Test 2 Review - University of South Carolinapeople.math.sc.edu/.../Math241Test2Review.pdf · f(x,y)...
Transcript of 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.