course Mth 271
Exercises 1-4 1. Repeat the introductory exercise for a beginning principle of $5000 and an annual interest rate of 5%. That is, calculate the principle at the end of each of the first 4 years, then calculate the principle at the end of 100 years.
P = $5000, r = .05, t = 4, 100
P(1.05)^4
5000(1.05)^4 = 6077.53
5000(1.05)^100 = 657506.30
2. Repeat the introductory exercise for a beginning principle of $500 and an annual interest rate of 12%. By what number would you multiply the amount at the beginning of the year to get the amount at the end of the year?
P = 500, r = .12
Multiply by 1.12
3. Give the expression for the 100-th year ending principle for an original principle of P0 and an interest rate of 6%.
P(t) = P0 (1.06)^100
4. What are the growth rate and growth factor for each of the following:
a) $500 is invested at 15% for 20 years
P0 = 500, growth rate = .15, growth factor = 1.15
P(t) = 500(1.1)^20
b) $30,000 is invested at 7% for 30 years
P0 = 30000, growth rate = .07, growth factor = 1.07
P(t) = 30000(1.07)^30
c) $2000 is invested at 5% for 40 years.
P(t) = 2000(1.05)^40
Exercises 5-7
5. For a $200 investment at a 10% annual rate, what are the growth rate and the growth factor? What therefore is the function P(t) that gives principle as a function of time?
For this function determine the principle at t = 0, t = 5, t = 10 and t = 20. Sketch an approximate graph of principle vs. time from t = 0 to t = 20.
growth rate = .1, growth factor = 1.1
P(t) = P0(1.1)^t t = 0, 5, 10, 20
P(t) = 200, 322.102, 518.75, 1345.5
How long does it take for the original $200 principle to double to $400?
400 = 200 (1.1)^t
2 = 1.1^t
ln2 = tln1.1
t = 7.27
At what approximate value of t does the principle first reach $300? Starting from that time, how long does it take the principle to double?
300 = 200(1.1)^t
1.5 = 1.1^t
ln1.5 = tln1.1
t = 4.25
600 = 300(1.1)^t
2 = 1.1^t
ln2 = tln1.1
t = 7.27
t = 7.27 + 4.25 = 11.52
At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
t = 20
P(t) = P0(1.1)^t
P(t) = 200(1.1)^20
P(t) = 1345.49
P(t) = 672.75
672.75 = 200(1.1)^t
3.36 = 1.1^t
ln3.36 = tln1.1
t = 12.7
6. Determine the doubling time for a $200 investment at a 20% annual rate and compare to the results of #1. How did a doubling of the rate affect the doubling time of the investment?
P0 = 200, growth rate = .20, growth factor = 1.2
400 = 200(1.2)^t
2 = 1.2^t
ln2= tln1.2
t = 3.802
7. Determine the doubling time for $2000 investment at a 10% annual rate and compare to the results of #1. How did a ten-fold increase in the initial principle affect the doubling time of the investment?
P0 = 2000, growth rate = .1, growth factor = 1.1
4000 = 2000(1.1)^t
2 = 1.1^t
ln2 = tln1.1
t = 7.3
The growth rate is different for the two problems, meaning even though there is a ten-fold increase, they can’t be compared. If they growth rates were the same, they would have the same doubling time.
Exercise 8-9
8. On a single set of coordinate axes, sketch principle vs. time for the first four years, using four different functions, each with an initial principle of $1. Let the rate the 10% for the first function, 20% for the second, 30% for the third and 40% for the fourth.
P0 = 1, r = .1, .2, .3, .4, t = 1, 2, 3, 4
P(t) = P0(1.1)^t
r = .1 r = .2 r = .3 r = .4
t P(t) t P(t) t P(t) t P(t)
1 1.1 1 1.2 1 1.3 1 1.4
2 1.21 2 1.44 2 1.69 2 1.96
3 1.33 3 1.73 3 2.20 3 2.74
4 1.46 4 2.07 4 2.86 4 3.84
Does the final principle increase by the same amount when the rate increases from 10% to 20% as it does between 20% and 30%, and is the change in final principle between the 30% and 40% rates the same as the other two? If not what kind of progression is there in the final amounts?
There is a difference in the increase for the different rates. The higher the rates, the more the principle increases.
Estimate for each rate the time required to double the principle from the initial $1 to $2. As the percent rate increases in increments of 10%, does the doubling time change by a consistent amount?
r = .1, .2, .3, .4
P0 = 1, P(t) = 2
P(t) = P0(1 + r)^t
2 = 1(1 + r)^t
r = 1.1 r = 1.2 r = 1.3 r = 1.4
t = 7.27 t = 3.80 t = 2.64 t = 2.06
There is no consistent change in the doubling time.
9. Repeat the preceding exercise for an initial principle of $5. You can do this very quickly if you think about how to do it efficiently.
P0 = 5, r = .1, .2, .3, .4
r = .1 r = .2 r = .3 r = .4
t P(t) t P(t) t P(t) t P(t)
1 5.5 1 6 1 6.5 1 7
2 6.05 2 7.2 2 8.45 2 9.8
3 6.65 3 8.64 3 10.98 3 13.7
4 7.32 4 10.37 4 14.28 4 19.21
Exercise 10-11
10. Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 0, for a $2000 investment at 10%. Simplify this equation as much as possible using valid operations on the equation.
t = 0, P0 = 2000, growth rate = .1, growth factor = 1.1
P(t) = 2000(1.1)^t
2000(1.1^(t0 + doublingTime)) = 2 (2000(1.1^t0))
Divide by 2000
1.1^(t0 +doublingTime) = 2(1.1^t0)
1.1^t0 + 1.1^doublingTime = 2(1.1^t0)
Divide by 1.1^t0
1.1^doublingTime = 2
11. Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. Simplify this equation as much as possible using valid operations on the equation.
t = 2, P0 = 5000, growth rate = .8, growth factor = 1.8
5000(1.8^(2 + doublingTime)) = 10000
1.8^(2 + doublingTime) = 2
1.8^2 + 1.8^doublingTime = 2
1.8^doublingTime = .61728
Exercises 12-14
12. Use your calculator to evaluate (1 + 1/n) ^ n for n = 2, 4, 10, 100, 1000, and 10000. For each value of n, write down the difference between 2.71828 and your result. Make a reasonable estimate of what the differences would be for n = 100,000 and for n = 1,000,000.
(1 + (1/n))^n n = 2, 4, 10, 100, 1000, 10000
n f(n) f(n) - e n = 100000, f(n) = 2.718268, f(n) – e = .0000138
2 2.25 .46828 n = 1000000, f(n) = 2.7182804, f(n) – e = .00000142
4 2.44 .27828
10 2.59 .12828
100 2.705 .01328
1000 2.717 .00328
10000 2.71814 .00014
13. As n continues to increase, (1 + 1/n) ^ n continues to approach 2.71828. However, your calculator will eventually begin to malfunction as you attempt to use larger and larger numbers for n. Most calculators will begin giving smaller and smaller results, and will finally give just 1. This is a result of the approximate nature of the calculator's binary approximation to base-10 arithmetic, and to the limits of its precision.
Determine the approximate value of n at which your calculator begins to give you bad answers. Suggestion: use n = 100,000, then 1,000,000, etc. (just add another 0 each time).
n = 1,000,000,000, f(n) = 1.10517
14. Use DERIVE to determine the approximate number n required to obtain the value 2.71828.
I don’t understand what you mean by this.
Exercise 15
15. If the amount of plutonium decreases by 7% per millennium, then how much of a 16-gram sample will remain after 10, 20, 30, 40 and 50 millenia?
r = -.07, P0 = 16, t = 0, 10 ,20, 30, 40, 50
Q(t) = 16(1.00 + (-.07))^t
t Q(t)
0 16
10 7.74
20 3.75
30 1.81
40 .878
50 .425
As t increases, the quantity of Q(t) decreases. The graph is an exponential line that approaches 0.
15.5. If we start with a quantity Q0 of plutonium, and if the amount of plutonium decreases by 7% per millennium, then what expression represents the quantity Q(t) of plutonium after t years (after t years, not after t millenia)?
Q(t) = Q0(1 + r)^t
Exercises 16-17
16. The quantity of a certain radioactive element decreases by 15% per day. The initial amount present is 30 grams.
What function Q(t) gives the quantity Q as a function of time t?
r = -.15, Q0 = 30g
Q(t) = Q0(1 + r)^t
Q(t) = 30(.85)^t
t 0 1 2 3 4 5 6 7 8 9 10
Q(t) 30 25.5 21.68 18.42 15.66 13.31 11.31 9.62 8.17 6.95 5.91
Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible.
15 = 30(.85)^’dt
½ = .85^’dt
‘dt = 4.265
17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour.
What function Q(t) gives the quantity Q of the antibiotic as a function of the time t since 10:00 a.m.?
r = -.11, Q0 = 550mg
Q(t) = Q0(1 + r)^t
Q(t) = 550(1 + (-.11))^t
How much antibiotic will be present at 3:00 p.m.?
Q(t) = 550(.89)^5
Q(t) = 307.12mg
t 0 1 2 3 4 5 6 7 8 9 10
Q(t) 550 489.5 435.65 387.73 345.08 307.12 273.34 243.27 216.51 192.69 171.50
Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible.
Q(t) = Q0(1 + r)^t
Q(t) = 550(1 -.11)^’dt
275 = 550(.89)^’dt
½ = .89^’dt
‘dt = 5.95
Exercises 17-20
17. For the function Q(t) = Q0 (.9 ^ t), if t = 3, then Q(t) = Q0 (.9^3) = .729 Q0.
Find a value of t for which Q(t) lies between .05 Q0 and .1 Q0.
t = 22, Q(t) = .098
t = 28, Q(t) = .0523
Find values of t for which Q(t) lies within each of the following ranges:
between .005 Q0 and .01 Q0
t = 40, Q(t) = .0147
t = 49, Q(t) = .0057
between .0005 Q0 and .001 Q0.
t = 59, Q(t) = .0019
t = 71, Q(t) = 00056
In terms of this exercise explain why the positive x axis is a horizontal asymptote for this function.
The value for n(Q0) approaches 0, but will never reach it
18. For each of the following functions find a value of t such that Q(t) lies between .05 Q0 and .1 Q0:
Q(t) = Q0 (.8 ^ t)
t = 8, Q(t) = .167
t = 13, Q(t) = .052
Q(t) = Q0 (.7 ^ t)
t = 5, Q(t) = .168
t = 8, Q(t) = .057
Q(t) = Q0 (.6 ^ t)
t = 4, Q(t) = .129
t = 5.75, Q(t) = .053
Q(t) = Q0 (.5 ^ t).
t = 3, Q(t) = .125
t = 4.25, Q(t) = .0525
19. For the function Q(t) = Q0 (1.1^ t), where we note that the growth rate is positive, find a value of t such that Q(t) lies between .05 Q0 and .1 Q0 (hint: the value of t will be negative).
t = -24, Q(t) = .101
t = -30, Q(t) = .057
Find values of t for which Q(t) lies within each of the following ranges:
between .005 Q0 and .01 Q0
t = -48, Q(t) = .0103
t = -54, Q(t) = .0058
between .0005 Q0 and .001 Q0.
t = -.68, Q(t) = .001532
t = -79, Q(t) = .000537
20. For each of the following functions find a value of t such that Q(t) lies between .05 Q0 and .1 Q0:
Q(t) = Q0 (.8 ^ t)
t = 8, Q(t) = .167
t = 13, Q(t) = .054
Q(t) = Q0 (.7 ^ t)
t = 5, Q(t) = .168
t = 8, Q(t) = .057
Q(t) = Q0 (.6 ^ t)
t = 4, Q(t) = .129
t = 5.75, Q(t) = .053
Q(t) = Q0 (.5 ^ t)
t = 3, Q(t) = .125
t = 4.25, Q(t) = .0525
Exercises 21-24
21. What value of b would we use to express each of the following functions in the form y = A b^x?
y = 12 ( 2^(-.5x) )
y = Ab^x, b = 2
y = 12(.707)^x
b = .707
y = .007 ( 2^(.71 x) )
y = Ab^x, b = 2
y = .007(1.635)^x
b = 1.635
y = -13 ( 2^(3.9 x) )
y = Ab^x, b = 2
y = -13(14.928)^x
b = 14.928
22. What value of b would we use to express each of the following functions in the form y = A b^x? (Note: You may use e = 2.718 as a reasonable approximation. Or you may use the e^x key on your calculator. You may even use DERIVE: you get the number e by holding down the ALT key and pressing the E key.)
y = 12 ( e^(-.5x) )
y = 12(.606^x)
b = .606
y = .007 ( e^(.71 x) )
y = .007(2.034^x)
b = 2.034
y = -13 ( e^(3.9 x) )
y = -13(49.402^x)
b = 49.402
23. Try to find a good approximation to the value of k for which the function y = 9 ( 2^(kx) ) is the same as y = 9 ( 13 ^ x ). You will have to use trial and error for now. Soon you will learn to get a precise answer using logarithms.
9(2^kx) = 9(13^x)
k = 3.72
24. Try to find a good approximation to the value of k for which the function y = 9 ( e^(kx) ) is the same as y = 9 ( 13 ^ x ). You will have to use trial and error for now. Soon you will learn to get a precise answer using logarithms.
y = 9(e^kx) = 9(13^x)
k = 2.57
Exercises 25-26
Obtain and simplify as far as possible, solving where possible, the system of equations corresponding to each of the following situations:
25. A bacteria culture grows exponentially, according to the form y = A b^x, where y is the area covered by the culture and x is time in hours. Find the model corresponding to a culture which originally covers 12 square centimeters and which, after 8 hours, covers 20 square centimeters.
y = Ab^x, (0,12)(8,20)
20 = Ab^8
12 = Ab^ 0 A = 12
20 = 12b^8
1.66 = b^8
b = 1.065
Repeat for the form y = A (2^(kx) ).
y = A(2^kx), (0,12)(8, 20)
12 = A(2^0k)
20 = A(2^8k) A = 12
20 = 12(2^8k)
1.66 = 2^8k
26. The amount of current flowing through a certain capacitor, as it discharges through the fingers of your left hand, obeys an exponential model of form y = A b^x, where y is current and x is time. The current is 4 microAmps after 2 seconds, and has decreased to 3 microAmps 5 seconds later.
y = Ab^x, (2, 4)(7,3)
4 = Ab^2
3 = Ab^7
4/3 = Ab^2/Ab^7
4/3 = b^-5
b = .944
4 = A(.944)^2
A = 4.487
Repeat for the form y = A (2^(kx) ).
y = A(2^kx)
3 = A(2^7k)
4 = A(2^2k)
4/3 = A(2^2k)/A(2^7k)
4/3 = 2^-5k
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