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?

- anonymous

- schrodinger

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- anonymous

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

- anonymous

differentiated would be p(t) = .5^t/30 ?

- anonymous

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
\[e^{30k}=\frac{1}{2}\]

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## More answers

- anonymous

you get
\[30k=\ln(.5)\] or
\[k=\frac{\ln(.5)}{30}\]

- anonymous

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

- anonymous

ok i believe i have confused youi
lets go slow

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

allright i guess I cant solve it, maybe they jst want the formulas.

- anonymous

that is weird

- anonymous

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

- anonymous

then you get \(k=\frac{\ln(\frac{1}{2})}{30}\) and leave it at taht

- anonymous

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

- anonymous

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

- anonymous

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 \(p(30)=100\)

- anonymous

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

- anonymous

solve via
\[30k=\ln(\frac{1}{2})\]
\[k=\frac{\ln(\frac{1}{2})}{30}\]

- anonymous

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

- anonymous

which is very strange to me, but if you say so then i believe you

- anonymous

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

- anonymous

to answer the next question
replact \(t\) by \(75\), that is, compute
\[p(75)\]

- anonymous

yeaa i go to city college in new york and they arent allowing any calcs for calc or precal

- anonymous

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
\[p(t)=200e^{\frac{\ln(.5)}{30}t}\]

- anonymous

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

- anonymous

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

- anonymous

hold on

- anonymous

they are not asking for the derivative, they are asking for the "differential equation"

- anonymous

the answer to the first question is
\[\frac{dP}{dt}=kP\]

- anonymous

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

- anonymous

this is the same as saying the instantaneous rate of change is proportional to the current amount

- anonymous

the solution to
\[\frac{dP}{dt}=-kP\] is
\[P(t)=P_0e^{-kt}\] which is what we solved above

- anonymous

i guess i should have said in your case \[\frac{dP}{dt}=-kP\] and \[P(0)=200\]

- anonymous

we cannot do 3 or 4 without a calculator, but i can show you how to do them if you like

- anonymous

wouldnt a be kt ? from the original decay equation?

- anonymous

ok nvm i got it ! thanks :)

- anonymous

yw

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