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: 13

Answer

\begin{aligned} \text { (a) } f(2,1) & =11 \\ f(2.1,1.05) & =10.4875\\ \Delta z &=-0.5125 \\ \text {( b) } dz &=-0.5 \end{aligned}

Work Step by Step

Given $$f(x, y)=16-(x^2+ y^2)$$ so, we have \begin{aligned} \text { (a) } f(2,1) &=16-(2^2+1^2)=16-5=11 \\ f(2.1,1.05) &=16-(2.1^2+1.05^2)=16-(4.41+1.1025)\\ &=16-5.5125=10.4875\\ \Delta z &=f(2.1,1.05)-f(2,1)=10.4875-11=5.5125\\ &=0.5125\\ \text {( b) } \text { since }d z &=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy,\\ &\frac{\partial f}{\partial x}= -2x=-2(2)=-4,\\ &\frac{\partial f}{\partial y} =-2y=-2(1)=-2,\\ & 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=-4 d x-2 d y=-4(0.1)-2(0.05) \\ &\ \ \ \ =0.4+0.1=-0.5 \end{aligned}
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