course Mth 173
Introduction to Exponential FunctionsRadioactive decay, compound interest, growth rates, growth factors, fitting exponential functions to data
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Introduction
Compound interest: $1000 at 10% Exercises 1-4
Compound interest: principle P at rate r Exercises 5-7
Graphing principle vs. time Exercises 8-9
Doubling time at rate r Exercises 10-11
Irreducible Required Knowledge
The Number e Exercises 12-14
Radioactive decay of plutonium Exercise 15
Radioactive decay: initial quantity Q0, rate r
Half-life Exercises 16-17
Asymptotes Exercises 17-20
The exponential function forms y = A b^x, y = A * 2^(kx) and y = A e^(kx) Exercises 21-24
Fitting an exponential function to data I: Obtaining a set of simultaneous (nonlinear) equations Exercises 25-28
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Introduction
The function f(x) = e ^ x is used most often as the basic form of the exponential function. Since e lies between 2 and 3 (e is approximately 2.71828), e^x lies somewhere between y = 2^x and y = 3^x. The function y = 2^x is also used as a basic form of the exponential function.
The general forms of the exponential function include
f(x) = A (2 ^ (kx)) + c,
f(x) = A b^x + c
and
f(x) = A e^(kx) + c.
The two fundamental examples through which we will begin to understand the behavior of exponential functions include compound interest and radioactive decay. We will later generalize exponential functions through the example of how the temperature of a hot or cold object approaches room temperature.
We will see that all these situations are governed by the fact that the rate of change in the quantity we are exploring (i.e., the principle of an investment, the radioactivity of a sample, the temperature of an object relative to room temperature) is always directly proportional to the quantity present.
These properties of exponential functions will be understood in terms of the laws of exponents, especially the following:
a^b * a^c = a^(b+c)
(a^b) ^ c = a ^ (bc).
We will also look at the doubling time for an increasing exponential (the time required for a given increasing exponential function to double is always the same, regardless of the starting time) and the half-life (the time required for a given decreasing exponential function to lose half its value is always the same regardless of the starting time) of an exponential function. These ideas, and the need to solve exponential equations, will later lead us to define and explore logarithmic functions.
Compound interest: $1000 at 10%
Introductory exercise: Suppose you are given $1000 to invest at 10% interest, with interest to be computed at the end of each year based on principle at the beginning of the year.
How much money will you have after each of the first four years?
Yr 1=1100
Yr2=1210
Yr 3=1331
Yr4 = 1464.10
How much will you have after 100 years? p(1+i)t or 1000(1+.1)100=$13780612.34 (I have taken several PhD level finance courses)
Complete this exercise, as best you can within 10 minutes, before you proceed.
You probably didn't have any trouble at all determining that after the first year you will have earned 10% of $1000, or $100, and that your principle will therefore be $1100.
There is a strong tendency to then figure that each year will see the addition of another $100, so that your totals will be $1200, $1300, $1400, etc.. This would especially be the case for anyone who had looked down the page and seen these numbers. However, this would not be correct. For example, during the second year your interest would be based on the $1100 present at the beginning of that year, so interest would be $110. This would bring the total after two years to $1100 + $110 = $1210.
The correct results could be computed as follows:
Year 1: start $1000, interest 10% ($1000) = .1* $1000 = $100, end $1000 + $100 = $1100.
Year 2: start $1100, interest 10% ($1100) = .1* $1100 = $110, end $1100 + $110 = $1210.
Year 3: start $1210, interest 10% ($1210) = .1* $1210 = $121, end $1210 + $121 = $1331.
Year 4: start $1331, interest 10% ($1331) = .1* $1331 = $133.1, end $1331 + $133.1 = $1464.1
We see that every year we add .1 of the beginning principle to that principle to get the ending principle.
Simplifying the Calculation Process
Expressing this yearly calculation in symbols, if P stands for the beginning principle, we add .1 P to P. This gives us
new principle = P + .1 P.
Of course we can apply the distributive law of multiplication over addition to obtain
new principle = P * (1 + .1).
It should be clear that P * (1 + .1) = P + .1 P. Of course 1 + .1 = 1.1, so
new principle = 1.1 P.
We see that the new principle is obtained by multiplying the old principle by 1.1. So our calculation scheme could have been simpler. We could have calculated as follows:
Year 1: Beginning principle $1000, ending principle $1000 * 1.1 = $1100.
Year 2: Beginning principle $1100, ending principle $1100 * 1.1 = $1210.
Year 3: Beginning principle $1210, ending principle $1210 * 1.1 = $1331.
Year 4: Beginning principle $1331, ending principle $1331 * 1.1 = $1464.1.
The 1.1 by which we multiply accounts for both our original principle P and our interest .1 P.
We note that the 4th-year ending principle has been obtained by multiplying by the factor 1.1 a total of 4 times. This is the same as multiplying our original $1000 by 1.1 * 1.1 * 1.1 * 1.1. Use your calculator to verify that the result of multiplying $1000 by 1.1 a total of four times is indeed $1464.1.
Of course $1000 * 1.1 * 1.1 * 1.1 * 1.1 = $1000 * (1.1 ^ 4). You can calculate this quantity directly using your calculator. Once more you should do so in order to validate the result.
This should give you a hint about how to calculate the 100th-year principle. See if you can figure out what to do, and then do the calculation, before you read further.
You probably realized that, over a period of 100 years we will add 10% a total of 100 times, and will therefore multiply by 1.1 a total of 100 times. This is equivalent to multiplying by (1.1) ^ 100. So our 100th-year total will be $1000 (1.1 ^ 100). You can find the number using your calculator.
Important Terminology: In this situation the 10%, which is the same as the proportion .1, is called the growth rate. This makes good sense because here money grows at the rate of 10% per year. The factor 1.1 by which we multiply beginning principle to obtain ending principle is called the growth factor. This also makes good sense, because in mathematics a number by which we multiply is called a factor, and this factor dictates how our principle grows.
The word 'rate' here is not used in the same sense as when we found average rates of changes. Here the 'rate' is a percent or proportion of the principle; the actual rate of change dy / dx depends on how much principle we have. This double use of the word 'rate' can be confusing, so think about the distinction carefully.
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.
1 5250
2 5512.50
3 5788.13
4 6077.53
100 657506.29
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?
1 560
2 627.20
3 702.46
4 786.76
100 41761132.86
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*(1.06)100
h factor for each of the following:
$500 is invested at 15% for 20 years 500*(1.15)^20=8183.27
$30,000 is invested at 7% for 30 years 30000*(1.07)^30=228367.65
$2000 is invested at 5% for 40 years. 2000*(1.05)^40=14079.98
For each situation, give an expression for the principle after t years.
Compound interest: principle P at rate r
The rate r at which principle grows determines the growth factor. For example, if the rate is r = 6% = .06, then the growth factor is 1 + .06 = 1.06. A rate of 15% would be r = .15, and the growth factor would be 1 + .15 = 1.15. It is clear then that the growth factor is 1 + r.
If interest is computed for t years, we multiply the original principle by the growth factor a total of t times. For example, if interest rate is 6% for 10 years, then we multiply by the growth factor 1.06 a total of 10 times; this is the same as muptiplying by 1.06 ^ 10. A 15% interest rate for 20 years would entail multiplying the original principle by 1.15 ^ 20. If the interest rate is r, then for the t-year amount we would multiply by (1+r) a total of t times; we would therefore multiply the original principle by (1+r) ^ t.
It is therefore easy to see that if the original principle is P0, then the t-year principle must be
t-year principle = P(t) = P0 (1 + r) ^ t.
We therefore see that the principle is a function P of time t. Using function notation we write
P(t) = P0 (1+r) ^ t.
For example, if $500 is invested at 15%, we have P0 = $500, r = .15, so growth rate is r = 15% = .15 and growth factor is (1+r) = 1.15. Thus
P(t) = P0 (1+r) ^ t = $500 (1.15) ^ t.
After 20 years we would have
P(20) = $500 (1.15) ^ 20.
You should calculate these amounts and sketch a reasonably accurate graph of P(t) vs. t.
As another example, if $30,000 is invested at 7%, we have P0 = $30000, r = .07, so growth factor r = 7% = .07 and growth rate is (1+r) = 1.07. Then
P(t) = P0 (1+r) ^ t = $30000 (1.07) ^ t.
After 30 years we would have
30-year amount: P(30) = $30000 (1.07) ^ 30.
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?growth rate is 10% and growth factor would be 1.1. P(t)=200*(1.1)t
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. P(0)=200, P(5)=322.10, P(10)=518.75
How long does it take for the original $200 principle to double. 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?Yr 4 p=~300 would take to year 11 to reach ~600 so the difference is 7 years.
At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result? P(20)=1345.50, of P(20)=672.75,
672.75=200*1.1^t
Divide both sides by 200:
3.3675=1.1^t
T=12.72915
So the doubling time is 20-12.72915 or 7.27085
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? P(20)=7667.50, of P(20)=3833.76
3833.76=200*1.2^t
Divide both sides by 200:
19.1688=1.2^t
T=16.1982
So the doubling time is 20-16.1982 or 3.8018
Approximately (52% , but rounding caused some error.)
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? P(20)=13455, of P(20)=6727.50
6727.50=2000*1.1^t
T=12.7275
So the doubling time is 20-12.7275=7.2725 or the same.
Graphing principle vs. time
Suppose the interest rate is 30%, and our original principle is $1. We can easily calculate the principle and interest for the first 4 years. You should quickly calculate these results. 1 1.3, 2 1.69, 3 2.20, 4 2.86
We find that the principle takes values $1, $1.3, $1.69, $2.20 and $2.86. The interest amounts are $.3, $.39, $.51and $.66.
These quantities are apparent in the graph shown below. Note that each principle exceeds the preceding principle by 30% (look at the blue lines), so that the amount by which the principle increases keeps increasing as the principle increases. This can be seen by the nonuniform increases in the lengths of the blue lines. It is worth noting that the semi-transparent red lines, which represent the actual interest amounts, also increase by 30% each year.
The approximate corresponding graph of P(t) = $1(1.3 ^ t) vs. t is shown below. This graph is approximate because points are joined by straight lines rather than by a smooth curve.
The expression P(t) = $1 (1.3 ^ t) can be recognized as an instance of the generalized form f(t) = A b^t + c of an exponential function, with A = $1, b = 1.3 and c = 0.
Exercise 8-9
8. On a single set of coordinate axes, sketch principle vs. 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.
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?No Increase in P appears to be increasing by ~$0.20 more then the preceding increase.
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?10% 7Yrs, 20% slightly < 4 Yrs, 30% around 2 Yrs, 40% just a tad over 2 yrs.
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. The increase appears to be $1.00 more then the preceding example (5*.2) and the answers to the periods of time would stay the same as I just changed the scale on my graph.
Challenge exercise for calculus-bound students: The graph shown above connects the points (0,1), (1,1.3), (2,1.69), etc. by straight line segments. Find the midpoint of each of these segments (halfway between the x values, halfway between the y values) and determine how far the midpoint actually deviates from the corresponding point on the graph of y = 1.3 ^ t, using the midpoint value of t. Then sketch an approximate graph using the function values at 0, .5, 1, 1.5, ..., connecting the points with straight line segments. Choose three of these segments, find their midpoints, and make the same comparison between the straight-line approximation and the actual function value. Did cutting the interval in half double the precision of the approximation, or did it do better (or worse) than that?
Midpoint of (0,1) and (1,1.3)=(.5, 1.15) y=1.3^t or 1.14 Diff .01
Midpoint of (1,1.3) and (2, 1.69)=(1.5, 1.5) y=1.3^t or 1.48 Diff .02
Midpoint of (2, 1.69) and (3,2.20)= (2.5, 1.945) y=1.3^t or 1.93 Diff .02
Midpoint of (3, 2.2) and (4, 2.86)=(3.5, 2.53) y=1.3^t or 2.50 Diff .03
Midpoint of (0,1) and (.5,1.15)=(.25, 1.08) y=1.3^t or 1.07 Diff .01
Midpoint of (1, 1.3) and (1.5, 1.5)=(1.25, 1.4) y=1.3^t or 1.39 Diff .01
Midpoint of (2,1.69) and (2.5, 1.945) = (2.25, 1.82) y=1.3^t or 1.80 Diff .02
Mispoint of (3,2.2) and (3.5, 2.53)=(3.25, 2.37) y=1.3^t or 2.35 Diff .02
Appears to be a smaller error in spite of difficulty with rounding on this example.
Doubling time at rate r
Preliminary exercise: Using the year-end values of the principle for the preceding example of a $1 initial investment at interest rate 30%, make your best estimate of the precise time at which the principle P(t) = $1 (1.3^t) first reaches $2, thereby doubling its initial value. Then determine how long it takes to double the 1-year principle of $1.3. Becomes $2.00 at 2.5 yrs. Doubles to $2.60 at 3.5 yrs.
Consider the above function P(t) = $1 (1.3 ^ t), which represents the changing principle on a $1 initial investment at an interest rate of 30%.
Recalling that the year-end values of this investment are $1.3, $1.69 and $2.20 at the ends of the first 3 years, we see that the investment reaches double its initial value at some time between t = 2 and t = 3 years, probably closer to t = 3 than to t = 2.
We see also that the approximate graph crosses the y = 2 line between year 2 and year 3, somewhat closer to year 3 (recall that the above graph is only approximate, using straight lines between data points; a smooth curve would clearly lie below this approximation and would therefore give us a somewhat later doubling time).
We might ask how the doubling time might differ if we started from some instant other than t = 0. For example, at time t = 1 the principle is $1.3. How long does it require to double to $2.6?
To see how long it takes to double the $1.3 principle, starting at t = 1, we can again consider the subsequent values $1.69, $2.20, $2.86 of the principle at the ends of succeeding years.
We see that between the end of the second succeeding year, when principle was $2.20, and the end of the third, when the principle was $2.86, the $1.3 had doubled to $2.6. The doubling time is again seen to be between 2 and 3 years, somewhat closer to 3 years.
The graph also shows that the $1.3 increases to $2.6, doubling its value, between t = 3 and t = 4, somewhat closer to t = 4; since the $1.3 value was reached at t = 1, then doubling time is between 'doublingTimeShort = 3-1 = 2 and 'doublingTimeLong = 4 - 1 = 3.
We can obtain an equation for the doubling time `doublingTime starting from t = 0.
The doubling time, starting at time t = 0 with principle P0 = $1, is the time 'doublingTime required to reach 2 P0 or $2, double the original amount.
We see then that the principle after time 'doublingTime, which is P('doublingTime), must therefore be 2 P0. So we can write the equation
P('doublingTime) = $2
to describe this condition. Since for this example P('doublingTime) = $1 (1.3 ^ 'doublingTime), we have
$1 (1.3 ^ 'doublingTime) = $2.
If we divide both sides by $1, we obtain
1.3 ^ 'doublingTime = 2.
This equation can be solved for 'doublingTime. However the solution requires logarithms. We will see soon what logarithms are and how they are used to solve such an equation.
We can also find an equation for the doubling time starting from t = 1:
The doubling time, starting at time t = 1 with principle P(1) = $1.3, is the additional time 'doublingTime required to reach 2 P(1) = $2.6.
At that time, 'doublingTime later than the starting time t = 1, t must be 1 + 'doublingTime. So the principle will be P(1 + 'doublingTime) and we can say that
P(1 + 'doublingTime) = 2 P(1)
or
P(1 + 'doublingTime) = $ 2.6.
Since P(1 + 'doublingTime) = $1 ( 1.3 ^ 'doublingTime), we can write
$1 (1.3 ^ (1 +'doublingTime)) = $2.6.
Dividing both sides by $1, we find that
1.3 ^ (1 + 'doublingTime) = 2.6.
This equation can be simplified using the rule a ^ (b + c) = a^b * a^c:
1.3^1 * 1.3^`doublingTime = 2.6.
Dividing through by 1.3, we obtain
1.3 ^ `doublingTime = 2,
the same equation we obtained for the doubling time starting with t = 0.
Once again this equation can be solved for 'doublingTime using logarithms.
We can symbolize the process to see why the doubling time doesn't depend on when we start:
If we calculate the doubling time starting at some general time t = t0, the condition for doubling is
P(t0 + 'doublingTime) = 2 P(t0).
This will be true for any function P(t).
For the specific function P(t) = $1 (1.3 ^ t), we have
$1 (1.3 ^ (t0 + 'doublingTime) ) = 2 ($1 (1.3 ^ t0) ).
Dividing by the $1 we obtain
1.3 ^ (t0 + 'doublingTime) = 2 (1.3 ^ t0).
This equation can be simplified as before:
1.3 ^ t0 * 1.3 ^ `doublingTime = 2 (1.3 ^ t0),
We can then divide both sides by 1.3 ^ t0 to obtain the familiar result
1.3 ^ `doublingTime = 2.
This shows that the doubling time is independent of the starting time.
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. 2000(1+.1)^doubling time = 4000
1.1^(doubling time)=2
Sketch a graph of principle vs. time and indicate on your graph how you obtain an estimate of the doubling time. Between 7 and 8 years
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. 2*5000(1+.08)^2=5000(1+.08)^(x-2)
2=1.08^(x-4)
Sketch a graph of principle vs. time and indicate on your graph how you obtain an estimate of the doubling time.
Irreducible Required Knowledge
You must know the laws of exponents as stated in your text.
You must know the meanings of growth rate and growth factor, and you must be able to explain in terms of these quantities how the form P(t) = P0 (1+r) ^ t arises naturally out of our consideration of compound interest.
The Number e
The number e appears in one of the standard forms f(t) = A e^(kt) + c of the exponential function. In many senses this is the most standard form, the one on which nearly every mathematician will agree. This number is arguably the most important number in mathematics.
We can understand e in terms of compound interest. Imagine that we start with some principle P0, and some hypothetical investment that yields 100% interest. Of course the growth rate is 100% = 1, so the growth factor is (1 + r) = 1 + 1 = 2. After one year we will have principle 2 P0.
Now suppose that instead of taking the entire 100% at the end of the year, we take the interest in two stages, taking 50% for the first half of the year and then applying the other 50% for the second half. In this case we have a growth factor of (1 + .5) = 1.5, which we get to apply twice. After the first application the principle will be 1.5 P0; applying the growth factor once more will give us
1.5 (1.5 P0) =1.5 ^ 2 P0 = 2.25 P0.
This is the principle at the end of the year. Notice that we now have more money than before, 2.25 P0 vs. 2 P0.
We might be encouraged to split the 100% into more than two parts, to attempt to obtain more money. If we split the 100% into four parts, 25% each, then we would have growth factor (1 + .25) = 1.25. We would be able to apply this factor four times. It should be easy to see that this will result in a year-end principle of
(1.25 ^ 4) P0 = 2.44 P0.
We see the following progression:
1 compounding: year-end principle is (1 + 1) P0 = 2 P0
2 compoundings: year-end principle is (1 + 1/2) ^ 2 P0 = 2 P0
4 compoundings: year-end principle is (1 + 1/4) ^ 4 P0 = 2 P0.
The fractions 1/2 and 1/4 indicate the splitting of the 100% interest into 2 and 4 parts. The exponents 2 and 4 indicate that the number of times we apply the interest is equal to the number of parts into which we divide the 100% interest.
If we were to divide the interest into n parts, then our growth rate for each application would be 1/n and the growth factor would be (1 + 1/n). We would therefore achieve year-end interest
(1 + 1/n) ^ n P0.
We can evaluate the quantity (1 + 1/n) ^ n for increasing values of n. Until we do so it is not clear whether the results will just keep getting larger and larger without limit, or whether they will remain bounded by some limit. It isn't even clear whether the results will continue increasing.
As you will see from the exercises, the numbers do continue increasing, but they also remain bounded.
We have already obtained 2, 2.25 and 2.44 for n = 1, 2 and 4. It turns out that as n gets larger and larger, the results approach a number a little greater than 2.7. This number cannot be represented exactly in decimal form, since it would continue indefinitely with no recurring pattern (such a number is called irrational; `pi and `sqrt(2) are other examples of irrational numbers). To 5 decimal places the number is 2.71828. The next four decimal places make it look like the number is repeating: we get 2.7818281828. But the apparent pattern is just a coincidence, and the number continues with no repeating pattern.
The number approached by (1 + 1/n) ^ n is called e:
e is the large-n limit of the expressioni (1 + 1/n) ^ n.
So if we divide the 100% interest into more and more parts, we approach the year-end principle e * P0, a bit more than 2.7 P0.
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.
At 2 dif .46828
At 4 dif .27687
At 10 dif .12454
At 100 dif .01347
At 1000 diff .00136
At 10000 Diff .00013
Estimate at 100,000 .000013
1000000 .0000013
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).It looks like about 10 million
14. Use DERIVE to determine the approximate number n required to obtain the value 2.71828. Abmy calculatorout 10 million on
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assignment #004
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Calculus I
07-09-2007
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10:19:53
query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?
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RESPONSE -->
rate=10%
factor=1.1
Double 7.27 yrs
$300 4.25 yrs
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10:20:58
** The t = 20 value is $200 * 1.1^20 = $1340, approx.
Half the t = 20 value is therefore $1340/2 = $670 approx..
By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx..
For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45). At 12.75=674.20 so it would probably be about12.72.
This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr.
This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **
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RESPONSE -->
ok
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10:21:16
query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE -->
ok
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10:21:29
** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double.
for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double.
Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double.
The final 4-year amount increases by more and more with each 10% increase in interest rate.
The doubling time decreases, but by less and less with each 10% increase in interest rate. **
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10:21:39
query #11. equation for doubling time
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10:21:53
** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore
P0 * (1+r)^t = 2 P0.
Note that this simplifies to
(1 + r)^ t = 2,
and that this result depends only on the interest rate, not on the initial amount P0. **
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10:24:50
Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.
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2*(5000*(1+.08)^2)=5000*(1+.08)^(t-2)
2=(1.08)^(t-4)
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10:26:21
**dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get
1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2].
This can be written as
1.08^2 * 1.08^doublingtime = 2 * 1.08^2.
Dividing both sides by 1.08^2 we obtain
1.08^doublingtime = 2.
We can then use trial and error to find the doubling time that works. We get something like 9 years. **
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oops
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10:27:24
Desribe how on your graph how you obtained an estimate of the doubling time.
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By moving between the amount at t=amount2 and double that
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10:27:34
**In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis.
The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **
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&#Very good responses. Let me know if you have questions. &#