Transcript 可微分與切平面

第二十七單元
切平面
Tangent Lines
The equation of the tangent line at x=a is
( y - f (a) ) = f ¢(a)( x - a)
Tangent Planes
Differentiability (One Variable)
One Variable Case
y= x
f is continuous at x0
f ( x0 + D x) - f ( x0 )
f+¢( x0 ) = lim+
D x® 0
Dx
f ( x0 + D x) - f ( x0 )
f- ¢( x0 ) = limD x® 0
Dx
y = x is not differentiable at x = 0
Differentiability
Two Variable Case
ìïï 1 ,
f ( x, y) = í
ïïî 0 ,
x> 0 , y> 0
otherwise
f (0 + D x, 0) - f (0, 0)
f x (0, 0) = lim
=0
D x® 0
Dx
f (0 , 0 + D y) - f (0, 0)
f y (0, 0) = lim
=0
D y® 0
Dy
f is not continuous at (0,0)
Along x = y (in the first quadrant)
Differentiability
lim
( x , y )® (0,0)
f ( x, y ) = 1
Along x = 0
lim
( x , y )® (0,0)
f ( x, y ) = 0
The existence of f x and f y is not guarantee
f is differentiable
Sufficient Condition for differentiability
Theorem :
Let f be a function of two variables, if
f , f x , f y are continuous in a disk centered
at ( x0 , y0 ), then f is differentiable at ( x0 , y0 )
f
All of us
are
continuous
fx
fy
D
( x0 , y0 )
Example
Show that f ( x, y) = x 2 y 2 + xy 3 is differentiable
for all ( x, y )
Solution :
¶ 2 2
f x ( x, y ) =
( x y + xy 3 ) = 2 xy 2 + y 3
¶x
¶ 2
f y ( x, y ) =
( x y + xy 3 ) = x 2 + 3xy 2
¶y
Q f , f x , f y are continuous for all ( x, y)
\ f is differentiable for all ( x, y)
Example
x
x
ax
d
d
x=
Þ
ax = a =
dx
x
dx
x
x
Show that f ( x, y) = xy is not differentiable
at point (0,0)
Solution :
yx
f x = xy x =
x
not continuous in the
disk centered at (0,0)
Graph of f (x)
The Direction Vector
The direction vector is
1 , 0 , f x (a, b)
The Direction Vector
The direction vector is
0 , 1 , f y ( a, b)
Normal Vector
v
a = 1 , 0 , f x (a, b)
v
b=
0 , 1 , f y ( a, b)
Normal Vector
v v
= a´ b
The Normal Vector of a Tangent Plane
v
v
a = 1 , 0 , f x ( a, b) , b =
0 , 1 , f y ( a, b)
1
0
fx
1
0
fx
0
1
fy
0
1
fy
< - fx , - f y , 1
>
Theorem
Z=f (x ,y)
< f x , f y , - 1>
Example
Find the equation of the tangent plane to
2
2
z = 6 - x - y at the point (1 , 2 ,1)
Solution :
v
N = < - 2 x, - 2 y , - 1 >
(1,2,1)
= < - 2, - 4, - 1 >
Plane Equation:
< - 2, - 4, - 1 > ×< x - 1, y - 2, z - 1 > = 0
- 2( x - 1) - 4( y - 2) - ( z - 1) = 0
Example(continued)
f ( x, y) = 6 - x2 - y 2
(1, 2,1)
Tangent plane to the level surface
Suppose S is a surface with F ( x, y, z) = k that
is, level surface of 3 variables
uv
v
ÑF
Suppose r (t ) = < x(t ), y(t ), z(t ) >
F ( x(t ), y(t ), z (t )) = k
¶ F dx ¶ F dy ¶ F dz
+
+
=0
¶ x dt ¶ y dt ¶ z dt
dx dy dz
¶F ¶F ¶F
,, ,
=0
,
,
×
dt dt dt
¶x ¶y ¶z
P( x0 , y0 , z0 )
Tangent Plane
dx dy dz
¶F ¶F ¶F
,, ,
=0
,
,
×
dt dt dt
¶x ¶y ¶z
uv
ÑF
uv v
Ñ F ×r ¢(t ) = 0
uv
v¢
Ñ F ( x0 , y0 , z0 ) ×r (t0 ) = 0
uv
\ Ñ F ( x0 , y0 , z0 ) ^ any curve passing through P
Example
Find the equation of the tangent plane and normal
x2 y 2 z 2
line to the ellipsoid
+
+
=3
4
1
9
at the point (- 2 ,1 , - 3)
Solution :
2
2
2
x
y
z
F ( x, y , z ) =
+
+
4
1
9
x
Fx ( x, y, z ) =
Fy ( x, y, z) = 2 y
2
2z
Fz ( x, y , z ) =
9
x
Fx ( x, y, z ) =
Fy ( x, y, z) = 2 y
2
Fx (- 2,1, - 3) = - 1
Example
Fy (- 2,1, - 3) = 2
Fz (- 2,1, - 3) = - 2/ 3
2z
Fz ( x, y , z ) =
9
< - 1, 2 , - 2 / 3 >
P(- 2,1, - 3)
2
- 1 ( x + 2) + 2 ( y - 1) - ( z + 3) = 0
3
graph
單元結語
單變數函數可微分的條件微左導數=右導數，
以幾何的角度來看，就是存在一條非鉛直
的切線。
雙變數函數可微分的條件就比較複雜了，
在此我們提出了可微分的充分條件，以幾
何的角度來看，就是切平面的存在性。
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