Precalculus (6th Edition) Blitzer

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

Chapter 3 - Section 3.5 - Exponential Growth and Decay; Modeling Data - Exercise Set - Page 507: 47


a. $T=45+25e^{-0.0916t}$ b. $51.3^\circ F$ c. $17.6min$

Work Step by Step

Using Newton's Law of Cooling $T=C+(T_0-C)e^{kt}$, we have: a. We have: $T_0=70^\circ F, C=45^\circ F, t=10min, T=55^\circ F$ Thus $55=45+(70-45)e^{10k}$ and $e^{10k}=\frac{10}{25}=0.4$ This gives $k=\frac{ln(0,4)}{10}\approx-0.0916$ The model equation is then $T=45+25e^{-0.0916t}$ b. With $t=15min$, we have $T=45+25e^{-0.0916(15)}\approx51.3^\circ F$ c. Letting $T=50$, we have $45+25e^{-0.0916t}=50$ and $e^{-0.0916t}=\frac{5}{25}=0.2$ Thus $t=-\frac{ln(0.2)}{0.0916}\approx17.6min$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.