Equation of the plane tangent to the given surface

Question

Find the equation of a plane tangent to the surface given by $$xyz+x^2-3y^2+z^3=14$$ at $$P=\left( 5,-2,3 \right)$$ In my opinion answer is: $$4x+27y+25z-41=0$$ If not please tell me what am i doing wrong.

Answer 1

In order to make the function a surface, take $ x(y, z) $ to be $ x $ as a function of $y, z $. Then your equation becomes,

$$ x(y, z)yz + x(y, z)^2 - 3y^3 + z^3 = 14 $$

Without solving for $ x $, we find the slope of $ x(y, z) $ in the y and z directions,

$$ \partial_y x(y,z) y z + xz + 2 x \partial_y x - 9y^2 = 0 \\ \partial_z x(y,z) y z + xy + 2 x \partial_z x + 3z^2 = 0 $$

Solving for the derivatives at the given point, $ P=(5, -2, 3) $

$$ \partial_y x(y, z) = \frac{21}{4} \\ \partial_z x(y, z) = \frac{17}{4} $$

We can then construct a tangent plane at $$ 4(x - 5) = 21(y + 2) + 17(z - 3) $$ by the standard formula. Notice that the derivatives match and the tangent touches at the given point.