Question #322866

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



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Expert's answer
2022-04-11T10:34:54-0400

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

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

For the function F(x,y,z)=2xz23xy4x7F(x,y,z)=2xz^2-3xy-4x-7 we have

Fx=2z23y4\frac{\partial F}{\partial x}=2z^2-3y-4 ; Fy=3x\frac{\partial F}{\partial y}=-3x ; Fz=4xz\frac{\partial F}{\partial z}=4xz, so the equation of the tangent plane at (1, -1, 2) is

(2223(1)4)(x1)+(31)(y(1))(2\cdot2^2-3\cdot(-1)-4)(x−1)+(-3\cdot1)(y−(-1))+(412)(z2)=0+(4\cdot 1\cdot 2)(z−2)=0

7(x1)3(y+1)+8(z2)=07(x−1)-3(y+1)+8(z−2)=0

7x3y+8z26=07x-3y+8z-26=0

Answer: 7x3y+8z26=07x-3y+8z-26=0 .


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