Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.4 Limits at Infinity; Horizontal Asymptotes - 3.4 Exercises - Page 243: 64

Answer

$a)$ see the proof. $b)$ the concentration is near to $30$ so for large time, the concentration of the solution in the tank will be the same as the concentration of the brine.

Work Step by Step

$a$- Since the variation of the mass in the reaction is zero it follows that: $$dm_s=0 \to m_b=m_{salt in the tank}$$ Before the tank we have: $$dm_b=C_b \cdot dV_b$$ $$dm_b=C_b \cdot Q_bdt$$ $$m_b=C_b \cdot Q_bt$$ In the tank: $$m_{salt in the tank}=C_s(t)\cdot V_T $$ $$m_{salt in the tank}=C_s(t)\cdot (V_w+V_b(t))$$ $$m_{salt in the tank}=C_s(t)\cdot (V_w+Q_bt)$$ so: $$C_b \cdot Q_bt=C_s(t)\cdot (V_w+Q_bt)$$ $$30 \cdot 25t=C_s(t)\cdot (5000+25t)$$ $$\frac{30 \cdot 25t}{5000+25t}=C_s(t)$$ $$\frac{30t}{200+t}=C_s(t)$$ ---------------------------------------------------------------- $b$- $$\lim\limits_{t \to \infty}\frac{30t}{200+t}=\lim\limits_{t \to \infty}\frac{30t}{t(\frac{200}{t}+1)}=\lim\limits_{t \to \infty}\frac{30}{\frac{200}{t}+1}=30$$ The concentration is near to $30$ so for large time, the concentration of the solution in the tank will be the same as the concentration of the brine.
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