Polytechnic Institute of NYUMA 2112 Worksheet 4

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Directions: Show all work for every problem. Most of Worksheet 4 questions arefrom the end of Sections 14.4 – 14.7, 15.1 – 15.3, and from Review Exercises andProblems for Chapter 14 and 15. Additional suggested Problems are: Section 14.4: 2,8, 14, 21, 24, 31, 35, 38, 42, 56, 64, 65, 69; Section 14.5: 4, 7, 15, 20, 23, 26, 38, 48, 51,57, 62; Section 14.6: 2, 6, 14, 19, 22, 25, 26, 31, 35, 38; Section 14.7: 2, 5, 12, 18, 20, 28,32, 40, 45; Section 15.1: 1, 6, 9, 13, 16, 21, 26, 28, 30; Section 15.2: 4, 7, 12, 16, 18, 21;Section 15.3: 6, 9, 14, 15, 20, 22, 30, 33. Please do not forget to practice True-Falsequestions from Check Your Understanding at the end of Chapters 14 and 15.

Problem Possible Points

1 10

2 10

3 10

4 10

5 10

6 10

7 10

8 10

9 10

10 10

Total 100

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(1) (a) (Section 15.1, Exercise 1, Page 798) The contour diagram of a function z =f(x, y) is given as shown in Figure 1a, which of the points A, B, C, D, E, Fand G appear to be critical points? Classify those that are critical points aslocal maxima, local minima, saddle points, or none of these.

Figure 1a

(b) (Section 15.1, Exercises 7–16, Page 798) For each of the following parts (i)-(iii),find the critical points and classify them as local maxima, local minima, saddlepoints, or none of these.

(i) f(x, y) = x3 + y2 − 3x2 + 10y + 6

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(ii) f(x, y) = (x+ y)(xy + 1)

(iii) f(x, y) = e2x2+y2

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(2) (a) (Section 14.4, Problem 73, Page 750) Consider the function f(x, y). If you startat the point (4, 5) and move toward the point (5, 6), the directional derivativeis 2. Starting at the point (4, 5) and moving toward the point (6, 6) gives adirectional derivative of 3. Find ∇f at the point (4, 5).

(b) (Section 14.6, Problem 31, Page 767) Suppose that x > 0, y > 0 and that zcan be expressed either as a function of Cartesian coordinates (x, y) or as afunction of polar coordinates (r, θ), so that z = f(x, y) = g(r, θ). (Recall that

x = r cos θ, y = r sin θ, r =√x2 + y2 , and, for x > 0, y > 0, θ = arctan(y/x).)

Show that (∂z

∂x

)2

+

(∂z

∂y

)2

=

(∂z

∂r

)2

+1

r2

(∂z

∂θ

)2

.

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(c) (i) (Section 14.7, Example 2, Page 769) Use the values of the function f(x, y)in Table 2 to estimate fxy(1, 2) and fyx(1, 2).

(ii) Find the Taylor polynomial of degree 1 for f about (1, 2).

(iii) Use the result of part 2(c)ii to estimate f(1.01, 1.99).

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(3) (a) (Section 14.5, Problem 57, Page 758) The temperature of a gas at the point(x, y, z) is given by G(x, y, z) = x2 − 5xy + y2z.

(i) What is the rate of change in temperature at the point (1, 2, 3) in the

direction ~v = 2i+ j − 4k?

(ii) What is the direction of maximum rate of change of temperature at thepoint (1, 2, 3)?

(iii) What is the maximum rate of change at the point (1, 2, 3)?

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(b) (Section 14.7, Example 5, Page 773) Find the Taylor polynomial of degree 2 at

the point (1, 7) for the function f(x, y) =1

xy.

(c) (Sample Exam) What is the shortest distance from the tangent plane of thegraph of f at (1, 7, 1/7) in part 3b to the point (4, 1, 7)?

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(4) (Section 14.7, Problem 46, Page 776)(a) Consider the function f(x, y) = (sinx)(sin y).

(i) Find the Taylor polynomials of degree 2 for f about the points (0, 0) and(π/2, π/2).

(ii) Use the Taylor polynomials to sketch the contours of f close to each ofthe points (0, 0) and (π/2, π/2).

(b) The maximum value of f(x, y) subject to the constraint g(x, y) = 240 is 6300.The method of Lagrange multipliers gives λ = 20. Find an approximate valuefor the maximum of f(x, y) subject to the constraint g(x, y) = 242.

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(5) (a) (Section 15.1, Problems 17, Page 799) Find A and B so that f(x, y) = x2 +Ax+ y2 +B has a local minimum value of 20 at (1, 0).

(b) (Section 15.1, Problems 18, Page 799) For f(x, y) = A− (x2 +Bx+ y2 +Cy),what values of A, B, and C give f a local maximum value of 15 at the point(−2, 1)?

(c) (Section 15.2, Example 6, Page 806) Does the function f have a global maxi-mum or minimum on the region R given by 0 < x2 + y2 ≤ 1?

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(6) (Section 15.3, Exercise 8, Page 815)(a) Sketch the contour of g(x, y) = x2−y2 = 1, and contours of f(x, y) = x2−y = c

for c = 0, 3/4, 1, 2, 5, 10, 20 on the same graph in the xy-plane.

(b) Use Lagrange multipliers to find global maxima, maximum value, global min-ima, minimum value of f(x, y) = x2 − y, subject to the equality constraintg(x, y) = x2 − y2 = 1.

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(7) (Sample Exam)(a) Sketch the contour of g(x, y) = x2 + y2 − 4 = 0, and contours of f(x, y) =

5x2−2xy+5y2 = c for c = 12, 16, 20, 24, 28 on the same graph in the xy-plane.

(b) Determine the global maxima, maximum value, global minima, minimum valueof f(x, y) = 5x2 − 2xy + 5y2, subject to the equality constraint g(x, y) =x2 + y2 − 4 = 0.

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(8) (Section 15.3, Exercises 12-17, Page 815) Drawn below is the contour diagram off(x, y) = x3 − 3x2 + y2.

(a) Label all contour lines with corresponding z-values.

(b) Determine the global maxima, maximum value, global minima, minimum valueof

f(x, y) = x3 − 3x2 + y2

in the constrained region

R =

{(x, y) ∈ R2

∣∣∣∣− 1

2≤ x ≤ 3,−2 ≤ y ≤ 2

}.

(Hint: inequality constraint.)

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Space for Question 8

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(9) (Section 15.3, Problem 30, Page 817) A firm manufactures a commodity at twodifferent factories. The total cost of manufacturing depends on the quantities, q1and q2, supplied by each factory, and is expressed by the joint cost function,

C = f(q1, q2) = 2q21 + q1q2 + q22 + 500.

The company’s objective is to produce 200 units, while minimizing production costs.How many units should be supplied by each factory?

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(10) (Section 15.3, Problem 33, Page 817)(a) Let f(x, y) = x2 +2y2. Find the minimum value m(c) of f on the line x+y = c

as a function of c.

(b) Give the value of the Lagrange multiplier λ at this minimum.

(c) What is the relation between your answers in parts (a) and (b)?

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Multivariable Functions

Points in 3-space are represented by a system of Cartesian coordinates. The

distance between (x, y, z) and (a, b, c) is√

(x− a)2 + (y − b)2 + (z − c)2.Functions of two variables can be represented by graphs, contour diagrams,

cross-sections, and tables.Functions of three variables can be represented by the family of level surfaces

f(x, y, z) = c for various values of the constant c.A linear function f(x, y) has equation

f(x, y) = z0 +m(x− x0) + n(y − y0) = c+mx+ ny, where c = z0 −mx0 − ny0.Its graph is a plane with slope m in the x-direction, slope n in the y-direction,

through (x0, y0, z0). Its table of values has linear rows (of same slope) and linearcolumns (of same slope). Its contour diagram is equally spaced parallel straightlines.

The limit of f at the point (a, b), written lim(x,y)→(a,b)

f(x, y), is the number L, if

one exists, such that f(x, y) is as close to L as we please whenever the distance fromthe point (x, y) to the point (a, b) is sufficiently small, but not zero.

A function f is continuous at the point (a, b) if lim(x,y)→(a,b)

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

function is continuous on a region R if it is continuous at each point of R.

Vectors

A vector ~v has magnitude (denoted ‖~v‖) and direction (denoted ~v/ ‖~v‖).Examples are displacement vectors, velocity and acceleration vectors andforce. We can add vectors, and multiply a vector by a scalar. Two non-zero vectors,~v and ~w, are parallel if one is a scalar multiple of the other.

A unit vector has magnitude 1. The vectors i, j, and k are unit vectors in thedirections of the coordinate axes. A unit vector in the direction of any nonzero vector

~v is v =~v

‖~v‖. We resolve ~v into components by writing ~v = v1i + v2j + v3k = v1

v2v3

=

v100

+

0v20

+

00v3

where i =

100

, j =

010

, k =

001

.

If ~v = v1i+ v2j + v3k =

v1v2v3

and ~w = w1i+ w2j + w3k =

w1

w2

w3

then

‖~v‖ =√v21 + v22 + v23

~v + ~w = (v1 + w1)i+ (v2 + w2)j + (v3 + w3)k v1v2v3

+

w1

w2

w3

=

v1 + w1

v2 + w2

v3 + w3

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α~v = (αv1)i+ (αv2)j + (αv3)k

α

v1v2v3

=

αv1αv2αv3

The displacement vector from P1 = (x1, y1, z1) to P2 = (x2, y2, z2) is

−−→P1P2 = (x2 − x1)i+ (y2 − y1)j + (z2 − z1)k

=

x2 − x1y2 − y1z2 − z1

.The position vector of P = (x, y, z) is

−→OP . A vector in n dimensions is a

string of numbers ~v = (v1, v2, . . . , vn) =

v1v2...vn

.

Dot Product (Inner or Scalar Product)Geometric definition: ~v · ~w = 〈~v, ~w〉 = ‖~v‖ ‖~w‖ cos θ where θ is the angle between~v and ~w and 0 ≤ θ ≤ π.

Algebraic definition: ~v·~w = 〈~v, ~w〉 =

⟨ v1v2v3

, w1

w2

w3

⟩ = v1w1+v2w2+v3w3.

Two nonzero vectors ~v and ~w are perpendicular if and only if ~v · ~w = 〈~v, ~w〉 = 0.Magnitude and dot product are related by ~v ·~v = 〈~v,~v〉 = ‖~v‖2. We can extend thisidea of dot product to n dimensions in a similar way.

The equation of the plane with normal vector ~n = ai+ bj+ ck =

abc

and

containing the point P0 = (x0, y0, z0) =

x0y0z0

is ~n ·(~r−~r0) = 〈~n,~r − ~r0〉 = 0 which

is

⟨ abc

, x− x0y − y0z − z0

⟩ = 0 where ~r =

xyz

and ~r0 =

x0y0z0

or ax+by+cz = d,

where d = ax0 + by0 + cz0.If ~vparallel and ~vperp are components of ~v which are parallel and perpendicular,

respectively, to a vector ~u, then ~vparallel =〈~v, ~u〉‖~u‖2

~u and ~vperp = ~v − ~vparallel.

The work, W , done by a force ~F acting on an object through a displacement ~d is

W =⟨~F , ~d⟩

.

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Cross Product (Vector Product)Geometric definition:

~v × ~w =

(Area of parallelogramwith edges ~v and ~w

n

)= (‖~v‖ ‖~w‖ sin θ)n,

where 0 ≤ θ ≤ π is the angle between ~v and ~w and n is the unit vector perpendicularto ~v and ~w pointing in the direction given by the right-hand rule.

Algebraic definition:

~v × ~w =

∣∣∣∣∣∣i j kv1 v2 v3w1 w2 w3

∣∣∣∣∣∣where ~v = v1i+ v2j + v3k =

v1v2v3

, ~w = w1i+ w2j + w3k =

w1

w2

w3

.

Differentiation of Multivariable Functions

Partial derivatives of f .

fx(a, b) = Rate of change of f with respect to x at the point (a, b)

= limh→0

f(a+ h, b)− f(a, b)

h,

fy(a, b) = Rate of change of f with respect to y at the point (a, b)

= limh→0

f(a, b+ h)− f(a, b)

h.

On the graph of f , the partial derivatives fx(a, b) and fy(a, b) give the slope inthe x and y directions, respectively. The tangent plane to z = f(x, y) at (a, b) is

z = f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b).Partial derivatives can be estimated from a contour diagram or table of values

using difference quotients, and can be computed algebraically using the same rulesof differentiation as for one-variable calculus. Partial derivatives for functions ofthree or more variables are defined and computed in the same way.

The tangent plane approximation to f(x, y) for (x, y) near the point (a, b) is

f(x, y) ≈ f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b).The right-hand side is the local linearization. The differential of z = f(x, y)

at (a, b) is the linear function of dx and dy

df = fx(a, b)dx+ fy(a, b)dy.

Local linearity with three or more variables follows the same pattern as for func-tions of two variables.

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The tangent plane to a level surface of a function of three variables f at(a, b, c) is

〈grad f(a, b, c), ~r − ~r0〉 = fx(a, b, c)(x− a) + fy(a, b, c)(y − b) + fz(a, b, c)(z − c) = 0,

where ~r =

xyz

and in this case ~r0 =

abc

.

The directional derivative of f at P = (a, b) in the direction of a unit vector u is

fu(P ) =Rate of change

of f in directionof u at P = (a, b)

= limh→0

f(a+ hu1, b+ hu2)− f(a, b)

h= 〈grad f(P ), u〉 =

⟨[fx(P )fy(P )

],

[u1u2

]⟩.

The Chain Rule for the partial derivative of one variable with respect to anotherin a chain of composed functions:

• Draw a diagram expressing the relationship between the variables, and labeleach link in the diagram with the derivative relating the variables at its ends.

• For each path between the two variables, multiply together the derivatives fromeach step along the path.

• Add the contributions from each path.

If z = f(x, y), and x = g(t), and y = h(t), then

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt.

If z = f(x, y), with x = g(u, v) and y = h(u, v), then

∂z

∂u=∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂u,

∂z

∂v=∂z

∂x

∂x

∂v+∂z

∂y

∂y

∂v.

Second-order partial derivatives

∂2z

∂x2= fxx = (fx)x,

∂2z

∂x∂y= fyx = (fy)x,

∂2z

∂y∂x= fxy = (fx)y,

∂2z

∂y2= fyy = (fy)y.

Theorem: Equality of Mixed Partial Derivatives. If fxy and fyx are con-tinuous at (a, b), an interior point of their domain, then fxy(a, b) = fyx(a, b).

Taylor Polynomial of Degree 1 Approximating f(x, y) for (x, y) near (a, b)

f(x, y) ≈ L(x, y) = f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b).

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Taylor Polynomial of Degree 2

f(x, y) ≈ Q(x, y)

= f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b)

+fxx(a, b)

2(x− a)2 + fxy(a, b)(x− a)(y − b)

+fyy(a, b)

2(y − b)2.

Definition of Differentiability. A function f(x, y) is differentiable at thepoint (a, b) if there is a linear function L(x, y) = f(a, b) +m(x− a) +n(y− b) suchthat if the error E(x, y) is defined by

f(x, y) = L(x, y) + E(x, y),

and if h = x − a, k = y − b, then the relative error E(a + h, b + k)/√h2 + k2

satisfies

limh→0k→0

E(a+ h, b+ k)√h2 + k2

= 0.

Theorem: Continuity of Partial Derivatives Implies Differentiability.If the partial derivatives, fx and fy, of a function f exist and are continuous on asmall disk centered at the point (a, b), then f is differentiable at (a, b).

Optimization

A function f has a local maximum at the point P0 if f(P0) ≥ f(P ) for allpoints P near P0, and local minimum at the point P0 if f(P0) ≤ f(P ) for allpoints P near P0. A critical point of a function f is a point where grad f iseither ~0 or undefined. If f has a local maximum or minimum at a point P0, noton the boundary of its domain, then P0 is a critical point. A quadratic functionf(x, y) = ax2 + bxy + cy2 generally has one critical point, which can be a localmaximum, a local minimum, or a saddle point.

Second derivative test for functions of two variables. Suppose grad f(x0, y0) =~0. Let D = fxx(x0, y0)fyy(x0, y0)− (fxy(x0, y0))

2.

• If D > 0 and fxx(x0, y0) > 0, then f has a local minimum at (x0, y0).

• If D > 0 and fxx(x0, y0) < 0, then f has a local maximum at (x0, y0).

• If D < 0, then f has a saddle point at (x0, y0).

• If D = 0, anything can happen.

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Unconstrained optimization

A function f defined on a region R has a global maximum on R at the point P0

if f(P0) ≥ f(P ) for all points P in R, and a global minimum on R at the pointP0 if f(P0) ≤ f(P ) for all points P in R. For an unconstrained optimizationproblem, find the critical points and investigate whether the critical points giveglobal maxima or minima.

A closed region is one which contains its boundary; a bounded region is onewhich does not stretch to infinity in any direction.

Extreme Value Theorem for Multivariable Functions. If f is a continuousfunction on a closed and bounded region R, then f has a global maximum at somepoint (x0, y0) in R and a global minimum at some point (x1, y1) in R.

Constrained optimizationSuppose P0 is a point satisfying the constraint g(x, y) = c. A function f has alocal maximum at P0 subject to the constraint if f(P0) ≥ f(P ) for all pointsP near P0 satisfying the constraint. It has a global maximum at P0 subject tothe constraint if f(P0) ≥ f(P ) for all points P satisfying the constraint. Localand global minima are defined similarly. A local maximum or minimum of f(x, y)subject to a constraint g(x, y) = c occurs at a point where the constraint is tangentto a level curve of f , and thus where grad g is parallel to grad f .

To optimize f subject to the constraint g = c, find the points satisfying theequation

grad f = λ grad g and g = c.

Then compare values of f at these points, at points on the constraint wheregrad g = ~0 or grad g undefined, and at the endpoints of the constraint. The numberλ is called the Lagrange multiplier.

To optimize f subject to the constraint g ≤ c, find all points in the interiorg(x, y) < c where grad f is zero or undefined; then use Lagrange multipliers to findthe local extrema of f on the boundary g(x, y) = c. Evaluate f at the points foundand compare the values.

The value of λ is the rate of change of the optimum value of f as c increases (whereg(x, y) = c). The Lagrangian function L(x, y, λ) = f(x, y) − λ(g(x, y) − c) canbe used to convert a constrained optimization problem for f subject the constraintg = c into an unconstrained problem for L.

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• nth degree Taylor Polynomial of f(x) centered at x = a:

f(x) = f(a) + f ′(a)(x− a) +f ′′(a)

2!(x− a)2 +

f ′′′(a)

3!(x− a)3 + · · ·+ f (n)(a)

n!(x− a)n

• Taylor series of f(x) centered at x = a:

f(x) = f(a) + f ′(a)(x− a) +f ′′(a)

2!(x− a)2 +

f ′′′(a)

3!(x− a)3 + · · ·

• Taylor Series of important functions:

sin(x) = x− x3

3!+x5

5!− x7

7!+ · · ·

cos(x) = 1− x2

2!+x4

4!− x6

6!+ · · ·

ex = 1 + x+x2

2!+x3

3!+x4

4!+ · · ·

1

1− x= 1 + x+ x2 + x3 + · · · for − 1 < x < 1

ln(1 + x) = x− x2

2+x3

3− x4

4+ · · · for − 1 < x ≤ 1

(1 + x)p = 1 + px+p(p− 1)

2!x2 +

p(p− 1)(p− 2)

3!x3 + · · · for − 1 < x < 1

• Finite Geometric Series:

a+ ax+ ax2 + · · ·+ axn−1 =a(1− xn)

1− x

• Infinite Geometric Series:

a+ ax+ ax2 + · · · = a

1− xfor |x| < 1

• Ratio Test:For the series

∑an, suppose,

limn→∞

|an+1||an|

= L.

– If L < 1, then the series converges.– If L > 1, then the series diverges.– If L = 1, then the test fails.