Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

ok i believe i have confused youi
lets go slow

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

that is weird

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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 \(p(30)=100\)

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

hold on

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

wouldnt a be kt ? from the original decay equation?

Answer 36

ok nvm i got it ! thanks :)

Answer 37

yw