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A tank contains $200\; gal$ of water of $40\; lb$ salt dissolves. $5\; gal$ of brine , each containing $2\; lb$ of dissolved salt, run into tank per minute and mixture kept uniform by stirring runs out at same rate. Find amount of salt $y(t)$ in tank at any time $(t)$

$(a)\;y(t)=300-360\;e^{-0.05t} \\ (b)\;y(t)=360-300 e^{-0.05t} \\ (c)\;y(t) =400 -360 e^{-0.025t} \\ (d)\;y(t)=360-400 e^{-0.025t}$

1 Answer

The time rate of charge $y'= \large\frac{dy}{dt}$
of $y(t) $ =the inflow of salt - outflow
The inflow $=10 \;lb/min$
(5 gal brive, each containing 2 lb)
$y(t) $ is total amount of salt.
tank is always contains 200 gal because 5 gal flow in and 5 gal flow out/ min
Thus 1 gal contains $\large\frac{y(t)}{200}$$lb$ of salt.
Hence 5 out flowing gal containing $\large\frac{5 y(t)}{200}=\frac{y(t)}{200}$
$\qquad=0.025\;y(t) \;lb$
This is outflow.
$y'= $ salt inflow-out flow rate
Initially $y(0)=40$
$y'= -0.025 (y-400)$
$\large\frac{dy}{y- 400}$$=-0.025\;at$
$y-400=ce^{-0.025 t}$
$y(t)=400-360e^{-0.025 \;t}\;lb$
Hence c is the correct answer.
answered Feb 7, 2014 by meena.p

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