Unit #14 - Integral Applications in Physics and Economics Section 8.6

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1 Unit #14 - Integral Applications in Physics and Economics Section 8.6 Some material from Calculus, Single and MultiVariable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 25 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. PRACTICE PROBLEMS 1. We are adding money at a constant rate of $1 per year. The amount deposited in the small time interval t at time t will be $1 t. This money will grow at 8% interest for the remaining 1 t years, so its future value will be ($1 t) e.8 (1 t) Adding up all the deposits from t = to t = 1, and letting t dt, we get Future value = 1 1e.8(1 t) dt.8(1 t) 1 = 1e 1.8 = 1 ( e +e.8(1)).8 = $15, This is noticeably more than the actual $1, deposited, and seems reasonable for the amount of interest that should have been earned. 3. Similar to #1, the future value will be given by Future value = 5 2e.8(1 t) dt.8(5 t) 1 = 2e 5.8 = 2 ( e +e.8(5)).8 = $12, Note that this value is less than for the 1 year case in #1, even though the total deposit is still $1,. This is because there is only five years for the interest to accumulate. The present value can be computed more easily using the simple relationship on page 435, that Future Value = e rm Present value where r is the interest rate and M is the number of years. 1

2 Future Value Present value = e rm = $12, e.8 5 $8,242. TEST PREPARATION PROBLEMS 7. (a) The future value of a continuous income stream is given by the formula on page 435 of the text. In this case, the income stream is constant, so P(t) = R, and we are looking for R such that the future value equals $1,. Future value = 1 1, = Re 1, = R.1 Re.1(1 t) dt.1(1 t) [ e +e.1(1)] R = 1, e 1 $5, per year e The parents would need to deposit approximately $5, per year to achieve their goal. 2

3 (b) A single lump sum of P, over 1 years, would grow to P e.1(1). To find an initial deposit that would grow to 1, in ten years, we solve P e 1 = 1, P = 1, e 1 $36, They would need to deposit $36, today to have $1, in ten years. 9. You should choose whichever payment schedule maximizes either the present or current value of the entire schedule. Since one of the options is a lump-sum payment of $2,8 right now (and so its present value is $2,8), it would be easier to compare present values. When we consider the three payments of $1,, we need to reduce each of their value from the future to their present value. One payment is immediate, so has a present value of $1,. The next payment is a year from now, so needs to be discounted by e.6 1. The last payment will be given in two years, so needs to be discounted by e.6 2. The net present value of all three payments then is 3

4 $1,+$1, e.6 1 +$1, e.6 2 $ From this analysis, you are marginally better off taking the payment in $1, installments, although the difference is relatively slight $28 on the $3, amount. With either option, you would still be taking a loss relative to the full $3, which you are owed. 13. (a) If the extraction rate is q(t) = 1.1t millions of barrels per year, the total amount extracted in N years will be N 1.1t dt 4

5 We want that amount to equal 1 ( millions of barrels is already in the units), and solve for N: 1 = N 1.1t dt 1 = 1t.1t2 2 1 = 1N.1N2 2 Set up as quadratic: =.5N 2 1N +1 Use quadratic formula: N N = 1± 1 4(.5)(1) 2(.5) 1.6, years Only the 1.6 year estimate makes sense, as the rate would be negative at 189 years. If the estimate of 1 million barrels is correct, it should take 1.6 years to exhaust the well. (b) If oil sale price is $2 per barrel, and it costs $1 to extract, then every barrel of oil will produce a profit of $2 - $1 = $1. This means that we can translate the rate of oil into the rate of money flowing in to an account, with P(t) = rate of deposit = $1 q(t) = $1(1.1t) = $(1 t) millions of dollars per year This income stream will be flowing as long as the well is pumping, which we found would be for 1.6 years in part (a). Using the present value formula on page 435 of the text, we can compute the present value of all the oil that will be produced over that time. Present value = = Split into two integrals: = P(t)e rt dt (1 t)e rt dt 1e rt dt 1.6 te rt dt The first integral is 1e rt /r. The second integral requires integration by parts: selecting: u = t dv = e rt dt so du = dt v = e rt /r Using the integration by parts formula, 5

6 te rt dt = te rt /r + 1 r e rt dt = te rt /r 1 r 2e rt so the Present value = 1 e rt r = e rt ( 1 r 2 + t 1 r ( t r e rt 1 r 2e rt ) 1.6 ( 1 with r = 1% =.1, = e ( 1 e.1 1 ).1 $624.9 million The present value is roughly $625 million dollars. This seems reasonable, since the total value of the oil is $1, million dollars ($1 for each of the 1 million barrels). However, the oil will only be pumped out later, so the present value is slightly discounted. ) ) 6