Gradients in Other Coordinates

Let us let Ñ_uv denote the gradient in the uv-plane and Ñ_xy denote the gradient in the xy-plane under a smooth transformation T( u,v) . For technical reasons, we think of a gradient as a row vector in matrix form, so that by the chain rule we have

Ñuvf = [ fu,fv] = é ë  ¶f ¶x  ¶x ¶u +  ¶f ¶y  ¶y ¶u ,  ¶f ¶x  ¶x ¶v +  ¶f ¶y  ¶y ¶v ù û

In terms of the Jacobian, we have

Ñuvf = [ fx,fy] é ê ë xu xv yu yv ù ú û = ( Ñxyf ) J( u,v)

If we define the inverse Jacobian matrix by

J-1( u,v) = æ è  ¶( x,y) ¶( u,v) ö ø -1   é ê ë yv -xv -yu xu ù ú û

then J^-1( u,v) cancels J( u,v) , so that

( Ñuvf)  J-1( u,v) = ( Ñxyf) J( u,v) J-1( u,v) = Ñxyf

As a result, we have

Ñxyf = æ è  ¶( x,y) ¶(u,v) ö ø -1   [ fu,fv] é ê ë yv -xv -yu xu ù ú û = æ è  ¶( x,y) ¶( u,v) ö ø -1   [ fuyv-fvyu,-fuxv+fvxu]

That is, the gradient of f( u,v) is

Ñf = æ è  ¶( x,y) ¶(u,v) ö ø -1   é ë á yv,-xv ñ  ¶f ¶u + á -yu,xu ñ  ¶f ¶v ù û

(2)

with respect to T( u,v) = á x( u,v), y( u,v) ñ .        

EXAMPLE 5    What is the gradient with respect to

T( u,v) = á u2-v2,2uv ñ

Solution: Since y_v = 2u, x_v = -2v, y_u = 2v, and x_u = 2u,formula (2) becomes

Ñf =  1 4u2+4v2 é ë á 2u,2v ñ  ¶f ¶u + á -2v,2u ñ  ¶f ¶v ù û

using the Jacobian determinant in example 3. Thus,

Ñf =  1 2u2+2v2  ¶f ¶u  á u,v ñ  +   1 2u2+2v2  ¶f ¶v  á -v,u ñ

It follows that (2) provides a formula for finding normal vectors to the images under T( u,v) of level curves of g( u,v) .
That is, the Jacobian also maps normal vectors to curves to normal vectors to the images of those curves.    

EXAMPLE 6    What is the gradient in polar coordinates? What is a normal to the level curve rsec( q) = 2 at q = p/4?    

Solution: In polar coordinates, formula (2) becomes

Ñg = æ è  ¶( x,y) ¶( r,q) ö ø -1   é ë á yq,-xq ñ  ¶g ¶r + á-yr,xr ñ  ¶g ¶q ù û

Since x = rcos( q) and y = rsin( q) , we have

yq = rcos( q) , xq = -rsin(q) , yr = sin( q) ,  xr = cos(q)

and we found the Jacobian determinant in example 4. Thus,

Ñg = æ è  ¶( x,y) ¶(r,q) ö ø -1   é ë á rcos( q) ,rsin( q) ñ  ¶g¶r + á -sin( q) ,cos( q) ñ  ¶g ¶q ù û = ( r) -1 é ë r á cos( q),sin( q) ñ  ¶g ¶r + á -sin( q) ,cos( q) ñ  ¶g ¶q ù û =  ¶g¶r á cos( q),sin( q) ñ +  1 r  ¶g ¶q á -sin( q) ,cos(q) ñ

However, e_r = á cos( q) ,sin( q) ñ and e_q = á-sin( q) ,cos( q) ñ , so that

Ñg =  ¶g ¶r er+  1 r  ¶g ¶q eq

If g(r,q) = rsec( q) , then g_r = sec( q) and g_q = rsec(q) tan( q) , so that

Ñg = sec( q) er+sec( q) tan( q) eq

The level curve rsec( q) = 2 is the same as r = 2cos( q) , r ¹ 0, which is a circle of radius 1 centered at ( 1,0) . At q = p/4, we have r = Ö2 and

Ñg æ è Ö2,  p 4 ö ø = sec æ è  p 4 ö ø er æ è  p 4 ö ø  + sec æ è  p 4 ö ø tan æ è  p 4 ö ø eq æ è  p 4 ö ø

Since sec( p/4) = Ö2, tan( p/4) = 1, e_r( p/4) = á 1/Ö2, 1/Ö2 ñ and e_q = á -1/Ö2, 1/Ö2 ñ , the result is

Ñg æ è Ö2,  p 4 ö ø = á 1,1 ñ+ á -1,1 ñ = á 0,2 ñ

which is shown along with the circle in the figure below: