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

Question #154504

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


1
Expert's answer
2021-01-11T15:37:17-0500

                "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"


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