Transcript 偏微分(三度空間)

數學軟體應用
Directional Derivatives and the
Gradient Vector
Functions of Three Variables
The directiona l derivative of f
at ( x0 , y0 , z0 ) in the direction of the
vector u  a, b, c  is Du f ( x0 , y0 , z0 )
f ( x0  ha, y0  hb, z0  hc)  f ( x0 , y0 , z0 )
 lim
h0
h
if this limit exists.
f ( x0  hu )  f ( x0 )
Du f ( x0 )  lim
h0
h
f
f
f
f  f x , f y , f z  i 
j k
x
y
z
Du f ( x,y,z)  f ( x,y,z)  u  f ( x,y,z) cos 
f

u
 x,y,z
 是 f 與單位向量 u 之夾角
Tangent Planes to Level Surfaces
Fx ( x0 , y0 , z0 )( x  x0 )  Fy ( x0 , y0 , z0 )( y  y0 )
 Fz ( x0 , y0 , z0 )( z  z0 )  0
Maximum and Minimum
Values
A func ti on of two variables has a
local maximum at (a , b). If f(x,y)  f (a, b)
when (x , y) is near (a , b), the number
f(a,b) is called 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. 圖
Thm :
If f has a local maximum or minimum at (a, b)
and the first partial derivative s of f exist there,
then f x(a,b)  0 and f y(a,b)  0.
用matlab:畫圖
x=linspace(-2,2,25);
y=linspace(-2,2,25);
F(x,y)=x^2+y^2
[xx,yy]=meshgrid(x,y);
在(0,0)處有相對極
小值
zz=xx.^2+yy.^2;
mesh(xx,yy,zz);
Second Derivatives Test
Suppose the second partial derivative s of f are
continuous on a disk with center (a, b) , and
suppose that f x(a,b)  0 and f y(a,b)  0.
Let D  D(a,b)  f xx(a,b)f yy(a.b)  [f xy(a,b)] 2
(a) If D  0 and f xy(a,b)  0, then f(a,b) is a local minimum.
(b) If D  0 and f xy(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.
the point (a, b) is called a saddle point of f .
圖
Extreme Value Theorem for
Functions of Two Variable
If f is continuos on closed , bounded
2
set D in R .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.
圖
To find the absolute maximum and minimum
values of a continuous function on a closed ,
bounded set D :
1. Find the values of f at the critical points of
f in D.
2. Find the extreme values of f on the
boundary of D.
3. The largest of the values from steps
1 and 2 is the absolute maximum
value , the smallest of these values
is the absolute minimum value .
Lagrange Multiplies
Find the extreme values of f(x,y,z)
subject t o the constraint g(x,y,z)  k .
Suppose that a function f has an extreme
value at a point ( x0 , y0 , z0 ) satisfy
g ( x0 , y0 , z0 )  0 , there is a number
 such that f ( x0 , y0 , z0 )   g ( x0 , y0 , z0 )
Method of Lagrange Multipliers
To find the maximum and minimum values of f(x,y,z)
subject to the constraint g(x,y,z)  k ( assuming that
these extreme values exit ) :
(a) Find all values of x, y, z, and  such that
f(x,y,z)   g(x,y,z) and g(x,y,z)  k
(b) Evaluate f at all the points (x,y,z) that result from
step (a) . The largest of these values is the maximum
value of f ; the smallest is the minimum value of f .
Two Constrains
Suppose now that we wnat to find the maximum and
minimum values of f(x,y,z) subject to two constrains
of the form g(x,y,z)  k and h(x,y,z)  c
It can be shown that if an extreme value occurs at
(x0 ,y0 ,z0 ).Then the gradient vector f ( x0 , y0 , z0 ) is in
the plane determined by g ( x0 , y0 , z0 ) and h( x0 , y0 , z0 ).
So, there are numbers  and  such that
f ( x0 , y0 , z0 )   g ( x0 , y0 , z0 )  h( x0 , y0 , z0 )
absolute maximum value
用matlab
畫個peaks圖 轉到平面 標
上點
local maximum value
local minimum
value
absolute minimum
value
back
用matlab
畫‘’鞍點‘’的圖
Z =y^2-x^2
f(x,y) =y^2-x^2 在(0,0)處有鞍點
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