Assignment 5

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course Mth 279

6/21 8:00

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) P = P_0(1 + (r/n))^(nt)

3000 = 1000(1 + (.04/1))^(1t)

log(3) = log(1.04^t)

log(3) = tlog(1.04)

t = log(3)/log(1.04) = 28.01 years

B) 3000 = 1000(1 + (.04/4))^(4t)

3 = 1.01^(4t)

log(3) = log(1.01^(4t))

log(3) = 4t(log(1.01))

4t = log(3)/log(1.01)

t= (log(3)/log(1.01))/4 = 27.6 years

C) 3000 = 1000e^.04t

3 = e^.04t

ln(3) = 0.04t

t = ln(3)/.04 = 27.4 years

confidence rating #$&*:

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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:

P = Ae^(kt)

3000 = 1000e^(15k)

3 = e^(15k)

ln(3) = 15k

k = ln(3)/15 = .07 = 7%

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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:

1. find rate of growth

100,000 = 40,000e^(72k)

2.5 = e^(72k)

ln(2.5) = 72k

k = ln(2.5)/72 = .012 = 1.2%

2. find time to grow to 200,000 population

200,000 = 100,000e^(.012t)

2 = e^(.012t)

ln(2) = .012t

t = 57.7 more hours

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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. dP/dt = P’ = kP + M

P’ - kp = M

p(t) = -k g(t) = M

int(-k) = -kt

P(t) = Ce^(kt) + e^(kt)*int(e^(-kt)*M)

P(t) = Ce^(kt) + e^(kt)*M((-e^(-kt))/k)

P(t) = Ce^(kt) - (M/k)

B. increase of the population e^(kt) must be greater than (M/k)

C. migration rate M must equal the rate of population increase kP

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Your solution to the equation is good, but you haven't applied the initial condition P = P_0.

If you do you will find that 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

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Given Solution:

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Self-critique (if necessary):

Really no idea what this question is getting at. I had to try and look up parts B and C.

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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:

A) The same number that migrate in

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This number is not given. It must be computed for the new parameters.

Doing so, we get the condition

M = P_0 ( e^k - 1 )

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B) The migration rate M, must equal the rate of population increase kP. They are the same because to keep a steady or constant population, these rates must equal

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Given Solution:

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Self-critique (if necessary):

I was able to answer the first part of the question. But the second question didn’t make sense. I again had to try search for help.

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The difference between this migration rate and the preceding is thus

P_0 * ( e^k - 1 - k)

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:

Half life = ln(2)/k

120 = ln(2)/k

k = ln(2)/120 = 0.006

dQ/dt = (-kQ)

dQ/dt = (-.006 * 3)

dQ/dt = -.018

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Given Solution:

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Self-critique (if necessary):

Not really sure if this is correct. I found that the rate = -0.018, which means it is decreasing at that rate?

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That is the rate at which new material must be added, per day.

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&#Good responses. See my notes and let me know if you have questions. &#