The linearization of ƒ ( x , y ) is a tangent-plane

Chapter 13, Problem 55E

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The linearization of \(f(x,y)\) is a tangent-plane approximation Show that the tangent plane at the point \(P_{0}\left(x_{0}, y_{0}=f\left(x_{0}, y_{0}\right)\right)\) on the surface \(z=f(x,y)\) defined by a differentiable function \(f\) is the plane

                             \(f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)-\left(z-f\left(x_{0}, y_{0}\right)\right)=0\)

or

\(z=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)\)

Thus, the tangent plane at \(P_{0}\) is the graph of the linearization of \(f\) at \(P_{0}\) (see accompanying figure).

Equation Transcription:

Text Transcription:

f(x,y)

P_0(x_0,y_0,f(x_0,y_0))

z=f(x,y)

f

f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)-(z-f(x_0,y_0))=0

z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

P_0