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course Mth 279
2/22
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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Using the equation form f = p * (1 + i )^n
We can get
3000 = 1000 (1 + 0.04)^n
3 = 1.04^n
ln(1.04) * n = ln(3)
n = ln(3) / ln(1.04) = approximatly 28 years
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How long will it take if compounded quarterly at the same annual rate?
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3000 = 1000 (1 + (0.04/4))^(4n)
3000 = 1000 (1.01)^(4n)
3 = 1.01^(4n)
4n ln(1.01) = ln(3)
n = ln(3) / (4 ln(1.01)) = 27.6 years approx.
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How long will it take if compounded continuously at the same annual rate?
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using formula A(t) = A_0 * e^(rt)
3000 = 1000 e^(o.o4t)
3 = e^(0.04t)
ln(3) = 0.04t
t = ln(3) / 0.04 = 27.47 years approx.
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Your solution:
Answered above after each question in the spaces provided.
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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:
Using f = P(1 + i)^n
3000 = 1000 (1 + i )^15
and solving for i,
3 = (1+i)^15
15th root of 3 = 1 + i
1.076 = 1 + i
0.076 = i
or i = 7.6%
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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:
Using the formula for exponential growth, x(t) = x e^(kt)
solving for k,
100,000 = 40,000 e^(k * 72)
2.5 = e^(k * 72)
ln(2.5) = k * 72
k = ln(2.5) / 72 = 0.013 approx.
solving for t,
200,000 = 100,000 e^(0.013t)
2 = e^(0.013t)
ln(2) = 0.013t
t = ln(2) / 0.013 = approximatly 54.5 more hours
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3
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Your solutions to the above three problems all relied on the solutions of differential equations.
You have already demonstrated enough knowledge of these equations that I believe you would be able to write and solve the related equations, and you have already done in other exercises. So I'm no concerned.
However to remeber that if a problem of this nature arises on a test, you will be expected to write and solve the equation.
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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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dP/dt = k P + M has the general solution of P(t) = c e^(kt) - (M/k)
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If P(0) = P_0 then at t = 0 we have e^(kt) = e^0 = 1 and
P_0 = - M / k + c
so
c = P_0 + M / k
and our solution is
P(t) = -M / k + (P_0 + M / k) e^(k 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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the increase of the population e^(kt) must be greater than (M/k)
at t = 0, P(t) = c - (M/k) which means that the minimum population must be greater than (M/k)
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What migration rate is required to achieve a constant population?
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In terms of the solution mentioned above:
Since e^(k t) is positive and increasing, P(t) will therefore be increasing as long as P_0 + M / k > 0.
Similarly P(t) will be decreasing if P_0 + M / k < 0.
So the threshold migration rate occurs when P_0 + M / k = 0, giving us M = - k P_0.
If M > - k P_0, population increases.
If M < - k P_0, population decreases
Remember that positive M indicates migrating into the population. Your solution assumed positive M to indicate migration out of the population, which is OK if you state it that way.
If the migration rate M is positive (into the population), of course, the population will increase. If M is negative and of sufficient magnitude then the population will decrease.
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Your solution:
The migration rate M, must equal the rate of population increase k P.
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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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The number of individuls that migrate away each year equals the number that migreates in each year.
which would keep the population the same as the previous 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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The migration rate M, must equal the rate of population increase k P.
They are the same, bc to keep a steady or constant population, these rates must be equal.
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You didn't solve this one.
Here we will let M be the number that migrates away, so positive M will correspond to a loss from the population.
During the course of a year the population increases from P_0 to P_0 * e^(k).
So to keep the population stable 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 the 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:
first the half life of the first substance is 120,
120 = ln(2) / k
and solving for k,
k = 0.006
using a form of the equation, Q(t) = C e^(-kt)
4 = 3e^(-0.006 * 120) + M
4 - M = 3e^(-0.006 * 120)
M = 4 - 3e^(-0.006 * 120) = 2.54
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If you were replentishing and adding to the element every 120 days this is a plausible amount.
If you are replentishing at a constant rate, however, the result is somewhat different. See if you agree with the following:
The decay rate is constant so
dQ/dt = -kQ,
where Q is quantity present.
This is easily solved. We obtain
Q(t) = Q_0 e^(-kt)
The half-life is the time required to reach half the original quantity. Since this requires 120 days we have Q(120) = .5 Q_0 so that
.5 Q_0 = Q_0 * e^(-k * 120)
which we solve for k. The result is
k = -ln(.5) / 120 = .006
so our model is
Q(t) = Q_0 e^(-.006 t).
Left to itself, our original 3 grams will therefore decay in such a way that
Q(t) = 3 e^(-.006 t).
If, however, we add material at an appropriate constant rate r, the quantity of material present may be kept constant.
The rate at which the material is lost is
dQ/dt = -.018 e^(-.006 t)
At the initial instant t = 0, this rate is -.018, meaning that material is being lost at the rate of .018 grams / day.
If we add material continuously at this constant rate, then none will be lost.
Thus r = .018, meaning that we must continuously add .018 grams of new material per day.
It's worth noting that if rather than adding material continuously we just add a chunk consisting of .018 grams of the material at the end of each day, the amount of material will increase over time, with the beginning-of-day amount increasing toward a limiting value which is in excess of the original 3 grams. It would be interesting to calculate this limiting value.
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Your work is good, but I believe you do have some errors. In any case I've added some notes. Check them out and let me know if you have questions.
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