CHAPTER 43
Directional Derivatives and the
Gradient. Extreme Values
43.1
Given a function/(je, y) and a unit vector u, define the derivative of/in the direction u, at a point (*„, y 0 ), and
state its connection with the gradient Vf = (f x , / v ).
The ray in the direction u starting at the point (* 0 , y a ) is (x 0 , y 0 ) + tu (t > 0). The (directional) derivative, at
(x 0 , y 0 ), of the function / in the direction u is the rate of change, at (x a , y 0 ), of / along that ray, that is,
— /((*„, y 0 ) + tu), evaluated at t = 0. It is equal to Vf • u, the scalar projection of the gradient on u. (Similar
definitions and results apply for functions / of three or more variables.)
43.2
Show that the direction of the gradient Vfis the direction in which the derivative achieves its maximum, \Vf\, and
the direction of — Vf is the direction in which the derivative achieves its minimum, — \Vf\.
The derivative in the direction of a unit vector u is Vf • u = \Vf \ cos 0, where 6 is the angle between Vf and
u. Since cos 6 takes on its maximum value 1 when 0=0, the maximum value of Vf-v is obtained when u
is the unit vector in the direction of Vf. That maximum value is |V/|. Similarly, since cos 0 takes on its minimum
value —1 when 0 = rr, the minimum value of Vf-u is attained when u has the direction of —Vf. That
minimum value is — \Vf\.
43.3
Find the derivative of f(x, y) = 2x
2 - 3xy + 5y
2
at the point (1, 2) in the direction of the unit vector u making
an angle of 45° with the positive ;t-axis.
Hence, the derivative is V/-u = (-2,17)-(V2/2,V2/2) =
u = (cos45°, sin 45°) = (V2/2, V2/2). Vf = (4x -3y, -3x + lOy) = (-2,17) at (1,2).
43.4
Find the derivative of f(x, y) = x — sin xy at (1,77/2) in the direction of u =
V/=(l -ycosxy, -x cos xy) = (1,0). Hence, the derivative is V/-u= |(1,0)-(1, V5) = f.
43.5
Find the derivative of f(x, y) = xy
2
at (1, 3) in the direction toward (4, 5).
The indicated direction is that of the vector (4,5) - (1, 3) = (3, 2). The unit vector in that direction is
u = (l/vT3) (3,2); the gradient Vf=(y
2 ,2xy) = (9,6). So, V/-u = (9,6)43.6
Find the derivative of f(x, y, z) = X
2 y
2 z at (2,1,4) in the direction of the vector (1,2,2).
The unit vector in the indicated direction is u= 1(1,2,2). Vf= (2xy
2 z, 2x
2 yz, x
2 y
2 ) = (16,32, 64) =
16(1, 2,4). Hence, the derivative is Vf- u = 16(1,2,4) • $(1, 2,2) = ^(13) = ^.
43.7
Find the derivative of f(x, y, z) = x* + y
3 z at (-1,2,1) in the direction toward (0, 3, 3).
A vector in the indicated direction is (0,3,3) - (—1,2,1) = (1,1, 2). The unit vector in that direction is
Hence, is
43.8
On a hill represented by z = 8 - 4x
2 - 2y
2 , find the direction of the steepest grade at (1,1,2).
In view of Problem 43.2, we wish to find a vector v in the tangent plane to the surface at (1,1, 2) such that the
perpendicular projection of v onto the xy-plane is Vz = (-8*, -4y) = (-8, -4). Thus we must have v =
(-8, -4, c); and the component c may be determined from the orthogonality of v and the surface normal found
in Problem 42.105:
(-8, -4, c)-(-8, -4, -1) = 0
or
c = 80
Note that the direction of v is the direction of steepest ascent at (1,1, 2).
392
Directional Derivatives and the
Gradient. Extreme Values
43.1
Given a function/(je, y) and a unit vector u, define the derivative of/in the direction u, at a point (*„, y 0 ), and
state its connection with the gradient Vf = (f x , / v ).
The ray in the direction u starting at the point (* 0 , y a ) is (x 0 , y 0 ) + tu (t > 0). The (directional) derivative, at
(x 0 , y 0 ), of the function / in the direction u is the rate of change, at (x a , y 0 ), of / along that ray, that is,
— /((*„, y 0 ) + tu), evaluated at t = 0. It is equal to Vf • u, the scalar projection of the gradient on u. (Similar
definitions and results apply for functions / of three or more variables.)
43.2
Show that the direction of the gradient Vfis the direction in which the derivative achieves its maximum, \Vf\, and
the direction of — Vf is the direction in which the derivative achieves its minimum, — \Vf\.
The derivative in the direction of a unit vector u is Vf • u = \Vf \ cos 0, where 6 is the angle between Vf and
u. Since cos 6 takes on its maximum value 1 when 0=0, the maximum value of Vf-v is obtained when u
is the unit vector in the direction of Vf. That maximum value is |V/|. Similarly, since cos 0 takes on its minimum
value —1 when 0 = rr, the minimum value of Vf-u is attained when u has the direction of —Vf. That
minimum value is — \Vf\.
43.3
Find the derivative of f(x, y) = 2x
2 - 3xy + 5y
2
at the point (1, 2) in the direction of the unit vector u making
an angle of 45° with the positive ;t-axis.
Hence, the derivative is V/-u = (-2,17)-(V2/2,V2/2) =
u = (cos45°, sin 45°) = (V2/2, V2/2). Vf = (4x -3y, -3x + lOy) = (-2,17) at (1,2).
43.4
Find the derivative of f(x, y) = x — sin xy at (1,77/2) in the direction of u =
V/=(l -ycosxy, -x cos xy) = (1,0). Hence, the derivative is V/-u= |(1,0)-(1, V5) = f.
43.5
Find the derivative of f(x, y) = xy
2
at (1, 3) in the direction toward (4, 5).
The indicated direction is that of the vector (4,5) - (1, 3) = (3, 2). The unit vector in that direction is
u = (l/vT3) (3,2); the gradient Vf=(y
2 ,2xy) = (9,6). So, V/-u = (9,6)43.6
Find the derivative of f(x, y, z) = X
2 y
2 z at (2,1,4) in the direction of the vector (1,2,2).
The unit vector in the indicated direction is u= 1(1,2,2). Vf= (2xy
2 z, 2x
2 yz, x
2 y
2 ) = (16,32, 64) =
16(1, 2,4). Hence, the derivative is Vf- u = 16(1,2,4) • $(1, 2,2) = ^(13) = ^.
43.7
Find the derivative of f(x, y, z) = x* + y
3 z at (-1,2,1) in the direction toward (0, 3, 3).
A vector in the indicated direction is (0,3,3) - (—1,2,1) = (1,1, 2). The unit vector in that direction is
Hence, is
43.8
On a hill represented by z = 8 - 4x
2 - 2y
2 , find the direction of the steepest grade at (1,1,2).
In view of Problem 43.2, we wish to find a vector v in the tangent plane to the surface at (1,1, 2) such that the
perpendicular projection of v onto the xy-plane is Vz = (-8*, -4y) = (-8, -4). Thus we must have v =
(-8, -4, c); and the component c may be determined from the orthogonality of v and the surface normal found
in Problem 42.105:
(-8, -4, c)-(-8, -4, -1) = 0
or
c = 80
Note that the direction of v is the direction of steepest ascent at (1,1, 2).
392