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lovelymultani
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?
\[p(t)=200\left(\frac{1}{2}\right)^{\frac{t}{30}}\] is the quick and easy answer, but perhaps not the one your math teacher wants
differentiated would be p(t) = .5^t/30 ?
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 \[e^{30k}=\frac{1}{2}\]
you get \[30k=\ln(.5)\] or \[k=\frac{\ln(.5)}{30}\]
not sure what it means by "simplify" maybe write \[k=\frac{\ln(.5)}{30}=-.0231\] and so your function is \[p(t)=200e^{-.0231t}\]
ok i believe i have confused youi lets go slow
we can't use calculators on our final, so I have no way of getting -.0231t . Is there an alternative?
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
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
allright i guess I cant solve it, maybe they jst want the formulas.
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
then you get \(k=\frac{\ln(\frac{1}{2})}{30}\) and leave it at taht
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
therefore you function will look like \[p(t)=200e^{kt}\] but you do not yet know \(k\), we have to solve for it
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 \(p(30)=100\)
you get a) \[200e^{30k}=100\]b)\[e^{30k}=\frac{1}{2}\] we could have gone right to the second equation
solve via \[30k=\ln(\frac{1}{2})\] \[k=\frac{\ln(\frac{1}{2})}{30}\]
now if you are not allowed to use a calculator, then that is the best you can do you can write \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]
which is very strange to me, but if you say so then i believe you
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
to answer the next question replact \(t\) by \(75\), that is, compute \[p(75)\]
yeaa i go to city college in new york and they arent allowing any calcs for calc or precal
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 \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]
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
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
they are not asking for the derivative, they are asking for the "differential equation"
the answer to the first question is \[\frac{dP}{dt}=kP\]
or maybe your text uses \[\frac{dP}{dt}=-kP\] since this is for decay it makes no real different, since \(k\) is negative
this is the same as saying the instantaneous rate of change is proportional to the current amount
the solution to \[\frac{dP}{dt}=-kP\] is \[P(t)=P_0e^{-kt}\] which is what we solved above
i guess i should have said in your case \[\frac{dP}{dt}=-kP\] and \[P(0)=200\]
we cannot do 3 or 4 without a calculator, but i can show you how to do them if you like
wouldnt a be kt ? from the original decay equation?
ok nvm i got it ! thanks :)