On Monday 1feb99
Two homework sets were returned.
New homework was assigned, due Friday, 5feb:
P118 2, 3, 4, and (7) (8) for extra credit.
Old Business:
To cwoo@students.uiuc.edu
From gfrancis Sat Jan 30 20:10:15 1999
Subject: p101prob8
Christine
You wrote
>I was wondering if you could give help me out with problem 8 on page 101.
Let me use the notation C=cosh v and S=sinch v so that
C'=S and S'=C and C^2-S^2 = 1.
Note the similarity of the hyperbolic trig functions with the usual ones.
The left hand side of the equation to be verified requires no expansion.
For the right hand side, apply the chain rule to compute the partial w_u.
Namely
w_u = w_x x_u + w_y y_u = w_x C + w_y S
but when you derive by v there is a u-term in each
w_v = w_x x_v + w_y y_v = w_x u S + w_y u C.
When you do the algebra of expanding the RHS you'll get the LHS.
New Business
This week we cover sections 2.10, 2.12, and 2.13. In order to do this
we must simplify the essentials by using our differential notation and
you must learn how to expand the main formulas in various abstract and
concrete cases to see the correlation with the textbook.
For example, 2.10 in a nutshell looks like this:
Webnotation: You should write out on paper what I have to encode here
because .html does not yet support mathematical notation. Thus \d stands
for the partial symbol.
On the locus where F(x,u)= const the displacement must also vanish
\dF \dF
0 = dF = --- dx + --- du
\dx \du
Solving for du by multiplying by the inverse of a matrix
|\dF|-1 |\dF| |\dF|-1 |\dF|
du = - |---| |---| dx or <<*>> du + |---| |---| dx
|\du| |\dx| |\du| |\dx|
for which to make sense the matrix F_u must be square and invertible.
THEOREM
If, at a point (x*,u*) on 0=F(x*,u*), the Jacobian matrix of
partials F_u is square and invertible, then there is a function f(x) for
which u=f(x) satisfies F(x,f(x))=0, at least in neighborhood of (x*).
MOREOVER
Since du = f_x dx we get dim(x)*dim(u) differential equations by setting
the two matrices equal:
\df(x) |\dF|-1 |\dF|
----- = - |---| |---|
\dx |\du| |\dx|
Now, in a particular case, you have to expand this notation to see what
you need to check, what you need to compute, and what equations you must
solve to actually compute u=f(x), the implicitly defined function.
We did most of Problem 4 assigned above in class.