Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.4 Exercises - Page 946: 3

Answer

$x+y-2z=0$

Work Step by Step

Suppose $f$ has continuous partial derivatives. An equation of the tangent plane to the surface $z=f(x, y)$ at the point $P(x_{0}, y_{0}, z_{0})$ is $z-z_{0}=f_{x}(x_{0}, y_{0})(x-x_{0})+f_{y}(x_{0}, y_{0})(y-y_{0})$ ---- $f(x, y) =\sqrt{xy}$ $ \begin{array}{lllll} f_{x}(x, y) & =\sqrt{y}\cdot\dfrac{\partial}{\partial x}[x^{1/2}] & & f_{y}(x, y) & =\sqrt{x}\cdot\dfrac{\partial}{\partial y}[y^{1/2}]\\ & =\dfrac{\sqrt{y}}{2\sqrt{x}} & & & =\dfrac{\sqrt{x}}{2\sqrt{y}}\\ & & & & \\ f_{x}(1,1) & =\dfrac{1}{2} & & f_{y}(1,1) & =\dfrac{1}{2} \end{array} $ an equation of the tangent plane at $(x_{0}, y_{0}, z_{0})=(1,1,1) $ is $z -1=f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$ $z-1=\displaystyle \frac{1}{2}(x-1)+\frac{1}{2}(y-1) \qquad/\times 2$ $2z -2=x-1+y-1$ $0 =x+y-2z$
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