The Jacobian of a Transformation

In this section, we explore the concept of a "derivative" of a coordinate transformation, which is known as the Jacobian of the transformation. However, in this course, it is the determinant of the Jacobian that will be used most frequently.

If we let u = á u,v ñ , p = á p,q ñ , and x = áx,y ñ , then ( x,y) = T( u,v) is given in vector notation by

x = T( u)

This notation allows us to extend the concept of a total derivative to the total derivative of a coordinate transformation.    

Definition 5.1: A coordinate transformation T( u) is differentiable at a point p if there exists a matrix J( p) for which

lim u® p     || T( u) -T( p) -J( p) ( u-p) || || u-p || = 0

(1)

When it exists, J( p) is the total derivative of T( u) at p.

In non-vector notation, definition 5.1 says that the total derivative at a point ( p,q) of a coordinate transformation T( u,v) is a matrix J( u,v) evaluated at (p,q) .  In a manner analogous to that in section 2-5, it can be shown that this matrix is given by

J( u,v) = é ê ë   xu xv yu yv ù ú û

(see exercise 46). The total derivative is also known as the Jacobian Matrix of the transformation T( u,v) .        

EXAMPLE 1    What is the Jacobian matrix for the polar coordinate transformation?

   

Solution: Since x = rcos( q) and y = rsin( q) , the Jacobian matrix is

J( r,q) = é ê ë   xr xq yr yq ù ú û = é ê ë   cos( q) -rsin( q) sin( q) rcos( q) ù ú û

If u( t) = á u( t),v( t) ñ is a curve in the uv-plane, then x( t) = T( u( t) ,v( t)) is the image of u( t) in the xy-plane. Moreover,

dx dt = é ê ê ê ê ê ë      dx dt      dy dt   ù ú ú ú ú ú û = é ê ê ê ê ê ë   xu  du dt +xv  dv dt   yu  du dt +yv  dv dt   ù ú ú ú ú ú û = é ê ë   xu xv yu yv ù ú û   é ê ê ê ê ê ë      du dt      dv dt   ù ú ú ú ú ú û

The last vector is du/dt. Thus, we have shown that if x( t) = T( u( t) ) , then

dx dt = J( u)  du dt

That is, the Jacobian maps tangent vectors to curves in the uv-plane to tangent vectors to curves in the xy-plane.
In general, the Jacobian maps any tangent vector to a curve at a given point to a tangent vector to the image of the curve at the image of the point.        

EXAMPLE 2    Let T( u,v) = áu²-v²,2uv ñ

    a) Find the velocity of u( t) = át,t² ñ when t = 1.

    b) Find the Jacobian and apply it to the vector in a)

    c) Find x( t) = T( u( t)) in the xy-plane and then find its velocity vector at t = 1. Compare to the result in (b).        

Solution: a) Since u' ( t) = á 1,2t ñ , the velocity at t = 1 is u' ( 1) = á 1,2 ñ .

b) Since x( u,v) = u²-v² and y( u,v) = 2uv, the Jacobian of T( u,v) is

J( u,v) = é ê ë xu xv yu yv ù ú û = é ê ë 2u -2v 2v 2u ù ú û

which at the point ( 1,1) is given by

J( 1,1) = é ê ë 2 -2 2 2 ù ú û

Identifying u' ( 1) = á1,2 ñ with [ 1,2] ^t leads to

J( 1,1) u' ( 1) = é ê ë 2 -2 2 2 ù ú û é ê ë 1 2 ù ú û = é ê ë -2 6 ù ú û

c) Substituting u = t, v = t² into T( u,v) = áu²-v²,2uv ñ results in

x( t) = ( t2-t4,2t3)

which has a velocity of x' ( t) = á2t-4t³,6t² ñ . Moreover, x' (1) = á -2,6 ñ