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)=2x2e2x3y8f(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 Po=(xo,yo)P_o = (x_o,y_o) be a point in the domain of f. Then, the equation of the tangent plane to S at PoP_o  is given by

       z=f(xo,yo)+fx(xoyo)+fy(xo,yo)(yyo)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 fx(x,y)f_x(x,y) and fy(x,y)f_y(x, y), the we use 1-equation with x0=1x_0 = -1 and

yo=2y_o = 2

                       fx(x,y)=4x2e2xf_x(x,y) = 4x-2e^{2x}

                       fy(x,y)=3f_y(x,y) = -3

                       f(1,2)=2(1)2e2(1)328=111e2f(-1,2) = 2(-1)^2-e^{2*(-1)}-3*2-8=-11-\frac{1}{e^2}

                       fx(1,2)=4(1)2e2(1)=42e2f_x(-1,2) = 4*(-1)-2e^{2*(-1)}=-4-\frac{2}{e^2}

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

Then 1-equation becomes

            z = 111e2-11-\frac{1}{e^2} + (42e2)(x+1)(-4-\frac{2}{e^2})(x+1) + (3)(y+3)(-3)(y+3)

            z=111e24x2xe241e23y9z= -11-\frac{1}{e^2} - 4x-\frac{2x}{e^2}-4-\frac{1}{e^2}-3y-9

             z = 4x2xe23y1e224-4x-\frac{2x}{e^2}-3y-\frac{1}{e^2}-24


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