# Thread:

1. ## Quadratic Approximation

Hi!
Any help?

Q:
Determine the quadratic approximation surface at the point (0,0) on the surface of
z=sqrt(x+1)/(y+1)

2. Start by defining the "quadratic surface" by the standard form $q(x,y)=ax^2+bx+cxy+dy+ey^2+f$ and we'll find values for the coefficients such that $q(0,0)=z(0,0),q_x(0,0)=z_x(0,0),q_y(0,0)=z_y(0,0), q_{xx}(0,0)=$ $z_{xx}(0,0),q_{xy}(0,0)=z_{xy}(0,0),q_{yy}(0,0)=z_ {yy}(0,0)$. Since this is a system of six equations and six unknowns, we are guaranteed a unique answer.

Derivatives of z:
$z(x,y)=(x+1)^{1/2}(y+1)^{-1}, z(0,0)=+1$
$z_x(x,y)=\frac12(x+1)^{-1/2}(y+1)^{-1}, z_x(0,0)=+\frac12$
$z_y(x,y)=-(x+1)^{1/2}(y+1)^{-2}, z_y(0,0)=-1$
$z_{xx}(x,y)=-\frac14(x+1)^{-3/2}(y+1)^{-1}, z_{xx}(0,0)=-\frac14$
$z_{xy}(x,y)=-\frac12(x+1)^{-1/2}(y+1)^{-1}, z_{xy}(0,0)=-\frac12$
$z_{yy}(x,y)=+2(x+1)^{1/2}(y+1)^{-3}, z_{yy}(0,0)=+2$

Derivatives of q:
$q(x,y)=ax^2+bx+cxy+dy+ey^2+f, q(0,0)=f=z(0,0)=1, f=1$
$q_x(x,y)=2ax+b+cy, q_x(0,0)=b=z_x(0,0)=\frac12, b=\frac12$
$q_y(x,y)=cx+d+2ey, q_y(0,0)=d=z_y(0,0)=-1, d=-1$
$q_{xx}(x,y)=2a, q_{xx}(0,0)=2a=z_{xx}(0,0)=-\frac14, a=-\frac18$
$q_{xy}(x,y)=c, q_{xy}(0,0)=c=z_{xy}(0,0)=-\frac12, c=-\frac12$
$q_{yy}(x,y)=2e, q_{yy}(0,0)=2e=z_{yy}(0,0)=2, e=1$

So, $q(x,y)=-\frac18x^2+\frac12x-\frac12xy-y+y^2+1$ is a quadratic function that shares its six derivatives with $z(x,y)$. Notice that this is always possible, as long as all six derivatives exist and are defined at the point of interest.
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