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Mathematics
OpenStudy (anonymous):

can someone help me with my problem?

OpenStudy (anonymous):

and what is your problem jmfranco?

OpenStudy (anonymous):

2) We often use radio-isotopes to estimate the age of things (such as the Earth). Assuming the half-life of some isotope is 1000 years, and you found a rock of unknown age near the Mid Atlantic Ridge. You measured the amount of isotope in the rock and determined that 12.5% of the original amount when the rock formed was still present. How old is the rock

OpenStudy (anonymous):

Let's have a chat here franco. This is a growth and decay problem. Is that the topic you're studying?

OpenStudy (anonymous):

ok..yes it is

OpenStudy (anonymous):

great. now, sometimes growth and decay have been modeled by exponential functions. This is the model common model and the one you'll be using here. Do you know the form of the exponential function that is used to model growth and decay? is so, type it in here...

OpenStudy (anonymous):

I hope you don't mind that I'm not just giving you the answer, but you'll be better able to solve all the other problems you'll have this way. it will only take a few minutes.

OpenStudy (anonymous):

okay i don't really know exponential. i understand

OpenStudy (anonymous):

r us still there?

OpenStudy (anonymous):

yes, i'm here

OpenStudy (anonymous):

my connection just failed me for a bit, but i'm back

OpenStudy (anonymous):

this website is having a lot of traffic and seems slow, check out this site has a good explanation: http://serc.carleton.edu/quantskills/methods/quantlit/expGandD.html

OpenStudy (anonymous):

take a look and let me know if you have any questions

OpenStudy (anonymous):

please let me know once you have briefly looked at the page...

OpenStudy (anonymous):

then we can choose between those two equations and use the information you were given to solve the problem. it's not hard at all and you'll see.

OpenStudy (anonymous):

ok looked at it its sort of complicated can u go step by step

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

so the two functions are n(t) = n0*e^(kt)

OpenStudy (anonymous):

now, the exponent of e is that decides whether we have a growth or decay. If it is negative, then the graph starts high and comes down (decay) and the opposite is true if the exponent is positive.

OpenStudy (anonymous):

with me so far?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

by the way, what grade you in? so I know what you sort of should know...

OpenStudy (anonymous):

college lol

OpenStudy (anonymous):

hehe, no worries. I just graduated a year ago myself. any how...i hope you're not rushed cuz I'm definitely not moving very quickly...let me know if you're rushed.

OpenStudy (anonymous):

so let's go over the equation in more detail...

OpenStudy (anonymous):

the site gives good description for each variable involved.

OpenStudy (anonymous):

lol cool. well i got this question and another, these questions r complicated and due at 4

OpenStudy (anonymous):

today sorry don't rush its okay.

OpenStudy (anonymous):

I'm gonna ask you to rephrase them for me here: what does N(t) mean? and the N_0(t) (that is the subscript zero), the k? the t?

OpenStudy (anonymous):

once you have those, we can replace the variables with the information in your problem. I just wanna make sure you're with me...

OpenStudy (anonymous):

Nt is quatity in time, t is time,n0 is initial quantity, k is a constant and eX is exponential function

OpenStudy (anonymous):

good

OpenStudy (anonymous):

so, it's N(t) and it is a function of time. It's very important that you understand that. and eX is e^(x) which is the mathematically constant e (approx 2.71..) to power x.

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

so what the formulate says is N(t) [the amount as a function of time] = n0*e^(k*t) [the original amount times the exponential function to power k times t]

OpenStudy (anonymous):

k here is the rate of change. the larger is it, the faster the amount grows or, in this case, decays.

OpenStudy (anonymous):

What is the question asking of us? what do you think will be the variable that will be the final answer?

OpenStudy (anonymous):

stay with me franco, we're almost there

OpenStudy (anonymous):

not sure ive never done a prblem like this one

OpenStudy (anonymous):

the question asks you how old is the rock when only 12.5% of it is left..

OpenStudy (anonymous):

so we're looking for time, which will be contained in our variable t.

OpenStudy (anonymous):

a clue that is give to you without knowing it is the term "half-life"

OpenStudy (anonymous):

how long is the half-life of this rock?

OpenStudy (anonymous):

Franco?

OpenStudy (anonymous):

1000

OpenStudy (anonymous):

thank you.

OpenStudy (anonymous):

so that means that in a 1000 years, we'll only have 50% of the original material left. that IS what half life means.

OpenStudy (anonymous):

okay so what do i do??

OpenStudy (anonymous):

so let's set up the equation

OpenStudy (anonymous):

actually, I'll let you do it: in 1000 years, we will only have 50% of the original material. can you set that up?

OpenStudy (anonymous):

i'll prompt with good questions: first: what is t?

OpenStudy (anonymous):

nooo :( i don't kno how this is complicateddd!

OpenStudy (anonymous):

okay: time in this case is 1000 years. second: notice how you can divide both sides by n0 (the original material) - once we do this, what do we have on the left side of the equation?

OpenStudy (anonymous):

this is only solving the first component of the problem.any way, when we divide both sides by n0, we get the equation: N(t)/n0 = e^(kt)

OpenStudy (anonymous):

are you with me?

OpenStudy (anonymous):

in a thousand years, we know we will only have half left, so what will N(t)/n0 equal? think about this one cause it solves 50% of the problem!

OpenStudy (anonymous):

Franco? are you still with me?

OpenStudy (anonymous):

is n0 1000

OpenStudy (anonymous):

i didvide both sides by that?

OpenStudy (anonymous):

n0 is the original amount. was that given in the probem? 1000 is the number of years...

OpenStudy (anonymous):

before i loose you here, we divide the variable n0 by both sides so that we can remove it from the right side and put it on the left. you understand what this results in?

OpenStudy (anonymous):

I miss half life problems, what is that calculus 1

OpenStudy (anonymous):

no what is the asnwer/

OpenStudy (anonymous):

now the equation looks like this: \[n(t) \div n _{0} = e ^{k*t}\]

OpenStudy (anonymous):

no, there is no calculus involved. this is still pre-calc

OpenStudy (anonymous):

do you understand how we get that equation?

OpenStudy (amanda):

how do you know so much about math thierryu

OpenStudy (anonymous):

I studied engineering in college, i just graduated last year

OpenStudy (anonymous):

are you here to help us Amanda?

OpenStudy (anonymous):

Franco, do you see how we got that last equation I put in?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

good

OpenStudy (anonymous):

Now, if it has been a 1000 years and we only have 50% of the original materia.

OpenStudy (anonymous):

so now if t=1000, we're gonna look at n0/N(1000). All we know about this fraction is that N(t) at t=1000 is 50% of n0, the original amount.

OpenStudy (anonymous):

digest that.

OpenStudy (anonymous):

therefore, N(t)/n0 when t=1000 years should give us 1/2. do you agree?

OpenStudy (anonymous):

I meant to say N(t)/n0 all along. sorry for the confusion.

OpenStudy (anonymous):

bradley, you found me here!

OpenStudy (anonymous):

should just got to wolfram mathmatica, and type in halflife

OpenStudy (anonymous):

Franco, are you with me? do you see why N(t)/n0 at the half-time (t=1000) will give us 1/2 ?

OpenStudy (anonymous):

so wats the answer??? yea i see

OpenStudy (anonymous):

i did but i will let you help him first

OpenStudy (anonymous):

alright. so now we have 1/2 = e^(k*t)

OpenStudy (anonymous):

im a girl

OpenStudy (anonymous):

and t= 1000 so we really have .5=e^(1000k)

OpenStudy (anonymous):

franco plug in your time and your values for n and No

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

now we solve for k by first takiing the natural log of both sides: ln(.5) = ln(e^(1000k))

OpenStudy (anonymous):

we do this because we know that the ln(e^(x)) equals x.

OpenStudy (anonymous):

so ln(e^(1000k)) reduces to simply 1000k!

OpenStudy (anonymous):

now you can use your calculator to solve for k from teh equation: \[\ln(.5) = 1000*k\]

OpenStudy (anonymous):

what is k?

OpenStudy (anonymous):

once you find k, we go back to the board, but now we have everything we need expect time which is exactly what we're looking for!

OpenStudy (anonymous):

umm not sure..

OpenStudy (anonymous):

why? because we only have 12.5% left. so the ratio of N(t)/n0 = .125

OpenStudy (anonymous):

you're not sure about k?

OpenStudy (anonymous):

there you'd just use a calculator; get the natural log of .5 and then divide it by 1000

OpenStudy (anonymous):

are you with me now?

OpenStudy (anonymous):

i know this was elongated by if you go over this chat and solve the same or a similar problem, you'll get a hang of it. don't be discouraged.

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