Modeling a graph that shows age in relation to depth of an ice sample

[hey123]

Homework Statement

The thickness of the annually deposited ice layers in the ice core can be determined by examining chemical impurities and isotope ratios. Under pressure, ice exhibits the properties of a liquid. Therefore, the ice in the Greenland ice sheet is not only compressed, but also moves from the ice shelf towards the coasts. As a result, the layers of ice deposited each year become thinner with increasing depth. The following table shows the thickness of the ice layers in meters per year for the NEEM drill core:

Depth z in m 0 500 1000 1200 1400 1500 1600 Thickness per year λ in m a-1 0.25 0.20 0.13 0.10 0.037 0.018 0.010

Using the data from the table, create a graph that approximately shows the age t in a of an ice sample as a function of the depth z at which the sample was taken.

Relevant Equations

a= m* a/m, i.e. t=z/λ.

To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.

 

[pasmith]

You can use linear interpolation between the data points. Between $z_i$ and $z_{i+1}$ that gives you $$\int_{z_i}^z \frac{1}{\lambda(z)}\,dz = \int_{z_i}^z \frac{1}{A_i + B_iz}\,dz$$ which you can do analytically.

 

[haruspex]

A smoother result can be had by interpolating with splines, e.g. cubic splines: https://en.wikipedia.org/wiki/Spline_interpolation

 

[DEvens]

I don't have Excel installed right now so I can't graph your data. But here's what I would do.

Plot your data. Try to guess what a good shape for λ as a function of z. That is, find λ(z). Just eyeballing the data it looks like a straight line fit is not hopeless but not particularly good.

Then note that λ is the increase per year of z. That means $\frac{dz}{dt} = \lambda$ . That is, the rate of change of depth is just λ. Hopefully the shape of the curve you get is one that you can solve the differential equation for.

If you can't solve the differential equation, here is what you can do to approximate it. It's called Euler's methods. (Look it up if you are using it.) Start with z=0 at t=0 and your λ for z=0. Then note that one year later, by assuming a constant speed for the first year, you will have a z1 of λ(0). Then for year 1 to 2 assume a constant speed of λ(z1) where z1 is the depth at the end of year 1. So you get λ(0)+λ(z1) at the end of year 2 for z2. λ(0)+ λ(z1)+λ(z2) at the end of year 3 for z3. And so on. If you have Excel (or any spreadsheet prog) you should be able to sum that pretty quickly. And graph it at every 100 years or so. Or you could write a short program in your favorite computer program language.

 

[SebasHerbert]

 

[berkeman]

View attachment 370987

Keep in mind that we generally do not allow folks to do the OP's schoolwork for them, but in this case the thread is old enough that the OP has moved on and no longer has this as points-paying homework, so you posting that work is okay in this case.

 

[pasmith]

Keep in mind that we generally do not allow folks to do the OP's schoolwork for them, but in this case the thread is old enough that the OP has moved on and no longer has this as points-paying homework, so you posting that work is okay in this case.

Responses, like the original post, should be typed and make use of LaTeX rather than screenshots.