A community for students.
Here's the question you clicked on:
 0 viewing
lovelymultani
 3 years 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?
lovelymultani
 3 years 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?

This Question is Open

satellite73
 3 years ago
Best 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

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0differentiated would be p(t) = .5^t/30 ?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2you 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 \[e^{30k}=\frac{1}{2}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2you get \[30k=\ln(.5)\] or \[k=\frac{\ln(.5)}{30}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2not sure what it means by "simplify" maybe write \[k=\frac{\ln(.5)}{30}=.0231\] and so your function is \[p(t)=200e^{.0231t}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2ok i believe i have confused youi lets go slow

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2you 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2in 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

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0allright i guess I cant solve it, maybe they jst want the formulas.

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2then you get \(k=\frac{\ln(\frac{1}{2})}{30}\) and leave it at taht

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2ok lets make sure you understand it you know that the initial value is \(200\) so you know \(p_0=200\) what you start with

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2therefore you function will look like \[p(t)=200e^{kt}\] but you do not yet know \(k\), we have to solve for it

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2but 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 \(p(30)=100\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2you get a) \[200e^{30k}=100\]b)\[e^{30k}=\frac{1}{2}\] we could have gone right to the second equation

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2solve via \[30k=\ln(\frac{1}{2})\] \[k=\frac{\ln(\frac{1}{2})}{30}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2now if you are not allowed to use a calculator, then that is the best you can do you can write \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2which is very strange to me, but if you say so then i believe you

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0That 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2to answer the next question replact \(t\) by \(75\), that is, compute \[p(75)\]

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2the 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 \[p(t)=200e^{\frac{\ln(.5)}{30}t}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2now 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

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0allright 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2they are not asking for the derivative, they are asking for the "differential equation"

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2the answer to the first question is \[\frac{dP}{dt}=kP\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2or maybe your text uses \[\frac{dP}{dt}=kP\] since this is for decay it makes no real different, since \(k\) is negative

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2this is the same as saying the instantaneous rate of change is proportional to the current amount

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2the solution to \[\frac{dP}{dt}=kP\] is \[P(t)=P_0e^{kt}\] which is what we solved above

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2i guess i should have said in your case \[\frac{dP}{dt}=kP\] and \[P(0)=200\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2we cannot do 3 or 4 without a calculator, but i can show you how to do them if you like

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0wouldnt a be kt ? from the original decay equation?

lovelymultani
 3 years ago
Best ResponseYou've already chosen the best response.0ok nvm i got it ! thanks :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.