I'm practicing for a calculus exam and I've found this problem:
Prove that the equation $\frac{1}{8} z^3 x^2 + y^2 z - xy \ln(y) = -3$ defines an implicit function $y = \varphi (x, z)$ around the point $(1,1,-2)$. For this $\varphi$ find the equation of the tangent plane to the surface define by $y^2 \varphi (x,z) = 16$ in the point $(1,4,-2)$
Setting $F(x,y,z)=\frac{1}{8} z^3 x^2 + y^2 z - xy \ln(y) + 3 = 0$. By the implicit function theorem if the partial derivative of $F$ with respect to $y$ is different from $0$ at the given point then there exists such a function.
In order to find the equation of the plane I should evaluate the surface's equation at the desired point and use the gradient of $F$ in the formula $\Pi : <\nabla F (x,y,z), (x,y,z) - (1,4,-2)>$ ? I'm a little bit confused on how to use the information of the implicit function.
I've came up with this... I hope it's right.
The function is $C^1$ in the neighborhood of $(1,1,-2)$ and the partial derivative $F_y(1,1,-2) \neq 0$. So by the implicit function theorem there exists such a function.
To find the tangent plane to the given surface at the point $(1,4,-2)$ I set $G(x,y,z)=16\varphi(1,-2)=16$ and I look for $G$'s gradient.
$G_x = y^2 \varphi_x(x,z)$, $G_y = 2y \varphi(x,z)$, $G_z = y^2 \varphi_z(x,z)$
Using the formula for the implicit function's partial derivatives:
$$\frac{\delta \varphi}{\delta x_i} = -\frac{\frac{\delta f}{\delta x_i}}{\frac{\delta f}{\delta x_k}}(x, \varphi (x,z), z)$$
So I have: $\varphi_x = -\frac{2}{5}$ and $\varphi_z = \frac{1}{2}$
Substituting in $\nabla G(x,y,z)$: $G_x = - \frac{32}{5}$, $G_y = 8$ and $G_z = - \frac{8}{8}$
Finally the tangent plane to $G(x,y,z)$ at $(1,4,-2)$ is: $-\frac{32}{5}+8y+8z= \frac{48}{5}$