Answer to Question #154504 in Differential Geometry | Topology for SUNITA MUHAL

Question

Find the equation of the tangent to the plane f(x,y) at the point (-1,2)

F(X,y )= 2x^2 -e^2x-3y-8

Answer 1

                $f(x,y) = 2x^2-e^{2x} - 3y-8$       (-1,2)

Definition:

Let S be a surface defined by a differentiable function z = f(x, y), and let $P_o = (x_o,y_o)$ be a point in the domain of f. Then, the equation of the tangent plane to S at $P_o$  is given by

       $z = f(x_o, y_o) + f_x(x_o - y_o) +f_y(x_o, y_o)(y-y_o)$ 1-equation

First, we must calculate $f_x(x,y)$ and $f_y(x, y)$, the we use 1-equation with $x_0 = -1$ and

$y_o = 2$

                       $f_x(x,y) = 4x-2e^{2x}$

                       $f_y(x,y) = -3$

                       $f(-1,2) = 2(-1)^2-e^{2*(-1)}-3*2-8=-11-\frac{1}{e^2}$

                       $f_x(-1,2) = 4*(-1)-2e^{2*(-1)}=-4-\frac{2}{e^2}$

                       $f_y(-1,2) = -3$

Then 1-equation becomes

            z = $-11-\frac{1}{e^2}$ + $(-4-\frac{2}{e^2})(x+1)$ + $(-3)(y+3)$

            $z= -11-\frac{1}{e^2} - 4x-\frac{2x}{e^2}-4-\frac{1}{e^2}-3y-9$

             z = $-4x-\frac{2x}{e^2}-3y-\frac{1}{e^2}-24$