Directional Derivatives - (9.5)
1. Gradient of a Scalar Function:
The Vector Differential Operator:
, x y
i j y
x
or i j k x
y
z
Let fŸx, y and FŸx, y, z be differentiable functions. Then
fŸx, y f , f , and FŸx, y, z x y
, , .
x y z
F , F , F
x y z
.
Vectors fŸx, y and FŸx, y, z are called gradientvectors of f and F, respectively.
Notations: f, grad f.
Example Let FŸx, y, z 3e x
FŸx, y, z 6xe x
2
y
2
y
cos 2x
z . Find F and FŸ=, 1, 2 .
cos 2x
z
2
" 6z e x
y
3x 2 x 2
sin 2x
z , 2 y e
y
6x x 2
cos 2x
z , z2 e
y
sin 2x
z
2
2
2
FŸ=, 1, 2 "6=e = , " 3= e = , 0
2
Use the Scientific Notebook: define FŸx, y, z first and then type FŸx, y, z and click on Evaluate. Here is
what you get:
sin 2 xz 3 x 2 x 2 y
2
2
2
6x y e x y cos 2 xz " 6e x y
cos 2 xz , 6e x y sin 2 xz x2
z ,2 y e
z
To evaluate FŸ=, 1, 2 , first define gŸx, y, z FŸx, y, z , that is,
sin 2 xz 3 x 2 x 2
2
2
gŸx, y, z 6x y e x y cos 2 xz " 6e x y
z ,2 y e
y
cos 2 xz , 6e x
2
y
sin 2 xz
x
z2
and then type in gŸ=, 1, 2 and click on Evaluate. Here is what you see:
2
2
gŸ=, 1, 2 "6=e = , " 3 = 2 e = , 0
2
2. Directional Derivatives:
Let FŸx, y, z be differentiable. Partial derivatives F , F and F measure the rates of change of
x y
z
FŸx, y, z along the x "axis, y "axis and z "axis or in the direction u ¡1, 0, 0¢, v ¡0, 1, 0¢ and
¡0, 0, 1¢, respectively. How to evaluate the rate of change of F in a given direction w
d?
Definition Let u cosŸ2 , sinŸ2 be a unit vector in xy "plane and let z fŸx, y be a differentiable
function. The directional derivative of f in the direction u is defined as:
fŸx h cosŸ2 , y h sinŸ2 " fŸx, y D u fŸx, y lim
hv0
h
provided the limit exists.
Note that when 2 0, D u fŸx, y f , and when 2 =, D u fŸx, y f .
x
y
Observe that the following.
a.
1
fŸx h cosŸ2 , y h sinŸ2 " fŸx, y h
fŸx h cosŸ2 , y h sinŸ2 " fŸx, y h sinŸ2 fŸx, y h sinŸ2 " fŸx, y h
fŸx h cosŸ2 , y h sinŸ2 " fŸx, y h sinŸ2 fŸx, y h sinŸ2 " fŸx, y h
h
where the first term involves the change of f along the x "axis and the second terms involves the change
U
gŸb " gŸa of f along the y " axis. By the Mean Value Theorem for differentiation (g Ÿc , we
b"a
know there exist x 1 between x and x h cosŸ2 , and y 1 between y and y h sinŸ2 such that
fŸx h cosŸ2 , y h sinŸ2 " fŸx, y f x 1 , y h sinŸ2 Ÿh cosŸ2 f Ÿx, y 1 Ÿh sinŸ2 .
h
x
y
Then as h v 0, x 1 v x and y 1 v y, and
f x 1 , y h sinŸ2 Ÿh cosŸ2 f Ÿx, y 1 Ÿh sinŸ2 x
y
D u fŸx, y lim
hv0
h
f x 1 , y h sinŸ2 ŸcosŸ2 f Ÿx, y 1 ŸsinŸ2 x
y
f x, y ŸcosŸ2 f Ÿx, y ŸsinŸ2 fŸx, y u.
x
y
lim
hv0
u.
In a similar way, we can derive a similar formula for computing D u FŸx, y, z FŸx, y, z b. Let 2 be the angle between vectors u and FŸx, y, z . Then
|| cosŸ2 ||FŸx, y, z ||||u
|| cosŸ2 .
u ||FŸx, y, z ||||u
D u FŸx, y, z FŸx, y, z Since "1 t cosŸ2 t 1,
" ||FŸx, y, z || t D u FŸx, y, z t ||FŸx, y, z ||.
The maximum value of the directional derivative is ||FŸx, y, z || and it occurs when u FŸx, y, z and
the maximum value of the directional derivative is "||FŸx, y, z || and it occurs when u "FŸx, y, z .
So, the gradient vector F points in the direction in which F increases most rapidly whereas "FŸx, y, z points in the direction of the most rapid decrease of F.
x2 " y2
. Find the directional derivatives of F in the directions u i " 2j 3k
z2
and v FŸx, y, z at the point Ÿ2, 4, "1 .
Example Let FŸx, y, z FŸx, y, z 2
2
2x , "2y , "2Ÿx " y z2
z3
z2
D u FŸ2, 4, "1 4, " 8, " 24 Dv FŸ2, 4, "1 4, " 8, " 24 1
¡1, "2, 3¢ " 26 14 "13. 90
7
149
1
4 8 2 24 2
2
, FŸ2, 4, "1 4, " 8, " 24
4, " 8, " 24
4 2 8 2 Ÿ24 2 4 41 25. 61
xy
Example Find a vector that gives the direction in which FŸx, y, z ln z decreases most rapidly at
1, 1, 1 .
2 6 3
xy
The u "F 1 , 1 , 1 is a direction in in which FŸx, y, z ln z decreases most rapidly.
2 6 3
2
1, 1, " 1 ,
z
x y
FŸx, y, z lnŸx lnŸy " lnŸz , FŸx, y, z F 1 , 1 , 1
2 6 3
u "2, " 6, 3 .
2, 6, " 3 , u so that Ÿa D u fŸa, b 0, and
Example Suppose that fŸa, b 2, " 3 . Find a unit vector Ÿb D u fŸa, b is a maximum.
a. Let u cosŸ2 , sinŸ2 . Then
D u fŸa, b fŸa, b u 2 cosŸ2 " 3 sinŸ2 0 ® 2 cosŸ2 3 sinŸ2 , tanŸ2 2 ,
3
2
2
"1 2
"1
"1
, and u cos tan
, sin tan
2 tan
¡0. 832 05, 0. 554 7¢.
3
3
3
b.
u
1
fŸa, b ||fŸa, b ||
1
49
2, " 3
¡0. 554 7, "0. 832 05¢
Example The temperature T at a point Ÿx, y, z in space is inversely proportional to the square of the
distance from Ÿx, y, z to the origin. It is known that TŸ0, 0, 1 500. Find the rate of change of
T at Ÿ2, 3, 3 in the direction of Ÿ3, 1, 1 . In which direction from Ÿ2, 3, 3 does the temperature T
increases most rapidly and what is the maximum rate of change of T?
The function T is
TŸx, y, z C
. TŸ0, 0, 1 C 500 ® C 500.
1
x2 y2 z2
"1000y
"1000x
"1000z
,
,
2
2
2
2 2
2
Ÿx y z Ÿx y 2 z 2 Ÿx 2 y 2 z 2 TŸ2, 3, 3 gŸ2, 3, 3 " 500 , " 750 , " 750
121 121 121
The direction from Ÿ2, 3, 3 in the direction of Ÿ3, 1, 1 is u ¡1, "2, "2¢. So the rate of change is
1
D u TŸ2, 3, 3 " 500 , " 750 , " 750 ¡1, "2, "2¢ 2500 .
121 121 121
363
144
TŸx, y, z gŸx, y, z 2
The temperature T increases most rapidly in the direction v TŸ2, 3, 3 and the maximum rate of change
of T is
Dv TŸ2, 3, 3 ||TŸ2, 3, 3 || PwP 9. 690 94.
Example Find a function f if possible such that fŸx, y 2x y e x
2
y
,
x 2 e x2
2 y
y
f 2x y e x 2 y ® fŸx, y ; 2x y e x 2 y dx e x 2 y CŸy x
f 1 x 2 e x 2 y C U Ÿy x 2 e x 2 y 2y cosŸy 2 2 y
2 y
y
C U Ÿy 2y cosŸy 2 , CŸy fŸx, y e x
3
2
y
sinŸy 2 C
; 2y cosŸy 2 dy sinŸy 2 C
2y cosŸy 2 .