H
Heater!
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The problem is as follows:
"Plutonium-239 has a decay rate of approximately 0.003% per year. Suppose that plutonium-239 is released into the atmosphere each year for 20 years at a constant rate of 1 lb per year. How much plutonium-239 will remain in the atmosphere after 20 years?"
My answer:
The amount of plutonium remaining is given exactly by the definite integral shown, where the chemical is released into the atmosphere at a rate of R(t) pounds per year for T years at a decay rate of k.
∫_(t to 0)(R(t) e^kt (dt)
Substitute the given values in the definition for the future amount. The value of k will be negative because it is decaying.
∫(20 to 0) 1 e^(-0.003t)(dt)
Now, we integrate with respect to t.
[-333.33e^(-0.003(t)) ](20 to 0)
Lastly, we evaluate the result over the integral from t = 0 from t = 20.
-333.33(e^(-0.003(20))-e^(-0.003(0))) ≈ 19.41162802
Therefore, the approximate amount of plutonium-239 in the atmosphere after 20 years is 19.412 lbs.
I'm coming up with the answer of 19.412 lbs, but the book says the answer is 19.994 lbs.
"Plutonium-239 has a decay rate of approximately 0.003% per year. Suppose that plutonium-239 is released into the atmosphere each year for 20 years at a constant rate of 1 lb per year. How much plutonium-239 will remain in the atmosphere after 20 years?"
My answer:
The amount of plutonium remaining is given exactly by the definite integral shown, where the chemical is released into the atmosphere at a rate of R(t) pounds per year for T years at a decay rate of k.
∫_(t to 0)(R(t) e^kt (dt)
Substitute the given values in the definition for the future amount. The value of k will be negative because it is decaying.
∫(20 to 0) 1 e^(-0.003t)(dt)
Now, we integrate with respect to t.
[-333.33e^(-0.003(t)) ](20 to 0)
Lastly, we evaluate the result over the integral from t = 0 from t = 20.
-333.33(e^(-0.003(20))-e^(-0.003(0))) ≈ 19.41162802
Therefore, the approximate amount of plutonium-239 in the atmosphere after 20 years is 19.412 lbs.
I'm coming up with the answer of 19.412 lbs, but the book says the answer is 19.994 lbs.