Assignment 5

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course Mth 279

6/24 10:00

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

confidence rating #$&*:

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Given Solution:

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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. The increase of the population, e^(kt), must be greater than (M/k) to achieve a minimum population for long term grown.

C. The migration rate M must equal and opposite the rate of population increase kP, to achieve a constant population.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Really no idea what this question is getting at.

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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

B) |m| must equal |kP|

confidence rating #$&*:

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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. Not sure if -.018 is every 120 days or every day because when we found k the half life was 120 days.

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&#This looks good. See my notes. Let me know if you have any questions. &#