Mathematical Tools for Physics

8—Multivariable Calculus 216

8.5 Gradient
The equation ( 5 ) for the differential has another geometric interpretation. For a function such asf(x,y) =
x^2 + 4y^2 , the equations representing constant values offdescribe curves in thex-yplane. In this example, they
are ellipses. If you start from any fixed point in the plane and start to move away from it, the rate at which
the value offchanges will depend on the direction in which you move. If you move along the curve defined by
f=constant thenfwon’t change at all. If you move perpendicular to that direction thenfmay change a lot.

The gradient offat a point is the vector pointing in the direction in which

f is increasing most rapidly and the component of the gradient along that

direction is the derivative offwith respect to the distance in that direction.

To relate this to the partial derivatives that we’ve been using, and to understand how to compute and to
use the gradient, return to Eq. ( 5 ) and write it in vector form. Use the common notation for the basis:ˆxandˆy.
Then let

d~r=dxˆx+dyyˆ and G~=

(

∂f

∂x

)

y

ˆx+

(

∂f

∂y

)

x

ˆy (9)

The equation for the differential is now

df=df(x,y,dx,dy) =G~.d~r (10)

G~

d~r

Because you know the properties of the dot product, you know that this isGdrcosθand it is largest when
the directions ofd~rand ofG~are the same. It’s zero when they are perpendicular. You also know thatdfis zero
whend~ris in the direction along the curve wherefis constant. The vectorG~is therefore perpendicular to this
curve. It is in the direction in whichfis changing most rapidly. Also becausedf=Gdrcos 0, you see thatGis
the derivative offwith respect to distance along that direction.G~is the gradient.