Math 1020, Fall 2015, Activity #09

Monday, September 21(/Wednesday, Sept. 23?): Beyond the linear model…

Learning Outcomes:

• Length-to-volume relationships and power functions

• Log-Log plots and their relationship to power function models

• Logarithm base ten as a measure of the “size” (“order of magnitude”) of a number

• Reinforcing the practical interpretation of variables in a modeling equation

NAME(s)______

The linear model was a good fit for the original data because the lake levels covered a small range of values. This is a phenomenon known as “local linearity,” which says that most quantities which can be modeled with a mathematical formula can have that formula simplified to a straight-line equation over small value ranges. (An example of this can be seen in, say, North Dakota or eastern Montana; looking at the small patch of earth you can see from the ground, the earth looks flat - - “linear” - - even though it’s clearly curved when seen from space). So, if we only wanted to predict lake volumes for the levels we’re likely to see under ordinary circumstances, the linear model would be just fine. (In technical terms, our model is valid for interpolation.) But it sometimes happens that lake levels drop or rise precipitously, to values outside our model’s valid range. This can be due to natural phenomena (extended periods of drought or unusually heavy rainfall), or to human interference (such as extensive groundwater pumping, which is suspected as a contributing cause of the recent drop in White Bear Lake). Recently, the Anderson Lakes (in western Bloomington and eastern Eden Prairie) chain was completely drained dry to try to kill off an invasive weed species! The engineers involved in this project needed an accurate prediction of the volume of water they’d be dealing with at each stage of the operation.

For a model which is valid for a broader range of depth (x) values, a more realistic equation would be a power function equation: , where C and r are specific numerical values (called “constants”) which we’ll need to work out. Don’t worry, EXCEL® will do most of the tedious number-crunching.

To help convince you that a power function would make sense when finding a volume from a depth:

•If a cube has sides of length x, its volume is

•If a sphere has radius r, its volume is .

Both of these formulas are power functions, which say that the volume of some object is proportional to a length raised to the third power.

Question 1.We are trying to find a formula giving the lake volume, y, in terms of the lake depth, x. Based on the two volume formulas above, take a guess at the power to which x might be raised in such a formula.

OK. In the LakeLevelData.xls spreadsheet, find “sheet 3.” This has columns entitled “logarithm of…”. We’re going to graph the data from columns D and E, logarithms of depth and volume. Why logarithms? you ask…

Note: This class is not going to get very deeply into what a logarithm is, from a technical standpoint. For our purposes, all you need to know is that the logarithm of a number is a way to measure its order of magnitude.

For example, 1 million is 1,000,000, which is . Thus, since the lake volumes in our data are measured in millions of cubic meters, they are all said to be “on the order of” , which is a way of saying what scale the measurements are on, without worrying about being too precise. The logarithm of a number which is on the order of will be “6 point something,” where the 6 tells us the number is “on the order of” a million or so, and the decimal part tells more precisely what value the number has.

Logarithms also have some nice algebraic properties which we’re not going to get into. The only thing about that we’re going to use is that, if you make a graph using the logarithms of x and y, and it gives a linear equation of log y = m·log x + b, then it follows that the variables x and y satisfy the equation .

OK, got that? No??? Just follow the instructions, it’s really not that bad, I promise.

Back to columns D and E from Sheet 3. Make a scatterplot of the data and have EXCEL® calculate the linear regression equation. Print the graph, with the trendline.

Question 2. Just from looking at the graph you just made, does the linear model seem to be a good “fit” for the data points? Justify your answer based on earlier discussions from the course.

Question 3. Write the linear equation you get, exactly as it appears on the graph, in the space below:

Question 4. Now in the equation you just wrote, replace “y” with “log y” and replace “x” with “log x,” and write the result in the space below:

Compare what you just wrote (question 4) with the equation which is boxed, in the 2nd paragraph on the preceding page.

Question 5. What is the value of m in the equation you just wrote?

Question 6. What is the value of b in the equation you just wrote?

Question 7.Use a calculator to find , to 2 places after the decimal.

Question 8. Finally, use the pattern to write our new, more sophisticated (and hopefully more reliable for extrapolated values) “power rule” model for lake volume:

Question 9. Look two pages back, Question 1, where you guessed what the power of x might be in our power - function volume model. Was your guess close to the actual power we ended up with?

We’ll analyze the accuracy and “reasonableness” of this model next, with an eye to the difficulties our linear model had when trying (and failing) to accurately predict the depth of Bush Lake by seeing what the model said to expect when the volume, y, of the lake was 0 .

Since the linear model was pretty good as far as interpolation went, we want to make sure that the new, power function model is, at least, no worse, so we’ll start by finding some relative errors from using the new model to predict volumes from depths for which we currently have data.

First a re-cap of what we’ve done so far:

Question 10. Write the power-rule equation we derived in the space below (this is the answer to Question 8 on the previous page). Note: This model gives volumes in cubic meters, not millions of cubic meters.

Question 11. Which variable represents lake depth, x or y?

Question 12. Which variable represents lake volume, x or y?

Now let’s check the accuracy of this new model in matching the actual values we had from our original data (as given in Sheet 2 of the spreadsheet LakeLevelData.xls; printout provided below;“maximum depth” and “volume”).

maximum depth (feet) / est. lake volume (millions of m^3)
33.46 / 1.7903
33.9 / 1.8719
33.87 / 1.8664
34.03 / 1.896
34.28 / 1.9423
34.85 / 2.048
34.11 / 1.9108
35.08 / 2.0906
37.41 / 2.5217
37.35 / 2.5113

Question 13. Using our new, power-rule model for predicting lake volume from lake depth (see Question 10), what volume is predicted for a depth of 34.28 feet? Answer in cubic meters. Round to the nearest whole number.

Question 14.According to the table above, what is the actual, measured lake volume when the depth is 34.28feet? Answer in cubic meters. Round to the nearest whole number.

Question 15. What is the error when using the new model to predict the lake volume when the depth is 34.28 feet? Answer in cubic meters. (Seepage 4 of Activity 7 for the definition of “error.”)

Question 16. What is the relative error when using the new power rule model to predict the lake volume when the depth is 34.28 feet? Answer as a percentage. (See page 4 of Activity 7 for the definition of “relative error.”)

Question 17. What was the relative error from using the linear model to predict lake volume when the lake elevation was 832.28 feet above sea level? (This was Question 14, activity 7 – just write that answer here, as a percent.)

Question 18. A depth of 34.28 feet corresponds to the lake surface elevation of 832.28 feet, so the relative errors in the preceding two questions are from predictions of volume at the same lake level. Given that, which model seems more accurate, the linear model or the power rule model? Briefly explain your answer.

Question 19. Using our new model, what volume is predicted for a depth of 37.41 feet? Answer in cubic meters.

Question 20. According to the table on the preceding page, what is the lake volume when the depth is 37.41feet? Answer in cubic meters.

Question 21. What is the error when using the new model to predict the lake volume when the depth is 37.41feet? Answer in cubic meters.

Question 22. What is the relative error when using the new model to predict the lake volume when the depth is 37.41feet? Answer as a percentage.

Question 23. What was the relative error from using the linear model to predict lake volume when the lake elevation was 835.41 feet above sea level? (This was Question 19, activity 7 – just write that answer here, as a percent.)

Question 24. A depth of 37.41feet corresponds to the lake surface elevation of 835.41 feet, so the relative errors in the preceding two questions are from predictions of volume at the same lake level. Given that, which model seems more accurate, the linear model or the power rule model? Briefly explain your answer.

So, that’s interpolation, which the linear model did just fine. Next time we’ll try some extrapolation.