Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Review - Exercises - Page 942: 36

Answer

(a) $P=\frac{100}{T}$ (b) $-\frac{1}{900}$

Work Step by Step

(a) Given $PT=100$, we have $P=\frac{100}{T}$ which shows $P$ as a function of $T$. (b) The instantaneous rate of change of P with respect to T when $T=300K$ is the derivative of $P$ with respect to $T$: $P'=\lim_{h\to 0} \frac{P(300+h)-P(300)}{h}=\lim_{h\to 0} \frac{\frac{100}{300+h}-\frac{100}{300}}{h}=\lim_{h\to 0} \frac{100(300-300-h)}{300h(300+h)}=\lim_{h\to 0} \frac{-1}{3(300+h)}=-\frac{1}{900}$ $lb/in^2K$
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