MODEL ESSAY

 

DIRECT AND INVERSE VARIATION

 

            The purpose of this essay is to show how the concepts of direct variation and inverse variation help people make decisions in their personal or professional lives.   Direct variation and inverse variation are mathematical models that show the relationship between variables.  In direct variation, when the “x” variable increases by a factor, the “y” variable increases by the same factor.  In inverse variation, when one variable increases by a factor, the other variable increases by the inverse or reciprocal of that factor.   In this essay I will critically think about two examples of these models through a discussion of the elements of thought.  One example is about a police investigation and the other example is about the demand and price of a laptop.

 

DIRECT VARIATION

 

PROBLEM 1:  A police investigator knows that the stopping distance, d, of an automobile is directly proportional to the square of the speed, s,  of the car.  On a particular road surface, a car requires 75 feet to stop when its speed is 30 mph.  The investigator measures the skid marks at an accident scene and finds them to be 125 feet long.  What was the speed of the car?

 

            Problem 1 is an example of direct variation because the stopping distance is directly proportional to the square of the speed.  The equations for direct variation are:  y = kx , where k is the constant of variation and therefore, the formula for the constant of variation, k, is k = y/x.  In this problem the information that is given are that a car requires 75 feet to stop when its speed is 30 mph and that the police investigator measured skid marks at an accident that are 125 feet long.   The constant of variation, k can be calculated as follows:

X = square of the speed, s²

Y = stopping distance, d

k = y/x = d/s ² = 75/(30²) = 0.08

 

The steps to solve the question, “What was the speed of the car?”:

y = kx = 0.08(x)

125 = 0.08 x

125/0.08 = x

1500 = x = s²

38.7 mp  s, the speed of the car that produces a skid mark of 125 ft.

 

            The algebraic model for this problem: y = 0.08x (d = 0.08s² ).  The verbal model for this problem:  The stopping distance, d is equal to the constant of variation, 0.8 times the square of the speed.  An example of a numerical model with 5 values for x, the square of the speed:

 

 

X, the square of the speed, s²:                     50       100     150     200     250     300

Y, the stopping distance, d:                          4.15    8.3       12.45  16.6    20.75  24.9   

 

 

The graphic model:

 

 

            The mathematical model helps one predict the speed of the car at the time of the accident and also, helps an investigator plan an investigation based on the predicted speed.  The purpose for using the concept “direct variation” if one is a police investigator is that its use can help an investigator make a decision in regards to whether or not a driver was speeding at the time of the accident.  The conclusion that one can make is that the skid mark would have been longer if the driver did not hit an object.  The assumption that one can make is that the direct variation model applies to this type of road surface.  The implication of using this model is that the police investigator can use the length of the skid marks in his investigation.  The point of view to be considered would be that of the police investigator and that of the driver who might feel that the length of the skid marks would not be sufficient to prove that he/she was speeding.        

 

INVERSE VARIATION

 

PROBLEM 2: You have a computer store and after conducting a study you find that the demand for a particular laptop varies inversely as the price of the product.  When the price is $800 the demand is 1200 units.  What is the approximate expected demand if you drop the price to $750?

 

Problem 2 is an example of inverse variation because the demand of a particular laptop varies inversely as the price of the product.  The equations for inverse variation are: y = k/x, where k is the constant of variation and therefore, the formula for the constant of variation, k, is k = xy. In this problem the information that is given are that when the price is $800 the demand is 1200 units.  The constant of variation, k, can be calculated as follows:

X = price of a particular laptop,p

Y = expected demand, in units, of laptop, d

K = xy = p(d) = 800(1200) = 960000

 

The steps to solve the question, “What is the approximate expected demand if you drop the price to $750?

Y = k/x = 960000/x

D = 960000/p

D = 960000/75 = 1280; the expected demand is 1280 units when the price is lowered to $750.

 

The algebraic model for this problem: y = 960000x (d = 960000/p).  The verbal model for this problem:  The expected demand, d, is equal to the constant of variation, 960000 divided by the unit price.  An example of a numerical model with 5 values for x, the price of a particular laptop:

X, price of a particular laptop, p:                             700     800     900     1000   1100

Y, expected demand of laptop(in units), d:              1371   1200   1067   960     873

 

The graphic model:

 

The mathematical model helps one predict the expected demand depending on the price of the unit and plan for this expected demand in areas such as inventory and back orders if demand exceeds supply.  The purpose for using the concept “inverse variation” if one is a business person is that this concept allows one to determine the expected demand in the future and make decisions depending on this prediction.  The conclusion that one can make is that one can expect the demand to vary inversely to the drop in price.  The assumption that the business person must make is that no other factor would dramatically affect the demand besides the price.   The implication of using the inverse variation model is that it allows a business person to plan for different sales scenarios depending on price of the unit.   The point of view of those who would be making decisions based on this model would be that of the business person and possibly the consumer who might feel that a severe drop in price would cause a sellout of that item or that item might soon be replaced by a better model.

 

CONCLUSION

 

            In conclusion, I have shown how variation is used to help people make decisions in their professional or private lives.  Variation equations help people plan and predict.  X and y variation relationships were shown using four models:  algebraic, numerical, verbal, and graphic models.  Also, I was able to critically think about each example by using the eight elements of thought: purpose, concept, information, question, conclusion, consequences, assumptions, and point of view.

 

 

 

MODEL ESSAY

 

 MATH MODELING

 

            The purpose of this essay is to show how mathematical modeling helps people make decisions in their personal and professional lives.  A mathematical model shows the relationship between two or more variables.  In this essay I will critically think about seven mathematical model problems using eight elements of thought: purpose, concept, information, question, conclusion, assumption, implication or consequence, and point of view.  The problems that I will discuss in this essay relate to pressure at a particular depth, the weight of an object based on the distance from the center of the earth, poverty income levels, the amount of medicine in the bloodstream, the value of a car over time, future value of a deposit in a bank, and the decay of a radioactive isotope.

 

PRESSURE AT A PARTICULAR DEPTH

 

PROBLEM 1:  When you swim underwater, the pressure “p” varies directly with the depth “d” at which you swim.  At a depth of 20 feet, the pressure is 8.6 pounds per square inch.  What is the pressure at 65 feet deep?

 

SOLUTION: 

P/65 = 8.6/20 ; P = 27.95 pounds per square inch.

At a depth of 20 feet, the pressure is 8.6 pounds per square inch.

 

            In this problem, the information given are: (1) the pressure varies directly with the depth,  and (2) at 20 feet the pressure is 8.6 pounds per square inch.  The verbal model is: pressure is equal to constant of variation (.43) times depth.  A numerical model for every 20 feet up to 100 feet:

 

Depth (ft)                  20       40       60       80       100

Pressure (p)              8.6       17.2    25.8    34.4    43

 

            The purpose of using the concept “direct variation” is to help those who need to dive make decisions about the pressure that is to be expected.  The algebraic model is p = .43 d and the conclusion that one can make is that when the depth increases by 1 foot, the pressure increase by .43.  The assumption that can be made is that constant of variation does not change as one travels deeper into the water.  The implications to an individual or submarine are that steps must be taken to withstand the increase in pressure as a submarine or individual dives deeper and deeper into the water.   From the point of view, safety is an issue as one dives deeper due to pressure and also, this is a concern for anyone travelling by submarine into deep waters where pressure can affect the submarine itself.

 

 

 

 

THE WEIGHT OF AN OBJECT BASED ON DISTANCE FROM EARTH’S CENTER

 

PROBLEM 2:  The weight of an object varies inversely with the square of its distance from the center of the earth.   Assume that the radius of the earth is 3960 miles.  If an astronaut weighs 220 pounds on the surface of the earth, what does he weigh when traveling 350 miles above the surface of the earth?

 

SOLUTION: 

K = xy

K = (3960^2)220

(3960^2)220 = (4310^2)weight

(3960^2)220 / (4310^2) = weight

185.72 lbs. = weight

An astronaut weighs approximately 185.72 pounds when traveling 350 miles above the surface of the earth. 

 

            In this problem, the information given: (1) the weight of the astronaut on the surface of the earth 3960 miles from the center of the earth is 220 pounds and (2) the weight of an object varies inversely with the square of its distance from the center of the earth. The verbal model is the weight of an individual is equal to the constant of variation (3960^2 x 220) divided by the distance from the center of the earth squared.  The numerical model for every 100 feet up to 500 feet:

 

X, distance^2    3960^2     4060^2     4160^2      4260^2             4360^2      4460^2         

Y, weight            220           209.3       199.35        190.11             186.48       173.44

 

            The purpose of using the concept “inverse variation” to help people make decisions regarding their weight when traveling above the surface of the earth in which case adaptions and training may be needed to adjust to the change in weight.

Based on the algebraic model, y = (3960^2)(200)/x^2,  one can conclude that since the weight varies inversely with the distance squared, the further one travels above the earth, the lighter the weight.   The assumption one can make is that no other factors will occur to affect the relationship between weight and the distance from the earth.  The consequences of this relationship from the point of view of an astronaut or NASA engineer/scientist is that one must train the astronaut to work under these conditions.

 

 

POVERTY INCOME LEVEL

 

PROBLEM 3:  The following table from the U.S. Census Bureau shows the poverty level for family of four in selected years (families whose income is below this level are considered to be in poverty).

 

 

Year                                 1994        1996        1997         1998       1999       2000    2001

 

Poverty income

Level (in thousands)         15.141    16.036    16.400    16.660    17.029    17.603    18.267

 

            Inputting the data into a TI-83, by creating 2 lists that correspond to the year and poverty income level, the linear regression equation is y = .423x + 13.413 (when x = 0 corresponds to 1990).  By replacing x in the equation with 5 for 1995, one gets a predicted poverty income level of $15,526 and by replacing x in the equation with 15, one gets a predicted poverty income level of  $19,753.  The verbal model for this equation:  the poverty income level is equal to a rate of change (slope) of .423 times x, the year when x=0 for 1990 plus a y-intercept equal to the poverty income level of $13,413.  The purpose of using the concept “linear regression” by the government is to make decisions and plans based on future predicted poverty income levels.  Based on the linear regression model, one can conclude that the poverty income level will increase at a constant rate.  Therefore, the assumption is that this rate of change (slope) will not change over time.  The implications of being able to predict future poverty income levels from the point of view of the government is that plans could be made to assist those in need.  From the point of view of families, they would be aware that even if they made more money in the future, their income could still be considered at the poverty income level or below.

 

 

EXPONENTIAL FUNCTIONS (DECAY OR NEGATIVE GROWTH)

 

PROBLEM 4:  Exponential functions are useful for modeling situations in which a quantity increases by a fixed factor.  (Exponential growth or decay.)  The table shows the amount of medicine for treating a disease in the bloodstream over the 9 hours following a dose of 10 mg.  It seems that the rate of decrease of the drug is approximately proportional to the amount remaining.

 

Time (hrs.)                 0          1          2          3          4          5          6          7          8         

Drug amount (mg)    10       8.3       7.2       6.0       5.0       4.4       3.7       2.8       2.5

 

a.     Use this information to find the exponential regression model (ExpReg: y = ab^x) for this data.

SOLUTION:  Using the TI-83, by inputting 2 lists, and calculating the exponential regression,  I conclude that the exponential regression model is y= (10.08339351).8402025505^x.

b.     Using your model, when will there be less than 1 mg. of the medicine in the patient’s bloodstream?

SOLUTION:

y=(10.08339351).8402025505^x

1/10.08339351 = .8402025505^x

log (1/10.08339351) = x log.8402025505

log (1/10.08339351)/log.8402025505 = x

13.27241135 = x

Therefore, in approximately 13.272 hours there will be less than 1 mg. of the medicine in the patient’s bloodstream.

c.      If the initial dose was 15 mg., when would the amount of the medicine in the bloodstream fall below 5 mg.?

SOLUTION:

5 = 15(.8402025505^X)

1/3 = .8402025505^X

log (1/3) = x log .8402025505

log(1/3) / log .8402025505 = x

6.309791917 = x

Therefore, in approximately 6.310 hours the amount of the medicine in the bloodstream will fall below 5 mg. if the initial dose was 15 mg.

 

In this problem, the information given: (1) time in hours, (2) drug amount over that time of 9 hours, and (3) that the rate of decrease of the drug is approximately proportional to the amount remaining.   The verbal model is that the drug amount (mg) is equal to the initial amount times the fixed factor raised to the power of time (hr).  The purpose of using the concept “exponential growth” is to help doctors predict the amount of medicine in the bloodstream over a given amount of time that allows decisions to be made in regards to the drug use.  Based on the algebraic model, one can conclude that the drug in the bloodstream decreases by a fixed factor.  One can assume that this fixed factor is not affected by any other factors within the body.  The consequences of being able to predict the amount of drugs in the bloodstream over time allows a doctor to decide the amount of any particular drug to be used during the day for a given patient.  From the point of view of the patients, he/she can feel confident that they are taking the right amount of medicine during the day if the doctor makes a decision based on these calculations and from the point of view of the doctors, they can feel confident that a safe dosage is being used by the patient.

 

 

 

DEPRECIATION OF A CAR

 

PROBLEM 5:  A Ford Taurus car depreciates at 8.5% per year.  What is the value of the car after 6 years if the car was purchased for $48,000? [Y=a(1-r)^t]

 

SOLUTION: 

Y= 48000(1-.085)^6  = 28168.76468

Therefore, the value of the car after 6 years if the car was originally purchased for $48000 would be approximately $28168.76.

 

            In this problem, the information given: (1) the depreciation rate, 8.5% per year and (2) the original purchase price, $48,000.  The algebraic model is y=48000(1-.085)^x.  The verbal model is that the future value of the car is equal to the original value times one minus the depreciation rate raised to the power of time in years.  The numerical model for every year up to the 6^th year is as follows:

 

Year, x                              1              2              3            4                5             6

 

Future value

Car                              43920         40187     36771     33645       30786      28169

 

            The purpose of using the concept “depreciation” (the loss of value due to use) is to help people make decisions about the future value of assets such as cars and office equipment and to plan accordingly based the calculations of their future value.  The assumption is that the depreciation rate will not change over time.   The implication or consequences of estimating the future value of an asset like a car is that one can plan for its sale knowing approximately how much money the item could be sold for in the future.  Also, the calculated value based on its age can be used from the point of view of a seller or buyer to negotiate a price that is agreeable to both parties.

 

 

 

COMPOUND INTEREST

 

PROBLEM 6:  If a bank offers 1.2% interest and compounds daily and you deposit $5000 and leave it untouched, how much money would you have after 25 years? [A=P(1+r/n)^(nt)]

 

SOLUTION: 

A = 5000(1 + 0.012/365)^(365X25) = 6749.260753

Therefore, after 25 years one would have $6749.26 in the bank if an initial deposit of $5000 was made and left untouched.

 

           

 

 

 

 

In this problem, the information given: (1) $5000 was deposited, (2) left untouched for 25 years, and (3) the interest rate is 1.2%. The numerical model for every 5 years up to the 25^th year is as follows:

 

Years, t                             5                     10                15                 20               25

 

Future value

Of deposit                  5309.20         5367.50         5986.10         6356.2           6749.30

 

            The purpose of using the concept “compound interest”  is to help people make decisions about the future value of their bank account if the interest was compounded and the original deposit was left untouched.   The conclusion that one can make regarding a $5000 deposit, if left untouched, is that in 25 years the deposit would be worth approximately $6749.30.  However, the assumption that is made is that the interest rate would not change over time.  From the point of view of the person depositing the money, the consequences of knowing what you might have in the future is that one can make plans for the money one expects or if one has a goal in mind they would know whether or not the money that is expected in the future is adequate.  From the point of a bank, the bank might try to have the highest interest rate to attract individuals who are looking to put their money in the bank for the long term.  This money can then be used to lend to others at a much higher rate so that the bank can in return make a profit.

 

 

 

 

 

ELEMENT X AND EXPONENTIAL DECAY (HALF-LIFE)

 

PROBLEM 7:  Element X is an extremely toxic radioactive isotope found int environment after a chemical plant exploded, [Y = a(.5)^(t/h).  Solve for the missing variable:

 

Future Amt.                Initial Amt.                 Time               Half-life

         65                          2048                           ?                    20 yrs.

 

SOLUTION:

65 = 2048(.5)^(t/20)

65/2048 = 2048(.5)^(t/20)  /  2048

log (65/2048 )= t/20 log .5

log (65/2048 )/ log .5  = t/20

4.977632187 = t/20

20(4.977632187) = t

99.55264374 = t

Therefore, in approximately 100 years the initial amount, 2048 grams of element X would decay to 65 grams.

 

            The information given: (1) the future amount after an unknown number of years is 64 grams, (2) the initial amount, 2048 grams, and the half-life of the element is 20 years.  The numerical model for every 20 years up to 100 years is as follows:

 

years of decay, t                20             40                  60                  80                  100

 

Future amount, gms.      1024          512                 256                 128                 64

 

            The verbal model for this problem is that the future amount of a radioactive isotope is equal to the original amount times a fifty percent decay rate raised to the power of the quotient, the time passed divided by the time it takes for the element to decay by fifty percent. The purpose of using the concept “exponential decay” is to help people such as scientists and government officials to determine the time it takes for a radioactive isotope to decay.  Based on the algebraic model, y=a(.5)^(t/h), one can conclude that 2048 grams of this element would decay to 64 grams  in 100 years.  The assumption being that no other factor can affect the rate of decay.   The implication, from the point of view of the government and the citizens, of the fact that this element takes so long to decay into barely traceable amounts means that its radioactivity will affect the environment in a very negative way. For a very long time the radioactive element would possibly make the area uninhabitable and totally unsuitable for use of any kind.

 

 

 

 

 

 

CONCLUSION

 

            In conclusion, I have shown how mathematical models are used to help people make decisions in their personal and/or professional lives.  The problems that I used to show this related to real life situations such as the effect of depth on pressure, weight from a distance from the earth, poverty income levels, medicine in the bloodstream, the loss of value of a car over time, compound interest and banks, and the half-life of a radioactive material.  With each problem, I was able to critically think the scenario by using the eight elements of thought.  The elements of thought that were utilized were purpose, concept, question, information, conclusion, assumption, implications or consequences, and point of view.