall the defin ition of the derivative:

f ' ( ) - I' f(x o + h) - f (xo)Xo - 1m h . h -tO

By letting f(x ) = e"" and xo = 0, we get he" - 1 - eO d IeO+ lim - h- = lim = _ eX = 1.

h -t O "-to h dx x = o

6. (c) From one-variable calcul us recall that lirn.,-t 0 (sin :z: / x) = 1. This fact an al 0 be verified by using I'Hopital's rule . Using this fact and the properties of limits , we compute

. 2 (.)2 ( .)2sm x _ I' Sill X _ } ' sm x _ 12 - 1lim - 1m - 1m - - . x 2x-tO x -t O X x-t O X

g. (c) It is obvious that the limit of the numerator is 0, and the limit of the denominator is 2 # 0, so the limit of he quotient is 0/ 2 = O.

10. (b) First, hold y = 0 constant and let x approach O. Then use I'Hopital's ru le:

. cosx - 1 - (:z:2 / 2) . -sinx - x . - cosx - 1 hm = 11m = hm . x-tO X4 ""-to 4x3 x-tO 12:z:2

This last limit tends to -00 since the numerator tends to - 2 and the denominator tends to 0, Thus, the limit does not exist.

11, (a) Let t =xy and use continuity of compositions (theorem 5) to get

I sin xy }' sin t 11m -- = 1m -- = , (x ,y )-t(O, O) xy t -tO t

13. (c) Use the fact that the limit of a vector is the limit of each component (theorem 3(v)). So Xwe get limx -t l (x2 ,e ) = (l,e).

16. (a) We will use theorem 5 (continuity of compositions) . Note that f (:z: ) = (1- x)8 + cos (1+ x3 ) is the sum of two functions . The first is u8 with 'U = 1 - x. Since tJ is continuous, theorem 5 says that (1 - x)8 is continuous. The second function is cos v with v = 1 + x3, Again, since v

3is continuous, theorem 5 says that cos (l + x ) is continuous. T he sum of continuous functions is continuous, so f(x ) is continuous.

17. (a) We can make a function continuous by equating f (xo ) and limx-txo f(x) . As in exercise 11(a) , we can Ie t = x + y , so

sin(:z: + y) _ l' sin t - 111m - Im-- - . (x ,y )-t(o,O) x +Y t-tO t

Thus, we let

sin (x + y) -~----''-'- = 1 for x = y = 0,

x +y

"

27 DIFFERENTIATION

and we have a continuous function. (b) First , we note that if x = y , then

xy . x'.l 1 lim ') " = lim 2 ., = -2 ' (x .y ) ~ (o. O ) x- + y- x~o x-

On the other hand , if x = -y, then x y . _x 2 1

lim = hm -- = -(x.y) ~ (o.O ) x 2 + y2 x~o 2x 2 2 Since the Hmiting value depends on the direction of approach, the limit does not exist at the origin , so it is not possible to make the function continuous at the origin.

22. (b) We want to find a 6 for every N > 0 such that 0 < x < 6 implies that 1/1xl > N. Let 0 < 6 < l i N , then Ixl < liN implies that 1/1xl > N. This is not true if the absolute values are omitted , i.e., limx~o(llx) may be +ex> or -ex> depending on which side of 0 we are approaching from (see the figure below.)

y

x

x

y =l/lxl y =l/x 27. (a) By the triangle inequality, 1a.3 + 3a 2 + al < la2 1+ 31a 21+ lal, since lal < 1 (we assume it is

small, since 6 ,is small), this is less than (or equal to) 514 Choose 6 < 1/500 , then for lal < 6, la3 +3a2 +al ::; 1/100 (note that this is a very rough estimate; a bigger 6 would probably work if we work harder to improve the inequality.)

2.3: DIFFERENTIATION

GOALS 1. Be able to state the definition of partial derivat ives.

2. Be able to compute a partial derivative or a matrix of partial derivatives.

3. Be able to compute a gradient .

4. Be able to compute a tangent plane.

STUDY HIN TS

1. Nota-tion. Class en means that the nth derivative is continuous.

2. Partial derivatives . Know the definition

of l ' f( Xl, "" Xi + h, ... , x n) - f(Xl' ... , Xi, ... , Xn)

- = In1 =---..:........:....:.---''---'---'----'-~~-'--'----=-'--'--'-'--'----'-'-'-

OXi h~ O h

To compute ofIOXi in examples, consider all variables except Xi to be constant and differentiate by one-variable methods . Differen iation is performed with respect to the variable Xi.

28 CHAPTER 2

3. Notation lor partial derivatives. In many texts, Ix is used for 8f/8x. If we wish to evaluate at a given point, we write

81 fx l(xo ,yo) ' or ~: I if z = f (x , y).a ' x 1(xo,Yo) (:Co ,Yo)

4. Tangent plane. The tangent plane to the graph of a function f (x , y) at (x o, Yo, f(xo, Yo) ) is given by

8f l 81 1z= f(x o,yo)+ ax (x -xo) + {) (Y -Yo). (Xo,Yo ) y (Xo ,Yo)

This equation is also used to compute linear approximation . Compare this equation to the equation of a tangent line and the linear approximation in the one-variable case. See section 2.6 for a gener lization.

5. Differ ntiability. Equation (2) in the text tells you that f : R2 -+ is differentiable if t he tangent plane approaches f (xo,yo ) as (x,y) approaches (x o,yo ). Now, if f : U C ]Rn -+ ]Rffl, then

I Il f(x ) - I(xo ) - T(x - xo) 11 01m - (2)X--+Xo Il x - xoll - ,

where T : ]Rn -+ ]Rm is the derivative. You should be able to get equat ion (2) from this definition . T his definition of differentiab.ility is most important for theortical work.

6. Gradient. T he gradient is a vector whose components are the partial derivatives of f , with 8f/8x; in the ,th position. Here, fis a real-valued function. Thi operation is denoted by the symbol 'V . Someti, es , it is denoted "grad." For a function I : ]R3 -+ JR. , you should remember the form ula

7. Derivative of vector-valued fu.nctions. Con ider a function I : Rn -+ ]R m. The derivative is an m x n matrix of partial derivativ s. The range consists of vectors with m components . Think of t he components as real-valued vector fu nctions; th n each row of the derivative matrix is a gradient. The derivative matrix of f , evaluated at Xo, is denoted Df(xo) .

8. Important facts . Differentiability implies continuity of the fun ct.ion, but continuity does not imply differen tiability. The existence of continuous partial derivatives implies differentiability but the converse is not true. If a funct ion is differentiable, then its partial derivatives exist , but the converse is, again , not true.

SOLUTIONS TO SELECTED EXERCISES

1. (b) Holding y constant and differentiating with respect to x, we get 81/8x = yexy . By symmetry, 8f /8y = xeXY In this problem, all we did to compute fJ!I fJy was to switch x and y. T his is what we mean by "symmetry" ; it only works for functions that are unchanged when x and yare swapped.

2. (b) Hold y con taut and use the chain rule to differentiate wi th respect to x. We get f}z 1 1 1 y x

. '1 1 8z-- . - 'y - . SImI ar y, 8yox - J1+ xy 2y'1 + xy - 2{1 + xy) ' 2(1 + xy) .

At (1,2), 8z/8x =1/3 and 8z/ oy =1/6. At (0,0), az/ax =az/ 8y = o. 3. (b) Holding y constant and using the quotient rule, we calculate:

8w _ 2x(x2 - y2 ) - 2x(x + 2 + y2) 8x (x 2 _ y2 )2

29 DIFFERENTIATION

Holding x constant and differentiating with the quotient rule , we get:

ow _ 2y(x2 - y2) - (-2y)(x + 2 + y2) _ 4x2y ay (x 2 - y2)2 - (x2 - y2)2 .

4. (b) We must show that the partials are continuous in the domain : of/ax = l/y - y/x 2 = (x 2- y2)/ x2y, which is continuous for x f. 0 and y f. OJ a f/ ay = _X/y2 + l/x :;;:: (y2 - x2 )/xy2, which is continuous for x f. O. T hus, f(x) is C l since its partial derivatives are continuous.

6. (b) The equation of the tangent plane is given by

Z = Zo + [fx(xo, yo)](x - xo) + [fy(x o, yo)](y - Yo).

Using .the result of exercise l(b) , we compute:

afl - l' afl =0', f( 0,1 ) = 1. ax (0 ,1) - , oy (0,1)

Therefore, the tangent plane is z = 1 + l(x - 0) - O(y - 1) or z - 1 + x. 7. (b) The first row contains the partial derivatives of xeY + cosy. The second roW contains

those of x, and the third row contains those of x + eY The fi rst column contains the partial derivatives with respect to x, and the second column contains those wi th respect to y. Thus, the matrix of partial derivatives is

YeY xe - sin y 1 1 0 . [ eY1

8. (b) The function f is a mapping from JR3 to JR 2, so the matrix of the partials is 2 x 3. Let It (x, y, z) =x - y, the first component of f. Similarly, let Iz(x, y, z) = y + z. Then

ay Df(x, y , z) = [ a;:

af l %1 ] _[1 -1 ()]olz - 0 1 1 .aiz alz

ax ay az

xy11. Using t.he result of exercise 1(b), x(of/ax ) = xye = y(af/ay) . 12. (b) Use the linear approximation formula, which is the same as the equation of the tangent

plane: z = Zo + [fx(xo, yo)](x - xo) + [f y(xo , yo)](y - Yo). Let z = f(x, y) = x3 + y3 - 6xy, IO = 1, I = 0.99, Yo = 2 and y = 2.01. We compute:

8flax =3x 2 - 6y, so fx(1 ,2) =-9. a f /ay=3 y2_6x, so / y(1, 2) =6.

Therefore, our linear approximation is z ~ - 3 + (-9)( -0.01) + 6(0.01) = -2.85. The actual value is -2.8485.

13. (c) The gradient is defined as the vector (of/ox , of/oy , a f/az) . Thus,

14. (c) The tangent plane is defined by \If(xo) . (x - xo) = O. Exercise 13 (c) gives us \l f (1, 0,1) = ei + 2ek,

so the tangent plane is e(x - 1) + 2e(z - 1) =0 or x + 2z = 3.

30 CHAPTER 2

17. We compute 'V/ (0,0 , 1) = (2:1:, 2y , - 2z ) 1(0 ,0,1) = (0, 0, - 2) = - 2k. 20. We want to find T in equation (4). By linearity, / (x) - f( x o) = f{ x - xo ). Denoting x - Xo

by h , we want to find T so that

I II / (h) - Thll - hl:To Ilhll -. If we choose T = f, the numerator vanishes for all h, so this T sati ties the condition; that is, t he deriv tive of a linear map is the map itself. For exam ple, in one variable , consider f (x) = ax . From one-variable calculus, T = f ' (xo) = a for all Xo.

2.4: INTROD UCTION T O P ATHS AND CURVES

G OALS

1. Given a path, be abl to compute the velocity vector.

2. Be able to find a tangent line for a given path.

STUDY HINTS

1. Paths . A path is a "formula" that describes a curve in space . The picture of the path, which we can draw on paper, is called the image of the path or the curve of the path.

2. Path images. Often, it is convenient to express a path in terms of x and y when you want to know the image of a path. This is done by elim inating the parameter. For example

t4 2(x, y) = (t2,t4) means t =..;x,so y = = (-J x)4 = x . Caution: In this example, x = t 2, so x is always non-negative.

3. Circular functions. If a path is parametr ized by an expression involving cos t nd sin t , the parameter can usually be eli minated by squaring and adding. Use the identity cos2 t+sin2 t = 1 and 0 her trigonometric identities.

4. Velocity. T he velocity vector 's components are first derivat ives of the components of the path. The velocity vector is tangent to the path.

5. Tangent lines. It is easy to find a tangent line if you remember that a line can be described as x + tv. The vector x is chosen to be a point on the path at to and v is the velocity vector c' (t o). Thus, the tangent line to a path i

l(t ) = c(to ) + (t - to )c'(to).

SOLUTIONS TO SELECT ED EXERC ISES

1. From y = 4 cost , we get y/4 = cost. Use the fact that

and substitution to obtain

1 x

Since 0 ~ t ::; 211" , th curve is an ellipse with y intercepts at 4 and x intercepts at l .

-4

31DIF FERENTIATION

3. As in example 1, c(t) has the form (xo, Yo , zo) + tv,

where (xo, Yo , zo) = (-1,2,0) and v = (2, 1,1). T hus, z c(t) is a line in R3. Specifically, it is the line t hrough (- 1, 2,0) with direction (2, 1,1) . 2

(-1,2,0)

x

7. A path's velocity vector is found by differentiating the individual components. In this case,

d 2 d 3 d )1" (t) ( dt (cos t ), dt (3t - t ), dt (t)

(- 2costsint,3 - 3t2 , 1).

12. T he tangent vector to a curve c(t) = (x(t), y(t is c' (t ) = (x'(t),lI(t. In this case, the tangent vector is (2t, 0).

15. T he equation for the tangent line is l (t) = c(to) + (t - to) c'(to). Here, to = 1 and c'(t) = (3 cos 3t, - 3 sin 3t, 5t3 / 2). T herefore , the tangent line is

l(t) = c(l) + (t - l) c'(I) (sin 3, co 3, 2) + (t - 1)(3 cos 3, -3 sin 3, 5) (sin 3 - 3 cos 3, cos 3 + 3sin 3, -3) + t (3 cos 3, - 3 sin 3, 5).

17. F irst , we need to find the tangent lin at to. We compute c (2) = (4,0 ,0) and c'(t) (2t , 3t2 - 4,0) , so c' (2) = (4,8, 0). T hus, the tangent line is l (t) = (4,0,0) + (t - 2)(4 ,8, 0). T he position of the particle a t t l = 3 is 1(3) = (4,0,0) + (3 - 2)(4 ,8, 0) = (8 , 8, 0).

2.5: PROPERTIES OF THE DERIVATIVE

GOALS

1. Be able to state the chain rule.

2. Be able to compute a partial derivative by using the chain rule .

STUDY HINTS

1. Chain rule. Suppose f is a funct ion of Yl, Y2, ... , Yn and each Yi is a function of x. Then df of dYl of dY2 ' of dYn-=-.-+- .-+ ... +- .dx aYI dx aY2 dx aYn dx '

Notice how each term appears to be dJl dx with ay, a.nd dYi "cane ling." However, bewar that the "su m" on the right-hand side is df / dx, not n t imes df / dx. Also note the different d's: "a" is for a function of many variables, while "d' is for a function of one variable.

2. Multivariable chain rule. T he multi var iable chain rule st ates that

D(f 0 g)(xo ) =::; Df(yo) Dg(x o) , where Yo = g(xo) . This is he product of two deriv tive m atrices , so any desired partial m y be obtained by multi plica ion.

32 CHAPTER 2

3. Chain rule, gradient relationship. Know that if fis r ai-valued and h(t ) == f( c (t)) , then

dh

dt = \7f(c(t)) . c' (t ) .

SOLUTIONS TO SELECTED EXERCISES

2. (b) The function f is differentiable by the sum rule. Its derivative is

af af][az' ay = [1 , 1] .

(f ) The function is differentiable by the chain rule. We know that z2 and y2 are differentiable by he product rule and that 1- Z2 - y2 is differentiable by the sum rule. Also, the square root function is differentiable (where its argument is positive), so th entire function is differentiable . Its derivative is

3. (b) This is a spe ial case of the first special case of the chain rule: dh af dx af du af dv dx = ax dz + au . dx + av . dx

af al du af dv az + au . dx + av . dx

5. (b) First, we compute I(c (t)) = exp(3t2 t3 ) = exp(3t5 ), so f' (t) =15t4 exp(3t5).

Next, by the chain ruie,

dl a1 dx a f dy '" '" 2- = - . - + - .- = ye . 6t + :te Y 3tY . dt ax dt ay dx

Subst itute x = 3t2 and y = t 3 to get

dl j dt = t3 exp(3t5) 6t + 3t2 exp(3t5 ) 3t2 = 15t4 exp(3t6 ),

which is the same as we got from a direct computation.

6. (b) Take th derivatives of each component to get c'(t) = (6t,3t 2) . 9. Substitute u = e"'- Y and v = x - y to get

By the chain rule, D(f 0 9)( X, y) = D/(u, v)D9(X, y).

First, we calculat

Df(u, v) = [ ~~~ ~~: ~~~~~~] = [ sec\uu- 1) =;:] . 1When (x, y) = (1 , I ), we have g(l , 1) = (e1- , 1 - 1) = (1, 0) . Hence,

D/(I, O) = [12 -1 ].

33DIFFERENTIATION

Next , we calculate

1 -1 ]so Dg(I, 1) = [ 1 -1 .

Therefore

D(f 0 g)(I, 1) == [~ ~1] [~ =~] = [~ ~2]' Alternatively, we may calculate D (f 0 g)(x, y) directly from (f 0 g)(x, V) .

13. (a) By the chain rule, dT =V'T(c(t)) . c'(t),dt

where c( t) = (cos t, sin t). Differentiate:

Substituting x =cos t, Y =sin t gives

2 tesinV'T(c(t)) (2 cos tesin t - sin3 t, cos t - 3 cos t sin 2 t),

c' (t) ( - sin t, cos t) .

Thus,

dT . / 4 3 ' 2 ')

- = -2sintcosteSm +sin t+cos tesmt -3cos tsin~t .dt

(b) Plug in x =cos t, Y =sin t into the expression for T and get

2 tesinT(t) = cos t - cost sin3 t.

Using techniques from one-variable calculus, we have

dT . . t 3 it 4 2' ? - = -2 S1l1 t cos team + cos teS n + S1l1 t - 3 cos t S1l1~ t ~ ,

which is the same as the answer obtained by the chain rule in part (a). 17. (a) If y(x) and G are differentiable, then by the chain rule, and the fact that G(x, y(x)) is

constant, dC = aC . dx + aC . dy = aC + oG . dy = 0 dx ax dx ay dx ox oy dx .

Solve for dy/dx: If oC/ay 0, then dy oC/ox dx - oC/oy'

(b) As in part (a), we differentiate C I and G2 by the chain rule:

dCI = aCI + aCI . ciYI + oCI . dY2 = 0 and dC2 = aC2 + aC2 . dYI + oC2 . dY2 = O.

dx ox aYI dx 8Y2 dx dx ox OYI dx oY2 dx

Assuming YI (x), Y2 (x) and G are differentiable and

aCt/aYI aC I /aY2 11=0 for all x , oCdOYI 8C2/ aY2I

then we can solve for dyt/dx and dY2/dx. Rewrite the two equations as

oCI dYI oGI dY2 oC I

-.-+_.- (1)oYI dx oY2 dx ox oG2 dYI oG2 dY2 OC2-. - +_.- (2)OYI dx aY2 dx ax

34

Mul tiply (1) by OG2 / OYl and (2) by - 8GI/OYl . Add th two together to get : 8G1 8G2 8G2 8Gl-- .-+ _ .

dY2 8x 8Yl 8x 8Yl dx = - 8" G=-l ---;

35DIFFERENTIATION

so , ab2 (J 0 g) (t) = a2 + b2 .

On the other hand, from part (a) , we have "V f(O, 0) =(of /ax, of/o y)(O ,0) = (0 , 0). Aliso, we compute g'(t) = (a, b), so "V f (O, 0) . g'(O) = (0,0) . (a, b) = O. This verifies, as required, that the chain rule does no t apply to this function.

24. The t.erm ow/ax on the left-hand side means the partial derivative of w(x , y,g(x , y)) with respect to x, holding y constant. T he term ow/ox on the right-hand side means the partial derivative of w(x, y , z ) with respect to x holding yand z constant . These two terms are not equal because different independent variables are held constant. Thus, the reasoning is indeed flawed .

2.6: GRADIENTS AND D IRE CTION AL D ERIVATIVES

GOALS

1. Be able to define a directional derivative.

2. Be able to compute a directional derivat ive.

3. Be able to explain the significance of the gradient.

4. Be able to understand the relationships among the direct ional derivatives, the gradient , the tangent plane and level sets.

STUDY HINTS

1. Important example. Many ex mples in this book use the fact that "V1' = r/1', where r = (x , y , z ) and r = Ilrll = Jx 2 + y 2 + z2. T his is derived in example 1. Much time can be saved by remembering this result .

2. Definition. The directional derivative is defined to be . f(x + hv) - f(x)

or I1m h .! f (x + tv) It=o 11-+0 The directional derivative gives a rate of change of f in the direction of v at x .

3. Geometric interpretation. Suppose v = ai + bj is a unit vector , (xo , YO) is a given point and z := f(x , y) is a urface. T he directional derivative tells us the "slope" of a curve at (xo ,Yo) in the direction of v . T he curve is formed by the intersection of the surface with the plane described by the set of points sv + tk. If v is not a unit vec tor , then the "slope" may be determined by normalizing v to be a unit vector .

4. Relation to partial derivatives . T he partials of/ox , of/ oy and of/ OZ are t he directional derivatives of f in the directions i , j and k , respec ively.

5. Computing directional derivatives. To compute t he direct ional derivative of fin the direction of v , the easiest formula to use is "Vf (x) v . The directional derivative is a scalar , not a vector . The vector v is often chosen to be a unit vector .

6. Gradient pmperties. Recall that "V f = (af/ ox )i + (of/oy) j + (af/oz)k. You should know that "Vf points in the direction in which f is increasing fastest and - \lf points in the direction of fastest decrease. The gradient of f is always orthogonal to a level surface of f

7. Tangent plane. In terms of the gradient , the equation of the tangent plane at Xo is "V f(xo) . (x - xo) = O. This generalizes the formula given is section 2.3.

36 CHAPTEP

SOLUTIONS TO SELECTED EXERCISES

1. The directional derivative of / (x) at Xo in the direction v is "V/ (xo) . v . In thi case,

V' / (1,1,2) (1/ V5,2/ V5, 0) = (z2,3y2, 2xz )I(1 ,1,2)' (l/V5,2/ v'5,0) (4, 3, 4) (1/ vf:5,2/vI5,0)

= 1O/v15 =2v'5.

2. (b) Given / (x,y) = \n(lx2 + y2 ), we compute

"VI(x,y) =

At (1 , 0) , V'/ (1 , 0) = (1 , 0) and the directional derivative i V' /(1 ,0) v =2/ vI5. 3. (b) Given /(x, y, z ) = eX + yz, we calculate "V /(x, y, z) = eXi + zj + yk and thus V' / (1 ,1 , 1) =

ei+ j + k. The unit vector parallel to (1, -1 , 1) is (1, - 1, 1)/..)3. Thus, the directional derivati is V' /( 1, 1, 1) . (1, - 1, 1)/..)3 = e/ ..j3.

4. (c) For a function of three variables, I (x, y , z), the tangent plane to the surface is V' /(xo, Yo , zo) . (x - xo, y - Yo, z - zo) = 0.

To use this formula, we need to describe the surface f(x, y, z) = constant. In this Cal: f (x,y,z) = xyz = 1, so V'/(x,y,z) = (yz,xz ,xy) and at (1,1 1), V'f = (1, ] , 1) . Therefor the desired tangent plane is (1,1 , ] ) . (x - 1, y - 1, z - 1) = 0, i.e., x + y + z =3.

5. (b) We have z = (cosx)(cosy), so Zx = - sinxeosy and, by symmetry, Zy = - sinycosx . . (0 , rr/ ,0), we compute zX = 0 and zy = - 1. Therefore, the equation of the tangent plane i_

z = Zo + [zx (xo, yo)]( x - xo) + [Zy(xo . yo)]( y - yo ) =0+ O(x - 0) - l(y - rr/2) , i.e., z + y = 7r/ 2.

6. (c) Given f (x, y, z) = 1/(x2 + y2 + z2 ) = 1/1,2, we have Ix = - 2x / (x2 + y2 + z2)2 = - 2x/r 4By symmetry, I y = - 2Y/ I and 10 = - 2z/ r4 . Then "VI = (- 2/ r4 )(xi + yj + zk) = - 2r/r

where r = xi + yj + zk and r = l x2 + y2 + z2. 7. ( ) The direction of fastest increase is along the gradient vector. U ing the result of exerci:

6(c), we get the direction of fastest increase as 1(1,1,1 ) = (- 2/9)(i + j + k). 8. The gradient vector is normal to a surface. Here, we have I(x, y, z) = x3 y'3 + y - z + 2 = _

so Ix = 3x2y'3, Iy = 3x3 y2 + 1 and /z = -1. At the point (0 , 0,2), we compute Ix = 0, Iy = and f. = -1. Therefore, a normal vector is V' / (O, 0,2) = j - k. Normalize this to get the un normal: (l/.../2)(j - k) .

13. (b) By definition, "VI = (81/ 8x,8// 8y,8f/8z), so "VI = (yzexyz,xzexY"xyexyO). Giv g(t) = (6t,3t2 ,t3 ) , we differentiate each componen to get g' (t) = (6 ,6t,3t2) _ From Seetio 2_5, we know that

(lo g)' (I ) = V'/(g(I)) . g'( I) = exp(18t6 )(3t5 ,6t4 ,18t3 ) . (6 ,6t ,3t2 )lt=1 = e18 (18 + 36 + 54) = 108e1S _

17. By definition, "V I = (Ix , Iy ). Since f is independent of y, I y = 0 and given that I ( x, y) = g(z we have Ix = 81/ 8x = g' (x ). Th refore, "V I(x, y) = (g'(x ), 0).

37DIFFERENTIATION

20. The direction in which the alt it ud is increasing most rapidly at the point (x, y) is \7z(x, y ) = (-2ax , - 2by).

At the point (1,1) , \7z(l , 1) = -2(a, b) , so the desired direction is - (ai + bj )/Ja2 + b2 If a marble were released at (1, 1) , it will roll in the di rection at which the altit ude i decreasing most rapidly, so the marble will roll down in the direction -\7z, i . . , (ai + bj )/Ja2 + b2 .

24 . (a) We want to maximize I (c(t)) = (cost )(sin t ). Set the first derivative equal to 0: df/dt :::: -(sint)(sin t ) + (cost )(cos t) = 0, so we get cos2 t = sin2 t or t :::: (7r/ 4 + n7r ), where n = 0, 1, 2, .. .. Since .::; t .::; 211' , we only want t = 7r/ 4,37r/4 , 511'/ 4, 711'/4. Evaluating at these points, we get l (c{7r/4)) = l (c(57r/4)) = 1/2 and l (c(37r/ 4)) :::: l(c(77r/ 4)) :::: - 1/ 2. T herefore, the maximum value of lalong the curve c(t ) is 1/ 2 and the minimum value is - 1/ 2.

SOLUTIONS TO SELECTED REVIEW EXERCISES FOR CHAPTER 2 1. (a) Since 3x2 and y2 are non-negative, there is no level curve if z < O. If z :::: 0, then the I vel

curve is the origin. If z > 0, then the level curve is an ellipse wit h the major axi parallel to the y axis and the minor axis parallel to the x axis . The ellipses get larger as z increase . Put all of these level curves together to get an elliptical paraboloid .

2. (c) First, consider the surface x yz :::: O. The surface consists of the planes x :::: 0, y :::: 0 and z :::: O. Now consider x yz :::: 1. When z :::: k , a positi ve constant, th I vel curve is xy :::: l/ k . Thus we get a hyperbola in the fi rst and third quadrants with asymptotes on the x and y axes. The hyperbolas get closer to the origin as z gets larger. Thus, the surface in the first octant has the plane x :::: 0, y :::: 0 and z :::: 0 as asymptotes. Similarly, there is a surface in t he octant where z > 0, x < 0 and y < O. Now, if z < 0, then there is a similar surface in either of the octants where x < 0 and y > 0 or x > 0 and y < 0. The level surfaces for xyz :::: c, where c is a positive constant , consist of four similar surfa.ces which move further from the origin as c gets larger. If c < 0, the level surface is positioned in the other four 0 tants.

3. (b) The function takes a point from ~1 to ~ 2 , so the size of the derivative matrix is 2 x 1. The derivative is

D / (x) :::: [ ~ ] . 5. We need to show the vector normal to the tangent plane of I (x , y) at the point (xo, Yo , I(xo, Yo) )

is parallel to the vector (xo, Yo , zo). The partial derivatives of I are: 01 - 1 2 2 / ! -Xoox (xo, yo) 9 2x (1 - x - Y )-1 2 :::: 2 2 . ~ (xa,ya) vI - Xo - Yo 8I ( ) -1 2Y(I_X2_y2)-1/2! :::: -Yo oy xo, Yo 2 . / 1 2 2 ' (x a,Ya) V - Xo - Yo

So the normal to the tangent plane is - Xo . -Yo .

. ,1 + J - k.VI - X6 - Y6 VI - X6 - Y6 Multiply the above through by -(1 - x5 - y6)1 / 2. We get (xo, Yo ,/(xo ,Yo)) = (xo, Yo, zo). Geomet rically, we are looking at the sphere : x 2 + y 2 + z2 :::: 1. The v ctors normal to the tangent planes are precisely the vectors r = (x, y, z). Those vectors that have the direction of e p (see exercise 7(a) in section 1.4) are perpendicular to the sphere.

7. (c) The equation of the tangent plane is z :::: I (xo, Yo) + [(01 / fJ x )(xo , Yo) ]( x ~ xo) + [( 0I / oy)(xo, yo)](y - Yo).

In this case, we have 1(-1, -1) :::: 1; ol / ox :::: y, (of/8x )( - 1, -1) :::: -1; of/oy :::: x , (of/oy)( - I , -1) :::: - 1. T herefor , the equation of the tangent plane is z :::: 1 - l (x + 1)l(y+ 1), i.e., x +y+ z + 1:::: O.

. .

38 CHAPTER.

8. (b) If f (x , Y, z ) = constant, then the equation of the tangent plane is "il f(xo) . (x - xo) = 3where x = (x , y,z). In this case, f(x,y,z ) = x - 2y3 + z3, so "ilf(x) = (3x2,-6y2,3;

and "il f (l, 1, 1) = (3, -5 , 3). Therefore, the tangent plane is (3, -6, 3) . (x - 1, Y - 1, z - 1) = 3x - 5y + 3z = 0, i.e. , x - 2y + z = O.

11. (b) The strategy here is to find a few "paths," compute the limit along those paths. Lx = 2y; then the limi t as y goes to 0 is

lim FX+YI- r f3Y I- v'3(x, II)~(o ,O) y';-=YI- y~Y1-;1- . On the other hand, take the path x = 4y. Then the limit as y goes to 0 is

lim JI x + y 1= lim f5YI= f. =/: v'3. (x, y )-+ (O ,O ) x - y y~O Y13"Y 1 V3

Since the limiting vain s depend upon the path taken, the limit does not exist .

12. (b) Hold y and z constant and use the hain rule to differentiate with respect to x. Th ofl ax = 10(x +y +z)9 . By symmetry, ofl ay = ofl az = 10(x +y + z )9. T he gradient is t vector (of/ox, of/oy, of/oz) = 10(x + y + z)9(i +j + k).

16. We compute

~z (1, - 2) =2x l =2; ~z (1, - 2) =2y l = - 4. x (1,- 2) Y (1,-2 )

Then, the tangent plane is z = f(l, - 2) + (8 zl ox )I(1, _2) (x - 1) + (az l ay)I(1,_2) (y + 2), z =5+ 2(x-1) - 4(y+ 2) or 2z -4y - z = 5. Geometrically, the gradient of f (x , y) = x 2 + at (1, - 2) is perpendicular to the level urve z = 5. The tangent plane of the graph of f the plane that contains the line perpendicular to the gradient of f at (1, - 2) and lying in t horizontal plane z = 5, and the tangent plane has slope .../22 +42 = 2v5 relative to the plane.

18. (b) The directional derivative in the direction ofv is "ilf(xo) , v /llv il. We compute "ilf(x) = (y + z , x + z , x + y) , so "il f(l, 1,2) = (3, 3,2) . Also, we have

Ilvll = ) 102 + (-1)2 + 22 = .../105. Thus, the directional derivative in the direction of v is (3, 3, 2) . (10 , - 1,2)I v105 =31/ M

21. The bug should move in the direction of - "ilT(x,y), since this is the direction of fast decrease. We compute - "ilT(x, y) =- 4xi + 8yj and -"ilT(1 , 2) =4i + 16j, so the bug shou move in the direction 4i + 16j.

24. Let u = :z: - y, then

a! du of01 01 and 81 ou ay au . dy - AU .ox

Holding y constant, we get

8z = (!:.) 81 = (!) 01 .

ax y ax y au

Holding x constant and using the quotient rule , we calculate

az =~ [al .y _ 1(1)] =~ [_ of . y - I] . ay y2 oy y2 au

Therefore,

39 DIFFERENTIATION

27. (ii) (a) The sum rule tells us that x2 + y4 is differentiable. Also , x2y2 is differentiable by the product rule. Finally, x 2 y2 l(x 2 + y4) is differentiable by the quotient rule since (x , y) 1- (0 , 0), and so x2+ y4 i- O. Holding y constant, we get

of _ 2xy2(X2 + y4) - 2X(x2 y2) _ 2x y6 ax (X2+y4)2 (x 2 +y4)2

Holding x constant , we get

o f 2x2y(x2+ y4) _ 4y.3 (X2 y2) 2x4y - 2x 2y5 oy (x 2 + y4)2 (x2 + y4)2

At the origin, we use the definition of the partial derivative:

o o f (0 0) ::= lim f(O + h, 0) - f(O, 0) ::= lim h2 - 0 = lim 0 = 0, ox' h-+ Qo h h-+O h h-+O

and similarly,

o of (0 , 0) = lim f(O ,O + h) - f(O ,0) = lim }l-~ = lim 0 =o. By "-+0 h "-+0 h "-+0

(b) The partials exist at (0,0) and

lim f( x, y) =0, (x ,y)-+(O,o) II(x, y) - (0,0)11

so by the definition of differentiability (equation (2), section 2.3), f is differentiable at (0,0). The function f is differentiable at all other points because the partials are continuous. Thus f is differentiable. However, as (x,y) approaches the origin, aflox and o f loy do not approach o (take, for example, the path x = y); thus the partial derivatives are not continuous .

28. (b) The gradient vector is 'V f = (oflox, ofloy). If (x, y) i- (G, 0), then 2

of . ( 1) 2x y (1)- =YSID - cos ox x2+y2 (X 2+y2)2 x 2 +y2'

and by symmetry

of . ( 1) 2x y2 (1)- = X Sill - cos 1 oy X2+y2 (X 2+y2)2 x2+y2

Now, if (x, y) = (0 , 0), then the definition of the partial derivative gives us

of _ . f(O + h, 0) - f(O ,0) _. (0 + h)(O)sin ((0 + h) ~ + (0)2) I. 0 0 (0 , 0) - hm h - hm h = 1m - =o. 0X - "-+ 0 h-+O h-+O h

Similarly, (ofl oy)(O , O) =O. Therefore, if (x,y) i- (0,0), then 2[ . ( 1) y ( [ )] 2x'Vf(x,y) = YSID - ') cos Jx2 + y2 (x2 + y2 )~ x2 + y2

[ . ( 1) (1)] .2xy2+ XSIl1 x 2+ y2 - (x 2+ y2)2 cos . x2+ y2 J, and thus 'V f(O, 0) ~ Oi + OJ.

40 CHAPTER _

29. (b) The directional derivative is 'VJ (xo) . v. Here,

aJ =_ x sin(v x2 + y2 ), ax V x2 + y2

and by symm try, 2~~ vxZY+ y2 sin( v x + y2 ) .

Therefore, 'V J (1, 1) = (( - 1/ V2) sin (V2), (- 1/.;2) sin (.;2)) and the directional derivative is (( -1/v2) sin (v2), (- 1/ v2) sin (h )) . (1 / Y2, 1/ 0 ) = - sin (v2).

33. (b) Direct ly, we first compute g(u) = J (h (u) ) = sin2 31.1 + cos 8u . Then

!~ =6(sin 3u)(cos3u) - 8sinSu. When u = 0, dg/ du = O. By the chai n rule,

dg [ 3cos 3U ] . [ 3 cos 31.1 ]-d = DJ(x,y) Dh(u) = [2x, 1] 8 ' S = [2sm3u, 1] S ' 8 u - sm u - sm u

= 6(sin 3u)(cos 3u ) - 8 sin Su.

Again, wh n u = 0, dg / du = O.

35 . The normal to the surface J(x, y, z) = x2 + 2y2 + 3z2 = 6 is

aJ aJ aJ )( ax' oy' oz = (2x, 4y, 6z ) = (x, 2y, 3z ).

At (1, 1, 1), the normal is (1,2,3), so he unit normal is (1,2, 3)/Jf4, the direc ion offiight . T veloci y of the particle is the speed times the direction , or 10(1,2, 3)/J14. The position of particle at any time can be found by finding the equation of t he line through (1 ,1, 1) with direction 10( 1,2, 3) / V14, and it is (1, 1, 1)+10t (l , 2, 3)/Jf4. This impli s tha.t x = 1+10t/ K y = 1 + 20t/ v1A and z = 1 + 30t/ ..;u. At som t ime T, the particle is on he sph x2+ y2 + z2 = 103, which means that (1+ 10T;./f4) 2 + (1 + 20T / ..;u)2 +( 1+ 30T/J14)2 = 1 Simplifying, we get 3 + (120/ v'14)T + 100T2 - 103 = O. Solving for T using the quadra formula., and taking th positive T only, we get T = (-3 + V359)/5V14.

3S. (a) By substitution, Z = (x + y)(x - y) = x2 - y2, so az/ ax = 2x and az/oy = - 2y. (b) By the chain rule, we have

az oz aU. oz av , ax = au' ax + ov' ox = vl1) + u(l) =v+ u = 2x.

Also, we have oz az ou az ov ay = au . ay + av . By = ve l ) + u( -1) = v - u = - 2y.

41. Let u(t) = J(t )g(t) and h(u) = ell. By the chain rule, we have

dh dh du [ dJ d9 ] [dJ d9] .dt = du . dt =e U dtg (t) + J(t) dt = dtg(t) + J(t) dt exp[J(t)g(t)] .

46. The velocity is d fined as the derivative of displacement. Therefore, we want to compu au/ot = - 6cos(x - 6t ) + 5cos(x + 5t ). When t = 1/3, x = 1 and the velocity is au./at = -6 cos(-1) + 6cos(3). Sine cos (- x) = cos x, the velocity is 6(cos (3) - cos( l )).

41 DIFFERENTIATION

49. (a) As given, P is a function of T and V. We can also write

so T is a function of P and V. Finally, we can write

p = RTV2 - a (V - {3) V2(V - jJ) .

Upon rearranging, we get 3pv - (PjJ - RT)V2 + o:V - 0:{3 = O.

Since this is a cubic equation in V, it is theoretically possible to write Yin terms of P and T. Therefore, any two of V, P or T determines the third variable. (b) From the equation for T , we hold V constant and get

(aT/aP ) = (V - {3)/R . From the equation for P, we hold T constant and get

(fJP/fJV) = -RT/(V - jJ)2 + 2Vo:/ V 4 = - RT/(V - jJ)2 + 20:/V3. Now, hold P constant and differentiate the equation for P by implici t differentiat.ion . We get

aV aV . _ R(V - jJ) - 7jf(R T) 2V7jf0: _ ~ _ aV [ RT _ 20:]

o - (V - {3)2 + V4 - V - {3 aT (V - jJp V3' or equivalently,

aV R aT - (RT 20:) . (V - jJ) (V _ {3)2 - V3

(c) Using the results of part (b) , we get

( aT) (BP) (BV) (V - {3) (-R T 20: ) [ R 1 aP BV fJT = ~ (V - jJ)2 + V3 (V _ jJ) ((v~~p _ ~~) =-1. 50. (a) The question is asking for the directional deriva ive in the direction of a unit vector. Here,

our unit vector is (1,1)/v'2. Also V h(x , y) = (-0.00130x, -0.00048y) and Vh( -2, -4) == (0.00260,0.00196). Therefore, the directional derivative at (-2, -4) in the direction of (1,1)/0 is 'Vh( -2, -4) . (1, 1)/v'2 = 0.00456//2. This means that the height increases (0 .00456//2) miles per horizontal mile traveled. (b) The direction of the steepest upward path is V h( - 2, -4) = (0.00260,0.00196) .

TEST FOR CHAPTER 2

1. True or false. If false, explain why.

(a) If a 3-variable function ! (x , y, z) has partial derivatives l x, Iy and Iz at the origin, then the function is differentiable at the origin.

(b) A gradient vector for a function fin IR 2 is parallel to level curves of f (c) A function f is continuous at a point Xo if

lim I(x} = I(xo).

X---+Xo

42 CHAPTER _

(d) In general , a ontinuous function is differentiable. (e) Given a function !(w, x, y, z), th directional derivative in the direction of (0 , 0,1,0)

the same as 8f18y. 22. Let u = x y + z and v = xyz. Also, let x = 2a + h, y = ab2 and z = a + 2a2b. Furthermor~

let a = d3 and b = c - d. Compute 8(1., v)18(c ,d) at (c, d) = (1,1) . 3. Use the linear approximation to estimate th distance between the origin and the poi..

(8.01,3.9 1.04). 4. Let c(t ) be a mapping from ]R to ]R3 and let ! (x, y , z ) be a mapping from ]R3 to R. 10::

h(t ) = !(c(t )). When t = 1, we have (z, y, z) = (-1,2,4) , c'( I ) = (3,4, -2), dyldt = 1 ar dzl dt = 2. Is the rate of decrease fastest in the positive :c di r ction , the negative x directi the positive 11 direct ion r the negative y direction?

5. Let z = (z + y) 2 - 5x3 + 2ye X be the equat ion of a surface in space. Find the equation f t tangent plane at x = 0, y = 3.

6. Let Y.., if x i= 0

g(x, y) = x . { o if x = 0 Compute 8g18z and 8g 18y at the origin if they exist there.

7. Let u and v be functions of x and V, and let x and y be functions of s and t. Furthermore. is known t hat

8(u, v) 8( z, y) = [2 1] 0 - 1]8(s, t) 0 -1 and 8(z, y) [ 3 2 .

Calcul ate 8u.18 sand 8v 18t.

8. (a) Evaluate the following limit for !(w, z, y, z) = w - z2y3z : lim !(5, 2, -1, 1) - !(5, 2, - 1+ h, 1). h -+O h

(b) Is it possible to define g(O, 0) so that g(x , y) = (x2 + y2) /( x2 + y) is continuous on all 1R2? Explain why or why not .

9. Let u (x, y ) = 2x + y2 + eX. At (1,1) show that u increases faster in the direction parallel

the x axis than in th direction parallel to the 11 axis.

10. A recent survey showed that patient satisfaction, s, at a pharmacy depended mainly up.. three factors - the waiting time in minutes (t) , the patient's percei v d degree of illness (i) a scale from 0 to 10, and the dollar cost of the prescription (c) . T h patient satisfaction ind is given by S(t, i, c) = (1000 - c) / it2 . (a) In what direction does patient satisfaction increase most rapidly at the point (1 0, 0.5, 10C (b) At the point (10, 0.5, 100) how fast and in what direction (positive or negative) d

patient satisfaction change for each extra minute of waiting time? (c) The administrators do not want s to decrease by more than 1 per un it change of t, i

c. Is this goal met with a price decrease of 3/ 5 dollars and an increase in waiting time 4/ 5 minutes? Explain .

.. --------------------------~----------------------~------------------

43

3 HIGHER-ORDER DERIVATIVES; MAXIMA AND MINIMA

3.1: ITERATED PARTIAL DERIVATIVES

GOALS

1. Be able to compute iterated partial derivatives.

2. Be able to explain when mixed partials are equal.

STUDY H INT S

1. Iterated partial deri vatives. T hese are high r-order derivatives, such as second and th ird derivatives. With several variables, higher-order derivat ives may be taken with respect to different

variables. The notation :::y means (:x ) (~~) , which is also denoted fy x . 2. EquaLity of m ixed partials. If the second partial derivatives are cont inuous, then an iterated

partial derivative may be computed in any order.

3. Warning. Note that the theorem on equality of mixed partials requires continuous partial derivatives. If this requirement is viola ed , different orders of differe ntiation may yield different results.

4. EvaLuating partials at a given point. Always remember to differentiate com pletely before substi tuting given values. With mixed partials , you may substitute for a variable only after you have completed differentiating in that variable.

5. A pplications. The heat equation, the wave equation and the potential (Laplace) equation are famous examples of how higher-order derivatives occur in nature. T here is normally no need to memorize these equations in a vector calculus course.

SOLUTIONS TO SELECTED EXERCISES

2. Note that when we differentiate wi th re pect to either x or y, eZ is "constant ," so the fi rst partial derivatives are of/ox = -1/x 2 + e- Y and of/oy = - xe-Y Finding the partial derivatives of the fi rst partial derivatives gives us these second partial derivatives:

7. (b) Rewrite the func ion as 2X2 +7x 2y 2x 7x

z = =-+3x y 3y 3 ' provided x #-O and y #- O. W compute:

o z 2 7 fJz 2x fJ2z 0 (Oz ) o x = 3y + 3; fJy = - 3y2 ; ox2 = ox ox =0;

-

44 ( HAPTE 2 2/)2z a ( oZ ) 2 a (az ) a z a z 8 ( az ) 4x

8xay = 8x 8y = - 3y2 = ay ax = ay8x; ay'.! = ay ay = 3y3 The function is not continuous if either x =0 or y = O. Hence, the funct ion is not differentia whenever x =0 or y =O.

3 . . a f 8 af oa f a ( Oh )(0 (8 f ))11. By defi rutlOn, oxayaz = ax ay az . Let h = az ' so axayoz = Ox oy and

8 (8h) ah a ( a f ) 0 ( aftheorem 1, this is als equal to ay ax . Also, by theorem 1, ax = ox oz = 8z ax and therefore

o3f a (8 (8 f ) ) oaf 8xayaz = ay 8z 8x = 8yozax

15. (a) We are given f(x , y) = x arctan (x/ y), so we compute:

fx arctan (~) + ::' 1 ( \ / ~ ) = 2 xy 2 + arctan ( ~) . y y + xy~ x+ y y x - x _ x 2 fy = 1 + (x 2 / y2 ) . --:;J2 = x2 + y2

y(x2+y2)_ 2x2y 1 1 y3 _ x2y Y 2~ fx x = (x2+ y2)2 + Y1 + (x 2/ y2 ) - (x2+ y2 )2 + x2+ y2 - (x2+ y2)2 -2x(x 2 + y2) + 2x3 _ 2xy2

/:ey fyx = (x2+ y2 )2 = (x2 + y2)2 x2 2x 2yfyy --=-~"7. 2y = .= (x2 + y2 )2 (x2 + y2 )2

19. We have U x =3x2- 6xy, so U xx = 6x - 6x = O. Also, uy =3x2, so U yy = o. Substitution gi us 'Uxx + 'U yy = 0+ 0 =o. Thus, u(x, y) satisfies Laplace's equation and so u(x, y) is harmon

20. (b) For u = ;c2 + y2 , we get U x =2x, so U xx =2. Also, u y = 2y, so U yy = 2. Substitution in Laplace's equation gives us 'Ux ., +Uy y = 2 + 2 :f. 0, so x 2 + y2 is not harmonic. (d) For U = 11 + 3x2y, we get u., = 6xy, so Uxx = 6y. Also, U y = 3y2 + 3x2 , so U yy = Substitution into Laplace's equation gives us Uxx + U yy = 6y + 6y "I 0, so y3 + 3x2 y is r harmonic.

23. Given V (x, y, z ) = - GmM/Jx2 + y2 + z2 = - GmM/ r, we compute:

GmMx

24. (a) If (x, y) "I (0,0), we can compute the first partial derivatives in the usual way: a f = (y(x2 - y2) + 2x2y)( x2 + y2 ) _ 2x2y(x2 _ y2 ) _ x4 y _ yS + 4x2y3 ax (x2+ y2 )2 (x2 + y2 )2 8f (x(x2 - y2 ) _ 2xy2 )(x2+ y2 ) _ 2xy2(x2 _ y2) x5 _ 4x3y2 _ xy4 ay (x2+ y2 )2 (x2 + y2) 2

45 HIGHER-ORDER DERIVATIVES; MAXIMA AND MINIMA

(b) To compute the part ial derivatives at (0,0), we need to use the definition of the partial derivati es. First, hold y constant at and differentiate with respect to x at Xo = 0. ate that f(O,O) is defined to be 0. Then

h(0)(h2 - 02 ) fj l (0 0) = lim 1(0 + h, 0) - 1(0,0) = lim --..:hc::....2---'+h,--0,,-2_ = lim Q= lim 0= 0. fjx ' 11-+0 h h-+O h-+ O h 11-+0

Similarly, we hold x constant at and differentiate with respect to y at Yo = 0. Then 0(h)( 02 - h2 )

{)I (OO) -I' I (O , O+ h)- I (O,O) _ l' 02 + h2 , - 1m - 1m = lim Q=0.

fjy 11 -+ 0 h h-+O h " -+ 0 h

(c) By definition, :::y = :x ( ~: ). First, we use ~~ from part (a) and then perform di ffe rentiation as in part (b) . By definition,

h5{)f fj f - 4h3(0)2 - h(0)4 {P I . {)y (O+h,O)- {)y (O,O) (h2 + 02 ) 2 -0 . h5/h4 -{) (0,0) = hm h = lim __'---_ ..,-!.... = hm -- = l.{) J ____

x Y h-+O h -+O h 11 -+ 0 h Similarly,

fj f of (0) 4h - h5 + 4(0 )2h3 02 f - (0, + h) - -(0,0) (02 + h2 )2 - --(0,0) = lim {)% fjx = lim --'----~----oyox h-+ O h h -+ O h

(d) The mixed partials are not equal, which is consistent wi th the fact that the second partials are not continuous at (0 , 0) .

3.2: TAYLOR'S THEOREM

GOALS 1. Be able to write down the firs t few terms of Taylor's formula for a given function.

STUDY HINTS 3

1. Notation. The summation symbol L: means to sum all possible combinations of (i, j) with i ,j= l

i and j ranging from 1 to 3, i.e., (i, j ) = (1,1), (1 ,2), (1 , 3), (2, 1), (2,2), (2, 3) , (3, 1), (3, 2), and (3,3) . In general , if m indices are summed from 1 to n, there will be nm terms; in our case there are 32 = 9 terms.

2. Review. Before continuing, you may wish to review Taylor's formula from your one-variable calculus text. Recall that the Taylor series can be used to approximate the values of functions.

3. Taylor 's fo rmula. You should know the pat tern for the general formula. As a reminder, Taylor's form ula is

The second term, which involves the second partial derivatives , will become important in the coming sections. The term involving the third partials sums up 3n terms, so it may be unreasonable to ask you to compute all of these terms unless n = 2.

46 CHAPTER

4. Computing Ta ylor 's formula. Remember that you will n ed to compute all of the partl derivatives of the same order. For example, when computing second partials, on must compu 82 f 82 f 82 f .. 82 f 82 f (j2 f -82' -82' ... , a 2 as well as all of the mixed partials 8 8 '8 8 ' 8 a ,etc. Note tha

Xl X n X l X2 X2 X l Xl X3X 2

we do not need to compute !.l 82! ,i f. j, twice because mixed partials are equal (assuminz

U XiUX j continuity) .

5. Taylor's formu la remainder. Recall that in one-variable calcul us, the remainder is determina. at some point between X o and X o + h. Now the r mainder is det rmined at some point on th line between the vectors Xo and Xo + h, where h = (hl ' h2 , ... , hn ).

SOLUTIONS TO SELECTED EXERCISES

1. Recall that Taylor's formula is a polynomial which approximates a function. If our function is itself a polynomial, then this function must be its own Taylor series as well. Hence the secondorder Taylor formula for f is f (h1 + h2 ) = (hi + h2 )2 =