Computing the gradient in polar coordinates using the Chain rule

Suppose we are given g(x, y), a function of two variables. If (r, θ) are the usual

polar coordinates related to (x,y) by x = r cos θ, y = r sin θ then by substituting

these formulas for x, y, g “becomes a function of r, θ”, i.e g(x, y) = f(r, θ). We want

to compute ∇g in terms of fr and fθ. We have to be more precise since ∇g is a vector

at each point (x, y). We introduce the unit vectors er in the r direction and eθ in the

θ direction which are orthogonal and positively oriented. That is,

(0.1) er =< cos θ, sin θ >, eθ =< − sin θ, cos θ >

in terms of the standard basis vectors i, j of cartesian coordinates. Then we want to

write

(0.2) ∇g = a(r, θ)er + b(r, θ)eθ ,

where a and b are computed in terms of fr and fθ. Notice that the positively oriented

basis vectors er, eθ vary from point to point.

We use the chain rule:

gx = frrx + fθθx,(0.3)

gy = frry + fθθy .(0.4)

It remains to compute rx, ry, θx, θy. Using r2 = x2 + y2, we find rx = xr

= cos θ

and similarly ry = sin θ. Likewise using θ = arctan yx, we find θx = − y

x2+y2= sin θ

rand

θy = xx2+y2

= cos θr

. Putting everything together gives:

gx = cos θfr − sin θfθr,(0.5)

gy = sin θfr + cos θfθr.(0.6)

This last equation means that ∇g = frer + fθreθ . (Note carefully that ∇g is not

equal to frer + fθeθ .)

1