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 - Review - Exercises - Page 992: 18

Answer

$\frac{\partial C}{\partial T}=3.587$ $\frac{\partial C}{\partial S}=1.24$ and $\frac{\partial C}{\partial D}=0.016$

Work Step by Step

$\frac{\partial C}{\partial T}=4.6-2\times 0.055\times T+3\times 0.00029\times T^2-0.01(S-35)=3.587$ Substitute $T=10$ and $S=35$ $\frac{\partial C}{\partial T}=4.6-2\times 0.055\times 10+3\times 0.00029\times 10^2-0.01(35-35)=3.587$ $\frac{\partial C}{\partial S}=1.34-0.01(10)=1.24$ and $\frac{\partial C}{\partial D}=0.016$
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