Ben LoBrutto

Footnote 18 Project

12:40 Class


11.7 DISCOVERY PROJECT: Quadratic Approximations and Critical Points

In Section 11.4 the linearization of a function f of two variables at point (a, b)  is defined as:


The graph of L is the tangent plane to the surface z = f(x, y) at (a, b, f(a, b) and the corresponding linear approximation is f(x, y) = L(x, y)The Linearization L is also called the first degree Taylor polynomial of f at (a, b).


Step 1.   
If f has continuous second-order partial derivatives at (a, b) , then the second-degree Taylor polynomial of f at (a, b) is


And the approximation

is called the quadratic approximation to f  at (a, b).


Verify that Q has the same first- and second- order partial derivatives as
f at (a, b ).

Since f (a, b) is a constant:

=





Now,



Qx = fx(a, b) + fxx(a, b) x 0 + fyy(a, b) x 0




EQUAL


 

 

EQUAL


Step 2.
In this problem the behavior of the polynomial  f(x, y) = ax2 + bxy + cy2 is analyzed by identifying the graph as a paraboloid.


a)      Show that if  a does not equal 0 then



EQUAL


b)      Let  D = 4ac - b2 Show that if D > 0 and a > 0, then f  has a local minimum at (0,0).




Remember:



Local min when:



So, if  a > 0, and 4ac - b2 > 0, then fxx > 0, and D > 0 , so  f  has a local minimum at (0,0)


c) Show that if D > 0 and a < 0, then  f   has a local maximum at (0,0)

Local max when D > 0, 0

If  D > 0, then 4ac - b2   is positive. So,


(+


So, f has a local max at (0,0).


d)
     
Show that if D < 0, then (0,0) is a saddle point. 

Saddle point when D < 0


If 4ac - b2 < 0, then two scenarios can exist:


a
is positive, in which case:



Or, a is negative, in which case:



So, if  D < 0, then D is negative, and is a saddle point.  



Step 3.  a)
Suppose is any function with continuous second-order partial derivatives such that  f (0, 0) = 0, and (0,0) is a critical point of  f .

Write an expression for the second-degree Taylor polynomial  Q , of  f  at (0,0).





b) What can you conclude about  Q from Problem 2?




From Problem 2 we can conclude that:

If  fxx(0, 0) is positive, and is positive, then f has a local min at (0,0).

If  fxx(0, 0) is negative, and is positive, then f has a local max at (0,0).

If   is negative, then  (0, 0)  is a saddle point.


c) In view of the quadratic approximation , what does part (b) suggest about  f ?

Part (b) suggests that f has either a local min, local max, or saddle point at (0,0)