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 2005 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 $1000 per year. The amount deposited in the small time interval ∆t at time t will be $1000 · ∆t. This money will grow at 8% interest for the remaining 10 − t years, so its future value will be
($1000∆t) · e0.08·(10−t)
Adding up all the deposits from t = 0 to t = 10, and letting ∆t → dt, we get
Future value = Z 10 0 1000e0.08(10−t)dt = 1000e0.08(10−t) −1 0.08 10 0 = 1000 0.08 −e0+ e0.08(10) = $15, 319.26
This is noticeably more than the actual $10,000 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 = Z 5 0 2000e0.08(10−t)dt = 2000e0.08(5−t) −1 0.08 5 0 = 2000 0.08 −e0+ e0.08(5) = $12, 295.62
Note that this value is less than for the 10 year case in #1, even though the total deposit is still $10,000. 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 = erM
·Present value where r is the interest rate and M is the number of years.
Present value = Future Value erM
= $12, 295.62 e0.08·5
≈$8, 242.00
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 $100,000.
Future value = Z 10 0 Re0.10(10−t) dt 100, 000 = Re0.10(10−t) −1 0.10 10 0 100, 000 = R 0.10 h −e0+ e0.1(10) i R = 10, 000 e1−e0 ≈$5, 819.77 per year
(b) A single lump sum of P0, over 10 years, would grow to P0e0.10(10). To find an initial
deposit that would grow to 100, 000 in ten years, we solve
P0e1= 10, 000
P0=
10, 000
e1 ≈$36, 787.94
They would need to deposit $36,787.94 today to have $100,000 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 $2,800 right now (and so its present value is $2,800), it would be easier to compare present values.
When we consider the three payments of $1,000, 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 $1,000. The next payment is a year from now, so needs to be discounted by e−0.06·1. The last payment
$1, 000 + $1, 000 · e−0.06·1+ $1, 000 · e−0.06·2≈$2828.68
From this analysis, you are marginally better off taking the payment in $1,000 installments, although the difference is relatively slight ∼ $28 on the $3,000 amount. With either option, you would still be taking a loss relative to the full $3,000 which you are owed.
13.
(a) If the extraction rate is q(t) = 10 − 0.1t millions of barrels per year, the total amount extracted in N years will be
Z N
0
We want that amount to equal 100 (“millions of barrels” is already in the units), and solve for N : 100 = Z N 0 10 − 0.1t dt 100 = 10t −0.1t 2 2 N 0 100 = 10N −0.1N 2 2 Set up as quadratic: 0 = 0.05N2−10N + 100
Use quadratic formula: N = 10 ±p100 − 4(0.05)(100) 2(0.05)
≈10.6, 189.4 years
Only the 10.6 year estimate makes sense, as the rate would be negative at 189 years. If the estimate of 100 million barrels is correct, it should take ≈ 10.6 years to exhaust the well.
(b) If oil sale price is $20 per barrel, and it costs $10 to extract, then every barrel of oil will produce a profit of $20 - $10 = $10.
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 = $10 × q(t) = $10(10 − 0.1t) = $(100 − 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 10.6 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 = Z 10.6 0 P (t)e−rt dt = Z 10.6 0 (100 − t)e−rt dt
Split into two integrals: = Z 10.6 0 100e−rt dt − Z 10.6 0 te−rt dt
The first integral is −100e−rt/r. The second integral requires integration by parts:
selecting: u = t dv = e−rt dt
so du = dt v = −e−rt/r
Z te−rt dt = −te−rt/r +1 r Z e−rt dt = −te−rt/r − 1 r2e− rt
so the Present value = −100e−
rt r − −t re −rt− 1 r2e− rt 10.6 0 = e−rt 1 r2 + t − 100 r with r = 10% = 0.1, = e−1.06 1 0.01 − 10.6 − 100 0.1 −e0 1 0.01− 0 − 100 0.1 ≈$624.9 million
The present value is roughly $625 million dollars. This seems reasonable, since the total value of the oil is $1,000 million dollars ($10 for each of the 100 million barrels). However, the oil will only be pumped out later, so the present value is slightly discounted.
