Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Review Exercises - Page 514: 89


a. $T=65+120e^{-0.1438t}$ b. after about 8 minutes.

Work Step by Step

a. We are given $T_0=185^{\circ}F, C=65^{\circ}F, T(2)=155^{\circ}F$ We have $155=65+(185-65)e^{2k}$ which gives $k=\frac{ln(90/120)}{2}\approx-0.1438$ Thus the model equation is $T=65+120e^{-0.1438t}$ b. Letting $T=105$, we have $65+120e^{-0.1438t}=105$ which gives $t=-\frac{ln(40/120)}{0.1438}\approx8min$ that is, after about 8 minutes.
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