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anonymous

  • one year ago

The rate of decay in the mass, M, of a radioactive substance is given by the differential equation dM /dt = -kM, where k is a positive constant. If the initial mass was 200g, then find the expression for the mass, M, at any time t.

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  1. anonymous
    • one year ago
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    @Astrophysics @ganeshie8 @nincompoop

  2. anonymous
    • one year ago
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    \[\frac{ 1 }{ -k }\]\[\int\limits \frac{ dM }{ M } = \int\limits dt\]

  3. anonymous
    • one year ago
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    is that right?

  4. anonymous
    • one year ago
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    Yea you mean \[\frac{ -1 }{ k} \times \int\limits \frac{ dM }{ M}=\int\limits dt\] R Right?

  5. anonymous
    • one year ago
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    integrate both side and it is \[\frac{ -1 }{ k}( \ln M)=t\]

  6. anonymous
    • one year ago
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    okay how did you integrate the left side

  7. anonymous
    • one year ago
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    okay nvm i see what you did

  8. anonymous
    • one year ago
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    Cross multiply so you can get M by it self. \[-1(\ln M)=kt\]

  9. anonymous
    • one year ago
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    now i just solve for M right?

  10. anonymous
    • one year ago
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    ln M=-kt

  11. anonymous
    • one year ago
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    M = e ^-kt

  12. anonymous
    • one year ago
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    e both sides and it should look \[M=e ^{-kt}\]

  13. Astrophysics
    • one year ago
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    Don't forget the constant you will need it for your initial amount

  14. anonymous
    • one year ago
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    Yea integrate both side and solve for M because he want to find an expression for M

  15. anonymous
    • one year ago
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    200 = e^-k(0) ...?

  16. anonymous
    • one year ago
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    Sorry had trouble with equation table.

  17. anonymous
    • one year ago
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    these are my answer choices by the way M = 200ln(kt) M = 2e−kt M = 200 ekt M = 200 e−kt

  18. anonymous
    • one year ago
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    is it D

  19. Astrophysics
    • one year ago
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    yes

  20. anonymous
    • one year ago
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    Yes, because if it is M=200e^-kt

  21. anonymous
    • one year ago
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    okay thanks for the help guys :D

  22. anonymous
    • one year ago
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    if it is M=200e^-kt*

  23. anonymous
    • one year ago
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    you multiply it by 200 since it is decaying and it is the initial principal.

  24. Astrophysics
    • one year ago
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    \[\frac{ dM }{ dt } = -kM \implies \int\limits \frac{ d M}{ M } = \int\limits - k dt \] \[\ln(M) = -k t+C\] \[\large e^{\ln(M)} = e^{-kt+C}\] \[M = e^{-kt}e^C\] let \[e^c = C\] so we have \[M = Ce^{-kt}\] after putting our initial conditions we get \[M = 200e^{-kt}\]

  25. Astrophysics
    • one year ago
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    Just to note initial conditions is t = 0 meaning M = 200, so C = 200

  26. anonymous
    • one year ago
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    This method works too, but I just multiply the initial mass amount by the expression.

  27. anonymous
    • one year ago
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    Since he probably is taking a calculus class with some basic differential equation problems. From my experience, it is similar to when I took Calculus. Doubt this is a Differential Equation class questions. I never took DE.

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