#$&*
course MTH 173
7/11 (I knowwww)
Exercises 1-41. 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.
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Year 1: start $5000, interest 5% ($5000) = .05* $5000 = $250, end $5000 + $250 = $5250.
Year 2: start $5250, interest 5% ($5250) = .05* $5250 = $262.50, end $5250 + $262.5 = $5512.50.
Year 3: start $5512.50, interest 5% ($5512.5) = .05* $5512.5 = $275.63, end $5512.5 + $275.63 = $5788.13.
Year 4: start $5788.13, interest 5% ($5788.13) = .05* $5788.13 = $289.41, end $5788.13 + $289.41 = $6077.54
P(t) = 5000*(1.05)^t
so
P(100) = $657,506.29
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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?
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Yr 1: $500*1.12 = $600
Yr 2: $600*1.12 = $672
Yr 3: $672*1.12 = $752.64
Yr 4: $752.64*1.12 = $842.96
You multiply it by 1.12.
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3. Give the expression for the 100-th year ending principle for an original principle of P0 and an interest rate of 6%.
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P(100) = P0(1.06)^t
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4. What are the growth rate and growth factor for each of the following:
$500 is invested at 15% for 20 years
$30,000 is invested at 7% for 30 years
$2000 is invested at 5% for 40 years.
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growth rate = .15, growth factor = 1.15
P(t) = 500(1.15)^t
P(20) = 500(1.15^20) = $8183.27
growth rate = .07, growth factor = 1.07
P(t) = 30,000(1.07)^t
P(30) = 30,000(1.07)^30 = $228,367.65
growth rate = .05, growth factor 1.05
P(t) = $2000(1.05)^t
P(40) = $2000(1.05^40) = $14,079.98
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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.
How long does it take for the original $200 principle to double to $400?
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?
At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
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Growth rate will be .1, growth factor 1.1, so P(t) = 200*1.1^t
P(0) = $200
P(5) = $322.1
P(10) = $518.75
P(20) = $1345.50
So then 2 = 1.1^t
ln(2) = t*ln(1.1)
ln(2)/ln(1.1) = t
t = 7.27 yrs for doubling
300 = 200*1.1^t
3/2 = 1.1^t
ln(3/2) = t(ln(1.1))
t = 4.25 yrs to reach $300
It's still going to take 7.27 yrs to double, since the ratio is still going to be 2 on the left side of the equation.
Or
you could plug into the equation again and get 600 = 200*1.1^t, where t =~11.53 yrs, so 11.53 - 4.26 = 7.27yrs.
then
half of $1345.50 is $672.75 and
672.75 = 200*1.1^t
ln(672.75/200) = t(ln(1.1))
t = 12.73 yrs. from the beginning
the doubling time will still be 7.27yrs since 2 = 1.1^t will give the same result as before.
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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?
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2 = 1.2^t
ln(2) = t(ln(1.2))
t = 3.8yrs
It nearly cut it in half.
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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?
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we would still end up with 2 = 1.2^t if the interest rate isn't changing b/c $4000 = $2000(1.2^t), so the ratio on the left side of the equation
will always be 2 for doubling time.
so the doubling rate would still be 7.27yrs.
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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.
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?
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?
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P(t) = 1(1.1)^t
P(t) = 1(1.2)^t
P(t) = 1(1.3)^t
P(t) = 1(1.4)^t
No, it doesn't:
2 = 1.1^t
ln(2) = t(ln(1.1)
t = 7.27yrs
ln(2) = t(ln(1.2))
t = 3.8yrs
ln(2) = t(ln(1.3))
t = 2.64yrs
ln(2) = t(ln(1.4))
t = 2.06yrs
The final principle will be greater for the 40% since the doubling time is lower.
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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.
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Initial principle should have no difference on doubling times, so the final principle will still be greater since the doubling time is still
lower.
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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?
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(.5, 1.15), (1.5, 1.495), (2.5, 1.9435), (3.5, 2.52655)
y = 1.3^.5 = 1.140
y = 1.3^1.5 = 1.482
y = 1.3^2.5 = 1.927
y = 1.3^3.5 = 2.505
1.15-1.14 = .01
1.495-1.482 = .013
1.9435-1.927 = .0165
2.52655-2.505 = .02155
(.25, 1.07), (.75, 1.22), (1.25, 1.391)
y = 1.3^.25 = 1.068
y = 1.3^.75 = 1.217
y = 1.3^1.25 = 1.388
1.07-1.068 = .002
1.22-1.217 = .003
1.391-1.388 = .003
looks like it did better to me, since comparatively .01/1.14 is about .86% error and .002/1.068 is about .19% error
also .013/1.482 is about .88% error and .003/1.217 is about .25% error
and .0165/1.927 is about .86% and .003/1.388 is about .22%, etc.
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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.
Sketch a graph of principle vs. time and indicate on your graph how you obtain an estimate of the doubling time.
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2P(t_initial) = P(t_initial + doubling time)
2P(0) = P(0+d.t.)
$4000 = $2000*1.1^(0+d.t.)
2 = 1.1^d.t.
ln(2)/ln(1.1) = d.t.
d.t. = 7.27yrs
or you could sketch the graph and find $2000 and $4000 on the y-axis. The change in the x-axis in terms of time will be the doubling time.
in this case, $2000 = $2000*1.1^t1 would give us t1 = 0, and $4000 = $2000*1.1^t2, and t2= 7.27, and 7.27yrs - 0yrs = 7.27yrs. You can calculate
both t values and find the difference.
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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.
Sketch a graph of principle vs. time and indicate on your graph how you obtain an estimate of the doubling time.
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P(2+doub.time) = 2P(2)
$5000(1.08^(2+d.t.)) = 2(5832)
1.08^(2+d.t.) = 2.3328
(2+d.t.)ln(1.08) = ln(2.3328)
2 + d.t. = 11
d.t. = 9 years
or we could have solved
2 = 1.08^t
and gotten 9 years, since d.t. is independent of t.
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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.
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(1 + 1/2) ^ 2 = 2.25
(1 + 1/4) ^ 4 = 2.44140625
(1 + 1/10) ^ 10 = 2.59374246
(1 + 1/100) ^ 100 = 2.704813829
(1 + 1/1000) ^ 1000 = 2.716923932
(1 + 1/10000) ^ 10000 = 2.718145927
differences .46828, .27687375, .12453754, .013466171, .001356068, .00013407
estimate for n = 100,000 would be 2.7182666
estimate for n = 1,000,000 would be 2.71827866
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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).
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Let me see if I can count this correctly. Mine started popping out the same answer at 100,000,000,000. It gave me 2.718281828 for that one and
for 10,000,000,000.
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14. Use DERIVE to determine the approximate number n required to obtain the value 2.71828.
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I'll come back to this one to do the DERIVE program, but the closest I can get without it is 50050017.
(1 + 1/n) ^ n = 2.71828
n =~ 50050017
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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?
Sketch a graph of the amount of plutonium vs. the number of millennia, for t = 0 to 50 millennia.
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Q(t) = 16(.93)^t
P(10) = 7.744g
P(20) = 3.748g
P(30) = 1.814g
P(40) = .8779g
P(50) = .4249g
My graph looks like it's decreasing at a decreasing rate.
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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)?
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since 1 millenium = 1000yrs, with t in yrs.,
Q(t) = Q0(.93)^(t/1000)
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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?
Sketch a graph of Q(t) vs. t for 0 <= t <= 10 days. Use this graph to estimate the half-life of the element.
Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible.
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Q(t) = 30(.85)^t
around 4
the graphs of y=30(.85)^t and y=15 intersect around 4.27.
.5 = .85^t
ln(.5) = t(ln(.85))
t = 4.27
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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.?
How much antibiotic will be present at 3:00 p.m.?
Sketch a graph of Q(t) vs. t for 0 <= t <= 10 hours. Use this graph to estimate the half-life of the element.
Write the equation you would solve to find the half-life of this element. Simplify the equation as much as possible.
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Q(t) = 550(.89)^t
Q(5) = 550(.89)^5 =~ 307.12mg
around 6 hrs.
.5 = .89^t
ln(.5) = t(ln(.89))
t =~ 5.95 hrs.
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Suggested Exercise: The value of the half-life of plutonium given here might or might not be very accurate. Find out what the actual half-life
of plutonium really is, and determine how long it would take to reduce an initial quantity to 1% of its original activity.
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I'm not seeing where the half-life of Plutonium was given here to compare, but Plutonium-239 has a half-life of 24360 yrs according to google, so
I'll use that.
.01 = .5^(t/24360)
ln(.01) = (t/24360yrs)(ln(.5))
t/24360yrs = 6.64386
t =~ 161844 yrs
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Challenge Exercise for Calculus-bound students: Suppose that the activity of a sample of plutonium is initially 5 decays per second. Using the
model above, determine the activity after 5, 10, 15, and 20 millenia. Then find the approximate average number of decays per second from 0-5
millenia, 5-10 millenia, 10-15 millenia and 15-20 millenia. Use these averages to determine the total approximate number of decays during each
of these time intervals, and the total for the entire 20 millenia. Speculate on how many decays there will be from the initial time to
eternity.
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I don't think I did the plutonium thing right, so I'm just going to use the antibiotic equation and pretend it's plutonium.
I mean, who wouldn't want plutonium instead of medicine?
@& Plutonium will put you out of your misery faster and more effectively than most antibiotics.*@
Q(t) = 5dec/sec(.94)^t
Q(5) = 3.67 d/sec
Q(10) = 2.69 d/s
Q(15) = 1.98 d/s
Q(20) = 1.45 d/s
0-5 ave: 4.335d/s
5-10 ave: 3.18d/s
10-15 ave: 2.335d/s
15-20 ave: 1.715d/s
so 5millenia(1000yrs/millenia)(365days/yr)(24hr/day)(60min/hr)(60sec/min) = 1.5768*10^11sec
mult. that by ave rate, and we get 6.835*10^11 decays, 5.01*10^11 decays, 3.682*10^11 decays, and 2.7*10^11 decays respectively
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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.
Find values of t for which Q(t) lies within each of the following ranges:
between .005 Q0 and .01 Q0
between .0005 Q0 and .001 Q0.
Sketch a reasonably accurate graph of Q(t) vs. t, with the range of t sufficient to allow Q(t) to fall below .01 Q0.
In terms of this exercise explain why the positive x axis is a horizontal asymptote for this function.
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for t = 25, Q(t) = .0718Q0
for t = 48, Q(t) = .0064Q0
for t = 70, Q(t) = .00063Q0
because no matter how high your t gets, your line will only ever approach a value of 0 in terms of quantity.
the graph will decrease at a decreasing rate, so the magnitude of the slope will just decrease (approaching left to right);it'll never actually
hit or cross the line.
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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)
Q(t) = Q0 (.7 ^ t)
Q(t) = Q0 (.6 ^ t)
Q(t) = Q0 (.5 ^ t).
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1. Q(12) = .0687 Q0
2. Q(8) = .0576 Q0
3. Q(5) = .0778 Q0
4. Q(4) = .0625 Q0
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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).
Find values of t for which Q(t) lies within each of the following ranges:
between .005 Q0 and .01 Q0
between .0005 Q0 and .001 Q0.
Sketch a reasonably accurate graph of Q(t) vs. t, with the range of t sufficient to allow Q(t) to fall below .01 Q0.
In terms of this exercise explain why the negative x axis is a horizontal asymptote for this function.
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Q(-25) = .092*Q0
Q(-50) = .0085*Q0
Q(-75) = .000786*Q0
this time it's approaching from the right to the left, and as the t value increases in negativity/magnitude, the Q value approaches 0, but never
reaches it.
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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)
Q(t) = Q0 (.7 ^ t)
Q(t) = Q0 (.6 ^ t)
Q(t) = Q0 (.5 ^ t).
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Same as #18?
1. Q(12) = .0687 Q0
2. Q(8) = .0576 Q0
3. Q(5) = .0778 Q0
4. Q(4) = .0625 Q0
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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 = .007 ( 2^(.71 x) )
y = -13 ( 2^(3.9 x) )
Write each of these functions in the form y = A b^x.
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y = 12(.7071)^x
y = .007(1.636)^x
y = -13(14.93)^x
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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 = .007 ( e^(.71 x) )
y = -13 ( e^(3.9 x) )
Write each of these functions in the form y = A b^x.
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y = 12(.6065)^x
y = .007(2.03)^x
y = -13(49.4)^x
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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.
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2^(kx) = 13^x
so 2^k = 13
k=~ 3.7
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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.
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9(e^(kx)) = 9(13^x)
e^(kx) = 13^x
e^k = 13
k =~ 2.56
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Preliminary exercise: What two simultaneous equations do we obtain when we substitute the data points (3,9) and (7,2) into the form y = A ( 2^
(kx) )?
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9 = A(2^(k*3))
2 = A(2^(k*7))
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Exercises 25-28
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.
Graph the resulting function.
Repeat for the form y = A (2^(kx) ).
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20cm^2 = 12(rate^8)
20/12 = rate^8
eighth_root(20/12) = rate
rate = 1.066 cm^2/hr or 6.6% increase every hour
1.066^x = 2^(kx)
so 1.066 = 2^k
k =~ .092207
so
y = 12(2^(.092207x)
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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.
Graph the resulting function.
Repeat for the form y = A (2^(kx) ).
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3 = 4(rate)^(3 seconds)
3/4 = r^3
cubic_root(3/4) = r
r=~.9085603
so
y = 4(.9085603)^(t-2)
annnd
4(.9085603)^(t-2) = 4(2)^(k*(t-2))
.9085603 = 2^k
k=~ -.13835
so
y = 4(2^(-.13835*(t-2)))
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I don't see 27-28?
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"
&#Your work looks good. See my notes. Let me know if you have any questions. &#
#*&!