15.6 Directional Derivatives and the Gradient Vector
Definition (Directional derivative)
Given a function f (x, y) and a unit vector u =< a, b >, we
define the directional derivative of f at v = (xo , yo ) in the
direction u as
Du f (xo , yo )
=
lim
h→0
f (v + hu) − f (v)
h
f (xo + ha, yo + hb) − f (xo , yo )
h→0
h
=
lim
provided that the limit exists.
Theorem If f is differentiable at (xo , yo ), then f has the directional derivative in the direction of any unit vector
u =< a, b > at (xo , yo ) and
Du f (xo , yo )
= fx (xo , yo )a + fy (xo , yo )b
= < fx (xo , yo ), fy (xo , yo ) > · < a, b >
(Proof) Let φ(h) = f (xo + ha, yo + hb). Then
Du f (xo , yo )
=
lim
h→0
φ(h) − φ(0)
h
=
φ0 (0)
=
a
=
afx (xo , yo ) + bfy (xo , yo )
∂f
∂f
(xo + ha, yo + hb) + b
(xo + ha, yo + hb)
∂x
∂y
h=0
Definition (The gradient vector ∇f ) Given a function z = f (x, y), we define the gradient of f to be the vector
∇f =< fx (x, y), fy (x, y) >= fx i + fy j
Property Du f (x, y) = ∇f (x, y) · u
Definition (Directional derivative and gradient vector of a function with three variables)
Consider a function f (x, y, z).
(a) When u =< a, b, c > is a unit vector, we define the directional derivative of f at v =< xo , yo , zo > to
be
f (v + hu) − f (v)
f (xo + ah, yo + bh, zo + ch) − f (xo , yo , zo )
= lim
h→0
h→0
h
h
lim
provided that the limit exists.
(b) We also define the gradient vector of f by
1
∇f =< fx , fy , fz >= fx i + fy j + fz k
Property Du f (xo , yo , zo ) = ∇f (xo , yo , zo ) · u
Example Find the directional derivative of f at the given point in the indicated direction.
(a) f (x, y) = x sin(xy) ;
(2, 0), in the direction θ = π/3
(b) f (x, y) = ln(x2 + xy + y 2 ) ;
(2, 1), in the direction v =< −1, 2 >
(Answers)
(a)

 ∇f =< sin(xy) + xy cos(xy), x2 cos(xy) >
:
 u =< cos π/3, sin π/3 >
:



 ∇f =
2x + y
x + 2y
,
2 x2 + xy + y 2
+
xy
+
y
(b)


√
√

u =< −1/ 5, 2/ 5 >
x2
∇f (2, 0) =< 0, 4 >
√
1
< 1, 3 >
2
√
Du f (2, 0) = ∇f (2, 0) · u = 2 3
u=
:
∇f (2, 1) = h5/7, 4/7i
Du f (2, 1) = ∇f (2, 1) · u =
3
√
7 5
Theorem Suppose that f is a differentiable function with two(or three) variables. Let x be a point in the domain of
f and u be a unit vector in R2 (or R3 ).
(a) Then
|Du f (x)| = |∇f (x) · u| = k∇f (x)k| cos θ| ≤ k∇f (x)k
where θ is the angle between ∇f (x) and u.
(b) Particularly, notice that Du f (x) is maximized when ∇f (x) and u are in the same direction and the
maximum value is k∇f (x)k.
Example Find the maximum rate of change of f and the direction in which it occurs at the given point.
(a) f (p, q) = qe−p + pe−q , (0, 0)
(b) f (x, y, z) = tan−1 (x + 2y + 3z), (1, 1, 1)
(Answers) At (a,b), maximum rate occurs in the direction of ∇f (a, b) and the maximum rate is k∇f (a, b)k.
(a) ∇f =< fp , fq >=< −qe−p + e−q , e−p − pe−q > ; ∇f (0, 0) =< 1, 1 >.
√
Maximum rate is 2 and it occurs in the direction of ∇f (0, 0) =< 1, 1 >
1
2
3
,
,
1 + (x + 2y + 3z)2 1 + (x + 2y + 3z)2 1 + (x + 2y + 3z)2
√
Maximum rate is 14/37 and it occurs in the direction of < 1, 2, 3 >
(b) ∇f =
;
∇f (1, 1, 1) =< 1/37, 2/37, 3/37 >.
Example Find all points at which the direction of the fastest change of the function f (x, y) = x2 + y 2 − 2x − 4y is
< 1, 1 >.
(Answer) Look for (x, y) such that < fx , fy > / / < 1, 1 >. From < 2x − 2, 2y − 4 >= k < 1, 1 >, 2x − 2 = k = 2y − 4, and the answer is
{(x, y) : x − y = −1}.
Theorem
(a) Given a function y = f (x), let g(x, y) = f (x) − y. Then
2
∇g(xo , yo ) ⊥ Tangent line to y = f (x) at (xo , yo )
(b) Given a function z = f (x, y), let g(x, y, z) = f (x, y) − y. Then
∇g(xo , yo , zo ) ⊥ Tangent plane to z = f (x, y) at (xo , yo , zo )
(Proofs)
(a) The slope of the tangent line to y = f (x) at (xo , yo ) is f 0 (xo ). Hence the vector < 1, f 0 (xo ) > is parallel to the tangent line. On
the other hand, ∇g(xo , yo ) =< gx (xo , yo ), gy (xo , yo ) >=< f 0 (xo ), −1 >. Because < 1, f 0 (xo ) > · < f 0 (xo ), −1 >= 0, (a) is
proven.
(b) It suffices to show that ∇g(xo , yo , zo ) is perpendicular to any curve on the surface z = f (x, y) through (xo , yo , zo ). Suppose
< x(t), y(t), z(t) > is a curve on the surface z = f (x, y) such that (xo , yo , zo ) = (x(to ), y(to ), z(to )).
⇒ z(t) = f (x(t), y(t))
⇒ g(x(t), y(t), z(t)) = f (x(t), y(t)) − z(t) = 0
dg
=
dt
∂g
⇒
,
∂x
∂g
⇒
,
∂x
⇒
∂g dx
∂g dy
∂g dz
+
+
=0
∂x dt
∂y dt
∂z dt
∂g ∂g
dx dy dz
,
·
,
,
=0
∂y ∂z
dt dt dt
∂g ∂g
dx dy dz
,
⊥
,
,
That is, ∇g(xo , yo , zo ) ⊥ < x0 (to ), y 0 (to ), z 0 (to ) >.
∂y ∂z
dt dt dt
Example Use gradient vector to find (a)an equation of the tangent plane and (b)an equation of the normal line to
the surface
x − z = 4 arcsin(yz)
at (x, y, z) = (1, 0, 1)
(Answers) Let g(x, y, z) = 4 arcsin(yz) − x + z. Then the vector ∇g(1, 0, 1) =< −1, 4, 1 > is perpendicular to the tangent plane to the surface
x − z = 4 arcsin(yz) at (1, 0, 1). Hence
(a) tangent plane : −1(x − 1) + 4(y − 0) + (z − 1) = 0
x−1
y−0
z−1
(b) normal line :
=
=
−1
4
1
3