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course Mth 279
7/17 at 11:30pm
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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Your solution:
####a.) For the equation, k = .04 and P_0 = $1000
It would work for the equation P(t) = Ae^(.04t). At p(0) = A(1) = 1000, so the equation is P(t)= 1000e^(.04t).
1000e^(.04t) = 3000, making .04t = ln(3), and t = ln(3)/ .04. Making t = approximately 28 years.
b.) It would take approximately 28 quarters to earn the amount of money, which is roughly 7 years.
c.) It would have to run through the continuous compound cycle 28 times to reach the desired level of $3000.
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Annual interest is applied once a year to the previous principle.
If you compound quarterly you divide the annual interest rate by 4, and apply it four times each year, each time applying it to the amount that resulted from the previous application.
Compounding continuously the function would be P(t) = 1000 e^(.04 t), which is the function you used.
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confidence rating #$&*:#2
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Given Solution:
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Self-critique (if necessary):
####I am not sure if there is a different formula for yearly, quarterly, and continuously.
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Self-critique rating:####2
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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:
####The formula for compounding interest is P(t) = Ae^kt, using the formula dP/dt = KP and solving it using the first order, linear homogeneous equation.
To find the annual return rate, k, to reach $3000 in 15 years, you plug in numbers to get y(15) = 1000e^k(15) = 3000.
You then get 15k = ln(3). THen k = ln(3) / 15, which approximately gives k = .073 or 7.3%.
confidence rating #$&*:#3
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Given Solution:
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Self-critique (if necessary):
####Ok
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Self-critique rating:####3
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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:
#### Use the formula p(t) = Ae^kt to solve the problem, and I am puting t in units of hours. p(0) = Ae^0 = 40 000 and p(72) = Ae^(72k) = 100 000.
Divide the larger function by the smaller to get (Ae^(72k) = 100 000) / (Ae^0 = 40 000), giving the equation e^(72k) = 2.5.
From there solving for k, it gives 72k = ln(2.5), and k = ln(2.5)/72, making k approximately equal to .013.
To find how long it will take to get to 200 000, you need to use 40 000 e^(.013t) = 200 000 and solve for t.
It gives .013t = ln(5), making t = ln(5) / .013, making t equal approximately 129 hours to reach the population size of 200 000.
confidence rating #$&*:#3
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Given Solution:
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Self-critique (if necessary):
####ok
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Self-critique rating:####3
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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:
#### a) the form dP/dt = kP + m, can be written as P' - kP = m, in the form of a linear, nonhomogeneous equation.
Thus multiplying both sides by e^-kt giving you (e^(-kt) *P)' = me^-kt.
Then after taking the integral of both sides and solving for P you get the equation p(t) = -(m/k) + C / e^(-kt).
To find C, p(0) = -(m/k) + C = 0, making C = m/k. Thus the equation is p(t) = -(m/k) + m / (k e^-kt).
b) P(t) = (P + (M/k))e^(-kt) - (M/k)
As t approaches infintity P(t) (P + (M/k))*1 - (M/k), because e will approach e^0 = 1
For (P + (M/k))*1 - (M/k)
P + (M/k) - (M/k) , Population in terms of M/k is
P = -((M/k) - (M/k)) = 0.
Thus the population must be zero to have an exponential growth.
c) Constant Population means dP/dt = 0 so using
dP/dt = kP + M, setting dP/dt equal to zero
0 = kP + M
M = -kP, which means the amount of population leaving is equal to the amount that it gorws
Confidence ####2
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Given Solution:
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Self-critique (if necessary):
####I am not sure how to find the minimum population for exponential growth??????
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The question asked about population growth, not exponential growth.
The population will grow if the M > -k P_0.
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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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Your solution:
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a) If the migration occured all at once, you could solve the equation as a linear homogeneous using the equation p(t) = Ae^kt.
Therefore making t = 1 year, the growth of the population would be Ae^k, thus the migration m, would equal Ae^k.
b)The migration rate to keep the population the same would have to equal the growth, where, in the previous the migration rate would be less than the growth rate to keep the population exponentially growing.
confidence rating #$&*:2
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Given Solution:
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Self-critique (if necessary):
####Again, not sure if I have done the previous one correctly.
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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 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:
####We know that dP/dt = -kP, and solving we get dP/P = -kdt, and after taking the integral of both sides it solves to ln(P) = -kt + C.
P = e^(-kt + C), which goes to P = Ae^-kt. SInce its a half life, P(t) / A = .5, so .5 = e^-kt.
Solving for k, with t = 120, ln(.5) = -k(120), and k = -ln(.5)/120, which means k = .006.
This gives us the form P(t) = Ae^(-.006t).
With the intial condition that P(0) = A = 3, it comes out to be P(t) = 3e^(-.006t).
TO find how much must be added to get to 4 grams of material,
4 = 3e^(-0.006 * 120) + M
4 - M = 3e^(-0.006 * 120)
M = 4 - 3e^(-0.006 * 120)
Meaning that M = 2.54g must be added to reach the desired amount.
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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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confidence rating #$&*:#2
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Given Solution:
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Self-critique (if necessary):
####Not sure if I did this correctly.
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Self-critique rating:####2"
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&#Good work. See my notes and let me know if you have questions. &#