Respuesta :
Answer:
Step-by-step explanation:
Let
[tex]f(x,y,z) = z-x^2-y^4-e^{xy}[/tex].
The partial derivatives of this function are
[tex]\frac{\partial f}{\partial x} (x,y,z) = ye^{xy} - 2x \\ \frac{\partial f}{\partial y} (x,y,z) = xe^{xy} - 4y^3 \\\frac{\partial f}{\partial z} (x,y,z) = 1[/tex]
The tangent plane equation through a point [tex]A(x_1,y_1,z_1)[/tex] is given by [tex]f'_x (x_1,y_1,z_1)(x-x_1) + f'_y(x_1,y_1,z_1)(y-y_1) + f'_z(x_1,y_1,z_1)(z-z_1) = 0[/tex]
In this case, we have
[tex]x_1 = 1, y_1 = 0, z_1 = 2.[/tex]
The values of the partial derivatives in this point are
[tex]\frac{\partial f}{\partial x} (1,0,2) = 0 \cdot e^{0} - 2\cdot 1 = -2 \\ \frac{\partial f}{\partial y} (1,0,2) = 1 \cdot e^{0} - 0 = 1 \\ \frac{\partial f}{\partial y} (1,0,2) = 1[/tex]
So, the equation is
[tex]-2(x-1) + 1 \cdot (y-0) + 1\cdot (z-2) = 0[/tex]
Therefore, the equation for the plane tangent to the surface at the point [tex](1,0,2)[/tex] is given by
[tex]2x-y-z = 0[/tex]
