University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 411: 43



Work Step by Step

Here, $T-T_s=(T_0-T_s)e^{-kt}$ This implies that $39^{\circ}-T_s=(46^{\circ}-T_s)e^{-10k}$ and $33^{\circ}-T_s=(46^{\circ}-T_s)e^{-20k}$ This implies that $\dfrac{39^{\circ}-T_s}{46^{\circ}-T_s}=(\dfrac{39^{\circ}-T_s}{46^{\circ}-T_s})^2$ Thus, $T_s=-3^{\circ}$
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