Answer to Question #240073 in Calculus for sai

Question

Can the surface x⁵y³ + x²y + z = 3 be expressed as the graph of a function x = f(y, z) near the point (2, 1/2, −3)? If yes, compute f_y(1/2, −3) and f_z(1/2, −3).

Answer 1

Let $F(x,y,z)=x^5y^3 + x^2y + z- 3$, and $P=(2,1/2, -3)$. Then

$dF(x,y,z)=(5x^4y^3 + 2xy)dx +(3x^5y^2 + x^2)dy + dz$

$dF(P)=12dx +28dy + dz$

Equation $dF(P)=12dx +28dy + dz=0$ is the equation of the tangent plane to the surface F=0 at the point P.

Since $\frac{\partial F}{\partial x}(P)=12\ne 0$ is invertible, then, by the implicit function theorem, the surface F=0 locally, in a neiborhood of (yz)-projection of the point P, can be expressed as a graph of differentiable function $x=f(y,z)$.

The tangent plane to the graph $x=f(y,z)$ at the point $P'=(1/2,-3)$ has equation

$dx-\frac{\partial f}{\partial y}(P')dy-\frac{\partial f}{\partial z}(P')dz=0$

But we know that the tangent plane to the surface F=0 at the point P has equation

$12dx +28dy + dz=0$

We have two formulas of the same plane. Therefore, they must be proportional:

${1}:{12}=-\frac{\partial f}{\partial y}(P'):28=-\frac{\partial f}{\partial z}(P'):1$

Therefore,

$\frac{\partial f}{\partial y}(P')=-28/12=-7/3$

$\frac{\partial f}{\partial z}(P')=-1/12$