```Math 280 Supplement to Friday, 5 Feb 99.

The next homework due on 10feb is p.129 ##1,2,10,11.

On the Inverse Function Theorem and Applications.

Suppose we apply the IFT to a special case that
F(x,u) := x - f(u), in other words, that we wish
to solve x = f(u) (i.e. where F(x,y)=0) for u as
a function of x. Since F_x = I (the identity matrix)
and F_u = - f'(u) (the Jacobian matrix of all partials
\dx/\du) we need only to check that det f'(u) != 0.

For the f'(u) is invertible, and, by the IFT, so is
f at least in a neighborhood of the point in question.
Thus (whether you can find it explicitly or have to
use numerical approximations), there is a differentiable
function u = g(x), for which the two items predicted by
the IFT are true:
-1
(i) 0 = F(x, g(x)) => x = f(g(x)) i.e. g = f  .

-1                        -1
(ii) g'(x)dx = du = -(F_u)  (F_x)dx => g'(x) = f'(u)  .

In particular, since the Jacobian of the inverse is the
(matrix) inverse of the Jacobian, the Jacobian determinant
of the inverse is the reciprocal of the determinant of the
Jacobian. This all is most useful when dealing with two
or more coordinate systems on the same geometrical space.

We will have plenty of opportunities to use the IFT in
the context of changing coordinates. It is impractical to
dwell on the details here and now.

When x = f(u) but dim x != dim u the machinery of differentials,
Jacobian matrices and determinants, and the IFT can also be applied
but in a more circumspect manner. When you are not preserving
the dimension, it matters first of all if dim x > dim u or
if dim x < dim u. (In the latter case, we prefer to write the
functional relation as u = f(x), so that x , or rather _x,
is generically a point or vector in 3-space. Secondly, it
matters whether the smaller dimension is 1 or 2.

Case 1. _x = (x,y,z) and dim u = 1.
Here we speak of a parametrized curve, the parameter being u,
and a point _x moving along that curve. We thus are dealing with three
functions of one variable, often as not called time t. The displacement
along the curve measured by d _x = _v dt, where _v = (x',y',z') is
the velocity vector of the moving point, is a set of three linear
equations in 4 variables, and hance one of the, dt, can be "eliminated"
by "solving" for in terms ot the others.  I.e. dx:dy:dz= x':y':z' or

dx    dy   dz        x - x_0   y - y_0   x - y_0
dt = --  = -- = --    =>  ------- = ------- = -------
x'    y'   z'          x'         y'       z'

Thus, we have derived the equation of the tangent line to a curve at a point.

Case 2.  u = f(x,y,z) and dim u =1.
Here we speace of a scalar field which measures a real number
at each point in space. The locus of points where u = u_0 is constant,
is a level-surface for f. From du = f_x dx + f_y dy +f_z dz we observe
that the Jacobian f' = [f_x,f_y,f_z] is the transpose of the gradient

0 =  du =  f_x dx + f_y dy +f_z dz = [f_x,f_y,f_z]|dx|
|dy|                                                                            |dz|

on the surface says that the gradient is perpendicular to every
displacement (dx,dy,dz) which is "along" (i.e.  tangent to) the surface.

Therefore, the equation of the tangent plane immediately follows as:

f_x(x_0,y_0,z_0)(x-x_0) + (f_y)(y-y_0) + (f_z)(z-z_0) = 0.

Note, that it may be an old fashioned, but still very practical trick of
the mind to consider the tangent of a curve to be the line through two
"consecutive" points on a curve. Then an infinitesimal displacement on
a curve or surface is a line through two consecutive points on the curve
or surface, hence has to be tangent to it.

Note that if the surface in a nhd of a point is not tangent to a principal
direction, e.g.  if f_z != 0, the we can solve

dz = - (f_x/f_z)dx - (f_y/f_z) dy

and the IFT says the surface is, locally, the graph z = g(x,y). Although
we may never be able to solve for g explicitly, we already know its partials,

g_x = - (f_x/f_z) , g_y - (f_y/f_z).

If you understand the above discussion the every explicit problem becomes
a matter of "expanding" the compressed notation to fit the context of the
problem. It is therefore more efficient to remember the compressed notation,
and learn how to unravel it as needed.

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