Let fx y 3x y1 x2 y2 Compute the two critical points P1a

Let f(x, y) = (3x - y)/(1 + x^2 + y^2). Compute the two critical points P_1(a_1, b_1]) and P_2(a_2, b_2) of f(x, y), and compute the two quadratic approximations Q_k(x, y) of f(x, y) at (a_k, b_k) for k = 1, 2. Complete the square of the two quadratic approximations Q_k(x, y), and determine if they are ellipic paraboloids or hyperbolic paraboloids.

Solution

Ans:

we to find out the derivative

f_x =[ (1+x^2+y^2) (3)-(3x-y)(2x)] / (1+x^2+y^2)^2

= (-3x^2 +3y^2+3xy+3)/ (1+x^2+y^2)^2

f_y = [(1+x^2+y^2)(-1) + (3x-y)(2y)]/ (1+x^2+y^2)^2

= (-x^2-3y^2+6xy-1)/ (1+x^2+y^2)^2

now we need to solve it

let

  f_x =0 and   f_y = o

(-3x^2 +3y^2+3xy+3)/ (1+x^2+y^2)^2 =0   

(-3x^2 +3y^2+3xy+3) = 0 ....................................1

(-x^2-3y^2+6xy-1)/ (1+x^2+y^2)^2 = 0

(-x^2-3y^2+6xy-1) = 0 .............................................2

solving 1 and 2 we get (adding 1 and 2)

-4x^2 + 9xy +2 =0 ...........................................3

multiplying 2 by 3 and substracting from 1 we get

12 y^2 - 16xy +6 = 0 ...........................................4

we get these 3 and 4 are quadratic are irredicuible there are no roots and no more solution

Thus there is just one critical point (0,0)

b) from these 3 and 4 equation

-4x^2 = -9xy-2

x^2 = 9/4 xy +1/2

and 12y^2 = 16xy -6

y^2 = 4/3 xy -1/2

from these two equation we conclude that it is elliptic paraboloids