anonymous
  • anonymous
The rate of decay in the mass, M, of a radioactive substance is given by the differential equation dM dt equals negative 1 times k times M, where k is a positive constant. If the initial mass was 150g, then find the expression for the mass, M, at any time t.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
https://gyazo.com/8d62d90df17bfc829809548f15163bd1
anonymous
  • anonymous
@Zarkon
anonymous
  • anonymous
Can you confirm my answer?

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More answers

anonymous
  • anonymous
@baru Can you confirm for me please?
anonymous
  • anonymous
@jim_thompson5910 will you be my hero?
jim_thompson5910
  • jim_thompson5910
If \(\Large M = 150e^{-kt}\) then what is \(\Large \frac{dM}{dt}\) equal to?
anonymous
  • anonymous
-kM
anonymous
  • anonymous
@jim_thompson5910 when I integrated I got M=e^-kt
anonymous
  • anonymous
I wasn't sure what to do with the initial mass, so using the choices I multiplied it.
anonymous
  • anonymous
Would you like to see my work?
jim_thompson5910
  • jim_thompson5910
sure, go ahead and post your work
anonymous
  • anonymous
anonymous
  • anonymous
Please ignore the 100-60 crap, that was for another question.
jim_thompson5910
  • jim_thompson5910
You should have M in absolute values. And don't forget the +C \[\Large \ln(|M|) = -kt+C\] \[\Large |M| = e^{-kt+C}\] \[\Large |M| = e^{C}*e^{-kt}\]
anonymous
  • anonymous
Why in absolute values?
jim_thompson5910
  • jim_thompson5910
because the domain of ln(x) is x > 0 the absolute values ensure that the argument is not negative
anonymous
  • anonymous
Ahh ok.
anonymous
  • anonymous
Would e^c be equal to 150?
jim_thompson5910
  • jim_thompson5910
now you'll use the information that 150 is the initial mass so M = 150 when t = 0 \[\Large |M| = e^{C}*e^{-kt}\] \[\Large |150| = e^{C}*e^{-k*0}\] \[\Large |150| = e^{C}*e^{0}\] \[\Large |150| = e^{C}*1\] \[\Large |150| = e^{C}\] \[\Large 150 = e^{C}\] \[\Large e^{C} = 150\] you are correct
anonymous
  • anonymous
Thank you good sir.
jim_thompson5910
  • jim_thompson5910
Now we make the claim that \[\Large M = 150e^{-kt}\] here is how to check that claim If \[\Large M = 150e^{-kt}\] then \[\Large \frac{dM}{dt} = -150ke^{-kt}\] -------------------------------- Start with the initial differential equation and perform substitutions \[\Large \frac{dM}{dt} = -k*M\] \[\Large -150ke^{-kt} = -k*150e^{-kt}\] \[\Large -150ke^{-kt} = -150ke^{-kt} \ \ \ \color{green}{\checkmark}\] So the solution has been confirmed
jim_thompson5910
  • jim_thompson5910
Of course, the more general solution is \[\Large M = Ce^{-kt}\] but we don't need to worry about that
anonymous
  • anonymous
Got it. Thanks

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