Jump to content
Sign in to follow this  

math hl IA type II radiometric dating

Recommended Posts

Guest mr_whatsisname

i am doing the radiometric dating IA in maths hl and i must say i am in quite a tough spot... i havent got the slightest idea how to conjecture an equation...i dont even have a sample of a Math Portfolio...can i please get some help here? i know there is a sample one but i need to be a vip member to view it...i just need an example....in kindah desperate... thx guys :wub: (math is definitely my weakest subject :D )

Share this post


Link to post
Share on other sites

1) Thought about switching to SL math? I did it and since you have 4 HLs you could too

2) Have you asked for help from your teacher on this?

Share this post


Link to post
Share on other sites
Guest simon_vargas

Hello, my teacher just gave me also a portfolio on radiometric dating, can anyone please help me, Im kind of lost here

Share this post


Link to post
Share on other sites

You know, we can't help you unless you tell us what you need to be helped in. So any specific question you'd like to ask?

Share this post


Link to post
Share on other sites
You know, we can't help you unless you tell us what you need to be helped in. So any specific question you'd like to ask?

Hey im new here i have an example that maybe can be usefull to u but i dont know how to send it if someone could teach me, it can be very helpful at times.

Share this post


Link to post
Share on other sites

im doing the same portfolio, and i have no clue where to start

i entered the data from part 1 into microsoft excel and created a scatter plot. the trendline tool gave me an equation for the data: y = 0.0005x4 - 0.0485x3 + 1.8934x2 - 32.269x + 209.8 but when i graph it, most of the points dont work

Share this post


Link to post
Share on other sites

Try removing the last line so that you can get an exponential decay curve, which is exactly how the points behave when they are modeled.

Share this post


Link to post
Share on other sites

Hi everyone :P I'm new here :P

I'm doing the Radiometric thing too :)

My guess is that for each roll it is expected that 1/6 of the dice will be 6, thus leaving 5/6 of them.

So perhaps the model would follow a simple decay equation which is Nn = No x (5/6)^n :)

I tried and it worked. The only prob is that sometimes I happened to be too lucky (or unlucky) to have to throw the dice much more fewer time than expected. It made the graph a little bit messy :)

But so far I'm stuck with the stuff relating to radiocarbon decay. Can anybody explain to me what is the e doing in the equation? :) I got quite confused with calculating this thing, let alone the fact that I've assumed too much from wikipedia... Would it do any fatal to my marks?

Share this post


Link to post
Share on other sites

http://mathworld.wolfram.com/ExponentialDecay.html

This site gives a good explanation on how they obtained the equation from the initial rate of decay. The 'e' is Euler's number, and in this case is used to 'cancel out' the natural logarithm of Ln(N/N0) which was obtained in the final step of the Mathworld website (integration of separated differential equation). If you remember, natural logarithms are logarithms with a base of e, Euler's number.

To brush up on integration of exponents/logarithms: http://ltcconline.net/greenl/courses/116/I...Integration.htm

To brush up on natural logarithms: http://en.wikipedia.org/wiki/Natural_logarithm

http://mathworld.wolfram.com/NaturalLogarithm.html

The Mathworld website helped clarify some linguistic issues with the IA and our non-IB calculus book helped while learning it for the first time.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.