The Directional Derivative   

Let p = ( x0,y0) be fixed and let u = á m,n ñ be a unit vector. Then a vertical slice of z = f( x,y) is the curve formed by the intersection of the vertical plane through the line r(t) = p+ut and the surface z = f( x,y) . In particular, z( t) is the height of the surface z = f(x,y) at the point r( t) .

Since u is a unit vector, the point r( h) is a distance h from r( 0) . Thus, a run of h causes a rise of z( h) -z( 0)

It follows that the slope of the tangent line at r( 0) is

slope =
lim
h® 0 
 z( h) -z( 0)
h
which is simply z' ( 0) . However, r' ( 0) = u and the chain rule yields
z' ( 0) =  dz
dt
ê
ê


t = 0 
= Ñf·r' ( 0) = Ñf·u
That is, the slope of the tangent line is Ñf·u, where u is a unit vector in the direction of r( t) .       

Definition 6.1    The directional derivative of f(x,y) in the direction of a unit vector u is denoted by Duf and is defined to be
Duf = Ñf·u
In particular, Duf is the slope of the tangent line to the curve formed by the intersection of z = f( x,y) and the vertical plane through a point p parallel to a unit vector u.

       

For example, the ordinary partial derivatives are special cases of Duf:
 f
x
= Dif        and        f
y
= Djf
That is, fx yields slopes of vertical slices in the xz-plane and fy yields slopes of vertical slices in the xy-plane.       

EXAMPLE 3    Find the derivative of f( x,y) = 1.1x2 - 0.1xy in the direction of v = á 3,4 ñ.       

Solution: Since v is not a unit vector, we first finds its direction vector:
u =  1
v
v =  1
5
á 3,4 ñ =
 3
5
,  4
5
  = á 0.6, 0.8 ñ  
The gradient of f is Ñf = á 2.2x - 0.1y, -0.1x ñ, so that
Du  f   =   á 2.2x - 0.1y, -0.1x ñ · á 0.6, 0.8 ñ
=   0.6( 2.2x - 0.1y) + 0.8 (-0.1x )
=   1.24x - 0.06y

LiveGraphics3d Applet

  Check your Reading: What is Du f at the point ( 1,0) ?