CALCULUS 265 Honors
Fall Semester 2000
8:00 to 8:50 MTTF
Carver 0002

Instructor: Hentzel
Office: 432 Carver
Office Hours: 9:00-10:00 MTTF

E-mail  hentzel@iastate.edu
phone  294-8141
Website

http://www.public.iastate.edu/~mathclasses/honors.265

August 25:  Section 12.5         
p863:1,18,41     
p864:47          


(* This is a Mathematic Program to study how to solve two functions    *)
(* in two variables f(x,y) = 0 and g(x,y) = 0  for simultaneous zeros. *)

(* Here are two complicated functions *)

f[{x_,y_}] :=   x Cos[y^2] + y^2 Sin[x] + 5;

g[{x_,y_}] :=   x^2 + 3 x y + Tan[y];

(* find x and y such that f(x,y) = 0 and g(x,y) = 0  *)

p = f[{x,y}];          (* make a copy of f(x,y) to differentiate *)

fx[{x_,y_}] = D[p,x];

fy[{x_,y_}] = D[p,y];

q = g[{x,y}];          (* make a copy of g(x,y) to differentiate *)

gx[{x_,y_}] = D[q,x];

gy[{x_,y_}] = D[q,y];
                       (* this is a 2x2 matrix of the partials *)

m[{x_,y_}] := {{fx[{x,y}],fy[{x,y}]},{gx[{x,y}],gy[{x,y}]}};

                       (* this updates the point to a better point *)

w[{x_,y_}] := {x,y} - Inverse[m[{x,y}]].{f[{x,y}],g[{x,y}]};

{x,y} = {8.0,2.0};          (* starting value INCLUDE DECIMAL POINT   *)

Do[ {x,y} = w[{x,y}], {i,1,100}];    (* iterates the expression 100 times *)

Print[" point ",{x,y},"  f-value ",f[{x,y}]," g-value  ",g[{x,y}] ]

-------------------------------------------------------------------------
-------------------------------------------------------------------------
Use the above program to find simultaneous zeros starting at various {x,y}.

Suggested starting places should include 1. {0.0,0.0}, 2. {1.0,1.0}, 3. {3.0,5.0} 

Not all starting places will work.   Start the program at 10 different points

and write down the starting point and the zero if it finds a simultaneous zero.  

If not, write down the starting point and that it did not converge to a zero.



Main idea:  The differentials are approximations to the function.

Key words:  Chain Rule, Implicit Differentiation

Goal:  Know the 3 (variations) of the chain rule.

dx  = fx dx + fy dy

dx/dt = fx dx/dt + fy dy/dt       first chain rule.


f(x,y,z)   and x = x(t), y = y(t)   z = z(t);

find df/dt.


f(x,y,z)  and x = r(u,v), y = s(u,v), z = t(u,v) .   

Write out the formula for     df/du      df/dv


w = x^3 + y^2 + 3 z      x = uv     y = u+v     z = u^2 + v^2.

Find dw/du       dw/dv  at (u,v) = (1,2).

So at (u,v) = (1,2)


Answer:

      (x,y,z) = (2,3,5)

dw/du = dw/dx dx/du + dw/dy dy/du + dw/dz dz/du

      =  12    2    +    6    1   +   3    2   

      = 36 


dw/dv = 30


Implicit differentiation

x^2 -3 x y + y^2 -2x + y - 5 = 0.

Find dy/dx.

Fx = 2x -3y -2

Fy = -3x + 2y + 1
                      2x -3y -2
dy/dx = -Fx/Fy =  - -------------
                    -3x + 2y + 1


The reasoning goes like this.

let w = x^2 -3 x y + y^2 -2x + y - 5 ,

Then dw = Fx dx + Fy dy.

If dw = 0 then  dy/dx = -Fx/Fy.


Question:  How would you use the above formula for the derivative to

plot the curve?