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Review for Midterm II
Math 20
December 4, 2007
Announcements
Midterm I 12/6, Hall A 78:30pm
ML Office Hours Wednesday 13 (SC 323)
Old exams and solutions on website
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Outline
Rank and other Linear AlgebraLinear dependenceRank
EigenbusinessEigenvector and EigenvalueDiagonalizationThe Spectral Theorem
Functions of several variables
Graphing/Contour PlotsPartial Derivatives
Differentiation
The Chain RuleImplicit Differentiation
OptimizationUnconstrained Optimization
Constrained Optimization
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Rank and other Linear AlgebraLearning Objectives
Determine whether a set of vectors is linearly independent Find the rank of a matrix
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Linear Independence
DefinitionLet{a1, a2, . . . , an} be a set of vectors in Rm. We say they arelinearly dependent if there exist constants c1, c2, . . . , ck R, notall zero, such that
c1a1+c2a2+ +ckan =0.
If the equation only holds when all c1 =c2 = =cn = 0, thenthe vectors are said to be linearly independent.
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Deciding linear dependence
We showed
a1, . . . , an LD c1a1+ +cnan=0 has a nonzero soln
a1 . . . an A
c1...
cn
c
=0 has a nonzero soln
system has some free variables rref(A) has a column with no leading entry to it
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Example
Determine if the vectors
101 ,
3
22 ,
0
21
are linearly dependent.
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Example
Determine if the vectors
101 ,
3
22 ,
0
21
are linearly dependent.
Solution
1 3 0
0 2 2
1 2 1
1
+
1 3 00 2 2
0 1 1
1 3 0
0 1 10 0 0
3
+
1 0 3
0 1 10 0 0
So the vectors are linearly dependent.
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Example
Determine if the vectors
101 ,
3
22 ,
0
21
are linearly dependent.
Solution
1 3 0
0 2 2
1 2 1
1
+
1 3 00 2 20
1 1
1 3 0
0 1 10 0 0
3
+
1 0 3
0 1 10 0 0
So the vectors are linearly dependent.
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Deciding linear independence
So
a1, . . . , an LI
every column of rref(A) has a leading entry to it
A
InO
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Example
Determine if the vectors
1
011
,3
220
,0
211
are linearly dependent.
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Solution
1 3 0
0 2 21 2 1
1 0 1
1 0 3
0 1 10 0 0
1 0 1
1
+
1 0 3
0 1 10 0 0
0 0 2
1 0 0
0 1 0
0 0 1
0 0 0
So the vectors are linearly independent.
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Rank
Definition
The rank of a matrix A, written r(A) is the maximum number oflinearly independent column vectors in A.
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Rank
Definition
The rank of a matrix A, written r(A) is the maximum number oflinearly independent column vectors in A. IfA is a zero matrix, wesayr(A) = 0.
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Example
Since
rref1 3 0
0 2 21 2 1 = 1 0 3
0 1 10 0 0 this matrix has rank 2.
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Example
Since
rref
1 3 00
2 2
1 2 11 0 1
=
1 0 00 1 0
0 0 10 0 0
this matrix has rank 3.
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Another way to compute rank
Theorem
Book Theorem 14.1 The rank ofA is the size of the largestnonvanishing minor ofA.
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Rank and consistency
FactLetA be an m
n matrix, b an n
1vector, andAb the matrixA
augmented byb.Then the system of linear equationsAx= b has a solution (isconsistent) if and only if r(A) =r(Ab).
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Rank and redundancy
FactLetA be an m n matrix, b an n 1vector, andAb the matrixAaugmented byb. Suppose that r(A) =r(Ab) =k
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Rank and freedom
FactLetA be an m n matrix, b an n 1vector, andAb the matrixAaugmented byb. Suppose that r(A) =r(Ab) =k
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Outline
Rank and other Linear AlgebraLinear dependenceRank
EigenbusinessEigenvector and EigenvalueDiagonalizationThe Spectral Theorem
Functions of several variables
Graphing/Contour PlotsPartial Derivatives
Differentiation
The Chain RuleImplicit Differentiation
OptimizationUnconstrained OptimizationConstrained Optimization
Ei b i
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EigenbusinessLearning Objectives
Determine if a vector is an eigenvalue of a matrix
Determine if a scalar is an eigenvalue of a matrix
Find all the eigenvalues of a matrix Find all the eigenvectors of a matrix for a given eigenvalue
Diagonalize a matrix
Know when a matrix is diagonalizable
Ei b i
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Eigenbusiness
DefinitionLet A be an n n matrix. The number is called an eigenvalueofA if there exists a nonzero vector x
R
n such that
Ax= x. (1)
Every nonzero vector satisfying (1) is called an eigenvectorofAassociated with the eigenvalue .
E a le
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ExampleMidterm II, Fall 2006, Problem 4
Let A=4 2
1 1
Problem
Is21 an eigenvector for A?
Example
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ExampleMidterm II, Fall 2006, Problem 4
Let A=4 2
1 1
Problem
Is21 an eigenvector for A?
SolutionUse the definition of eigenvector:
4 2
1 1 2
1 = 6
3 = 32
1So the vector is an eigenvector corresponding to the eigenvalue3.
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Let A= 4 21 1
ProblemIs0 an eigenvalue forA?
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Let A= 4 21 1
ProblemIs0 an eigenvalue forA?
Solution
The number0 is an eigenvalue forA if and only if the determinantofA 0I=A is zero. But
det A= 4 1 1 (2) = 6.
So its not.
Methods
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Methods
To find the eigenvalues of a matrix A, find the determinant ofA I. This will be a polynomial in (called thecharacteristic polynomial ofA, and its roots are theeigenvalues.
To find the eigenvector(s) of a matrix corresponding to aneigenvalue , do Gaussian Elimination on A I.
Diagonalization Procedure
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Diagonalization Procedure
Find the eigenvalues and eigenvectors.
Arrange the eigenvectors in a matrix P and the corresponding
eigenvalues in a diagonal matrix D. If you have enough eigenvectors so that the matrix P is
square and invertible, the original matrix is diagonalizable andequal to PDP1.
Example
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Example
ProblemLetA=
2 32 1
. Diagonalize.
Example
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Example
ProblemLetA=
2 32 1
. Diagonalize.
Solution
To find the eigenvalues, find the characteristic polynomial and itsroots:
|A I| =2 3
2 1
= (2 )(1 ) 6
=2
3 4 = ( + 1)( 4)So the eigenvalues are1 and4.
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To find an eigenvector corresponding to the eigenvalue1,
A+I=
3 32 2
1 10 0
So
11
is an eigenvector.
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To find an eigenvector corresponding to the eigenvalue 4,
A 4I=2 3
2 3
1 3/20 0
So
32
is an eigenvector.
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LetP=
1 31 2
so
P1 =1
5 2 31 1
Then
A=
1 31 2
1 00 4
1
5
2 31 1
The Spectral Theorem
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p
Theorem (Baby Spectral Theorem)SupposeAnn has n distinct real eigenvalues. Then A isdiagonalizable.
The Spectral Theorem
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p
Theorem (Baby Spectral Theorem)SupposeAnn has n distinct real eigenvalues. Then A isdiagonalizable.
Theorem (Spectral Theorem for Symmetric Matrices)
SupposeAnn is symmetric, that is, A =A. Then A is
diagonalizable. In fact, the eigenvectors can be chosen to bepairwise orthogonal with length one, which means thatP1 =P.Thus a symmetric matrix can be diagonalized as
A= PDP,
Outline
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Rank and other Linear AlgebraLinear dependenceRank
EigenbusinessEigenvector and EigenvalueDiagonalizationThe Spectral Theorem
Functions of several variables
Graphing/Contour PlotsPartial Derivatives
Differentiation
The Chain RuleImplicit Differentiation
OptimizationUnconstrained OptimizationConstrained Optimization
Functions of several variables
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Learning Objectives
identify functions, graphs, and contour plots
find partial derivatives of functions of several variables
Types of functions
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linear
polynomial
rational Cobb-Douglas
etc.
Examples
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ProblemIn each of the following, find the domain and range of the function.Is it linear? polynomial? rational? algebraic? Cobb-Douglas?
(a) f(x, y) =y x(b) f(x, y) = y x(c) f(x, y) = 4x2 + 9y2
(d) f(x, y) =x2 y2(e) f(x, y) =xy
(f) f(x, y) =y/x2
(g) f(x, y) = 116 x2 y2
(h) f(x, y) =
9 x2 y2
(i) f(x, y) = ln(x2 +y2)
(j) f(x, y) =e(x2+y2)
(k) f(x, y) = arcsin(y
x)
(l) f(x, y) = arctanyx
Graphing/Contour Plots
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A function of two variables can be visualized by
its graph: the surface (x, y, f(x, y) in R3
a contour plot: a collection of level curves
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Example
Graph and contour plot off(x, y) =y
x
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Example
Graph and contour plot off(x, y) =y
x
2
1
0
1
2 2
1
0
1
2
4
2
0
2
4
2 1 0 1 2
2
1
0
1
2
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Example
Graph and contour plot off(x, y) =
y
x
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Example
Graph and contour plot off(x, y) =
y
x
2
1
0
1
22
1
0
1
2
0.0
0.5
1.0
1.5
2.0
2 1 0 1 2
2
1
0
1
2
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Example
Graph and contour plot off(x, y) = 4x2 + 9y2
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Example
Graph and contour plot off(x, y) = 4x2 + 9y2
2
1
0
1
2 2
1
0
1
2
0
20
40
2 1 0 1 2
2
1
0
1
2
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Example
Graph and contour plot off(x, y) =x2
y2
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Example
Graph and contour plot off(x, y) =x2
y2
2
1
0
1
2 2
1
0
1
2
4
2
0
2
4
2 1 0 1 2
2
1
0
1
2
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Example
Graph and contour plot off(x, y) =xy
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Example
Graph and contour plot off(x, y) =xy
2
1
0
1
2 2
1
0
1
2
4
2
0
2
4
2 1 0 1 2
2
1
0
1
2
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Example
Graph and contour plot off(x, y) = yx2
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Example
Graph and contour plot off(x, y) = yx2
2
1
0
1
2 2
1
0
1
2
5
0
5
2 1 0 1 2
2
1
0
1
2
E ample
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Example
Graph and contour plot off(x, y) = 1
16 x2
y2
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Example
Graph and contour plot off(x, y) =
9 x2 y2
E l
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Example
Graph and contour plot off(x, y) =
9 x2 y2
2
0
2
2
0
2
0
1
2
3
3 2 1 0 1 2 3
3
2
1
0
1
2
3
E l
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Example
Graph and contour plot off(x, y) = ln(x2 +y2)
E l
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Example
Graph and contour plot off(x, y) = ln(x2 +y2)
2
0
2
2
0
2
1
0
1
2
3 2 1 0 1 2 3
3
2
1
0
1
2
3
E l
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Example
Graph and contour plot off(x, y) =e(x2+y2)
Example
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Example
Graph and contour plot off(x, y) =e(x2+y2)
2
0
2
2
0
2
0.0
0.5
1.0
3 2 1 0 1 2 3
3
2
1
0
1
2
3
Example
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Example
Graph and contour plot off(x, y) = arcsin(y x)
Example
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Example
Graph and contour plot off(x, y) = arcsin(y x)
2
1
0
1
22
1
0
1
2
1
0
1
2 1 0 1 2
2
1
0
1
2
Example
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Example
Graph and contour plot off(x, y) = arctan
yx
Example
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Example
Graph and contour plot off(x, y) = arctan
yx
2
1
0
1
22
1
0
1
2
1
0
1
2 1 0 1 2
2
1
0
1
2
The process
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To differentiate a function of several variables with respect to one
of the variables, pretend that the others are constant.
Examples
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Example
Letf(x, y) = 3x+ 2xy2 2y4. Find both the partial derivatives off.
Examples
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Example
Letf(x, y) = 3x+ 2xy2 2y4. Find both the partial derivatives off.
SolutionWe have
f
x= 3 + 2y2
f
y = 4xy 8y3
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ExampleLet w= sin cos. Find both the partial derivatives ofw.
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ExampleLet w= sin cos. Find both the partial derivatives ofw.
SolutionWe have
w
= cos cos
w
= sin sin
Example
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Let f(u, v) = arctan(u/v). Find both the partial derivatives off.
Example
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Let f(u, v) = arctan(u/v). Find both the partial derivatives off.
SolutionFor this its important to remember the chain rule!
f
u =
1
1 + (u/v)2
u
u
v =
1
1 + (u/v)21
v
fu
= 1
1 + (u/v)2v
uv
= 1
1 + (u/v)2uv2
Another way to write this is
fu
= vu2 +v2
fv
= uu2 +v2
Example
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Example
Let u= x21 +x22 + +x2n . Find all the derivatives ofu.
Example
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Example
Let u= x21 +x22 + +x2n . Find all the derivatives ofu.
SolutionWe have a partial derivative for each index i, but luckily theyresymmetric. So each derivative is represented by:
uxi
= 1
2
x21 +x22 + +x2n
xi
(x21 +x22 + +x2n )
= xi
x21 +x22 + +x2n
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Example
Let f(x, y) = 3x+ 2xy2 2y4. Find all the second derivatives.
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Example
Let f(x, y) = 3x+ 2xy2 2y4. Find all the second derivatives.Solution
2
fx2
= 0 2
fxy
= 4y
2f
yx = 4y
2f
y2 = 24y2
Tangent Planes
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FactThe tangent plane to z=f(x, y) through (x0, y0, z0 =f(x0, y0))has normal vector(f1(x0, y0), f
2(x0, y0),1) and equation
f1(x0, y0)(x x0) +f2(x0, y0)(y y0) (z z0) = 0
Tangent Planes
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FactThe tangent plane to z=f(x, y) through (x0, y0, z0 =f(x0, y0))has normal vector(f1(x0, y0), f
2(x0, y0),1) and equation
f1(x0, y0)(x x0) +f2(x0, y0)(y y0) (z z0) = 0
orz=f(x0, y0) +f
1(x0, y0)(x x0) +f2(x0, y0)(y y0)
Tangent Planes
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FactThe tangent plane to z=f(x, y) through (x0, y0, z0 =f(x0, y0))has normal vector(f1(x0, y0), f
2(x0, y0),1) and equation
f1(x0, y0)(x x0) +f2(x0, y0)(y y0) (z z0) = 0
orz=f(x0, y0) +f
1(x0, y0)(x x0) +f2(x0, y0)(y y0)
This is the best linear approximation to f near (x0, y0).
is is the first-degree Taylor polynomial (in two variables) for f.
Outline
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Rank and other Linear AlgebraLinear dependenceRank
EigenbusinessEigenvector and EigenvalueDiagonalizationThe Spectral Theorem
Functions of several variables
Graphing/Contour PlotsPartial Derivatives
DifferentiationThe Chain RuleImplicit Differentiation
OptimizationUnconstrained OptimizationConstrained Optimization
Fact (The Chain Rule, version I)
When z = F (x , y ) with x = f (t) and y = g (t), then
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When z F(x, y) with x f(t) and y g(t), then
z
(t) =F
1(f(t), g(t))f
(t) +F
2(f(t), g(t))g
(t)
or
dz
dt =
F
x
dx
dt +
F
y
dy
dt
Fact (The Chain Rule, version I)
When z=F(x, y) with x=f(t) and y=g(t), then
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e ( , y ) t (t) a d y g (t), t e
z
(t) =F
1(f(t), g(t))f
(t) +F
2(f(t), g(t))g
(t)
or
dz
dt =
F
x
dx
dt +
F
y
dy
dt
We can generalize to more variables, too. IfF is a function ofx1, x2, . . . , xn, and each xi is a function oft, then
dzdt
= Fx1
dx1dt
+ Fx2
dx2dt
+ + Fxn
dxndt
Tree Diagrams for the Chain Rule
F
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F
x
t
dxdt
Fx
y
t
dydt
Fy
To differentiate with respect to t, find all leaves marked t.Going down each branch, chain (multiply) all the derivativestogether. Then add up the result from each branch.
dz
dt =
dF
dt =
F
x
dx
dt +
F
y
dy
dt
Fact (The Chain Rule, Version II)
When z=F(x, y) with x=f(t, s) and y=g(t, s), then
F F
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z
t
=F
x
x
t
+F
y
y
tz
s =
F
x
x
s +
F
y
y
s
F
x
t s
y
t s
ExampleSupposez=xy2, x=t+s and y=t s. Find z
t and z
s at
(t, z) = (1/2, 1) in two ways:
(i) B i di l i f d b f
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(i) By expressing zdirectly in terms oft and s before
differentiating.(ii) By using the chain rule.
ExampleSupposez=xy2, x=t+s and y=t s. Find z
t and z
s at
(t, z) = (1/2, 1) in two ways:
(i) B i di tl i t f t d b f
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(i) By expressing zdirectly in terms oft and s before
differentiating.(ii) By using the chain rule.
Solution (i)
We have
z= (t+s)(t s)2 =s3 ts2 t2s+t3
So
zt
= s2 2ts+ 3t2z
s = 3s2 2ts t2
Solution (ii)
W h
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We have
z=xy2
x=t+s y=t sSo
z
t =
z
x
x
t +
z
y
y
t
=y2 1 + 2xy 1 = (t s)2 + 2(t+s)(t s)z
s =
z
x
x
s +
z
y
y
s
=y2
1 + 2xy(
1) = (t
s)2
2(t+s)(t
s)
These should be the same as in the previous calculation.
Theorem (The Chain Rule, General Version)
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Suppose that u is a differentiable function of the n variables
x1, x2, . . . , xn, and each xi is a differentiable function of the mvariables t1, t2, . . . , tm. Then u is a function of t1, t2, . . . , tm and
u
ti= u
x1
x1ti
+ u
x2
x2ti
+ + uxn
xnti
In summation notation
u
ti=
n
j=1u
xj
xjti
Implicit DifferentiationThe Big Idea
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FactAlong the level curve F(x, y) =c, the slope of the tangent line is
given by dy
dx =
dy
dx
F
= F/xF/x
= F
1(x, y)
F2(x, y)
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More than two variables
The basic idea is to close your eyes and use the chain rule:
E l
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Example
Suppose a surface is given by F(x, y, z) =c. If this defines z as afunction ofx and y, find zx and z
y.
More than two variables
The basic idea is to close your eyes and use the chain rule:
E l
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Example
Suppose a surface is given by F(x, y, z) =c. If this defines z as afunction ofx and y, find zx and z
y.
SolutionSetting F(x, y, z) =c and remembering z is implicitly a function
of x and y, we get
F
x +
F
z
z
x
F
= 0 =z
x
F
= F
x
Fz
F
y +F
zzy
F
= 0 = zyF = FyFz
Tree diagram
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F
x y z
x
F
x
+F
z z
xF= 0 =
z
xF=
Fx
F
z
ExampleProblem 16.8.4
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ProblemLet D=f(r,P) denote the deman for an agricultural commoditywhen the price is P and r is the producers total advertising
expenditure. Let supply be given by S=g(w,P), where w is anindex for how favorable the weather has been. Assumegw(w,P)> 0. Equilibrium now requires f(r,P) =g(w,P).Assume that this equation defines P implicitly as a differentiablefunction of r and w. Compute Pwand comment on its sign.
Solution
We have
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f(r,P)
g(w,P)
0
f g
r w P
w
fP
Pw
f=g
gw g
PPw
f=g
Answer
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So P
w
f=g
=gw
f
P g
P
f
P0. We assumed that
g
w>0. So in this case,P
w
f=g
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Rank and other Linear AlgebraLinear dependenceRank
Eigenbusiness
Eigenvector and EigenvalueDiagonalizationThe Spectral Theorem
Functions of several variables
Graphing/Contour PlotsPartial Derivatives
DifferentiationThe Chain Rule
Implicit DifferentiationOptimization
Unconstrained OptimizationConstrained Optimization
OptimizationLearning Objectives
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Find the critical points of a function defined on an open set(so unconstrained)
Classify the critical points of a function Find the critical points of a function restricted to a surface
(constrained)
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Theorem (Fermats Theorem)Let f(x, y) be a function of two variables. If f has a localmaximum or minimum at(a, b), and is differentiable at (a,b), then
f
x(a, b) = 0
f
y(a, b) = 0
As in one variable, well call these points critical points.
Theorem (The Second Derivative Test)
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Let f(x, y) be a function of two variables, and let(a, b) be acritical point of f . Then
If 2f
x22fy2
2fxy
2>0 and
2fx2
>0, the critical point is a
local minimum.
If 2
fx22
fy2 2fxy2 >0 and 2fx2
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f(x, y) = 4xy x4 y4
Example
ProblemFind and classify the critical points of
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f(x, y) = 4xy x4 y4
SolutionWe have f
x = 4y
4x3 and f
y = 4x
4y3. Both of these are
zero when y=x3 and x=y3 So x9 =x. Since
x9 x=x(x8 1) =x(x4 + 1)(x2 + 1)(x+ 1)(x 1)
the real solutions are x= 0, x= 1, and x=
1. Thecorresponding y values are0, 1, and1. So the critical points are
(0, 0), (1, 1), (1,1)
The second derivatives are
2f
x2 = 12x2
2f
yx = 4
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2
fxy
= 4 2
fyx
= 12y2
So
H(x, y) = 43x2 11 3y2
At (0, 0), the matrix is
0 11 0
, which has determinant
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2
1
0
1
2 2
1
0
1
2
30
20
10
0
2 1 0 1 2
2
1
0
1
Theorem (The Method of Lagrange Multipliers)
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Let f(x1, x2, . . . , xn) and g(x1, x2, . . . , xn) be functions of severalvariables. The critical points of the function f restricted to the setg= 0 are solutions to the equations:
f
xi
(x1, x2, . . . , xn) =g
xi
(x1, x2, . . . , xn) for each i= 1, . . . , n
g(x1, x2, . . . , xn) = 0.
Note that this is n+ 1 equations in the n+ 1 variables.x1, . . . , xn, .
ProblemFind the critical points and values of
f(x, y) =ax2 + 2bxy+cy2
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( , y ) y y
subject to the constraint that x2 +y2 = 1.
ProblemFind the critical points and values of
f(x, y) =ax2 + 2bxy+cy2
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( )
subject to the constraint that x2 +y2 = 1.
SolutionWe have
fx =g
x = 2ax+ 2by=(2x)fy =g
y = 2bx+ 2cy=(2y)
So the critical points happen when
a bb c
xy
=
xy
The critical values are
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f(x, y) =
x ya b
b c
xy
=
x y
xy
=(x2 +y2) =
The critical values are
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f(x, y) =
x ya b
b c
xy
=
x y
xy
=(x2 +y2) =
So
The critical points are eigenvectors!
The critical values are eigenvalues!
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