anonymous
  • anonymous
can someone help me with my problem?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
and what is your problem jmfranco?
anonymous
  • 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
anonymous
  • anonymous
Let's have a chat here franco. This is a growth and decay problem. Is that the topic you're studying?

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anonymous
  • anonymous
ok..yes it is
anonymous
  • 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...
anonymous
  • 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.
anonymous
  • anonymous
okay i don't really know exponential. i understand
anonymous
  • anonymous
r us still there?
anonymous
  • anonymous
yes, i'm here
anonymous
  • anonymous
my connection just failed me for a bit, but i'm back
anonymous
  • 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
anonymous
  • anonymous
take a look and let me know if you have any questions
anonymous
  • anonymous
please let me know once you have briefly looked at the page...
anonymous
  • 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.
anonymous
  • anonymous
ok looked at it its sort of complicated can u go step by step
anonymous
  • anonymous
sure
anonymous
  • anonymous
so the two functions are n(t) = n0*e^(kt)
anonymous
  • 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.
anonymous
  • anonymous
with me so far?
anonymous
  • anonymous
yes
anonymous
  • anonymous
by the way, what grade you in? so I know what you sort of should know...
anonymous
  • anonymous
college lol
anonymous
  • 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.
anonymous
  • anonymous
so let's go over the equation in more detail...
anonymous
  • anonymous
the site gives good description for each variable involved.
anonymous
  • anonymous
lol cool. well i got this question and another, these questions r complicated and due at 4
anonymous
  • anonymous
today sorry don't rush its okay.
anonymous
  • 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?
anonymous
  • 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...
anonymous
  • anonymous
Nt is quatity in time, t is time,n0 is initial quantity, k is a constant and eX is exponential function
anonymous
  • anonymous
good
anonymous
  • 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.
anonymous
  • anonymous
okay
anonymous
  • 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]
anonymous
  • anonymous
k here is the rate of change. the larger is it, the faster the amount grows or, in this case, decays.
anonymous
  • anonymous
What is the question asking of us? what do you think will be the variable that will be the final answer?
anonymous
  • anonymous
stay with me franco, we're almost there
anonymous
  • anonymous
not sure ive never done a prblem like this one
anonymous
  • anonymous
the question asks you how old is the rock when only 12.5% of it is left..
anonymous
  • anonymous
so we're looking for time, which will be contained in our variable t.
anonymous
  • anonymous
a clue that is give to you without knowing it is the term "half-life"
anonymous
  • anonymous
how long is the half-life of this rock?
anonymous
  • anonymous
Franco?
anonymous
  • anonymous
1000
anonymous
  • anonymous
thank you.
anonymous
  • 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.
anonymous
  • anonymous
okay so what do i do??
anonymous
  • anonymous
so let's set up the equation
anonymous
  • 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?
anonymous
  • anonymous
i'll prompt with good questions: first: what is t?
anonymous
  • anonymous
nooo :( i don't kno how this is complicateddd!
anonymous
  • 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?
anonymous
  • 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)
anonymous
  • anonymous
are you with me?
anonymous
  • 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!
anonymous
  • anonymous
Franco? are you still with me?
anonymous
  • anonymous
is n0 1000
anonymous
  • anonymous
i didvide both sides by that?
anonymous
  • anonymous
n0 is the original amount. was that given in the probem? 1000 is the number of years...
anonymous
  • 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?
anonymous
  • anonymous
I miss half life problems, what is that calculus 1
anonymous
  • anonymous
no what is the asnwer/
anonymous
  • anonymous
now the equation looks like this: \[n(t) \div n _{0} = e ^{k*t}\]
anonymous
  • anonymous
no, there is no calculus involved. this is still pre-calc
anonymous
  • anonymous
do you understand how we get that equation?
Amanda
  • Amanda
how do you know so much about math thierryu
anonymous
  • anonymous
I studied engineering in college, i just graduated last year
anonymous
  • anonymous
are you here to help us Amanda?
anonymous
  • anonymous
Franco, do you see how we got that last equation I put in?
anonymous
  • anonymous
yes
anonymous
  • anonymous
good
anonymous
  • anonymous
Now, if it has been a 1000 years and we only have 50% of the original materia.
anonymous
  • 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.
anonymous
  • anonymous
digest that.
anonymous
  • anonymous
therefore, N(t)/n0 when t=1000 years should give us 1/2. do you agree?
anonymous
  • anonymous
I meant to say N(t)/n0 all along. sorry for the confusion.
anonymous
  • anonymous
bradley, you found me here!
anonymous
  • anonymous
should just got to wolfram mathmatica, and type in halflife
anonymous
  • anonymous
Franco, are you with me? do you see why N(t)/n0 at the half-time (t=1000) will give us 1/2 ?
anonymous
  • anonymous
so wats the answer??? yea i see
anonymous
  • anonymous
i did but i will let you help him first
anonymous
  • anonymous
alright. so now we have 1/2 = e^(k*t)
anonymous
  • anonymous
im a girl
anonymous
  • anonymous
and t= 1000 so we really have .5=e^(1000k)
anonymous
  • anonymous
franco plug in your time and your values for n and No
anonymous
  • anonymous
ok
anonymous
  • anonymous
now we solve for k by first takiing the natural log of both sides: ln(.5) = ln(e^(1000k))
anonymous
  • anonymous
we do this because we know that the ln(e^(x)) equals x.
anonymous
  • anonymous
so ln(e^(1000k)) reduces to simply 1000k!
anonymous
  • anonymous
now you can use your calculator to solve for k from teh equation: \[\ln(.5) = 1000*k\]
anonymous
  • anonymous
what is k?
anonymous
  • 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!
anonymous
  • anonymous
umm not sure..
anonymous
  • anonymous
why? because we only have 12.5% left. so the ratio of N(t)/n0 = .125
anonymous
  • anonymous
you're not sure about k?
anonymous
  • anonymous
there you'd just use a calculator; get the natural log of .5 and then divide it by 1000
anonymous
  • anonymous
are you with me now?
anonymous
  • 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.
anonymous
  • anonymous
is it negative .0003
anonymous
  • anonymous
great. i didn't check but it looks about right
anonymous
  • anonymous
so on to solving the problem
anonymous
  • anonymous
i really havve to go can u just solve for me?
anonymous
  • anonymous
we only have 12.5% left; which means the ratio of N(t)/n0 = 12.5% = .125.
anonymous
  • anonymous
so since we now have k
anonymous
  • anonymous
we end up with \[.125 = e^(-.0003t)\]
anonymous
  • anonymous
remember how we solved for k? we do the same for t now
anonymous
  • anonymous
ln(.125) = ln(e^(-.0003*t)) = -.0003*t
anonymous
  • anonymous
so \[\ln(.125) = -.0003*t\]
anonymous
  • anonymous
can you solve for t? and that's gonna be the number of years that has passed for us to only have 12.5% of the rock left.
anonymous
  • anonymous
save this chat and go over it when you get a chance and you should be able to do this types of problems on your own.
anonymous
  • anonymous
did you find t?
anonymous
  • anonymous
I, for one, would love to know what it is...
anonymous
  • anonymous
no
anonymous
  • anonymous
divde ln(.125) by .0003
anonymous
  • anonymous
yes, by negative .0003
anonymous
  • anonymous
so how long will have passed franco?
anonymous
  • anonymous
6900?
anonymous
  • anonymous
about that much
anonymous
  • anonymous
sounds about right. definitely way more than a thousand because it is way past the 50% mark. so that makes sense.
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
you're welcome.
anonymous
  • anonymous
bradley, did you solve yours?
anonymous
  • anonymous
nope, i am still lost
anonymous
  • anonymous
would you care to chat on the other thread
anonymous
  • anonymous
since you're not rushed and i just got home, give me about half an hour to grab a snack and start dinner and I'll be back. I'm putting on a timer for 30 minutes sharp.
anonymous
  • anonymous
I may also have to review a little calculus ;)
anonymous
  • anonymous
haha okay i will be around
anonymous
  • anonymous
could you post the entire problem for me here again please?
anonymous
  • anonymous
using lagrange multipliers find the maxium and minimum values of the function f(x,y,z)=xy^2z^3 subject to the contraints x^2+y^2+2z^2=25

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