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Related Rate Applications

Example: The tank shown below contains 200 L of a salt solution with a concentration of

1 gram/liter. The tank is subsequently rinsed with fresh water flowing in at a rate of

2 liters/min. The solution in the tank is continuously stirred providing complete mixing

while flowing out of the tank at the same rate. Determine the time elapsed before the

concentration of salt in the tank reaches 10 % of its original value.

Applications involving changes of solute in a Tank

dy/dt = Rate of inflow ˗ Rate of outflow = Rate of change of solute

of solute of solute in the tank

dy/dt = r_i C_i ˗ r_o C_o y(0) = 200 grams of salt

For given data:

Rate of inflow = r_i C_i , where r_i = rate of flow in = 2 liters/min

C_i = concentration of salt entering in grams/liter = 0 (fresh water)

Rate of outflow = r_o C_o , where r_o = rate of flow out = 2 liters/ min

C_o = concentration of solute exiting in grams/liter = y(t)/V(t)

= 2/200 grams/min Note: Volume at any time = 200 L

So the governing d.e. is: dy/dt = ˗ (1/100) y(t) with the initial condition y(0) = 200

Separate variables: dy/y = ˗ (1/100) dt or ln (y) = ˗ (1/100) t + C₁

Solve for y(t) y(t) = C exp[ ( ˗ 1/100) t ] and y(0) = 200 = C

So y(t) = 200 exp[ ( ˗1/100) t ] and when y(t*) = 20, 1/10 = exp[ (˗1/100) t* ]

(˗1/100) t* = ln(1/10) and t* = 100 ln(10) minutes (result)

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