1. The textbook develops a mathematical model called Newton's Law of Cooling to predict the temperature of an object that is initially warmer than its surroundings, and that over time cools down to match the temperature of its surroundings. This lab explores the model graphically, to complement the analytic approach that the text develops.
Read the problem, presented as Example 3 on page 171. The information provided that is used to build the model includes three facts that are given:
The initial temperature of the soda is 72 degrees
The refrigerator is 44 degrees
The soda has cooled to 61 degrees after a half hour.
Now study the graph presented on Lab page 1. Identify features of the graph that correspond to the three facts that are given.
[d] 1. A
[c] 2. B
[b] 3. C
[e] 4. x
[a] 5. y
a. temperature axis
b. refrigerator temperature
c. soda temperature after half hour
d. initial temperature of soda
e. time axis
2. The model developed for Newton's Law of Cooling says that the correct function to model the change of temperature over time is an exponential decay, towards the temperature of the surroundings. A "pure" decay should be modeled with a negative exponent. It could vary in rate according to the base of the exponential function chosen, or according to the coefficient of the variable in the exponent. That means we could write it as or as . We will use the second form. Note the alternative notation . Why is it necessary to use a negative exponent?
a. The temperature can take on negative values
*b. A positive exponent and a base bigger than one models growth, not decay
c. The time can take on negative values
d. Time is always increasing
e. Temperature may be decreasing or increasing
3. A pure exponential decay has as a limiting value . What is the limiting value?
a. exp(-a)
b. -a
*c. 0
d. 1
e. a
f. exp(a)
4. How does the model adjust for a limiting value to be the temperature Ts of the surroundings? (In this case, Ts = 44.)
a. A horizontal translation of the horizontal asymptote of 44 units
*b. A vertical translation of the horizontal asymptote of 44 units
c. A stretching of the vertical axis using a multiplier of 44
d. A stretching of the horizontal axis using a multiplier of 44
5. A pure exponential function exp(-ax) has a value of what when x is 0?
a. -a
b. -1
c. 0
*d. 1
e. a
6. The exponential function we are building has to have a horizontal asymptote of 44, so it is a "pure" exponential shifted up 44 units. It has to pass through the point (0,72). How do we adjust the shifted function so it has a y intercept at 72-44=28 units above the asymptote?
a. Scale vertically by multiplying by 72
b. Scale vertically by multiplying by 44
*c. Scale vertically by multiplying by 28
d. Shift vertically by adding 72 more
e. Shift vertically by adding 28 more
7. So far the model looks like the temperature at time x is 44+28*exp(-a*x). All that is left is determining a to make a best fit. We use the second point (30,61) to do this. You can study carefully the book's algebraic approach to solving an exponential equation to find a (the book uses the letter k), but here we will determine a experimentally. In the f(x)= text box enter the equation with a unspecified, as 44+28*exp(-a*x), and hit Graph. The result misses the second point entirely, but does pass through (0,72) with the correct horizontal asymptote. Adjust a with the slider to find two numbers that bound the correct value of a above and below.
a. -10 and -9.9
b. -5.1 and -5.0
c. -0.1 and 0
*d. 0 and 0.1
e. 1 and 1.1
8. Refine your answer to question 7 by changing the left hand value of the slider to be the smaller number and the right hand value of the slider to be the larger number. Now the slider adjustment is finer, and the correct value of a can be experimentally determined to the nearest thousandth. What is a?
Correct Answer(s):
a. .017
9. Now that the model is built we can use it to predict other values. The grapher has a black dot called trace point that is available. Use the up and down arrows to put the trace point on the model, then use the left and right arrows to move it to see points on the graph. What temperature does the model predict after 90 minutes? Answer to the nearest degree.
Correct Answer(s):
a. 50
10. Consider the graph on Lab page 2. Fit a Newton's Law of Cooling model to the data points and asymptote that are given, and use it to predict the temperature after 60 minutes. Answer to the nearest degree.
Correct Answer(s):
a. 47
11. Go to lab page 3. Figure out how to modify the model to account for an object warming up. What is the predicted temperature at time 60? Answer to the nearest degree
Correct Answer(s):
a. 57
b. 58