Follow the steps below and do your work in the provided GeoGebra windows. [br][i][br]NOTE to VSC students taking this class with me[/i]: be sure you are accessing this through a GeoGebra Classroom Link! The URL of this page should have the word “classroom” in it. If not, then go back to Canvas and be sure to access this page from the Project 1 page. Also: I strongly recommend you login to GeoGebra.org with a free account so your work is saved, and you can come back later to review or modify it. [br][br][i]For external readers[/i]: these "project" activities are meant to be taken as part of my course, so these sections of the GeoGebra book my not be as intelligible as others. My apologies. .

The primary goal of this project is to use [b]Newton's Law of Cooling[/b] to model heat dissipation. A secondary goal is that you will also gain deeper conceptual and procedural/algebraic knowledge of this process.

[list=1][*]In Geogebra declare the following numbers: [code]T_A=72[/code] and [code]k=0.2[br][/code] [/*][*]In Geogebra create the slope field for [math]\frac{dT}{dt}=-k\left(T-T_A\right)[/math] making reference to the numbers you declared in an earlier step. Remember: [i]T[/i] (temperature in degrees Fahrenheit) is the dependent variable, and [i]t[/i] (time in minutes) is independent. Therefore [i]T [/i]is playing the role of [i]y[/i], and [i]t[/i] is playing the role of [i]x[/i]. After you've entered the appropriate code you will have a visual representation of the solution space of Newton's law of cooling. Adjust the [i]y[/i]-axis so you can see a stable solution at [i]y[/i]=72. The slope field should be called [code]slopefield1[/code] in Geogebra. If not, rename it now. In a text box in Geogebra, describe the nature of the solutions.[/*][*]A cake is removed from an oven at minute 0 with a temperature measured at 300 degrees Fahrenheit. It is left to rest in a room with ambient temperature 72 degrees Fahrenheit. Plot a point in the plane that represents this initial measurement. Rename it [code]HotOutOfOven[/code][/*][*]Use the [code]Locus[/code] command to construct a numerical solution of the differential equation through the point [code]HotOutOfOven[/code]. The command is [code]Locus(slopefield1,HotOutOfOven)[/code]. When referencing objects in Geogebra commands, case matters. [/*][*]At minute 3 ([i]t[/i]=3) the cake is measured to be 200 degrees Fahrenheit. Plot a point in the plane that represents this measurement. Rename it [code]GettingCool[/code]. [/*][*]As you can see the numerical solution you created earlier does not go through [code]GettingCool[/code] exactly. The reason is not the numerical methods that are implemented by the [code]Locus[/code] command. The problem is that the value of [i]k[/i] that we are using isn't quite right, and doesn't reflect the actual transfer of heat from the cake into the ambient temperature of the room. As you know, [i]k[/i] is different for different objects because different objects dissipate heat differently. There's two methods to calculate [i]k[/i]: algebraic (precise) and numerical (systematically estimate). In the numerical method, you "guess and check" in a systematic "high/low" fashion until your numerical solution from above gets within a self-selected tolerance of [code]GettingCool[/code]. Do that now to get the numerical integral from [code]Locus[/code] within 0.01 (vertically) of [code]GettingCool[/code]. The other method to find [i]k[/i] is algebraic, and which we discussed in a previous lesson. The summary from that lesson is that we use the algebraic method for solving linear differential equations (sometimes called "the method of integrating factors") to set up a system of two equations to find the constant of integration [i]c[/i] that yields a specific solution to the differential equation that goes through [code]HotOutOfOven[/code], and also selects a value of [i]k[/i] so the solution goes through [code]GettingCool[/code]. You do not need to do that in this project (although it is listed as a challenge exercise below). [/*][*]Use your numerically estimated [i]k[/i] value and your specific solution to answer the following question: How long will it take until the cake has cooled to 85 degrees? Plot your answer as a point on your [code]Locus[/code] function.[/*][/list]

A small body of water is measured to be 85 degrees Fahrenheit at 8pm after having been heated up during a warm day. The ambient air temperature around the body of water is 55 degrees Fahrenheit. Later at 8:25pm the same body of water is measured to have cooled to 82 degrees Fahrenheit. Assume that Newton's Law of Cooling applies to the temperature of the small body of water, and that the ambient temperature around the lake stays at a constant 55 degrees Fahrenheit. Adjust and replicate what we did in Part 1 to predict what time the temperature of the water will cool below 75 degrees. Hint: If you need to extend a [code]Locus[/code] solution, plot a point on the [code]Locus[/code] solution near its right tail, and run [code]Locus[/code] again on the second point and on the slope field of the differential equation.[br]

Adjust the previous question but build into your model that the ambient temperature decreases by .4 degrees every minute. Answer the same question: when will the temperature cool below 75 degrees? Put your work in the GeoGebra applet below.

Use algebraic methods to find the specific solutions in Parts 1, 2 and Challenge 1. Plot your answers in the above GeoGebra windows. Name your specific solutions something that makes it clear what they are, such as [code]SpecificSolutionPart1(x)[/code], and SpecificSolutionPart2(x), and [code]SpecificSolutionChallenge1(x)[/code].