# Answer in Calculus for Anuj #238273

A tank having a capacity of 1000 liters, initially contains 400 liters of sugar water having a concen-

tration of 0.2 Kg of sugar for each liter of water. At time zero, sugar water with a concentration of

50 gm of sugar per liter begins pumped into the tank at a rate of 2 liter per minute. Simultaneously,

a drain is opened at the bottom of the tank so that the volume of the sugar-water solution in the

tank reduces 1 liter per minute. Determine the following:

Let s(t) = amount, in kg of sugar at time t. Then we have

“frac{dt}{ds}” = (rate of salt into tank) – (rate of salt out of tank)

“= (0.05 frac{kg}{L} u00b7 5 frac{L}{min}) + (0.04 frac{kg}{Min} u00b7 10 frac{L}{min}) u2212 frac{s kg}{1000 L} u00b7 15 frac{L}{min}\n= (0.25 frac{kg}{Min}) + (0.04 frac{kg}{Min} ) u2212 frac{15s kg}{1000 min} \”

So we get the differential equation

“frac{ds}{ndt} = 0.65 u2212nfrac{15s}{n1000}\nfrac{ds}{ndt} =frac{n130 u2212 3s}{n200}”

We separate s and t to get

“frac{1}{n130 u2212 3}nds = frac{1}{n200}ndt”

Integrate

“int frac{1}{130-30s}ds=-frac{1}{30}ln left|130-30sright|+C\nfrac{1}{200}t+C\nimplies s =nfrac{130 u2212 C_4e^{u22123t/200}}{n3}”

Since we begin with pure water, we have s(0) = 0. Substituting,

“C_4 = 130”

“s =nfrac{130 u2212 130e^{u22123t/200}}{n3}\ns(60) =nfrac{130 u2212 130e^{u22123t/200}}{n3} = 25.7153”

Thus, after one hour there is about 25.72 kg of sugar in the tank.

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