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The halflife of censium137 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?
 one year ago
 one year ago
The halflife of censium137 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?
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.2
\[p(t)=200\left(\frac{1}{2}\right)^{\frac{t}{30}}\] is the quick and easy answer, but perhaps not the one your math teacher wants
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
differentiated would be p(t) = .5^t/30 ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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 \[e^{30k}=\frac{1}{2}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
you get \[30k=\ln(.5)\] or \[k=\frac{\ln(.5)}{30}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
not sure what it means by "simplify" maybe write \[k=\frac{\ln(.5)}{30}=.0231\] and so your function is \[p(t)=200e^{.0231t}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
ok i believe i have confused youi lets go slow
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
we can't use calculators on our final, so I have no way of getting .0231t . Is there an alternative?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
allright i guess I cant solve it, maybe they jst want the formulas.
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
then you get \(k=\frac{\ln(\frac{1}{2})}{30}\) and leave it at taht
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
therefore you function will look like \[p(t)=200e^{kt}\] but you do not yet know \(k\), we have to solve for it
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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 \(p(30)=100\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
you get a) \[200e^{30k}=100\]b)\[e^{30k}=\frac{1}{2}\] we could have gone right to the second equation
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
solve via \[30k=\ln(\frac{1}{2})\] \[k=\frac{\ln(\frac{1}{2})}{30}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
now if you are not allowed to use a calculator, then that is the best you can do you can write \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
which is very strange to me, but if you say so then i believe you
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
to answer the next question replact \(t\) by \(75\), that is, compute \[p(75)\]
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
yeaa i go to city college in new york and they arent allowing any calcs for calc or precal
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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 \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
they are not asking for the derivative, they are asking for the "differential equation"
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
the answer to the first question is \[\frac{dP}{dt}=kP\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
or maybe your text uses \[\frac{dP}{dt}=kP\] since this is for decay it makes no real different, since \(k\) is negative
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
this is the same as saying the instantaneous rate of change is proportional to the current amount
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
the solution to \[\frac{dP}{dt}=kP\] is \[P(t)=P_0e^{kt}\] which is what we solved above
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i guess i should have said in your case \[\frac{dP}{dt}=kP\] and \[P(0)=200\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
we cannot do 3 or 4 without a calculator, but i can show you how to do them if you like
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
wouldnt a be kt ? from the original decay equation?
 one year ago

lovelymultaniBest ResponseYou've already chosen the best response.0
ok nvm i got it ! thanks :)
 one year ago
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