Transcript CALCULUS For Business, Economics, and the Social and life

Mathematics for Business
(Finance)
Instructor: Prof. Ken Tsang
Room E409-R11
Email: [email protected]
1
Chapter 6: Calculus of two
variables
In this Chapter:
Functions of 2 Variables
Limits and Continuity
Partial Derivatives
Tangent Planes and Linear Approximations
The Chain Rule
Maximum and Minimum Values
Double integrals and volume evaluation
DEFINITION:
A function f of two variables is a rule that
assigns to each ordered pair of real numbers (x, y)
in a set D a unique real number denoted by f (x, y).
The set D is the domain of f and its range is the
set of values that f takes on, that
is,
.
f ( x, y) ( x, y)  D
We write z=f (x, y) to make explicit the value taken on
by f at the general point (x, y) . The variables x and y
are independent variables and z is the dependent
variable.
Domain of
f ( x, y) 
x  y 1
x 1
Domain of
f ( x, y)  x ln(y  x)
2
Domain of
g ( x, y )  9  x  y
2
2
Graph of
z=f(x,y)
Graph of
g ( x, y )  9  x  y
2
2
DEFINITION:
The level curves of a function f of two
variables are the curves with equations f (x,
y)=k, where k is a constant (in the range of f).
Contour map of
g ( x, y )  9  x  y
2
2
The graph of h(x, y)=4x2+y2
is formed by lifting the level curves.
DEFINITION
Let f be a function of two variables whose domain D
includes points arbitrarily close to
(a, b). Then we say that the limit of f (x, y) as (x, y)
approaches (a ,b) is L and we write
lim
( x , y )  ( a ,b )
f ( x, y )  L
if for every number ε> 0 there is a corresponding
number δ> 0 such that
If
and
then
( x, y )  D
0  ( x  a ) 2  ( y  b) 2  
f ( x, y)  L  
DEFINITION
A function f of two variables is called continuous
at (a, b) if
lim
( x , y ) ( a ,b )
f ( x, y )  f ( a , b )
We say f is continuous on D if f is continuous at
every point (a, b) in D.
If f is a function of two variables, its partial
derivatives are the functions fx and fy defined
by
f ( x  h, y )  f ( x, y )
f x ( x, y )  lim
h 0
h
f ( x, y  h )  f ( x, y )
f y ( x, y )  lim
h 0
h
NOTATIONS FOR PARTIAL DERIVATIVES
If z=f (x, y) , we write
f

z
f x ( x, y )  f x 

f ( x, y ) 
 Dx f
x
x
x
f

z
f y ( x, y )  f y 

f ( x, y ) 
 Dy f
y
y
y
RULE FOR FINDING PARTIAL DERIVATIVES
OF z=f (x, y)
1.To find fx, regard y as a constant and
differentiate f (x, y) with respect to x.
2. To find fy, regard x as a constant and
differentiate f (x, y) with respect to y.
The partial derivatives of f at (a, b) are
the slopes of the tangents to C1 and C2.
The second partial derivatives of f.
If z=f (x, y), we use the following notation:
( f x ) x  f xx
  f   2 f
2 z




2
x  x 
x
x 2
( f x ) y  f xy
  f 
2 f
2 z




y  x  yx
yx
( f y ) x  f yx
  f 
2 f
2 z







x  y  xy
xy
( f y ) y  f yy
  f   f
 z





2
2


y  y 
y
y
2
2
CLAIRAUT’S THEOREM
Suppose f is defined on a disk D that contains the
point (a, b) . If the functions fxy and fyx are both
continuous on D, then
f xy (a, b)  f yx (a, b)
The tangent plane contains the
tangent lines T1 and T2
Suppose f has continuous partial
derivatives. An equation of the tangent
plane to the surface z=f (x, y) at the point
P(xo ,yo ,zo) is
z  z0  f x (x0 , y0 )(x  x0 )  f y (x0 , y0 )(y  y0 )
The differential of x is dx=△x, if y=f(x), then dy=f’(x)dx is
the differential of y.
For a differentiable function of two variables,
z= f (x ,y), we define the differentials dx
and dy (i.e. small increments in x & y
directions). Then the differential dz (total
differential), is defined by
z
z
dz  f x ( x, y)dx  f y ( x, y)dy  dx  dy
x
y
For such functions the linear approximation is:
f (x, y)  f (a, b)  f x (a, b)(x  a)  f y (a, b)(y  b)
THE CHAIN RULE Suppose that z=f (x, y) is
a differentiable function of x and y, where x=g
(t) and y=h (t) and are both differentiable
functions of t. Then z is a differentiable
function of t and
dz f dx f dy


dt x dt y dt
THE CHAIN RULE (GENERAL VERSION) Suppose
that u is a differentiable function of the n variables x1,
x2,‧‧‧,xn and each xj is a differentiable function of the
m variables t1, t2,‧‧‧,tm Then u is a function of t1, t2,‧‧‧,
tm and
u u dx1 u x2


‧‧‧
ti x1 dti x2 dti
for each i=1,2,‧‧‧,m.
u xn

xn ti
DEFINITION A function of two variables has a
local maximum at (a, b) if f (x, y) ≤ f (a, b)
when (x, y) is near (a, b). [This means that f
(x, y) ≤ f (a, b) for all points (x, y) in some disk
with center (a, b).]
The number f (a, b) is called a local
maximum value.
If f (x, y) ≥ f (a, b) when (x, y) is near (a, b),
then f (a, b) is a local minimum value.
35
THEOREM If f has a local maximum or
minimum at (a, b) and the first order partial
derivatives of f exist there, then fx(a, b)=0
and fy(a, b)=0.
A point (a, b) is called a critical point (or
stationary point) of f if fx (a, b)=0 and fy (a,
b)=0, or if one of these partial derivatives
does not exist.
SECOND DERIVATIVES TEST Suppose the second
partial derivatives of f are continuous on a disk with
center (a, b) , and suppose that both fx (a, b) and fy
(a, b)=0 [that is, (a, b) is a critical point of f]. Let
D  D(a, b)  f xx (a, b) f yy (a, b)  [ f xy (a, b)]
2
(a)If D>0 and fxx (a, b)>0 , then f (a, b) is a local
minimum.
(b)If D>0 and fxx (a, b)<0, then f (a, b) is a local
maximum.
(c) If D<0, then f (a, b) is not a local maximum or
minimum.
NOTE 1 In case (c) the point (a, b) is called a
saddle point of f and the graph of f crosses its
tangent plane at (a, b).
NOTE 2 If D=0, the test gives no information: f
could have a local maximum or local minimum at
(a, b), or (a, b) could be a saddle point of f.
NOTE 3 To remember the formula for D it’s helpful
to write it as a determinant:
D
f xy f xy
f yx f yy
 f xx f yy  ( f xy )
2
39
z  x  y  4xy  1
4
4
41
42
EXTREME VALUE THEOREM FOR
FUNCTIONS OF TWO VARIABLES
If f is continuous on a closed, bounded set D
in R2, then f attains an absolute maximum
value f(x1,y1) and an absolute minimum
value f(x2,y2) at some points (x1,y1) and
(x2,y2) in D.
43
EXTREME VALUE THEOREM FOR
FUNCTIONS OF TWO VARIABLES
If f is continuous on a closed, bounded set D
in R2, then f attains an absolute maximum
value f(x1,y1) and an absolute minimum
value f(x2,y2) at some points (x1,y1) and
(x2,y2) in D.
44
Double integrals over rectangles
Recall the definition of definite integrals of
functions of a single variable
Suppose f(x) is defined on a interval [a,b].
Taking a partition P of [a, b] into subintervals:
a  x  x  x  x  b
0
n 1
1
Choose the points in [ x , x ] and let
i 1
i
n
x  x  x
i
i
Using the areas of the small rectangles to
approximate the areas of the curve sided echelons
i 1
and summing them, we
have
n
 f ( x )x
(1)
i 1
*
i
i
P  max{x }
i
(2)
b
 f ( x)dx  lim  f ( x )x
a
n
P 0 i 1
*
i
i
Volume and Double Integral
z=f(x,y) ≥ 0
on rectangle
R=[a,b]×[c,d]
S={(x,y,z) in R3 | 0 ≤ z ≤ f(x,y), (x,y) in R}
Volume of S = ?
Double integral of a function of two variables
defined on a closed rectangle like the following
R  [a, b]  [c, d ]  {( x, y )  R | a  x  b, c  y  d }
2
Taking a partition of the rectangle
a  x  x  x  x  b
0
1
m 1
m
c  y  y  y  y  d
0
1
n 1
n
ij’s column:
y
z
(xi, yj)
f (xij*, yij*)
Rij
Sample point (xij*, yij*)
x
y
x
Area of Rij is Δ A = Δ x Δ y
*
*
f
(
x
,
y
Volume of ij’s column:
ij
ij )A
m
n
Total volume of all columns:  
i 1 j 1
f ( x , y )A
*
*
ij
ij
ij
double Riemann sum
m
n
V   f ( x , y ) A
*
ij
i 1 j 1
*
ij
Definition
V  lim
m
n
 f ( x , y
m,n  i 1 j 1
*
ij
*
ij
) A
Definition:

The double integral
of f over the rectangle R is
f ( x, y)dA
R
m
n
 f ( x, y)dA  lim  f ( x , y
R
m,n  i 1 j 1
*
ij
*
ij
)A
if the limit exists
m
Double Riemann sum:
n
*
*
f
(
x
,
y
 ij ij )A
i 1 j 1
Example 1
z=16-x2-2y2
0≤x≤2
0≤y≤2
Estimate the volume
of the solid above
the square and
below the graph
m=n=4
V≈41.5
m=n=8
V≈44.875
Exact volume?
V=48
m=n=16
V≈46.46875
Example 2
R  [1,1]  [ 2,2]

R
z
1  x dA  ?
2
Integrals over arbitrary regions


f (x,y)
A


0
R

A
A is a bounded plane region
f (x,y) is defined on A
Find a rectangle R containing
A
Define new function on R:
 f ( x, y) if ( x, y)  A
f ( x, y)  
0, otherwise
f ( x, y)dA  f ( x, y)dA
R
Properties
Linearity
 [ f ( x, y)  g( x, y)]dA   f ( x, y)dA  g( x, y)dA
A
A
A
 cf ( x, y)dA  c f ( x, y)dA
A
A
Comparison
If f(x,y)≥g(x,y) for all (x,y) in R, then
 f ( x, y)dA   g( x, y)dA
A
A
Additivity
A2
A1
If A1 and A2 are non-overlapping regions then

f ( x, y )dA   f ( x, y )dA   f ( x, y )dA
A1  A2
A1
Area
1dA   dA  area of A
A
A
A2
Computation

If f (x,y) is continuous on rectangle R=[a,b]×[c,d]
then double integral is equal to iterated integral
d b
b d
c a
a c
 f ( x, y)dA   f ( x, y)dxdy   f ( x, y)dydx
y
R
d
y
fixed
c
x
a
x
b
fixed
Fubini’s Theorem If f is continuous on the
rectangle R  {( x, y) | a  x  b, c  y  d }, then
 f ( x, y )dA    f ( x, y )dydx    f ( x, y )dxdy
b d
d
b
a c
c
a
R
More generally, this is true if we assume that f
is bounded on R , f is discontinuous only on
a finite number of smooth curves, and the iterated
integrals exist.
Note
If f (x, y) = g (x) h(y) then

R
b
 d

f ( x, y )dA    g ( x)h( y )dxdy    g ( x)dx    h( y )dy
c a
a
 c

d b
EXAMPLE 1 Evaluate the iterated integrals
(a)
3

2
0 1
2
x ydydx (b)
(See the blackboard)
2
3
  (x
1
0
2
 y)dxdy
EXAMPLE 2
Evaluate the double integral
where
 ( x  3 y )dA
2
R
R  {( x, y) | 0  x  2, 1  y  2}.
(See the blackboard)
EXAMPLE 4 Find the volume of the solid S that is
bounded by the elliptic paraboloid x  2 y  z  16
, the plane y  2 and x  2 , and three coordinate
planes.
2
2
Solution We first observe that S is the solid that
lies under the surface z  16  x 2  2 y 2
and the above the square
R  [0,2]  [0,2].
V   (16  x  2 y )dA
2
2
R
   (16  x  2 y )dxdy
2
2
0
0
2

  16 x 
2
0

 2y x
2

 4 y dy
88
y
3
 y
2
0

1 3
x
3
2

88
3
2
4
3
3



2
0
 48
x 2
x 0
dy
More general case

If f (x,y) is continuous on
A={(x,y) | x in [a,b] and h (x) ≤ y ≤ g (x)} then
double integral is equal to iterated integral
y
b g ( x)
g(x)
 f ( x, y)dA    f ( x, y )dydx
A
A
h(x)
a
x
x
b
a h( x)
Similarly

If f (x,y) is continuous on
A={(x,y) | y in [c,d] and h (y) ≤ x ≤ g (y)} then
double integral is equal to iterated integral
y
d g ( y)
 f ( x, y)dA   f ( x, y)dxdy
d
A
y
h(y)
c
R
c h( y )
g(y)
x
Type I regions
D  {( x, y) | a  x  b, g ( x)  y  g ( x)}
1
2
If f is continuous on a type I region D such that
D  {( x, y) | a  x  b, g ( x)  y  g ( x)}
1
then
 f ( x, y )dA   
b g2 ( x )
D
a g1 ( x )
2
f ( x, y )dydx
Type II regions
(4)
(5)
D  {( x, y) | c  y  d , h ( y)  x  h ( y)}
1
d
h2 ( y )
c
h1 ( y )
 f ( x, y)dA   
D
2
f ( x, y )dxdy
where D is a type II region given by Equation 4


   ( x  2 y)dy dx
it is Type I region!
1
1
1 x 2
2 x2
  xy  y
1
1


2

y 1 x 2
y 2 x 2
dx
   x(1  x )  (1  x )  2 x  4 x dx
1
2
2
2
3
4
1
   3x  x  2 x  x  1dx
1
4
3
2
1
1
x x
 x x

  3   2   x
5 4
3 2

1
32

15
5
4
3
Example 2 Find the volume of the solid under the
paraboloid z  x  y and above the region D in the
2
2
xy-plane bounded by the line y  2 x and the parabola
y  x2.




Type I
D  {( x, y ) | 0  x  2, x  y  2 x}
2
Type II
y
D  {( x, y ) | 0  y  4,  x  y}
2
Type I
Solution 1
 ( x  y )dA   ( ( x  y )dy)dx
8
1
dx
  (2 x  3 x  x  3 x )dx
2
y 2 x
D
2
 2 1 3
  x y  y 
0
0
3  y x2

2
1 6
14 3
4
 0 ( 3 x  x  3 x )dx
2
2
2x
0
x2
2
3
3

1 7
x
21
2
2
4
1 5
5x
6


216
7 4 2

x
6
35
0
Type II
Solution 2

4
0

1 3
x
3
 ( x  y )dA    ( x  y )dxdy
2
D
yx
2 5/ 2
 15 y
2

 y
2
7
2
x y
x y / 2
7/2
dy  
4
0

y
1
3
4
y
0
y/2
3/ 2
216
13 4 4
 96 y  35
0
2
y 
5/ 2
2
1 3
y
24

 y dy
1
2
3
Example 3 Evaluate D xydA , where D is the region
bounded by the line
and the parabola
2
y
 2 x  6.
y  x 1



D as a type I
D as a type II