## 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?

• This Question is Open
1. satellite73

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

2. lovelymultani

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

3. satellite73

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

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

5. satellite73

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

ok i believe i have confused youi lets go slow

7. lovelymultani

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

8. satellite73

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

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

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

11. satellite73

that is weird

12. lovelymultani

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

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

14. satellite73

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

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

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

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

18. satellite73

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

19. satellite73

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

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

21. lovelymultani

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

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

23. lovelymultani

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

24. satellite73

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

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

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

hold on

28. satellite73

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

29. satellite73

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

30. satellite73

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

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

32. satellite73

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

33. satellite73

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

34. satellite73

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

35. lovelymultani

wouldnt a be kt ? from the original decay equation?

36. lovelymultani

ok nvm i got it ! thanks :)

37. satellite73

yw