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 = (fx, / v ).
The ray in the direction u starting at the point (*0, ya) is (x0, y 0 ) + tu (t > 0). The (directional) derivative, at
(x0, y 0 ), of the function / in the direction u is the rate of change, at (xa, y0), of / along that ray, that is,
— /((*„, y0) + 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) = 2x2 - 3xy + 5y2 at the point (1, 2) in the direction of the unit vector u making
an angle of 45° with the positive ;t-axis.
u = (cos45°, sin 45°) = (V2/2, V2/2). Vf = (4x -3y, -3x + lOy) = (-2,17) at (1,2). Hence, the deriva-
tive is V/-u = (-2,17)-(V2/2,V2/2) =
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) = xy2 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=(y2,2xy) = (9,6). So, V/-u = (9,6)-
43.6 Find the derivative of f ( x , y, z) = X2y2z 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= (2xy2z, 2x2yz, x2y2) = (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* + y3z 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 - 4x2 - 2y2, 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).
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43.9 For the surface of Problem 43.8, find the direction of the level curve through (1,1) and show that it is
perpendicular to the direction of steepest grade.
The level curve z = 2 is 8 - 4x2 - 2v 2 = 2, or 2x2 + y2 = 3. By implicit differentiation, 4jc +
dyldx = -2x/y = -2. Hence, a tangent vector is (1, -2,0). A vector in the direction of steepest
grade is (-8,-4,80), by Problem 43.8. Because (-8,-4, 80) •(!,-2,0) = 0, the two directions are perpen-
dicular.
43.10 Show that the sum of the squares of the directional derivatives of z =/(*, y) at any point is constant for any
two mutually perpendicular directions and is equal to the square of the gradient.
This is simply the Pythagorean theorem: If u and v are mutually perpendicular unit vectors in the plane, then
Problem 33.4 shows that Vf=(Vf-u)u + ( V f - v ) v . A direct computation of Vf'Vf then gives |V/|2 =
(Vf-u)2 + (Vf-v)2.
43.11 Find the derivative of z = x In v at the point (1, 2) in the direction making an angle of 30° with the positive
x-axis.
The unit vector in the given direction is
So, the derivative is
43.12 If the electric potential V at any point (x, y) is V= In find the rate of change of V at (3,4) in the
direction toward the point (2,6).
A vector in the given direction is (—1,2), and a corresponding unit vector is
Hence, the rate of change of V in the specified direction is
43.13 If the temperature is given by f ( x , y, z) = 3x2 - 5y2 + 2z2 and you are located at ( 3 , |, |) and want to get
cool as soon as possible, in which direction should you set out?
V/= (6*, -Wy, 4z) = (2, -2, 2) = 2(1, -1,1). The direction in which/decreases the most rapidly is that of
-V/; thus, you should move in the direction of the vector (-1,1, -1).
43.14 Prove that the gradient VFof a function F(x, y, z) at a point P(x0, y0, z 0 ) is perpendicular to the level surface of F
going through P.
Let F(x0, y0, z 0 ) = k. Then the level surface through P is F(x, y, z) = k. By Problem 42.109, a normal
vector to that surface is (Fx, Fy, F2), which is just VF.
43.15 In what direction should one initially travel, starting at the origin, to obtain the most rapid rate of increase of the
function f ( x , y, z) = (3 - x + y)2 + (4x - y + z + 2)3?
The appropriate direction is that of V/= (/,, fy, f:) = (2(3 - x + y)(-l) + 3(4* - y + z + 2)2(4), 2(3 -
x + y) + 3(4* -y + z + 2)2(-1), 3(4* - v + z + 2)2) = (42, -6,12) = 6(7, -1,2).
43.16 If fix, y, z) = jc3 + y 3 - z, find the rate of change of/at the point (1,1, 2) along the line
in the direction of decreasing x.
A vector parallel to the given line is (3,2, -2). Since the first component, 3, is positive, the vector is pointing
in the direction of increasing x. Hence, we want the opposite vector (-3, -2,2). A unit vector in that direction
is Hence, the required rate of change is Since
we get
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 = (fx, / v ).
The ray in the direction u starting at the point (*0, ya) is (x0, y 0 ) + tu (t > 0). The (directional) derivative, at
(x0, y 0 ), of the function / in the direction u is the rate of change, at (xa, y0), of / along that ray, that is,
— /((*„, y0) + 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) = 2x2 - 3xy + 5y2 at the point (1, 2) in the direction of the unit vector u making
an angle of 45° with the positive ;t-axis.
u = (cos45°, sin 45°) = (V2/2, V2/2). Vf = (4x -3y, -3x + lOy) = (-2,17) at (1,2). Hence, the deriva-
tive is V/-u = (-2,17)-(V2/2,V2/2) =
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) = xy2 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=(y2,2xy) = (9,6). So, V/-u = (9,6)-
43.6 Find the derivative of f ( x , y, z) = X2y2z 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= (2xy2z, 2x2yz, x2y2) = (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* + y3z 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 - 4x2 - 2y2, 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).
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43.9 For the surface of Problem 43.8, find the direction of the level curve through (1,1) and show that it is
perpendicular to the direction of steepest grade.
The level curve z = 2 is 8 - 4x2 - 2v 2 = 2, or 2x2 + y2 = 3. By implicit differentiation, 4jc +
dyldx = -2x/y = -2. Hence, a tangent vector is (1, -2,0). A vector in the direction of steepest
grade is (-8,-4,80), by Problem 43.8. Because (-8,-4, 80) •(!,-2,0) = 0, the two directions are perpen-
dicular.
43.10 Show that the sum of the squares of the directional derivatives of z =/(*, y) at any point is constant for any
two mutually perpendicular directions and is equal to the square of the gradient.
This is simply the Pythagorean theorem: If u and v are mutually perpendicular unit vectors in the plane, then
Problem 33.4 shows that Vf=(Vf-u)u + ( V f - v ) v . A direct computation of Vf'Vf then gives |V/|2 =
(Vf-u)2 + (Vf-v)2.
43.11 Find the derivative of z = x In v at the point (1, 2) in the direction making an angle of 30° with the positive
x-axis.
The unit vector in the given direction is
So, the derivative is
43.12 If the electric potential V at any point (x, y) is V= In find the rate of change of V at (3,4) in the
direction toward the point (2,6).
A vector in the given direction is (—1,2), and a corresponding unit vector is
Hence, the rate of change of V in the specified direction is
43.13 If the temperature is given by f ( x , y, z) = 3x2 - 5y2 + 2z2 and you are located at ( 3 , |, |) and want to get
cool as soon as possible, in which direction should you set out?
V/= (6*, -Wy, 4z) = (2, -2, 2) = 2(1, -1,1). The direction in which/decreases the most rapidly is that of
-V/; thus, you should move in the direction of the vector (-1,1, -1).
43.14 Prove that the gradient VFof a function F(x, y, z) at a point P(x0, y0, z 0 ) is perpendicular to the level surface of F
going through P.
Let F(x0, y0, z 0 ) = k. Then the level surface through P is F(x, y, z) = k. By Problem 42.109, a normal
vector to that surface is (Fx, Fy, F2), which is just VF.
43.15 In what direction should one initially travel, starting at the origin, to obtain the most rapid rate of increase of the
function f ( x , y, z) = (3 - x + y)2 + (4x - y + z + 2)3?
The appropriate direction is that of V/= (/,, fy, f:) = (2(3 - x + y)(-l) + 3(4* - y + z + 2)2(4), 2(3 -
x + y) + 3(4* -y + z + 2)2(-1), 3(4* - v + z + 2)2) = (42, -6,12) = 6(7, -1,2).
43.16 If fix, y, z) = jc3 + y 3 - z, find the rate of change of/at the point (1,1, 2) along the line
in the direction of decreasing x.
A vector parallel to the given line is (3,2, -2). Since the first component, 3, is positive, the vector is pointing
in the direction of increasing x. Hence, we want the opposite vector (-3, -2,2). A unit vector in that direction
is Hence, the required rate of change is Since
we get
