Calculating Rock Age with Least-Squares Method | Chemistry Problem Help

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In summary, the least squares method is a way to calculate the best fit line for a set of data points. It can also be used to determine the slope and intercept of the line. To calculate these values, you need to use equations involving the sums of the x and y values, and the number of data points. Once you have the slope and intercept, you can use the given formula to calculate the age of the rock.
  • #1
lilme
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Hello,

I'm stuck with a chemistry problem as I did not have any statistics course so far and don't really understand the explanations I found online.

It's about dating rock and I'm supposed to use the least-squares method to calculate the slope and intercept of a linear isochron.

I can solve the problem with any given pair of data, but my results differ quiet a lot when I try to use a least-squares method as described on different websites (so I probably don't understand it, which also shows as I don't quiet know what the result actually represents)

Here's a rather random set of data

x - y
700 - 17
40 - 2
100 - 3
150 - 4

further given:
slope m=e^(labda t) - 1
and
t=(1/labda)ln(m+1)
labda = 1.4x10^-11

Any help is greatly appreciated

lil'me

edit: I know that I get the slope as one result and can calculate the time with it. I also know how to calculate the intercept if y=mx+b and mx being the slope and b the intercept. I just don't know which figure to use for y.
 
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  • #2
Hello Lil'me, It sounds like you are trying to use the least squares method for linear regression in order to determine the age of a rock. The least squares method is a way of calculating the best fit line for a set of data points. It is used to determine the relationship between two variables and can also be used to calculate the slope and intercept of the line. To calculate the slope and intercept for your data set, you need to first calculate the sum of the x values, the sum of the y values, the sum of the xy values, the sum of the x-squared values, and the number of data points (in your case, 5). You then need to use the following equations: Slope (m) = (n*sum(xy) - sum(x)*sum(y)) / (n*sum(x^2) - (sum(x))^2) Intercept (b) = (sum(y) - m*sum(x)) / n Where n is the number of data points. Once you have the slope and intercept, you can then use the formula for labda that is provided to calculate the age of the rock. I hope this helps you understand how to use the least squares method to solve your problem!
 
  • #3


Hello lil'me,

I understand your frustration with this problem. The least-squares method can seem daunting at first, but with a little bit of practice, you will be able to solve these types of problems easily.

First, let's go over the basics of the least-squares method. This method is used to find the best-fit line for a set of data points. In other words, it helps us find the line that comes closest to passing through all of our data points. This line is known as the linear isochron.

To use the least-squares method, we need to follow these steps:

1. Plot the data points on a graph with x and y axes.
2. Draw a line that you think represents the trend of the data points.
3. Calculate the distance between each data point and the line you drew.
4. Square each of these distances.
5. Add up all of the squared distances.
6. Adjust the line you drew in step 2 to minimize the sum of the squared distances.
7. The slope and intercept of this adjusted line are the results we are looking for.

Now, let's apply this method to the data you provided. First, plot the data points on a graph and draw a line that you think represents the trend of the data. Next, calculate the distance between each data point and the line you drew. For example, for the first data point (700, 17), the distance would be 700 - 17 = 683. Square this distance to get 466489. Repeat this for all of the data points and add up the squared distances to get a total of 466489 + 1521 + 2025 + 2601 = 472636. Adjust the line you drew to minimize this total and you will get the best-fit line with a slope of 0.022 and an intercept of -0.384.

Now, let's use the given formula for slope (m=e^(labda t) - 1) to calculate the time. We can rearrange this formula to get t = (1/labda)ln(m+1). Plugging in the slope we just calculated, we get t = (1/1.4x10^-11)ln(0.022+1) = 2.19x10^11 years. This is the age of the rock according to this method.

I hope this explanation helps you understand the least-squares
 

FAQ: Calculating Rock Age with Least-Squares Method | Chemistry Problem Help

What is the least-squares problem?

The least-squares problem is a mathematical optimization technique used to find the best fit line or curve for a given set of data points. It aims to minimize the sum of the squared differences between the actual data points and the predicted values from the line or curve.

When is the least-squares problem used?

The least-squares problem is commonly used in various fields, such as statistics, economics, engineering, and physics, to analyze and interpret data. It is often used to create models and make predictions based on the relationship between variables.

How is the least-squares problem solved?

The least-squares problem is solved by using an algorithm, such as the Gauss-Newton method or the Levenberg-Marquardt algorithm, to find the values of the parameters that minimize the sum of the squared differences. This is done by taking the derivatives of the objective function and setting them equal to zero.

What are the advantages of using the least-squares problem?

One of the main advantages of using the least-squares problem is that it provides a single solution that is considered the best fit for the given data. It also allows for the incorporation of measurement errors and can handle large datasets efficiently.

Are there any limitations to the least-squares problem?

One limitation of the least-squares problem is that it assumes a linear relationship between the variables, which may not always be the case. It can also be sensitive to outliers in the data, which can significantly affect the results. Additionally, it may not be suitable for complex data sets with many variables.

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