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course MTH 279
2/1 8 pm
Section 2.4.*********************************************
Question: 1. How long will it take an investment of $1000 to reach $3000 if it is compounded annually at 4%?
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How long will it take if compounded quarterly at the same annual rate?
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How long will it take if compounded continuously at the same annual rate?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
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28.0 years
Calculated using formula: P(t) = (1 + r)^t * A_0
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27.6 years
Calculated using the formula:P(t) = (1 + r/4)^(4t) * A_0
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27.5 years
Calculated using the formula: P(t) = A_0 * e^(rt)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: 2. What annual rate of return is required if an investment of $1000 is to reach $3000 in 15 years?
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Your solution:
7.6%
Calculated using the formula: P(t) = (1 + r)^t * A_0
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: 3. A bacteria colony has a constant growth rate. The population grows from 40 000 to 100 000 in 72 hours. How much longer will it take the population to grow to 200 000?
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Your solution:
54.5 MORE hours.
P(t) = Ce^(kt)
P(0) = 40,000 leads us to C = 40,000
P(72) = 100,000 leads us to k = 0.0127
P(t) = 200,000 leads us to t = 126.5 hours
The difference between 126.5 and 72 gives us our time of 54.5 hrs.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: 4. A population experiences growth rate k and migration rate M, meaning that when the population is P the rate at which new members are added is k P, but the rate at they enter or leave the population is M (positive M implies migration into the population, negative M implies migration out of the population). This results in the differential equation dP/dt = k P + M.
Given initial condition P = P_0, solve this equation for the population function P(t).
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In terms of k and M, determine the minimum population required to achieve long-term growth.
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What migration rate is required to achieve a constant population?
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Your solution:
P(t) = (P_0 - M/k )e^(kt) + M/k
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-M/k must be less than P
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M = -kP
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I'm not sure if my P(t) equation is correct. When I tried to solve it myself as a linear nonhomogenous equation, I got an extra k^(-2) term, so I simply modeled my equation after the equation found in the book.
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You haven't shown enough detail for me to evaluate your work, but your solution looks good.
The equation could be written
P ' - k P = M
Its solution would be
P = -M / k + c / e^(-k t) = - M / k + c e^(k t).
Using P_0 for P(0) the solution could be written
P(t) = -M / k + (P_0 + M / k) e^(k t).
Stable population occurs when P_0 + M/k = 0, so that M = -k P.
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Self-critique rating:
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Question: 5. Suppose that the migration in the preceding occurs all at once, annually, in such a way that at the end of the year, the population returns to the same level as that of the previous year.
How many individuals migrate away each year?
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How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
P - (P_0 - M/k)e^(kt) individuals leave
The rate would be the same as in the preceding question since the overall change in population over this particular span of time would be 0.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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For a year the population increases from P_0 to P_0 * e^(k).
So the number of migrating individuals must be the difference
M = P_0 e^k - P_0 = P_0 ( e^k - 1 ).
Previously the threshold migration rate was M = P_0 * k.
The difference is P_0 ( e^k - 1 ) - P_0 k = P_0 * ( e^k - 1 - k).
e^k - 1 is greater than k. There are a number of ways to see this, but the Taylor expansion of e^k is probably the most direct.
e^k = 1 + k + k^2 / 2! + k^3 / 3! + ..., so
e^k - 1 = k + k^2 / 2! + k^3 / 3! + ...
which is greater than k by k^2 / 2! + k^3 / 3! + ... .
Conceptually, if the migration is continuous throughout the year many of the the migrating individuals don't reproduce until they have left and so do not contribute to the population growth, whereas if they stick around until the end of the year they do contribute.
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Question: 6. A radioactive element decays with a half-life of 120 days. Another substance decays with a very long half-life producing the first element at what we can regard as a constant rate. We begin with 3 grams of the element, and wish to increase the amount present to 4 grams over a period of 360 days. At what constant rate must the decay of the second substance add the first?
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Your solution:
r_s = 0.006575
R = r_s - r_e where R would be the overall growth rate, r_s would be the rate of the substance adding to the element, and r_e would be the rate of the element decaying.
R = 7.99 e-4, calculated from the equation Q(t) = Ce^(kt)
r_e = 0.00577 calculated from the equation tao = ln(2)/k
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK"
&#Your work looks good. See my notes. Let me know if you have any questions. &#