test 2 review - university of south carolinapeople.math.sc.edu/.../math241test2review.pdf · f(x,y)...

Test 2 Review

Problem 1 (1999):

Problem 1 (1999):

f(x, y) = x2y2 + x + y

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u =〈0, 3〉|〈0, 3〉|

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u =〈0, 3〉|〈0, 3〉|

= 〈0, 1〉

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u = 〈0, 1〉

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u = 〈0, 1〉

∇f = 〈2xy2 + 1, 2x2y + 1〉

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u = 〈0, 1〉

∇f = 〈2xy2 + 1, 2x2y + 1〉 = 〈5, 9〉

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u = 〈0, 1〉

∇f = 〈2xy2 + 1, 2x2y + 1〉 = 〈5, 9〉

~u · ∇f = 〈0, 1〉 · 〈5, 9〉

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(a) Find the directional derivative off(x, y) at the point(2, 1) in the direction of〈0, 3〉.

~u = 〈0, 1〉

∇f = 〈2xy2 + 1, 2x2y + 1〉 = 〈5, 9〉

~u · ∇f = 〈0, 1〉 · 〈5, 9〉 = 9

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(b) There are infinitely many different values for the di-rectional derivative off(x, y) at the point(2, 1) (sincethere are infinitely many directions that can be used to com-pute the directional derivative). Which of these is maximal?In other words, what is the largest value of the directionalderivative off(x, y) at the point(2, 1)?

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(b) There are infinitely many different values for the di-rectional derivative off(x, y) at the point(2, 1) (sincethere are infinitely many directions that can be used to com-pute the directional derivative). Which of these is maximal?In other words, what is the largest value of the directionalderivative off(x, y) at the point(2, 1)?

|∇f(2, 1)|

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(b) There are infinitely many different values for the di-rectional derivative off(x, y) at the point(2, 1) (sincethere are infinitely many directions that can be used to com-pute the directional derivative). Which of these is maximal?In other words, what is the largest value of the directionalderivative off(x, y) at the point(2, 1)?

|∇f(2, 1)| = |〈5, 9〉|

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(b) There are infinitely many different values for the di-rectional derivative off(x, y) at the point(2, 1) (sincethere are infinitely many directions that can be used to com-pute the directional derivative). Which of these is maximal?In other words, what is the largest value of the directionalderivative off(x, y) at the point(2, 1)?

|∇f(2, 1)| = |〈5, 9〉| =√

25 + 81

Problem 1 (1999):

f(x, y) = x2y2 + x + y

(b) There are infinitely many different values for the di-rectional derivative off(x, y) at the point(2, 1) (sincethere are infinitely many directions that can be used to com-pute the directional derivative). Which of these is maximal?In other words, what is the largest value of the directionalderivative off(x, y) at the point(2, 1)?

|∇f(2, 1)| = |〈5, 9〉| =√

25 + 81 =√

106

Problem 3 (1999):

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

=

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

= (2x + yz + 1)

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

= (2x + yz + 1)(sin(√

s) − 2t)

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

= (2x + yz + 1)(sin(√

s) − 2t)

+ xz

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

= (2x + yz + 1)(sin(√

s) − 2t)

+ xz(s2 cos t)

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

= (2x + yz + 1)(sin(√

s) − 2t)

+ xz(s2 cos t) + (xy + 2)

Problem 3 (1999):

Using the Chain Rule, compute∂w

∂twhere

w = x2 + xyz + x + 2z, x = t sin(√

s) + 2s − t2

y = 2s + s2 sin(t), andz = t2s3 − 2t

You do not need to put your answer in terms ofs andt (thevariablesx, y, andz can appear in your answer).

∂w

∂t=

∂w

∂x·∂x

∂t+

∂w

∂y·∂y

∂t+

∂w

∂z·∂z

∂t

= (2x + yz + 1)(sin(√

s) − 2t)

+ xz(s2 cos t) + (xy + 2)(2ts3 − 2)

Problem 6 (1999):

Problem 6 (1999):

Find the critical points of the function

f(x, y) = 3x + xy2

where(x, y) is restricted to points in the set

S = {(x, y) : x2 + y2 ≤ 9}.

Also, determine the maximum and the minimum values off(x, y) in S as well as all points(x, y) where these ex-treme values occur.

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

fx = 3 + y2

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

fx = 3 + y2

STOP!!

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

fx = 3 + y2

DON’T COMPUTEfy!!

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

fx = 3 + y2

fx NEVER EQUALS ZERO!!

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

fx = 3 + y2

THE CRITICAL POINTS ARETHE POINTS ON THE BOUNDARY!!

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

fx = 3 + y2

Critical Points: all points(x, y) wherex2 + y2 = 9

f(x, y) = 3x + xy2, S = {(x, y) : x2 + y2 ≤ 9}

f(x, y) = 3x+xy2, x2+y2 = 9

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) =

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2)

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

−3 ≤ x ≤ 3

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

−3 ≤ x ≤ 3

g′(x) = 12 − 3x2

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

−3 ≤ x ≤ 3

g′(x) = 12 − 3x2

Checkx = −3, x = −2, x = 2, andx = 3.

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

−3 ≤ x ≤ 3

g′(x) = 12 − 3x2

Checkx = −3, x = −2, x = 2, andx = 3.

g(−3) = −9, g(−2) = −16, g(2) = 16, g(3) = 9

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

−3 ≤ x ≤ 3

g′(x) = 12 − 3x2

Checkx = −3, x = −2, x = 2, andx = 3.

g(−3) = −9, g(−2) = −16, g(2) = 16, g(3) = 9

f(x, y) = 3x+xy2, x2+y2 = 9

g(x) = 3x + x(9 − x2) = 12x − x3

−3 ≤ x ≤ 3

g′(x) = 12 − 3x2

Checkx = −3, x = −2, x = 2, andx = 3.

g(−3) = −9, g(−2) = −16, g(2) = 16, g(3) = 9

f(x, y) = 3x+xy2, x2+y2 = 9

g(−3) = −9, g(−2) = −16, g(2) = 16, g(3) = 9

f(x, y) = 3x+xy2, x2+y2 = 9

g(−3) = −9, g(−2) = −16, g(2) = 16, g(3) = 9

The maximum is16 and it occurs at(2, ±√

5) .

The minimum is −16 and it occurs at(−2, ±√

5) .

Problem 7 (1999):

Problem 7 (1999):

Letf(x, y) = x4 + 4xy + xy2.

The functionf(x, y) has 3 critical points. Calculate thethree critical points and indicate (with justification) whethereach determines a local maximum value off(x, y), a localminimum value off(x, y), or a saddle point off(x, y).

f(x, y) = x4 + 4xy + xy2

f(x, y) = x4 + 4xy + xy2

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

f(x, y) = x4 + 4xy + xy2

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

4x + 2xy = 0 implies x = 0 or y = −2

f(x, y) = x4 + 4xy + xy2

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

4x + 2xy = 0 implies x = 0 or y = −2

If x = 0, thenfx = 0 implies 4y + y2 = 0 so thaty = 0 or y = −4.

f(x, y) = x4 + 4xy + xy2

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

4x + 2xy = 0 implies x = 0 or y = −2

If x = 0, thenfx = 0 implies 4y + y2 = 0 so thaty = 0 or y = −4. If y = −2, thenfx = 0 implies4x3 − 4 = 0 so thatx = 1.

f(x, y) = x4 + 4xy + xy2

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

4x + 2xy = 0 implies x = 0 or y = −2

If x = 0, thenfx = 0 implies 4y + y2 = 0 so thaty = 0 or y = −4. If y = −2, thenfx = 0 implies4x3 − 4 = 0 so thatx = 1. The critical points are

(0, 0), (0, −4), and (1, −2).

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fx = 4x3 + 4y + y2 = 0

fy = 4x + 2xy = 0

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

D(0, 0) = 0 − 42 = −16 =⇒

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

D(0, 0) = 0 − 42 = −16 =⇒ saddle point at(0, 0)

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

D(0, 0) = 0 − 42 = −16 =⇒ saddle point at(0, 0)

D(0, −4) = 0 − (−4)2 = −16 =⇒ saddle point at(0, −4)

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

D(0, 0) = 0 − 42 = −16 =⇒ saddle point at(0, 0)

D(0, −4) = 0 − (−4)2 = −16 =⇒ saddle point at(0, −4)

D(1, −2) = 12 · 2 − 02 = 24 =⇒

f(x, y) = x4 + 4xy + xy2

(0, 0), (0, −4), and (1, −2).

fxx = 12x2, fyy = 2x, fxy = 4 + 2y

D = fxxfyy − f2xy

D(0, 0) = 0 − 42 = −16 =⇒ saddle point at(0, 0)

D(0, −4) = 0 − (−4)2 = −16 =⇒ saddle point at(0, −4)

D(1, −2) = 12 · 2 − 02 = 24 =⇒ local min at(1, −2)

Quick Overview:

Quick Overview:

• Computing a directional derivative requires aunit vector.

Quick Overview:

• Computing a directional derivative requires aunit vector.

• The largest value of the directional derivative at a pointis |∇f | at that point.

Quick Overview:

• Computing a directional derivative requires aunit vector.

• The largest value of the directional derivative at a pointis |∇f | at that point.

• The direction giving this largest value is in the directionof ∇f .

Quick Overview:

• Computing a directional derivative requires aunit vector.

• The largest value of the directional derivative at a pointis |∇f | at that point.

• The direction giving this largest value is in the directionof ∇f .

• The gradient at a point on a surfaceF = 0 is perpendic-ular to the tangent plane there.

Quick Overview:

• Computing a directional derivative requires aunit vector.

• The largest value of the directional derivative at a pointis |∇f | at that point.

• The direction giving this largest value is in the directionof ∇f .

• The gradient at a point on a surfaceF = 0 is perpendic-ular to the tangent plane there.

• When taking limits, every direction counts but some di-rections might count more than others.

Quick Overview:

• Computing a directional derivative requires aunit vector.

• The largest value of the directional derivative at a pointis |∇f | at that point.

• The direction giving this largest value is in the directionof ∇f .

• The gradient at a point on a surfaceF = 0 is perpendic-ular to the tangent plane there.

• When taking limits, every direction counts but some di-rections might count more than others.

• Think polar coordinates with limits (as(x, y) → (0, 0)).

Quick Overview:

• I promise to review the chain rule before taking this test.

Quick Overview:

• I promise to review the chain rule before taking this test.

• Handle boundary points for a max-min problem involv-ingf(x, y) by changing the problem to a single variablemax-min problem.

Quick Overview:

• I promise to review the chain rule before taking this test.

• Handle boundary points for a max-min problem involv-ingf(x, y) by changing the problem to a single variablemax-min problem.

• I promise not to forget endpoints in single variable max-min problems.

Quick Overview:

• I promise to review the chain rule before taking this test.

• Handle boundary points for a max-min problem involv-ingf(x, y) by changing the problem to a single variablemax-min problem.

• I promise not to forget endpoints in single variable max-min problems.

•D = fxxfyy − f2xy.

Quick Overview:

• I promise to review the chain rule before taking this test.

• Handle boundary points for a max-min problem involv-ingf(x, y) by changing the problem to a single variablemax-min problem.

• I promise not to forget endpoints in single variable max-min problems.

•D = fxxfyy − f2xy.

• I promise only to computeD at places where∇f = 0.

Quick Overview:

• I promise to review the chain rule before taking this test.

• Handle boundary points for a max-min problem involv-ingf(x, y) by changing the problem to a single variablemax-min problem.

• I promise not to forget endpoints in single variable max-min problems.

•D = fxxfyy − f2xy.

• I promise only to computeD at places where∇f = 0.

• If D > 0 andfxx > 0, then we’ve located a localminimum.

Quick Overview:

• If D > 0 andfxx < 0, then we’ve located a localmaximum.

Quick Overview:

• If D > 0 andfxx < 0, then we’ve located a localmaximum.

• If D < 0, then we’ve located a saddle point.

Quick Overview:

• If D > 0 andfxx < 0, then we’ve located a localmaximum.

• If D < 0, then we’ve located a saddle point.

• I promise to look at the first two sections of Chapter 16

Quick Overview:

• If D > 0 andfxx < 0, then we’ve located a localmaximum.

• If D < 0, then we’ve located a saddle point.

• I promise to look at the first two sections of Chapter 16(a little).

Quick Overview:

• If D > 0 andfxx < 0, then we’ve located a localmaximum.

• If D < 0, then we’ve located a saddle point.

• I promise to look at the first two sections of Chapter 16(a little).

• I will not study Lagrange multipliers.