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lovelymultani

  • 2 years ago

The half-life of censium-137 is 30 years. Suppose we have a 200 mg sample. Let P(t) be the mass remaining after t years. a) find differentiation equation satisfied by p(t) b) find p(t) and simplify c) how much mass remains after 75 years? d) after how many years is the madd reduced to 1 mg?

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  1. satellite73
    • 2 years ago
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    \[p(t)=200\left(\frac{1}{2}\right)^{\frac{t}{30}}\] is the quick and easy answer, but perhaps not the one your math teacher wants

  2. lovelymultani
    • 2 years ago
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    differentiated would be p(t) = .5^t/30 ?

  3. satellite73
    • 2 years ago
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    you are probably being asked for \[p(t)=p_0e^{kt}\] where \(p_0=200\) (the initial value) and you have to solve for \(k\) via \[e^{30k}=\frac{1}{2}\]

  4. satellite73
    • 2 years ago
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    you get \[30k=\ln(.5)\] or \[k=\frac{\ln(.5)}{30}\]

  5. satellite73
    • 2 years ago
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    not sure what it means by "simplify" maybe write \[k=\frac{\ln(.5)}{30}=-.0231\] and so your function is \[p(t)=200e^{-.0231t}\]

  6. satellite73
    • 2 years ago
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    ok i believe i have confused youi lets go slow

  7. lovelymultani
    • 2 years ago
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    we can't use calculators on our final, so I have no way of getting -.0231t . Is there an alternative?

  8. satellite73
    • 2 years ago
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    you are going to have a function that is an exponential function it is going to look like \[p(t)=p_0e^{kt}\] to answer your question, no, you cannot know what \(\ln(.5)\div 30\) is without a calculator

  9. satellite73
    • 2 years ago
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    in fact you can't do the third or fourth part without a calculator either you can write down an expression to solve, but you can't solve it

  10. lovelymultani
    • 2 years ago
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    allright i guess I cant solve it, maybe they jst want the formulas.

  11. satellite73
    • 2 years ago
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    that is weird

  12. lovelymultani
    • 2 years ago
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    Yes, the decay formula, I have that. Next i plugged in the intital amount. I am just confused about what to put in for kt and yes =( it sucks

  13. satellite73
    • 2 years ago
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    then you get \(k=\frac{\ln(\frac{1}{2})}{30}\) and leave it at taht

  14. satellite73
    • 2 years ago
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    ok lets make sure you understand it you know that the initial value is \(200\) so you know \(p_0=200\) what you start with

  15. satellite73
    • 2 years ago
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    therefore you function will look like \[p(t)=200e^{kt}\] but you do not yet know \(k\), we have to solve for it

  16. satellite73
    • 2 years ago
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    but you know that the half life is 30 years. i.e. you know if \(t=30\) you must have \(p(30)=100\) (since 100 is half of 200) so you can solve for \(k\) by replacing \(t\) by 30 and \(p(30)=100\)

  17. satellite73
    • 2 years ago
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    you get a) \[200e^{30k}=100\]b)\[e^{30k}=\frac{1}{2}\] we could have gone right to the second equation

  18. satellite73
    • 2 years ago
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    solve via \[30k=\ln(\frac{1}{2})\] \[k=\frac{\ln(\frac{1}{2})}{30}\]

  19. satellite73
    • 2 years ago
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    now if you are not allowed to use a calculator, then that is the best you can do you can write \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]

  20. satellite73
    • 2 years ago
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    which is very strange to me, but if you say so then i believe you

  21. lovelymultani
    • 2 years ago
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    That was a great walk through , thanks soo much. Just confused on one last thing. that entire p(t) is for b, correct. u also said b) e^30k = 1/2 up there

  22. satellite73
    • 2 years ago
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    to answer the next question replact \(t\) by \(75\), that is, compute \[p(75)\]

  23. lovelymultani
    • 2 years ago
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    yeaa i go to city college in new york and they arent allowing any calcs for calc or precal

  24. satellite73
    • 2 years ago
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    the work was finding \(k\) to use in the function \(p(t)=p_0e^{kt}\) the \(p_0=200\) part is easy , the hard part is finding \(k\) once you have it, the "final answer" is \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]

  25. satellite73
    • 2 years ago
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    now i am going to guess that if you cannot use a calculator on the final, there is no way they will ask you part 3 or 4 on a final, because you literally cannot do it

  26. lovelymultani
    • 2 years ago
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    allright so when differentiated p(t) = ln(.5)/30*t ? and they do ask but I am going to go through notes to see how it was solved

  27. satellite73
    • 2 years ago
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    hold on

  28. satellite73
    • 2 years ago
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    they are not asking for the derivative, they are asking for the "differential equation"

  29. satellite73
    • 2 years ago
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    the answer to the first question is \[\frac{dP}{dt}=kP\]

  30. satellite73
    • 2 years ago
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    or maybe your text uses \[\frac{dP}{dt}=-kP\] since this is for decay it makes no real different, since \(k\) is negative

  31. satellite73
    • 2 years ago
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    this is the same as saying the instantaneous rate of change is proportional to the current amount

  32. satellite73
    • 2 years ago
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    the solution to \[\frac{dP}{dt}=-kP\] is \[P(t)=P_0e^{-kt}\] which is what we solved above

  33. satellite73
    • 2 years ago
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    i guess i should have said in your case \[\frac{dP}{dt}=-kP\] and \[P(0)=200\]

  34. satellite73
    • 2 years ago
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    we cannot do 3 or 4 without a calculator, but i can show you how to do them if you like

  35. lovelymultani
    • 2 years ago
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    wouldnt a be kt ? from the original decay equation?

  36. lovelymultani
    • 2 years ago
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    ok nvm i got it ! thanks :)

  37. satellite73
    • 2 years ago
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    yw

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