Chapter 6 Analyzing Accumulated Change: Integrals in Action 6. Streams in Business and Biology You will find Excel very helpful when dealing with streams that are accumulated over finite intervals. Finding either the future or present value of a continuous income stream is simply finding the value of a definite integral. *6..1 as noted in the text is actually found in this guide as 6... 6..1 DETERMINING THE FLOW-RATE FUNCTION FOR AN INCOME STREAM Excel can often help you to find the equation for an income stream flow rate. Note that it is not necessary to use Excel to find such an equation we present this as a technique to use only if you find writing the equation from the word description difficult. We illustrate these ideas as they are given in Example 1 of Section 6. of Calculus Concepts. In Example 1, part a, we are told that the business s profit remains constant. Setup a worksheet table with two possible input values for the time involved. (You might use different years than the ones shown here.) In the output column, enter the amount invested: 10% of the constant profit. Use the techniques covered in Section 1.4.c to find the equation of the linear model. The model is y = 0x +57,900. In Example 1, part b, we are told that the business s profit grows by $50,000 each year. The first year s profit (which determines the initial investment at t = 0) is $579,000. Reason that if the profit grows by $50,000 each year, the next year s profit will be $579,000 + 50,000 = $69,000. You need to remember that a constant growth means a linear flow rate. Note that only 10% of the profit is invested. Fit a linear function to these two data points. Thus, the linear flow rate function is R(t) = 5,000t + 57,900 dollars per year t years after the first year of business. In Example 1, part c, we are told that the business s profit grows by 17% each year. The first year s profit (which determines the initial investment at t = 0) is $579,000. Reason that if the profit grows by 17% each year, the next year s profit will be $579,000 + 0.17(579,000) = $677,430. Note that only 10% of the profit is invested. Remember that a constant percentage growth means an exponential flow rate. Fit an exponential function to these two data points using the techniques covered in Section 1.3.. The exponential flow rate function is R(t) = 0.10(579,000)(1.17 t ) dollars per year t years after the first year of business. 69
70 Chapter 6 Part d of Example 1 gives data that describe the growth of the business s profit. Refer to the material in Section 1.3.3 of this Guide to review how to fit a log function to these data. Note: If you forget which type of growth gives which function, simply use what you are told in the problem and fill in the lists with approximately five data points. Draw a scatter plot and it should be obvious which function to fit to the data. 6.. FUTURE VALUE OF A DISCRETE INCOME STREAM The future value of a discrete r income stream is found by adding, as d increases, the terms of R( d ) 1 + where R(d) is n the value per period of the dth deposit, 100r% is the annual percentage rate at which interest is earned when the interest is compounded once in each deposit period, n is the number of times interest is compounded (and deposits are made) during the year, and D is the total number of deposit periods. It is assumed that initial deposits are made at time t = 0 unless it is otherwise stated. We use the situation in Example 4 in Section 6. of Calculus Concepts: When you graduate from college (say, in 3 years), you would like to purchase a car. You have a job and can put $75 into savings each month for this purchase. You choose a money market account that offers an APR of 6.% compounded quarterly. Part b of Example 4 asks for the future value of the deposits at the end of 3 years. Because the APR is 6.% compounded quarterly, r = 0.06, n = 4, and D = 3 years 4 deposits = 1 deposits. A constant year $75 each month is deposited, so R(d) = $75(3) = $5 deposited each quarter. The 3-year future value is given 11 1 d 0. 06 by 5 1 +. d = 0 4 Use the SUM command to create the summation. Note that the entries in column B contain the future value of each quarterly investment. For example, the third payment (#) is valued at $6.54 at the end of three years. D d NOTE: When you use the sequence command for discrete income streams, the first value is always 0. The last value is always D 1 because we start counting at 0, not 1. The increment will always be 1 because of the way the formula is designed.
Excel 000 Guide 71 Part a asks how much will you have deposited in 3 years. No interest is involved in this calculation. There are 1 months in each year and $75 is deposited each month for 3 years. A total of $700 is deposited. Part c of Example 4 asks for the present value of the amount you would have to deposit now to achieve the same future value as was found above. To find the present value P, use the Solver to solve the equation D r P 1 + = Future Value. n 1 0. 06 That is, solve P 1 + = 988. 10 for P. (Solver techniques are covered in Section 4 1.1.1j) We find that an initial investment of $484.48 will yield $988.10 after three years. Part d of Example 4 asks for the future value if the monthly interest rate is 0.5% instead of the previous APR compounded quarterly. Make the appropriate changes in the Future Value function. Remember that R(d) is the amount deposited per interest compounding period. The new Future Value formula for the d th monthly investment is 36 75 1 +. 005. We must calculate ( ) d 35 36 ( 1 +. 005) d = 0 d 75. Summing the terms of the sequence we find the future value of the investment is $964.96. 6.3 Integrals in Economics Consumers surplus (when defined by a definite integral) and producers surplus are easy to find using definite integrals. You should draw graphs of the demand and supply functions and think of the economic quantities in terms of area to better understand the questions being asked. 6.3.1 CONSUMER ECONOMICS We illustrate how to find the consumers surplus and other economic quantities when the demand function intersects the input axis as given in Example 1 of Section 6.3 of Calculus Concepts:
7 Chapter 6 Suppose the demand for a certain model of minivan in the United States can be described as D(p) = 14.1(0.933 p ) 0.5 million minivans when the market price is p thousand dollars per minivan. First draw a graph of the demand function. This is not asked for, but it will really help your understanding of the problem. Read the problem to see if there are any clues as to how to set the horizontal view for the graph. The price cannot be negative, so p 0. There is no price given in the remainder of the problem, so just guess a value with which to begin. Even though D is an exponential function, a constant has been subtracted from the exponential portion. So, D may cross the input axis. Zooming in we see that D has an x- intercept at x 58.17. Customers will not purchase this model minivan if the price per minivan is more than about $58. thousand. Note that the answer to Example 1, part c, is p $58. thousand. (Be sure to label the axes with variables and units of measure when you copy this graph to your paper.)
Excel 000 Guide 73 Part a of Example 1 asks at what price consumers will purchase.5 million minivans. Look at the labels on your graph and note that.5 is a value of D. You therefore need to find the price (an input). Use Solver to find the solution of the equation.5 = 14.1(0.933 p ) 0.5. At a price of $3,590,.5 million vans will be sold. Part b asks for the consumers expenditure when purchasing.5 million minivans. The consumers expenditure is price * quantity = area of the rectangle with height =.5 million minivans and width $3.59 thousand per minivan. The area is about $59 billion. The consumer s surplus in part d of Example 1 is the area under the demand curve to the right of p = 3.5903 and to the left of p = 58.1670. Calculate the definite integral using the techniques covered in Section 6.1.4. The surplus is about $7.4 billion. 6.3. PRODUCER ECONOMICS We illustrate how to find the producers surplus and other economic quantities as given in Example 4 of Section 6.3 of Calculus Concepts: The demand and supply functions for the gasoline example in the text are given by D(p) = 5.43(0.607 p ) million gallons and S(p) = 0 million gallons for p < 1 and S(p) = 0.79p 0.433p + 0.314 million gallons for p 1 when the market price of gas is p dollars per gallon. First draw graphs of the demand and supply functions. Part a of Example 4 asks for the market equilibrium point. You can find this by using the Solver or by using the intersection method on the graph. The graphs intersect at (1.831111,.176679). 7 6 5 4 3 1 0 0 1 3
74 Chapter 6 The total social gain is producers surplus + consumers surplus at the equilibrium price. 1.831111 Social gain = S ( p) dp + D( p) 1 3 1.831111 dp NOTE: The material in the rest of this Guide appears only in the complete version of the text. 6.4 Probability Distributions and Density Functions Most of the applications of probability distributions and density functions use technology techniques that have already been discussed. Probabilities are areas whose values can be found by integrating the appropriate density function. A cumulative density function is an accumulation function of a probability density function. Excel s robust statistical functions are especially useful for finding means and standard deviations of some probability distributions. Additionally, we will use our numerical integration techniques to work with expressions for which we have not developed algebraic techniques for finding antiderivatives. 6.4.1a NORMAL PROBABILITIES The normal density function is the most well known and widely used probability distribution. If you are told that a random variable x has a normal distribution N(x) with mean µ and standard deviation σ, the probability that x is between two real numbers a and b is given by ( x µ ) b b 1 N( x) dx = a e σ dx aσ π We illustrate these ideas with the situation in Example 8 of Section 6.4 of Calculus Concepts. In that example, we are given that the distribution of the life of the bulbs, with the life span measured in hundreds of hours, is modeled by a normal density function with µ = 900 hours and σ = 100 hours. Carefully use parentheses when entering the normal density function. The probability that a light bulb lasts between 9 and 10 hundred hours is P(9 x 10) = ( x 9) 10 1 e dx. 9 π The value of this integral must be between 0 and 1. There is approximately a 34% chance the bulb will last 900-1000 hours. 6.4.1b VIEWING NORMAL PROBABILITIES Graph the function for a normal probability using the same techniques discussed for graphing any function. (See Section 1.1.1.)