Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.4 Exercises - Page 905: 16

Answer

\begin{aligned} \text { (a) } f(2,1) & =1.0806 \\ f(2.1,1.05) & =1.0449 \\ \Delta z&= -0.0357 \\ \text {( b) } dz &=-0.0301169\end{aligned}

Work Step by Step

Given $$f(x, y)=x \cos y$$ so, we have \begin{aligned} \text { (a) } f(2,1)& = 2 \cos 1 =1.0806 \\ f(2.1,1.05) &= 2.1 \cos 1.05 =1.0449 \\ \Delta z &=f(2.1,1.05)-f(2,1)\\ &= 1.0449 – 1.0806\\ &=-0.0357 \\ \text {( b) } \text { since }d z &=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy,\\ &\frac{\partial f}{\partial x}= \cos y = \cos 1 =0.540302,\\ &\frac{\partial f}{\partial y} = -x \sin y =-2 \sin 1 =-1.68294,\\ & dx\approx\Delta x=2.1-2=0.1 \ \text{and} \\& dy\approx\Delta y=1.05-1=0.05 \\& \text{So we get},\\ &dz=\cos y\ d x-x \sin y \ d y\\ &\ \ \ \ =0.540302 (0.1)-1.68294 (0.05) \\ &\ \ \ \ =-0.0301169\\ \end{aligned}
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