A 40-gallon barrel is filled completely with pure water. Salt water with a concentration of 0.3 lb/gal is then pumped into the barrel, and the resulting mixture overflows at the same rate. The amount of salt in the barrel at time t is given byQ(t) = 12(1 − e−0.09t)where t is measured in minutes and Q(t) is measured in pounds.(a) How much salt is in the barrel after 3 min? (Round your answer to two decimal places.) lb(b) How much salt is in the barrel after 6 min? (Round your answer to two decimal places.)
Question
A 40-gallon barrel is filled completely with pure water. Salt water with a concentration of 0.3 lb/gal is then pumped into the barrel, and the resulting mixture overflows at the same rate. The amount of salt in the barrel at time t is given byQ(t) = 12(1 − e−0.09t)where t is measured in minutes and Q(t) is measured in pounds.(a) How much salt is in the barrel after 3 min? (Round your answer to two decimal places.) lb(b) How much salt is in the barrel after 6 min? (Round your answer to two decimal places.)
Solution
(a) To find out how much salt is in the barrel after 3 minutes, we need to substitute t = 3 into the given equation Q(t) = 12(1 − e^−0.09t).
Q(3) = 12(1 - e^(-0.09*3))
Calculate the exponent first: -0.09*3 = -0.27.
So, Q(3) = 12(1 - e^-0.27).
Now, calculate the value of e^-0.27. Using a calculator, you'll find that e^-0.27 ≈ 0.7634.
Substitute this value back into the equation: Q(3) = 12(1 - 0.7634) = 12*0.2366 ≈ 2.84 pounds.
So, after 3 minutes, there are approximately 2.84 pounds of salt in the barrel.
(b) Similarly, to find out how much salt is in the barrel after 6 minutes, we substitute t = 6 into the equation.
Q(6) = 12(1 - e^(-0.09*6))
Calculate the exponent first: -0.09*6 = -0.54.
So, Q(6) = 12(1 - e^-0.54).
Calculate the value of e^-0.54. Using a calculator, you'll find that e^-0.54 ≈ 0.5831.
Substitute this value back into the equation: Q(6) = 12(1 - 0.5831) = 12*0.4169 ≈ 5.00 pounds.
So, after 6 minutes, there are approximately 5.00 pounds of salt in the barrel.
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