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Chapter 2: Problem 24

In one of the early "Functions and Change" pilot courses at Oklahoma State University, the instructor asked the class to determine when in Example \(2.1\) terminal velocity would be reached. Three students gave the following three answers: Student 1: 58 seconds into the fall. Student 2: 147 seconds into the fall. Student 3: Never. Each student's answer was accompanied by what the instructor judged to be an appropriate supporting argument, and each student received full credit for the problem. What supporting arguments might the students have used to convince the instructor that these three different answers could all be deserving of full credit? (Hint: Consider the formula given in Example 2.1. For student 1, look at a table of values where the entries are rounded to two decimal places. For student 2, look at a table of values made by using all the digits beyond the decimal point that the calculator can handle. In this case that was nine. For student 3 , consider what value \(0.834^{t}\) must have to make \(176\left(1-0.834^{t}\right)\) equal to 176 .)

Short Answer

Expert verified
Each student used different assumptions: rounding, precision, and theoretical limits, to justify their answer.

Step by step solution

01

Understanding the Formula

In Example 2.1, terminal velocity is reached when the expression \(176 \times (1 - 0.834^t) = 176\). This implies that \(1 - 0.834^t = 1\). Solving for \(t\), we see that \(0.834^t\) needs to be 0.
02

Analyzing Student 1's Argument

Student 1 observed the table of values where entries are rounded to two decimal places. By examining values closely around 58 seconds, the value of \(0.834^t\) might round close enough to zero that \(176 \times (1 - 0.834^t)\) approximates 176 when noted at two decimal places.
03

Analyzing Student 2's Argument

Student 2 used more precise measurements, extending the decimals to all the digits possible by the calculator (up to nine). The value sequencing subtly shows \(176 \times (1 - 0.834^t)\) equaling 176 around 147 seconds, considering precision and more decimal places.
04

Analyzing Student 3's Argument

Student 3 pointed out mathematically that the expression \(0.834^t\) never equals 0, hence \(1 - 0.834^t\) can never reach exactly 1. Therefore, terminal velocity in the exact calculation is theoretically never reached.
05

Conclusion of Each Student's Justification

Student justification for full credit is supported by how values approach terminal velocity: Student 1 through rounding; Student 2 through precision; Student 3 through theoretical understanding of limits.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Modeling
Mathematical modeling is an essential process where real-world situations are represented using mathematical expressions or equations. This approach helps in analyzing, interpreting, and predicting outcomes by solving these equations. In the problem from the textbook, the scenario involves calculating terminal velocity, which means a velocity where an object's acceleration stops due to opposing forces like air resistance.
With the provided expression, the terminal velocity was to be determined through the equation:
  • The equation defines the behavior of velocity as a function over time.
  • It gives us an insight into how quickly or slowly an object reaches a steady speed.
Fulfilled by student's arguments, each student's reasoning stemmed from the interpretation of the equation at different levels of precision. Student 1's approach to interpreting solution models by rounding values aligns with practical and simplified modeling techniques used for rough estimates. In contrast, Student 2 delves deeper using full calculator precision, allowing a more exact mathematical model interpretation. Lastly, Student 3 uses a more abstract model, focusing on theoretical mathematics, showing that sometimes absolute conditions can never be met thoroughly.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a continuous variable, often denoted as the power or exponent. In their nature, exponential functions can represent rapid growth or decay, as seen in various natural phenomena. The function given in the exercise \[ 176 \times (1 - 0.834^t) \]depends on time \(t\), representing a decay because the base \(0.834\) is less than one.
Here's how it connects to the exercise:
  • Exponential decay models how the value decreases sharply at the beginning and slows over time.
  • This decrease implies that as more time passes, the velocity change becomes less significant.
Each student's conclusion about when terminal velocity was reached relied on the behavior of this function. While Student 1 and 2 looked at empirical evidence through observations and calculations, the exponential nature explains why Student 3 could argue using a theoretical lens that \(0.834^t \) will never precisely reach zero, signifying it's never actually reached.
Precision and Rounding
Precision and rounding deal with the degree of detail in numerical calculations. Precision refers to the exactness of a number, often defined by how many digits are used after a decimal point. Rounding, on the other hand, simplifies numbers by reducing the number of digits, often making calculations more manageable.
In analyzing the student responses, it鈥檚 clear how dramatically precision and rounding can affect outcomes.
  • Student 1's use of rounding to two decimal places results in finding terminal velocity at 58 seconds, suggesting that approximate values closely imitating real values can give practical insights.
  • Student 2 extended this concept to a more precise calculus, considering all possible digits to get a 'more correct' result of achieving terminal velocity after 147 seconds.
  • Student 3's standpoint shows perfect precision in that theoretical models never actually permit absolute conditions like reaching zero.
This shows that depending on the context, different levels of precision and methods of rounding can lead to varied, but valid conclusions.

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Most popular questions from this chapter

In reconstructing an automobile accident, investigators study the total momentum, both before and after the accident, of the vehicles involved. The total momentum of two vehicles moving in the same direction is found by multiplying the weight of each vehicle by its speed and then adding the results. \({ }^{19}\) For example, if one vehicle weighs 3000 pounds and is traveling at 35 miles per hour, and another weighs 2500 pounds and is traveling at 45 miles per hour in the same direction, then the total momentum is \(3000 \times 35+2500 \times 45=\) 217,500 . In this exercise we study a collision in which a vehicle weighing 3000 pounds ran into the rear of a vehicle weighing 2000 pounds. a. After the collision, the larger vehicle was traveling at 30 miles per hour, and the smaller vehicle was traveling at 45 miles per hour. Find the total momentum of the vehicles after the collision. b. The smaller vehicle was traveling at 30 miles per hour before the collision, but the speed \(V\), in miles per hour, of the larger vehicle before the collision is unknown. Find a formula expressing the total momentum of the vehicles before the collision as a function of \(V\). c. The principle of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. Using this principle with parts a and b, determine at what speed the larger vehicle was traveling before the collision.

Find a function given by a formula in one of your textbooks for another class or some other handy source. If the formula involves more than one variable, assign reasonable values for all the variables except one so that your formula involves only one variable. Now make a graph, using an appropriate horizontal and vertical span so that the graph shows some interesting aspect of the function, such as a trend, significant values, or concavity. Carefully describe the function, formula, variables (including units), and graph, and explain how the graph is useful.

City A lies on the north bank of a river that is 1 mile wide. You need to run a phone cable from City A to City B, which lies on the opposite bank 5 miles down the river. You will lay \(L\) miles of the cable along the north shore of the river, and from the end of that stretch of cable you will lay \(W\) miles of cable running under water directly toward City B. (See Figure \(2.111\) on the next page.) You will need the following fact about right triangles. A right triangle has two legs, which meet at the right angle, and the hypotenuse, which is the longest side. An ancient and beautiful formula, the Pythagorean theorem, relates the lengths of the three sides: Length of hypotenuse \(=\sqrt{\text { Length of one leg }}\) a. Find an appropriate right triangle that shows that \(W=\sqrt{1+(5-L)^{2}}\) b. Find a formula for the length of the total phone cable \(P\) from City A to City B as a function of \(L\). c. Make a graph of the total phone cable length \(P\) as a function of \(L\), and explain what the graph is showing. d. What value of \(L\) gives the least length for the total phone cable? Draw a picture showing the least-length total phone cable.

The amount of vegetation eaten in a day by a grazing animal is a function of the amount \(V\) of food available (measured as biomass, in units such as pounds per acre). \({ }^{21}\) This relationship is called the functional response. If there is little vegetation available, the daily intake will be small, since the animal will have difficulty finding and eating the food. As the food biomass increases, so does the daily intake. Clearly, though, there is a limit to the amount the animal will eat, regardless of the amount of food available. This maximum amount eaten is the satiation level. a. For the western grey kangaroo of Australia, the functional response is $$ G=2.5-4.8 e^{-0.004 V} $$ where \(G=G(V)\) is the daily intake (measured in pounds) and \(V\) is the vegetation biomass (measured in pounds per acre). i. Draw a graph of \(G\) against \(V\). Include vegetation biomass levels up to 2000 pounds per acre. ii. Is the graph you found in part i concave up or concave down? Explain in practical terms what your answer means about how this kangaroo feeds. iii. There is a minimal vegetation biomass level below which the western grey kangaroo will eat nothing. (Another way of expressing this is to say that the animal cannot reduce the food biomass below this level.) Find this minimal level. iv. Find the satiation level for the western grey kangaroo. b. For the red kangaroo of Australia, the functional response is $$ R=1.9-1.9 e^{-0.033 V} $$ where \(R\) is the daily intake (measured in pounds) and \(V\) is the vegetation biomass (measured in pounds per acre). i. Add the graph of \(R\) against \(V\) to the graph of \(G\) you drew in part a. ii. A simple measure of the grazing efficiency of an animal involves the minimal vegetation biomass level described above: The lower the minimal level for an animal, the more efficient it is at grazing. Which is more efficient at grazing, the western grey kangaroo or the red kangaroo?

An important model for commercial fisheries is that of Beverton and Holt. \({ }^{31}\) It begins with the study of a single cohort of fishthat is, all the fish in the study are born at the same time. For a cohort of the North Sea plaice (a type of flatfish), the number \(N=N(t)\) of fish in the population is given by $$ N=1000 e^{-0.1 t} $$ and the weight \(w=w(t)\) of each fish is given by $$ w=6.32\left(1-0.93 e^{-0.095 t}\right)^{3} $$ Here \(w\) is measured in pounds and \(t\) in years. The variable \(t\) measures the so-called recruitment age, which we refer to simply as the age. The biomass \(B=B(t)\) of the fish cohort is defined to be the total weight of the cohort, so it is obtained by multiplying the population size by the weight of a fish. a. If a plaice weighs 3 pounds, how old is it? b. Use the formulas for \(N\) and \(w\) given above to find a formula for \(B=B(t)\), and then make a graph of \(B\) against \(t\). (Include ages through 20 years.) c. At what age is the biomass the largest? d. In practice, fish below a certain size can't be caught, so the biomass function becomes relevant only at a certain age. i. Suppose we want to harvest the plaice population at the largest biomass possible, but a plaice has to weigh 3 pounds before we can catch it. At what age should we harvest? ii. Work part i under the assumption that we can catch plaice weighing at least 2 pounds.
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