Business Calculus problem, work included. Why am I coming up with the wrong answer?

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.

Update:

Cidyah, what are you doing to come to those answers?

2 Answers

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  • ?
    Lv 7
    1 decade ago
    Favorite Answer

    I got your answer, too..

  • cidyah
    Lv 7
    1 decade ago

    1----0.99997

    2----1.9999100009

    3----2.999820003599973

    4----3.999700008999865

    5----4.999550017999595

    6----5.999370031499055

    7----6.9991600503981095

    8----7.9989200755965975

    9----8.99865010799433

    10----9.99835014849109

    11----10.998020197986635

    12----11.997660257380696

    13----12.997270327572975

    14----13.996850409463148

    15----14.996400503950865

    16----15.995920611935746

    17----16.995410734317385

    18----17.994870871995357

    19----18.994301025869195

    20----19.99370119683842

    I am not sure whether this is a half-life problem. I computed the values exactly as the problem wanted me to.

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