Vector 365 Study Guide Unit 4

7/24/2019 Vector 365 Study Guide Unit 4

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Unit 4Partial Derivatives

Functions establish relationships of dependency: a function of one variable, f(x),indicates that the variablefdepends on the variablex; a function of severalvariables,f(x1, x2, , xn), indicates that the variablefdepends on all of thevariablesx1, x2, , xn. In this unit, we extend the basic calculus concepts welearned for functions of one real variable, such as limits, continuity, differentiation

and integration, to multivariable functions. You will learn to find the equations of

planes that are tangent to surfaces, and solve optimization problems for these

functions. The concepts for one variable functions extend naturally to functions of

several variables, but these functions are intrinsically more complicated than

functions of one variable, and we need to apply new skills and different ideas towork with such functions.

Objectives

When you have completed this unit, you should be able to

1. identify the graph of functions of two variables.

2. describe the level curves of functions of two variables.

3. establish the relationship between limits and limits along smooth curves.

4. identify discontinuities of functions of two variables.

5. obtain the partial derivative functions of functions of several variables.

6. obtain higher-order partial derivative functions of functions of several variables.

7. use differentials to approximate the value of functions of several variables.

8. use local linear approximations to estimate errors in approximation.

9. apply the Chain Rule for derivatives of functions of several variables.

10. obtain directional derivatives of functions of two and three variables.

11. give the geometrical interpretation of the gradient of a function of two variables.

12. give the equation of a plane tangent to a surface.

13. obtain the extreme values of functions of several variables.

14. obtain absolute extreme values of functions of several variables.

15. apply Lagrange multipliers to obtain the extreme values of functions of two and

three variables.

Mathematics 365: Multivariable Calculus Study Guide 89


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Functions of Two or More Variables

Indications

1. Read Section 13.1, Functions of Two or More Variables, pages 906-913 of the

textbook.

2. Read the Comments, below.

3. Complete the exercises assigned within and at the end of this section. If you have

difficulty, do not hesitate to contact your tutor to discuss the problem.

Comments

When we writef(x), we indicate that the variablefdepends on the independentvariablex. The distances of a particle depends on the time traveledt; hence,s(t).The costCof living depends on the taxest paid, hence,C(t). For a function ofseveral variablesf(x, y), the variablefdepends on the independent variables(x, y). The areaAof a rectangle depends on its length l and widthw; hence,A(l, w)andA(l, w) =lw.

Definition 4.1. Afunctionfof several variables is a rule that assign toeach(x1, x2, , xn)in Rn a real numberu = f(x1, x2, , xn), andwe write

f : Rn R.

Definition 4.2. Thedomainof a functionf : Rn R is the subsetDf ofR

n such thatf(x1, x2, , xn)is well defined for all(x1, x2, , xn)inDf.

Example 4.1. For the function

f(w,x,y,z) = wxy+z

to be well defined,y+z >0. Hence,

Df ={(w,x,y,z)R4 | y >z}.

Example 4.2. The domain of

F(x,y,z) =

x2 +y2 +z2 4

90 Study Guide Mathematics 365: Multivariable Calculus


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is the set

DF ={(x,y,z)R3 | x2 +y2 +z2 4}.

This set corresponds to the points in R3

on the surface and outside the spherex2 +y2 +z2 = 4.

Example 4.3. The domain of

g(x, y) = ln(xy)

is the set

Dg ={(x, y)R2 | xy >0}.

This set corresponds to the first and third quadrants of the Cartesian plane, without

thex andy axes.

The graph of a real-valued functionf : R R is represented in the Cartesianplane R2; it is the curve {(x, y) | y= f(x)}. Similarly, the graph of a two variablefunctionf : R2 R is represented in R3; that is, the graph of such a function is asurface in a three-dimensional space given by the set {(x,y,z) | z= f(x, y)}.Functions of more than three variables do not have a graph with a visual

representation.

We use level curves to identify quadratic surfaces. It is important that you

understand how we use these curves to visualize a surface. The use of technology is

very helpful, but not required in this course.

Observe that a level curvek of a two variable functionfis a curve in R2

. It is theintersection of a three-dimensional surfacez = f(x, y)and a two-dimensionalplanez(x, y) =k. A level curvek of a three variable functionfis a surface in R3.It is the intersection of a four dimensional surface u = f(x,y,z)and athree-dimensional surfaceu(x,y,z) =k.

Example 4.4. The level curves of the functionf(x, y) =x y2 are parabolasx= y2 +k for a constantk. This family of curves is shown in Figure 4.1,below.



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k= 2 k= 0 k= 2 k= 3k= 1

Figure 4.1:Level curves of the functionf(x, y) =x y2, Example 4.4

Exercises

1. Draw the surface that corresponds to the level curves considered in

Example 4.4.

Example 4.5. The level curves of the functionf(x,y,z) =z2 x2 y2 arehyperboloids of two sheets

z2

k x

2

k y

2

k = 1,

for nonzerok.

Iff : R2 R is a two variable function andr : R R2 is a vector-valuedfunction, then the two possible compositions of these functions are as listed below,

and as shown in Figures 4.2 and 4.3.

a. Iff(x, y)andr(t) = (u(t), v(t)), thenr f : R2 R2 and

(r f)(x, y) =r(f(x, y)) = (u(f(x, y)), v(f(x, y))= ((u f)(x, y), (v f)(x, y)).

This is a two variable vector-valued function.

b. Iff(x, y)andr(t) = (u(t), v(t)), thenf r: R R and

(f r)(t) =f(r(t)) =f(u(t), v(t)).

This is a single variable real-valued function.



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Df

f (x,y)

f r

(x,y)(r f )

Figure 4.2:Compositionr fof two variable function fand vector-valuedfunctionr, Example 4.5

r

r (t)

f

(t)(f r)

Figure 4.3:Compositionf rof vector-valued functionr and two variablefunctionf, Example 4.5

In three dimensions, these compositions are as shown in Figures 4.4 and 4.5, below.

r

(x,y)(r f )

Df

(x,y)

f (x,y)

.

.

r f

.

Figure 4.4:Three-dimensional view of compositionr fof two variable functionfand vector-valued functionr, Example 4.5

Example 4.6. Letf(x, y) =x2 xyand r(t) = (t2, t 3).

a. Two variable vector-valued function:

(r f)(x, y) =r(f(x, y))=r(x2 xy) = ((x2 xy)2, (x2 xy) 3)= (x4 2x3y x2y2, x2 xy 3).

Thus,(r f)(1, 2) = (1 2(2) 4, 1 2 3) = (7, 4).



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.

. (t)(f r)

r (t)r

(t)(f r)

t

Figure 4.5:Three-dimensional view of compositionf rof vector-valued functionrand two variable functionf, Example 4.5

b. Single variable real-valued function:

(f r)(t) =f(r(t)) =f(t2, t 3)= (t2)2 t2(t 3) =t4 t3 + 3t2.

Thus,(f r)(2) = 24 23 + 3(22) = 20.

Exercises

2. Do odd-numbered exercises 1-7, 21-41 and 51-67; on pages 914-916 of the

textbook.

3. Letf(x,y,z) =x+y+z and r(t) = (t, 3t 1, t). Finda. (r f)(x,y,z).b. (r f)(1, 2, 3).c. (f r)(t).d. (f r)(0).



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Limits and Continuity

Indications

1. Read Section 13.2, Limits and Continuity, pages 917-925 of the textbook.


3. Complete the exercises assigned at the end of this section. If you have difficulty,

do not hesitate to contact your tutor to discuss the problem.

Comments

IfCis a curve given by the vector-valued function

r(t) = (x(t), y(t))

and

limtt0

r(t) = (x0, y0),

then we say that the curveCapproaches the point(x0, y0)ast approachest0.

That is,

r(t)(x0, y0) as tt0.

Iff(x, y)is a two variable function, then the composition(f r)(t)is a singlevariable real-valued function, as we said in point(b) on page 92, and the limit of the

functionfalong the curveCat (x0, y0)is the limit

limtt0

(f r)(t) = limtt0

f(r(t)) = limtt0

f(x(t), y(t)) = lim(x,y)(x0,y0)

f(x, y).

Limits along a curveCare used to show that a limit at a point(x0

, y0

)does notexist. To do so, we need to choose two different curves, C1 and C2, which approach(x0, y0), and show that the limits along these curves are not equal. In Example 1 onpage 918 of the textbook, the limit

lim(x,y)(0,0)

xyx2 +y2

does not exist, because the limits along they-axis and along the line y = x are notequal.

We suspect that a limit does not exist at the points where the function is not defined.



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Example 4.7. The function

f(x, y) = xy

3x2 +y2

is not defined at(0, 0). To see if the limit exists at this point, we need to choose twocurves approaching(0, 0).

Let

r1(t) = (t, 0) and r2(t) = (t, t).

Then,

limt0

r1(t) = (0, 0) = limt0

r2(t),

and

limt0 f r1(t) = limt0 f(t, 0) = 0,

limt0

f r2(t) = limt0

f(t, t) = limt0

t2

3t2 +t2= lim

t0

1

3 + 1 =

1

4.

Hence, the limit lim(x,y)(0,0)

f(x, y)does not exist.

To define the limit of a function of several variables, we must define open and

closed disks.

Definition 4.3. Ann-dimensional open diskD with centre

c= (c1, c2, , cn)and radiusis the setD={x= (x1, x2, , xn) | x c< }.

Since x c= (x1 c1)2 + (x2 c2)2 + + (xn cn)2,D={x= (x1, x2, , xn) |

(x1 c1)2 + (x2 c2)2 + + (xn cn)2 < }.

Definition 4.4. Ann-dimensional closed diskD with centrec= (c1, c2, , cn)and radiusis the set

D={x= (x1, x2, , xn) | x c }.As in Definition 4.3,

D={x= (x1, x2, , xn) |

(x1 c1)2 + (x2 c2)2 + + (xn cn)2 }.

IfD is ann-dimensional open disk with centre

c= (c1, c2, , cn),then the set of all points inD exceptits centre is written as

D {(c1, c2, , cn)}.



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Thus,

D {(c1, c2, , cn)}=

{(x1, x2,

, xn)

|0< (x1

c1)2 + (x2

c2)2 +

+ (xn

cn)2 <

}.

Definition 4.5. Letf : Rn R, and assume thatfis defined at allpoints on a disk centred at (c1, c2, , cn), except possibly at(c1, c2, , cn).If, given any >0, there exists an open diskD with centre at(c1, c2, , cn)and radius >0 such that

|f(x1, x2, , xn) L|< for any(x1, x2, , xn)inD {(c1, c2, , cn)}, then

the limit of f at (c1, c2, , cn) is L,and we write

lim(x1,x2, ,xn)(c1,c2, ,cn)

f(x1, x2, , xn) =L.

Definition 4.5 is Definition 13.2.1 on page 920 of the textbook ifn= 2; it isDefinition 13.2.5 on page 924 ifn= 3.

From Definition 4.5, you can see that proving the existence of a limit is not trivial.

As in the case of single variable functions, the evaluation of limits is easy if the

function is continuous or if the function can be simplified into a continuous

function. By Theorem 13.2.4 on page 922, the functions considered in this course

are continuous on their domains.


f(x, y) = exy

x2 +y2

is continuous on R {(0, 0)}; hence,

lim(x,y)(0,1)

exy

x2 +y2 =

e0

1 = 1.

On the other hand, for the curves r1(t) = (t, 0)andr2(t) = (t, t), as inExample 4.7 on the facing page above, along the curve r1,

lim(x,y)(0,0)

exy

x2 +y2 = limt0 f(r1(t)) = limt0 f(t, 0) = limt0

1

t2 =;and along the curver2,

lim(x,y)(0,0)

exy

x2 +y2= lim

t0f(r2(t)) = lim

t0f(t, t)

= limt0

et2

2t2 =.

Hence, the function is not continuous at (0, 0).



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Exercises

4. Do odd-numbered exercises 1-23 and 41-51 on pages 925-926 of the

textbook.



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Partial Derivatives

Indications

1. Read Section 13.3, Partial Derivatives, pages 927-935 of the textbook.




Comments

In this section, you must appeal to your knowledge of single variable calculus,

because all definitions and results presented here are generalizations of topics in

single variable calculus to several variables.

Partial derivatives are the generalization of the derivative function studied in single

variable calculus. Since we have several variables, the partial derivative is defined

with respect to the direction of each variable.

Definition 4.6. Let

a. f(x1, x2, , xn)be a function ofn variables,b. (c1, c2, , cn)be a point in the domain off, andc. xi, for1in, be a differential in the direction of theith

variable.

Then, thepartial derivative offin the direction of the ithvariable atthe point(c1, c2, , cn)is the limit

limxi0

f(c1, c2, , ci+ xi, , cn) f(c1, c2, , ci, , cn)xi

Ifn= 2, Definition 4.6 is the partial derivative of a two variable function in thedirection of the

x-axis (first variable) if

i = 1, and in the direction of the

y-axis

(second variable) ifi = 2. See Definition 13.3.1 on page 927.

The notation of the partial derivative must indicate the function, the direction of the

derivative and the point of differentiation. Hence, for the partial derivative in the

direction of theith variable at(c1, c2, , cn), we write

fxi(c1, c2, , cn) = d

dxif(c1, c2, , xi, , cn)

xi=ci

.

The partial derivative functions are obtained from Definition 4.6 by replacingci byxi for i = 1, 2, , n.



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Definition 4.7. Let

a. f(x1, x2,

, xn)be a function ofn variables, and

b. xi, for1in, be a differential in the direction of theithvariable.

Then, thepartial derivative offin the direction of the ithvariableisthe limit

limxi0

f(x1, x2, , xi+ xi, , xn) f(x1, x2, , xi, , xn)xi

As before, the notation for the partial derivative function must indicate the direction

of the variable. Thus, the partial derivative function in the direction of (or with

respect to) theith variable is denoted by

fxi(x1, x2, , xn) =

xi f(x1, x2, , xn).

Ifu = f(x1, x2, , xn)we write

fxi = f

xi=

u

xi.

The value of the partial derivative at the point (c1, c2, , cn)is denoted as

fxi(c1, c2, , cn) = f

xi

(c1,c2, ,cn)

= u

xi

(c1,c2, ,cn)

.

The evaluation of derivative functions is relatively simple. If the derivative is with

respect to the variablexi, for1

i

n, the other variables are considered to be

constants, and the rules of differentiation for one single variable apply.


f(w,x,y,z) =w2x+x2y+y2z+z2w

has four partial derivative functions, one for each variable.

Forfw, we think ofx, y,z as constants, and

fw(w,x,y,z) = 2wx+z2.

Forfx, the constants arew , y , z; thus,

fx(w,x,y,z) =w2 + 2xy.

Similarly, forfy and fz , we have

fy(w,x,y,z) =x2 + 2yz and fz(w,x,y,z) =y

2 + 2zw.

Hence,

f

x

(1,2,3,4)

=w2 + 2xy(1,2,3,4)

= 13

and

f

z

(1,2,3,4)

=y2 + 2zx(1,2,3,4)

= 17.



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Pay attention to the two interpretations of the partial derivative: as a slope of a

tangent line and as a related rate.

Examine Figure 13.3.1 on page 929. The curveC1 is on the planey = y0, because

it is the intersection of this plane and the surface z = f(x, y). The curveC2is theintersection of the planex = x0and the surfacez = f(x, y). Hence,fx(x0, y0)isthe slope of the line tangent to the curve C1 at the point(x0, y0). This line is on theplaney = y0, and the line tangent to the curve C2is on the planex = x0.

To complete the implicit partial differentiation, we must be clear about the variables

that are considered to be constants.

Example 4.10. See Example 7 on pages 931 of the textbook.

We have the partial derivative

z

y

of the equation x2 +y2 +z2 = 1.

For the partial derivative

x

y,

we would need to solvex from the equation to obtain a function in terms ofy and z(i.e,x = f(y, z)) and to differentiate this function with respect to y, the variablezis a constant. Thus,

y[x2 +y2 +z2] =

y[1]

2x

x

y + 2y+ 0 = 0

x

y =y

x.

You can check that

y

z =z

y and

z

x =x

z.

Example 4.11. The equation

w2x+x2y+y2z+z2w= wxyz

has four variables.

To findw

x, we would solvew from the equation and differentiate the resulting

functionw = f(x,y,z)with respect tox.

Hence, the variable isx and the constants are y andz; thus,

x[w2x+x2y+y2z+z2w] =

x[wxyz],

2wxw

x +w2 + 2xy+ 0 +z2

w

x =xyz

w

x +wyz.



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Solving forw

x,we obtain

w

x

= wyz w2 2xy2wx+z

2

xyz.

The notation of higher-order partial derivatives is clear and consistent.

Example 4.12. In the notation

3f

x2y,

the power in the numerator indicates the number of times we must differentiate, and

the denominator gives the order of the variables in the differentiation, starting on

the right. Hence, the first derivative is with respect toy, the second and third withrespect tox.

Thus, iff(x,y,z) =x2y2z2, then

f

y = 2x2yz2,

2f

xy =

x

f

y

=

x[2x2yz2] = 4xyz2.

3f

x2y =

x

2f

xy

=

x[4xyz2] = 4yz2.

The subindex notation for partial derivative is read from left to right; hence,

fyxx = fyx2 = 3f

x2y.

Exercises

5. Explain the difference between Definition 4.6 above and the notation

fxi(c1, c2,

, cn) =

f

xi(c1,c2, ,cn)

.

6. Find the partial derivatives

w

y and

y

w

by implicit differentiation on the surface

w2x+x2y+y2z+z2w= wxyz.



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7. Considerf(x,y,z) =x2y2z2. Find the following higher-order partialderivatives:

a. 2f

zy

b. 3f

zxy

c. 3f

x2y.

8. Do odd-numbered exercises 1-9, 25-49, 59-63, and 85-99 on pages 936-939

of the textbook.



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Differentiability, Differentials and Local Linearity

Indications

1. Read Section 13.4, Differentiability, Differentials, and Local Linearity,

pages 940-946 of the textbook.




Comments

Observe that since

(x, y, z)=

x2 + y2 + z2,

and since

x= (x, y, z)(0, 0, 0) can be written as x0,the limit in Definition 14.4.2 on page 961 can be written as

limx0

f (fx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0)) xx .

The aim in the next sections will be to use vector notation to write this limit.

Definition 4.8. Thetotal differential(dw) ofw = fat(x0, y0, z0)is

dw = df=fx(x0, y0, z0)x+fy(x0, y0, z0)y +fz(x0, y0, z0)z.

Differentials are used to approximate values and estimate errors.

A. For each variable we have ameasurement errordefined as

x= x x0 and x= x0+ x.

B. Thepropagated or maximum errorinf is

f=f(x0+ x, y0+ y, z0+ z) f(x0, y0, z0).Thus,

f dw = fx(x0, y0, z0)x+fy(x0, y0, z0)y+fz(x0, y0, z0)z.



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C. Therelative errorin measurement is

f

f(x0, y0, z0).

Thus,

f

f(x0, y0, z0) df

f(x0, y0, z0).

It is always easier to evaluatedf thanf.

D. Thepercentage errorin measurement is

f

f(x0, y0, z0)(100).

Thus,

ff(x0, y0, z0)

(100) dff(x0, y0, z0)

(100).

E. Thelocal linear approximationoffat(x0, y0, z0)is

f(x0+ x, y0+ y, z0+ z)df+f(x0, y0, z0).Hence,

f(x,y,z)L(x,y,z)where

L(x,y,z) =fx(x0, y0, z0)(x

x) +fy(x0, y0, z0)(y

y)

+fz(x0, y0, z0)(z z) +f(x0, y0, z0).

Example 4.13. See Exercise 56 on page 948 of the textbook.

The legs of a right triangle are measured to be 3 cm and4 cm, with maximum errorof0.05cm in each measurement. Use differentials to approximate the maximumpossible error in the calculated value of the hypotenuse.

LetHbe the hypotenuse; thus,H=

a2 +b2, wherea andb are the legs of thetriangle. The errors in measurement area= 0.05andb= 0.05. The propagatederror in measurement isH. The differential at(3, 4)is

dH= aa2 +b2

a+ ba2 +b2

b(3,4)

=3(0.05)5

+4(0.05)5

= 0.07.

The maximum possible error is approximately 7%.

Example 4.14. We can find the exact value of the functionf(x, y) = sin x cos yat(30, 45), but we can only estimate the value off(32, 43) = sin(32) cos(43).

In this example, we use linear approximation. We have

x0= 30 =

6, y0= 45

=

4, x= 2 =

90, andy=2 =

90.



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So,

sin(32) cos(43) =f(32, 43) =f(30 + 2, 45 2)

=f

6+ x,

4+ y

.

We calculate the differential

df =fx

6

,

4

x+fy

6

,

4

y+f

6

,

4

.

fx

6

,

4

= cos

6

cos

4

=

6

4

fy

6

,

4

=sin

6

sin

4

=

2

4

and

f

6 ,

4

= 1

22 .

Hence,

sin(32) cos(43)

3

4

90

1

2

2

90

+

1

2

2

= (

3 +

2)

360 +

1

2

2.

Exercises

9. Give the definition of the total differentialdf of

f(x1, x2, , xn) at (c1, c2, , cn).

10. Do odd-numbered exercises 1-25, 31-41 and 49-55 on pages 947-949 of the

textbook.



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The Chain Rule

Indications

1. Read the section on matrices in Appendix A of thisStudy Guide.




Comments

We will use the Chain Rule to differentiate the composition of multivariable

functions. Our approach will make use of the differential matrix associated with

multivariable vector-valued functions.

Definition 4.9. Letf : Rn R withf(x1, x2, , xn)be afunction of several variables.

Thedifferential matrixDfoff is a 1 nmatrix in which the(1, i)entries are the partial derivative functions

f

xifor 1in.

Thus,

f

x1

f

x2. . .

f

xn1

f

xn

Example 4.15. The differential matrix of the functionf(w,x,y,z) =w2x2y2z2,is a 1 4matrix:

Df=

fx1

fx2

fx3

fx4

=

2wx2y2z2 2w2xy2z2 2w2x2yz2 2w2x2y2z

Definition 4.10. Letr : R Rm with

r(t) = (x1(t), x2(t), , xm(t))

be a vector-valued function.



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Thedifferential matrixDr is a m 1matrix in which the(i, 1)-entry isthe derivative

dxi

dt

for 1

i

m.

Hence,

Dr=

dx1dt

dx2dt...

dxmdt

Example 4.16. For the functionr(t) = (sin t, cos t, t3), we identify the componentfunctions

x1(t) = sin t, x2(t) = cos t and x3(t) =t3,

and its differential matrix is a 3 1matrix

dx1dt

dx2dt

dx3dt

=

cos t

sin t

3t2

.

Iff(x1, x2, , xn)is a multivariable function, and

r(t) = (x1(t), x2(t), , xn(t))

is a vector-valued function, then the composition F =f ris a real-valuedfunction. See Figure 4.6, below.

Thus,

F(t) = (f(r))(t) =f(x1(t), x2(t), , xn(t)),

and its derivativeF(t)is the product of the differential matrices offandr, asshown below.



21/48

n

f r

r f

Figure 4.6:Compositionf rof multivariable functionf(x1, x2, , xn)andvector-valued functionr(t) = (x1(t), x2(t), , xn(t))

dF

dt =

f

x1

f

x2. . .

f

xn1

f

xn

dx1dt

dx2dt...

dxmdt

=

f

x1

dx1dt

+ f

x2

dx2dt

+ + fxn

dxndt

Example 4.17. Let

f(x,y,z) =x2y2z2 and r(t) = (sin t, cos t, t3).

The derivative of the composition

F(t) =f(r(t)) =f(sin t, cos t, t3) = sin2 t cos2 t(t3)2 =t6 sin2 t cos2 t

is the product of the matricesDfandDr, as is shown below.

dF

dt =

2xy2z2 2x2yz2 2x2y2z

cos t

sin t

3t2

= 2xy2z2 cos t 2x2yz2 sin t+ 3(2x2y2z)t2

= 2xyz(yzcot t xzsin t+ 3xyt2).Since

F(t) =f(r)(t) =f(sin t, cos t, t3),

we havex = sin t,y = cos tandz = t3.



22/48

If we substitute these values, we have

2xyz(yzcos t xzsin t+ 3xyt2)= 2t3 cos t sin t(t3 cos2 t

t3 sin2 t+ 3t2 sin t cos t

= 2t5(t cos(2t) + 3 sin t cos t).

Differentiate the functionF(t) =t6 sin2 t cos2 tto see that

F(t) = 2t5(t cos(2t) + 3 sin t cos t).

Definition 4.11. A multi-variable vector-valued function is a function

f : Rn Rm,

with

f(x1, x2, , xn)= (f1(x1, x2, , xn), f2(x1, x2, , xn), , fm(x1, x2, , xn)).

The functionsfi for 1in are the component functions off.Itsdifferential matrixDf is a m nmatrix with(i, j)entry the partialderivative function

fixj

for 1im, 1jn.

Thus

Df=

f1x1

f1x2

. . . f1xn

f2x1

f2x2

. . . f2

xn...

fmx1

fmx2

. . . fm

xn

Example 4.18. For the function

f(x,y,z) = (x2

yz,xy2

z,xyz2

, x2

y2

z2

),

we have

f : R3 R4.

Hence, its differential matrix is a4 3matrix and its component functions are

f1(x,y,z) =x2yz, f 2(x,y,z) =xy

2z,

f3(x,y,z) =xyz2, f4(x,y,z) =x

2y2z2.



23/48

Thus,

Df=

f1x1

f1x2

f1x3

f2x1

f2x2

f2x3

f3x1

f3x2

f3x3

f4x1

f4x2

f4x3

=

2xyz x2z x2y

y2z 2xyz xy2

yz2 xz2 2xyz

2xy2z2 2x2yz2 2x2y2z

Theorem 4.12. Chain RuleIfr : Rn Rm and f : Rm Rk, then

f(x1, x2, , xm)= (f1(x1, x2, , xm), f2(x1, x2, , xm), , fk(x1, x2, , xm))

and

r(t1, t2, , tn)= (x1(t1, t2, , tn), x2(t1, t2, , tn), , xm(t1, t2, , tn)).

Its composition isF= f r, andF: Rn Rk . See Figure 4.7,below.

kn

r f

F = f r

m

Figure 4.7:CompositionF = f r, Chain Rule

The differential matrixDFof this composition is ank nmatrixequal to the product of thek

mandm

ndifferential matrices off

andr, as shown below.

DF=

f1x1

f1x2

. . . f1

xm

f2x1

f2x2

. . . f2

xm...

fkx1

fkx2

. . . fk

xm

x1t1

x1t2

. . . x1

tn

x2t1

x2t2

. . . x2

tn...

xmt1

xmt2

. . . xm

tn



24/48

The component functionsFiofF are the compositions Fi = fixi; thus,Fi : R

n R for 1ik.

By the matrix product,Fixj

is equal to the product of thei-th row of the matrixDfand thej-th column of thematrixDr.

Hence,

Fixj

= fix1

x1tj

+ fix2

x2tj

+ + fixm

xmtj

.

Example 4.19. If

f(x1, x2, , xn)is a multivariable function and

r(t) = (x1(t), x2(t), , xn(t))is a vector-valued function, then the composition

R= r fis a multivariable vector-valued function. See Figure 4.8, below.

nn

rf

r f

Figure 4.8:Compositionr fof multivariable functionf(x1, x2, , xn)andvector-valued functionr(t) = (x1(t), x2(t), , xn(t)), Example 4.19

The composition is

R(x1, x2, , xn) =

r(f(x1, x2, , xn))

= (x1(f(x1, x2, , xn)), x2(f(x1, x2, , xn)), , xn(f(x1, x2, , xn))).

Thus, the component functionRiofR are

Ri(x1, x2, , xn) =xi fi(x1, x2, , xn) =xi(fi(x1, x2, , xn))for1in.The relatedn ndifferential matrix is the product of the differential matricesDfandDr, as is shown below.



25/48

R1x1

R1x2

. . . R1

xn

R2x1

R2x2

. . . R2

xn...

Rnx1

Rnx2

. . . Rn

xn

=

dx1dt

dx2dt...

dxndt

f

x1

f

x2. . .

f

xn1

f

xn

=

f

x1

dx1dt

f

x2

dx1dt

. . . f

xn

dx1dt

f

x1

dx2dt

f

x2

dx2dt

. . . f

xn

dx2dt

...

f

x1

dxndt

f

x2

dxndt

. . . f

xn

dxndt

Example 4.20. Let

f(x,y,z) =exyz and r(t) = (t, t2, t3),

then

R(x,y,z) = (r f)(x,y,z) =r(f(x,y,z)) =r(exyz) = (exyz , e2xyz , e3xyz).

Hence,

R1(x,y,z) =exyz , R2(x,y,z) =e

2xyz and R3(x,y,z) =e3xyz .

Its differential matrix is

R1x

R1y

R1z

R2x

R2y

R2z

R3x

R3y

R3z

=

1

2t

3t2

yzexyz xzexyz xyexyz

=


yzexyz(2t) xzexyz(2t) xyexyz(2t)

yzexyz(3t2) xzexyz(3t2) xyexyz(3t2)



26/48

SinceR(x,y,z) =r(exyz), we havet = exyz , and we conclude that

R1x

R1y

R1z

R2x

R2y

R2z

R3x

R3y

R3z

=


2yze2xyz 2xze2xyz 2xye2xyz

3yze3xyz 3xze3xyz 3xye3xyz

You can check that this conclusion is correct by differentiating the component

functionsRi for 1i3.

Example 4.21. Let

r(u, v) = (sin u cos v, u2v2, u2 +v2) and f(x,y,z) =xyz.

The composition

F(u, v) = (f r)(u, v)=f(r(u, v)) =f(sin u cos v, u2v2, u2 +v2)

= sin u cos v(u2v2)(u2 +v2)

is a functionF : R2 R,and

x(u, v) = sin u cos v, y(u, v) =u

2

v

2

, z(u, v) =u

2

+v

2

.Thus,r(u, v) = (x(u, v), y(u, v), z(u, v)).

By the Chain Rule,

F

u

F

v

=

f

x

f

y

f

z

x

u

x

v

y

u

y

v

z

u

z

v

Since

x(u, v) = sin u cos v, y(u, v) =u2v2 and z(u, v) =u2 +v2,

we have

F

u

F

v

=

yz xz xy

cos u cos v sin u sin v2uv2 2u2v

2u 2v



27/48


28/48

We have

f(u, v) = (f1(u, v), f2(u, v)) = (uv,euv)

and

r(x,y,z) = (u(x,y,z), v(x,y,z)) = (xy+z, x+yz).

By the Chain Rule,

DF= (Df)(Dr) =

f1u

f1v

f2u

f2v

u

x

u

y

u

z

v

x

v

y

v

z

DF= v u

veuv ueuvy x 1

1 z y

=

yv+u vx+zu v+uy

euv(yv +u) euv(xv+zu) euv(v+yv)

Sinceu = xy +z and v = x+yz , we have, forF(x,y,z) = (F1(x,y,z), F2(x,y,z)),

F1x

= 2yx+yz +z,

F1y

=x2 + 2xyz +z2,

F1z

=x+ 2yz +xy2,

F2x

= (2yx+yz +z)e(xy+z)(x+yz),

F2y

= (x2 + 2xyz +z2)e(xy+z)(x+yz) and

F2z

= (x+ 2yz +yx)e(xy+z)(x+yz).

Example 4.23. Let

f(u, v) = (uv,euv, u+v) and r(x,y,z) = (xy+z, x+yz ).

Thus,f : R2 R3 and r: R3 R2.LetF : R2 R2 be the compositionF = r f.

IfF1and F2 are the component functions ofF, thenF1

v is the product of the first

row ofDrand the second column ofDf.



29/48

Hence,

F1

v =

y 1 1

u

ueuv

1

and

F1v

=uy +ueuv + 1.

Since

F(u, v) = (r f)(u, v) =r(f(u, v)) =r(uv,euv, u+v)),

we have

x= uv, y = euv and z= u+v.

Replacing these values, we obtain

F1v

= 2ueuv + 1.

LetR : R3 R3 be the compositionR = f r.

IfR1,R2 and R3are the component functions ofR, thenR2

z is equal to the

product of the second row ofDfand the third column ofDr.

Hence,

R2z

=

veuv ueuv1

y

=veuv +yueuv.

Since

R(x,y,z) = (f r)(x,y,z) =f(r(x,y,z)) =f(xy+z, x+yz),

we have

u= xy +z and v= x+yz.

Replacing these values, we obtain

R2z

= (x+yz)e(xy+z)(x+yz) +y(xy+z)e(xy+z)(x+yz).

Note: To do the exercises of Section 14.5, we must identify the functions involved

in the composition.


Letw = ln(3x2 2y+ 4z3),x = t1/2,y = t2/3 andz = t2.



30/48

We have

w(x,y,z) = ln(3x2 2y+ 4z3)and

r(t) = (x(t), y(t), z(t)) = (t1/2, t2/3, t2).

Thus,w : R3 R and r: R R3, and the composition isF = (w r)(t) =w(r(t)).

By the Chain Rule,

dF

dt

= w

x

w

y

w

z

x

t

y

t

z

t

and

dF

dt =

6x

3x2 2y+ 4z32

3x2 2y+ 4z312z2

3x2 2y+ 4z3

1

2

t

2

3 3

t

2t3

Multiplying and replacingx = t1/2,y = t2/3 andz = t2, we obtain

dF

dt =

3

3t 2t2/3 + 4t6 + 4

3(3t4/3 2t+ 4t17/3) 24

3t5 2t14/3 + 4t2 .


Letw = 3xy2z3,y = 3x2 + 2andz =

x 1.Then

w(x,y,z) = 3xy2z3 and r(x) = (x, 3x2 + 2,

x 1).The composition is

(r w)(x,y,z) =r(w(x,y,z)),a real-valued function.



31/48

Exercises

11. FindR1

y of the functionR in Example 4.23 above.

12. Do odd-numbered exercises 1-13 and 17-33 on page 957 of the textbook.

13. Read Theorem 13.5.3 and Example 7 on page 955.

14. Do exercises 41, 43,45, and 47 on page 958.

15. Read theorems 13.5.3, 13.5.4 on pages 955-956.

16. Do odd numbered exercises 51-57 on page 958.



32/48

Directional Derivatives and Gradients

Indications

1. Read Section 13.6, Directional Derivatives and Gradients, pages 960-967 of

the textbook.




Comments

Ifu =u1, u2, , un is a unit vector and c =c1, c2, , cn is a vector inVn ,then the vector-valued functionr(s) =c +svis a line in the direction of the vectoruand through the point(c1, c2, , cn).

Definition 4.13. Ifu is a unit vector andf : Rn R is a function ofseveral variables, then thedirectional derivative offin the direction ofuatc is the derivative of the compositionf

rat zero, if its exists, and

we write

Duf(c1, c2, , cn).

Thus,

Duf(c1, c2, , cn) = d(f r)ds

= df(c +su)

ds

s=0

.

Ifn= 3, then Definition 4.13, above, is Definition 13.6.2 on page 960.

Observe that

r(s) =c +sv= (c1+su1, c2+su2, , cn+sun);hence,Dr= (u1, u2, , un).By the Chain Rule we have

Duf= (Df)(Dr) = f

x1u1+

f

x2u2+ + f

xnun.



33/48

Definition 4.14. Iff : Rn R, then thegradient off, written f,is the differential matrix offas a vector. Hence

f(x1, x2,

, xn) = f

x1,

f

x2,

, f

xn .

If(c1, c2, , cn)is a point in Rn, then thegradient offat(c1, c2, , cn)is

f(c1, c2, , cn)

=

f

x1

c1,c2, ,cn

, f

x2

c1,c2, ,cn

, , fxn

c1,c2, ,cn

.

Fornequal to2and3, Definition 4.15 is Definition 13.6.4 on page 963. Weconclude that

Duf(c1, c2, , cn) =f(c1, c2, , cn) u.Using this notation we can write the limit on page 104 and in Definition 13.4.2 on

page 942, as

limx0

f f(x0, y0, z0) xx .

It is important that you understand the meaning and importance of Theorem 13.6.5Remark

on page 964.

a. For a functionf : Rn R, the gradient at a point(c1, c2, , cn)is a vectorinVn .

b. For each unitary vectoru, we have a directional derivative

Duf(c1, c2, , cn) =Duf=f u.

c. The unitary vector that gives the largest value is

ff .

That is, if

u= ff ,

then the functionfincreases most rapidly in the direction ofu, and themaximum value of the directional derivative is f.

d. The directional derivative in the direction of

u= ff

is

Duf =f u=f ff =f f

f =f.



34/48

e. If

u= ff ,

then the functionfdecreases most rapidly in the direction of u, and theminimum value of the directional derivative is f.

f. If

f(x0, y0)= 0,

then f(x0, y0)is normal to the level curvef(x, y) =c through(x0, y0).g. If

F(x0, y0, z0)= 0,

then F(x0, y0, z0)is normal to the level curveF(x,y,z) =c through(x0, y0, z0).

Exercises

17. Letf : Rn R be a several variable function and (c1, c2, , cn)a pointin Rn.

a. Give the definition of a functionfdifferentiable at(c1, c2, , cn).b. Give the definition of the total differential offat(c1, c2, , cn).68

18. Do odd-numbered exercises 1-27, 33-43 and 47-57, 61-65 and 85 on


19. (Optional) Do exercises 80 and 86 on page 970.



35/48

Tangent Planes and Normal Vectors

Indications

1. Read Section 13.7, Tangent Planes and Normal Vectors, pages 971-975 of the

textbook.




Comments

In this section, we want to find the equation of the plane tangent to two different

surfaces in a three-dimensional space:

a. the surface given by a two variable functionz = f(x, y), and

b. the surface given by the level curve of a three variable functionF(x,y,z) =c.

The equation of a plane is determined by a point, P, on the surface and a normal, n,to the plane.

The points on the surfacez = f(x, y)are of the form(x,y,f(x, y)). Thus, a fixedpoint on the surface is

P = (x0, y0, f(x0, y0)).

If

F(x,y,z) =f(x, y) z = 0,

then by implicit differentiation, the vector

F(P) =f(x0, y0), 1is normal to the tangent plane; hence, the equation of the tangent plane is

0 =F(P) xP=fx(x0, y0), fy(x0, y0), 1 x x0, y y0, z f(x0, y0)=fx(x0, y0)(x x0) +fy(x0, y0)(y y0) (z f(x0, y0)).

Thus,

z= f(x0, y0) +fx(x0, y0)(x x0) +fy(x0, y0)(y y0).



36/48

The line that is normal to the tangent plane is in the direction of the normal vector

F(P)and through the pointP = (x0, y0, f(x0, y0)). Hence, its parametricequation is

r(t) =x0, y0, f(x0, y0) +tf(x0, y0), 1.Thus.

x(t) =x0+ tfx(x0, y0) y(t) =y0+ tfy(x0, y0) z(t) =f(x0, y0) t.

Example 4.26. Let

z= ln

x2 +y2 and P(1, 0, 0).

Observe that the pointPis on the surface because0 = ln

(1)2 + 0.

The gradient is

f= xx2 +y2

, y

x2 +y2;

hence,

f(1, 0, 0) =1, 0.

The tangent plane has the equationz =(x+ 1).The parametric equations of the normal line are

x=1 t y= 0 z =t.

A pointP = (x0, y0, z0)is on the surfaceF(x,y,z) =c if

F(x0, y0, z0) =c,

and the vector F(x0, y0, z0)is normal to the surfaceF(x,y,z) =c.Hence, the tangent plane has equation

0 =F(x0, y0, z0) x (x0, y0, z0)=fx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0) x x0, y y0, z z0=fx(x0, y0, z0)(x x0) +fy(x0, y0, z0)(y y0) +fy(x0, y0, z0)(z z0)

The normal line to the tangent plane is in the direction ofF(x0, y0, z0)andthrough(x0, y0, z0). Hence, its parametric equation is

r(t) =x0, y0, z0 +tfx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0).

Thus,

x= x0+ tfx(x0, y0, z0) y = y0+ tfy(x0, y0, z0) z = z0+ tfz(x0, y0, z0).



37/48

Example 4.27. Let

xz yz3 +yz2 = 2.

The pointP(2, 1, 1)is on this surface because2(1) (1)(1)3 + (1)(1)2 = 2.

This surface is the level curve

F(x,y,z) =xz yz3 +yz2 = 2.

The gradient is

F =z, z3 +z2, x 3yz2 + 2yz;

thus,

F(2, 1, 1) =1, 0, 3.The equation of the tangent plane at P is

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

The parametric equations of the normal line at P are

x= 1 +t y=1 z = 1 + 3t.

If

z= f(x, y) and z = g(x, y)

are two different surfaces, and the pointP(x0, y0, z0)is in their intersection, then

z0= f(x0, y0) =g(x0, y0)

and the vectors

f(x0, y0), 1 and g(x0, y0), 1

are normal to their respective tangent planes.

Hence, by Formula (9) on page 816 of the textbook, the angle of intersection of

these planesand so their normal linesat this point is

cos = |f(x0, y0), 1 g(x0, y0), 1|f(x0, y0), 1g(x0, y0), 1 .

Therefore,

cos = |fx(x0, y0)gx(x0, y0) +fy(x0, y0)gy(x0, y0) + 1|fx(x0, y0)2 +fy(x0, y0)2 + 1

gx(x0, y0)2 +gy(x0, y0)2 + 1

.



38/48


Let

z=

x2 +y2 and z=

1

10 (x2

+y2

) +

5

2 .

The point of intersection of these two surfaces is (3, 4, 5), since

5 =

42 + 32 and 5 = 1

10(32 + 42) +

5

2.

The angle of intersection of their tangent planes is

cos =

3

5

3

5

+

4

5

4

5

+ 1

2

2= 1.

Hence, = 0, and they have a common tangent plane at this point.

The intersection of the surface

z= 1

10(x2 +y2) +

5

2

with theyz -plane is the curve

z= y2

10+

5

2.

The point(0, 1, 13/5)is in the intersection of the surface and the yz -plane. Theangle of intersection of the tangent plane of this surface and the yz -plane at thispoint is the angle between the normal f(0, 1), 1 and the vector 1, 0, 0.Thus,cos = 0and = 90.


Ifz = f(x, y)andz = g(x, y)intersect at the pointP(x0, y0, z0), then their normallines are perpendicular if

0 = cos

= |fx(x0, y0)gx(x0, y0) +fy(x0, y0)gy(x0, y0) + 1|fx(x0, y0)2 +fy(x0, y0)2 + 1

gx(x0, y0)2 +gy(x0, y0)2 + 1

.

Hence,

|fx(x0, y0)gx(x0, y0) +fy(x0, y0)gy(x0, y0) + 1|= 0.

IfF(x,y,z) =c andG(x,y,z) =d are two different surfaces, and the pointP(x0, y0, z0)is in the intersection of these two surfaces, then

F(x0, y0, z0) =c and G(x0, y0, z0) =d,

and the vectors

F(x0, y0, z0) and G(x0, y0, z0)



39/48

are normal to their respective tangent planes. Hence, as before, by Formula (9) on

page 834 of the textbook, the angle of intersection of their planes at this point is

cos = |F G|FG

.


IfF(x,y,z) =c andG(x,y,z) =d are two different surfaces, and the pointP(x0, y0, z0)is their intersection, then their normal lines are perpendicular if

0 = cos = |F G|FG .

Hence,

F

G= Fx

Gx

+Fy

Gy

+ Fz

Gz

= 0.

Exercises

20. Do odd-numbered exercises 11 and 25-31 on pages 975-976 of the textbook.



40/48

Maxima and Minima of Functions of Two Variables

Indications

1. Read Section 13.8, Maxima and Minima of Functions of Two Variables,





Comments

Note:For this section, you may find useful to review the definition of extreme local

maxima and minima, the Extreme Value Theorem, and the second derivative test for

single variable functions.

In this section, we discuss local maxima and minima for functions of two variables.

There are corresponding concepts for functions of several variables, but they are not

covered in this course.

Definition 4.15.

a. Anopen diskDr(C) =Dr(x0, y0)of radiusr >0 and centre atC= (x0, y0)is the set of all points psuch that pC< r(seethe left-hand side of Figure 4.9, below).

b. Theboundaryof an open diskDr(x0, y0)is the set of all pointspsuch that pC= r.

c. Aclosed diskDr(x0, y0)is the set of all pointspsuch thatpC r (see the right-hand side of Figure 4.9, below).Note that the closed discDr(x0, y0)is equal to the union of theopen discDr(x0, y0)and its boundary (see Figure 4.9, below).



41/48

(x0,y

0)

Dr(x

0,y

0)

r

Dr(x

0,y

0)

Figure 4.9:Open disk boundary=closed disk

In Section 14.8, bounded sets are defined as being inside a rectangle. The proper

definition of a bounded set is in terms of discs.

Definition 4.16. A setR is bounded if there is a pointC= (x0, y0)inRand a numberr >0 such thatR is included (contained) in the discDr(x0, y0).

That is, for anyqin R, qC< rfor somer >0.

Example 4.31. The line segment

L={(5, y)|1< y6}is bounded because it is included in the disc D4(5, 3).

Indeed, if(5, y)is inL, then

(5, y) (5, 3)=|y 3|.If3y , then

|y 3|= y 36 3 = 3 < 4;ify3, then

|y 3|= 3 y3 < 4.In either case, (5, y) (5, 3)< 4.

Definition 4.17. TheclosureR of a setR is the set of all pointsp such

that the intersection of any open discDr(p)andR is nonempty.That is,Dr(p) R=for anyr >0.

Example 4.32. The closure of the line segment

L={(5, y)|1< y6}is the set

L={(5, y)|1y6}.Observe that the point(5, 1)is in the closure ofL, and any other point inL is alsoin the closure.



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Definition 4.18. A setR is closedif it is equal to its closureR.

That is,R = R.

Example 4.33. The set of points

P ={(n, 0)|n is an integer}is not bounded and it is closed.

Observe that for anyr >0 and any integern, the point(n, 0)is in the intersectionDr(n, 0) P; hence, the intersection is nonempty. We leave to you to check thatany other point not in Pis not in the closure ofP.

Thus,P =P.

The Extreme-Value Theorem says that any continuous function f(x, y)on a closedand bounded setR has an absolute maximum and minimum value on the set R.That is, there are pointsp andqin R such thatf(p)is an absolute maximum valueonR andf(q)is an absolute minimum value onR.

As in the case of single variable functions, for two variable functions, local extreme

values occur at critical points. However, for these functions, not all critical points

correspond to local extreme points. A critical point may be a local maximum,

minimum or saddle point. We use the determinant

D(x0, y0) =fxx(x0, y0)fyy(x0, y0) f2xy(x0, y0)to determine if the critical point(x0, y0)is a maximum, minimum or saddle point.

Do not forget that ifD(x0, y0) = 0, the second partial test is inconclusive.

We use the second partial test to solve optimization problems of three variable

functions with one constraint equation. We use the constraint equation to obtain a

two variable function so that we can apply the second partial test.


Find the points on the surface x2 yz = 5that are closest to the origin.Letd(x,y,z) =

x2 +y2 +z2 be the function for the distance of any point

(x,y,z)from the origin.

From the equationx2

yz = 5, we havex2 = 5 +yz ; hence,

d(y, z) =

5 +yz +y2 +z2.

We want to minimize this function.

Then,

dy(y, z) = z+ 2y

2

5 +yz +y2 +z2and dz(y, z) =

y+ 2z

2

5 +yz +y2 +z2.

To find the critical points, we set

dy(y, z) = 0 =dz(y, z)



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and solve fory and z; thus,y = 0 =z.

The discriminant at(0, 0)is

D(0, 0) =dyy(0, 0)dzz(0, 0) d2yz(0, 0) = 15

15

1

20= 3

20 >0

and

dyy(0, 0) = 1

5>0.

The function has a relative minimum at(0, 0). For these values ofy andz, we havex2 = 5. The points on the surface are(5, 0, 0)and(5, 0, 0).

Exercises

21. Do odd-numbered exercises 1-19 and 31-47 on pages 985-987 of the

textbook.



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Lagrange Multipliers

Indications

1. Read Section 13.9, Lagrange Multipliers, pages 989-995 of the textbook.




Comments

To obtain the extreme values using the method in the previous section, we must

solve a variable from the constraint equation, which is not always feasible.

Therefore, another method is needed. The method of Lagrange multipliers is the

next possible option.

When we apply the method of Lagrange multipliers, we find points on the

constraint curveg(x, y) = 0at which the equation

f(x, y) =

g(x, y)

holds for some scalar.

Similarly, for functions of three variables, the Lagrange multipliers method yields

points on the constraint curveg(x,y,z) = 0at which the equation

f(x,y,z) =g(x,y,z)

holds for some scalar.

The scalar is called theLagrange multiplier.

This method requires that we solve several simultaneous equations. So, for two

variable functions,

fx(x, y) =gx(x, y),

fy(x, y) =gy(x, y)

and

g(x, y) = 0.

For three variable functions,

fx(x,y,z) =gx(x,y,z),



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fy(x,y,z) =gy(x,y,z),

fz(x,y,z) =gz(x,y,z)

and

g(x,y,z) = 0

The solution of these equations require several methods, and there is no one way of

solving them. Sometimes we find the value of, or when possible, we equalize theequations and solve them. You need to remember that division by zero is not

possible, and in some instances, we need to consider two cases: when a variable is

zero and when it is not zero.

Take note that g= 0at any point of the curveg(x, y) = 0org(x,y,z) = 0.


Use Lagrange multipliers to find the points on the surface x2 yz = 5that areclosest to the origin.

Letf(x,y,z) =

x2 +y2 +z2 be the distance function of any point (x,y,z)tothe origin. The functionsf andf2 have the same a minima at the same point(s).Hence we considerf2 instead, being the partial derivatives of this easier to obtain.

The constraint isg(x,y,z) =x2 yz 5 = 0.Thus,

f2 = 2xi + 2yj + 2zk g(x,y,z) = 2xi zj yk.

we have the equations

2x= 2x 2y =z 2z =y.From these equations2y+z = 2z+y, hence(2 )(y z) = 0. We have twopossibilities.

= 2, hence from2x= 2x= 4xwe havex = 0and by the constrainequationyz =5. Therefore2yz =2z2 =10andz =5. Thusy=5and we have the points(0, 5, 5). For these pointsf2 = 10.

y= z , hence from the second equationy(+ 2) = 0. If =2, thenz= 0 =y . By the constrain equationx =5and we have the points(5, 0, 0). For these pointsf

2

= 5.

We conclude that the function has a min at(5, 0, 0)and the minimum distanceis 5.


Find the point on the plane4x+ 3y+z = 2that is closest to(1, 1, 1).Letf(x,y,z) =

(x 1)2 + (y+ 1)2 + (z 1)2 be the distance function of any

point(x,y,z)to(1, 1, 1).



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The constraint isg(x,y,z) = 4x+ 3y+z 2 = 0.Thus,

g(x,y,z) =4, 3, 1 =0.The equations we need to solve are

fx(x,y,z) = x 1

(x 1)2 + (y+ 1)2 + (z 1)2 =4,

fy(x,y,z) = y+ 1

(x 1)2 + (y+ 1)2 + (z 1)2 =3,

fz(x,y,z) = z 1

(x 1)2 + (y+ 1)2 + (z 1)2 =

and

4x+ 3y+z = 2.

Hence,

(x 1)2 + (y+ 1)2 + (z 1)2 = x 14

= y+ 1

3 =z 1

Substituting in the last equation, we obtain

4x+ 3

3(x 1)

4 1

+

x 14

+ 1 = 2.

Solving yieldsx = 1; thus,y =

1andz = 1.

We could have solved this problem by seeing that

4(1) + 3(1) + 1 = 2;

that is, by noticing that the point(1, 1, 1)is on the given plane.

Lagrange multipliers can also be used for functions of two or three variables andRemark

two constraints.

Exercises

22. Do odd-numbered exercises 5-21, 25, 29 and 31 on pages 996-997 of the

textbook.



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Finishing This Unit

1. Review the objectives of this unit and make sure you are able to meet each of

them.

2. If there is concept, definition, example or exercise that it is not yet clear to you,

go back and re-read it. Contact your tutor if you need help.

3. You may want to do all or some of exercises 1-3, 5, 11-18, 23, 25-28 and 33-37

from the Chapter Review Exercises on pages 997-998 of the textbook.



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