Question

Consider the equation xyz = 4x² +y² −z². Use the Implicit Function Theorem to show that the given equation has a smooth unique local solution of the form z = g(x,y) about the point (2,0,4). Then find the local linearization of g about the point (2,0).

Answer 1

Solution:-

Assuming $s$ is an open subset of $R^{n+k}$ and that $F:S\to R^k$ is a function of class $C^1$ .

Assuming that $(a,b)$ is a point in that

$F(a,b)=0$ and det $D_yF(a,b)\not =0$

Consider a continuously differentiable function $F(x,y,z)=c$ such that

$dF\over dz$ $(x,y,z)\not=0$ then $F$ is $\alpha(x,y,z)$ in that $(x,y)$ is almost close to $(x,y)$ then

$F(x,y,z)=C$

 $xyz=4x^2+y^2-z^2$

$Let F=4x^2+y^2-z^2-xyz$

$∇F=[8x−yz,2y−xz,−2z−xy]$ $∇F(2,0,4)=[8×2−0×4,2×0−2×4,−2×4−2×0]$

$=[16,−8,−8]$

Linearization equation

$z=z o​ +F x​ (x−xy)+F_y​ (y−y_1 )$

$z o(x=2,y=0)\space then z=z_o$

$z o​ =4×(2) ^2 +0^2 −z_0^2 ​$

$2_o^2 =16^ 2$

$z_o=4$

$z=4+16(x−2)−8(y−0)$

$=4+16x−32−8y,$

$16x-8y-z-28=0$