Answer to Question #322866 in Calculus for Neyoclass

Question

Find an equation for the tangent plane to the surface 2xz^2-3xy-4x=7 at the point (1, -1, 2).

Select one:

A. 1

B. 0

Answer 1

In case if a surface is defined implicitly by an equation of the form $F(x,y,z)=0$ , then the tangent plane to the surface at a point $(a,b,c)$  is given by the equation:

$\frac{\partial F(a,b,c)}{\partial x}(x−a)+\frac{\partial F(a,b,c)}{\partial y}(y−b)$$+\frac{\partial F(a,b,c)}{\partial z}(z−c)=0$

For the function $F(x,y,z)=2xz^2-3xy-4x-7$ we have

$\frac{\partial F}{\partial x}=2z^2-3y-4$ ; $\frac{\partial F}{\partial y}=-3x$ ; $\frac{\partial F}{\partial z}=4xz$, so the equation of the tangent plane at (1, -1, 2) is

$(2\cdot2^2-3\cdot(-1)-4)(x−1)+(-3\cdot1)(y−(-1))$$+(4\cdot 1\cdot 2)(z−2)=0$

$7(x−1)-3(y+1)+8(z−2)=0$

$7x-3y+8z-26=0$

Answer: $7x-3y+8z-26=0$ .