Multivariate Calculus - Bruce E. Shapiro

LECTURE 8. FUNCTIONS OF TWO VARIABLES 57
Figure 8.5: Level curves for the function z = 2 − x − y 2 at z = −1, 0, 1.
Solution. Substituting z = k and cross-multiplying, we find
x 2 + y = k(x + y 2 ) = kx + ky 2
x 2 − kx = ky 2 − y
Completing the squares on the left hand side of the equation,
x 2 − kx = x 2 − kx + (k/2) 2 − (k/2) 2
= (x − k/2) 2 − k 2 /4
Doing a similar manipulation on the right hand side of the equation,
ky 2 − y = k(y 2 − y/k)
[
= k
= k
Equating the two expressions,
which is a hyperbola.
y 2 − (y/k) +
[ (
y − 1 ) ]
2
− 1
2k 4k 2
= k(y − 1/(2k)) 2 − 1/(4k)
( ) 1 2 ( ) ]
1
2
−
2k 2k
(x − k/2) 2 − k 2 /4 = k(y − (1/2k)) 2 − 1/4k
(x − k/2) 2 − k(y − (1/2k)) 2 = k2
4 − 1
4k = k3 − 1
4k
(x − k/2) 2 (y − (1/2k))2
(k 3 −
− 1)/4k (k 3 − 1)/(4k 2 = 1
)
Math 250, Fall 2006 Revised December 6, 2006.