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course MTH 279
1. How long will it take an investment of $1000 to reach $3000 if it is compounded annually at 4%?P = Ae^kt
3000 = 1000e^(.4)(t)
3 = e^ (.4t)
ln(3)/ .4 = t, when t is in years
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4% is .04, not .4.
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In any case the exponential function doesn't apply when compounding is not continuous.
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How long will it take if compounded quarterly at the same annual rate?
10ln(3) = t, when t is in years
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How long will it take if compounded continuously at the same annual rate?
No time because it can go from 1000 to infinity instanteaneously if constantly compounded by 4% ?
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2. What annual rate of return is required if an investment of $1000 is to reach $3000 in 15 years?
3000 = 1000 e^(15k)
ln(3) /15 = k
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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?
100,000 = 40,000 e^(72k)
ln(2.5) / 72 = k
200,000 = 100,000 e^(t * ln(2.5)/72)
ln (2) / ((ln(2.5)) /72) = t
t is approximately 54.47 hours
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The rate of population growth is proportional to the population, so
dP/dt = kP
which can be written
P ' - k P = 0,
a first-order linear homogeneous equation which we solve in the usual manner, obtaining general solution
P(t) = Ae^(kt), A > 0.
This is the point at which you started your solution. However your solution should start with this differential equation.
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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).
dP/dt -kP = M
p(t) = -k
P(t) = -kt
u = e^(-kt)
e^(-kt) dP/dt = ke^(-kt) P = e^(-kt)M
integrate both sides
e^(-kt) P(t) = -M/k e^(-kt) + C
P(t) = -M/k + Ce^(kt)
P(t) = -M/k + P_0 e^(kt)
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C is not equal to P_0. The equation
P(t) = -M/k + P_0 e^(kt)
does not yield P(0) = P_0, but rather P(0) = -M/k + P_0.
P(0) = P_0 so
-M/k + C = P_0
and
C = P_0 + M/k.
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You haven't addressed the question of threshold migration rates.
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In terms of k and M, determine the minimum population required to achieve long-term growth.
If M > - k P_0, population increases.
If M < - k P_0, population decreases
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Good, though this doesn't appear to follow from the solution you gave.
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What migration rate is required to achieve a constant population?
So the threshold migration rate occurs when P_0 + M / k = 0, giving us M = - k P_0
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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?
It would have to be the same as -kP_0
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How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question?
Steady and constant population are the same concept different words
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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?
0.5C=C*e^(-k*120)
0.5=e^(-k*120)
ln0.5=(-k*120)
-k=-0.00577
C(t)=Ce^(-0.00577*t)
C(t)=3e^(-0.00577*t)
dC/dt = -.018 e^(-.00577*t)
t = 0, rate is -.018, material is being lost at the rate of .018 grams / day.
r = .018, meaning that we must continuously add .018 grams of new material per day.
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&#This looks good. See my notes. Let me know if you have any questions. &#