Query_03

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course MTH-279

06/15 around 8:33PM.

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 f = p * (1 + I) ^ n

3000 = 1000 * (1 + 0.04) ^ n

3 = 1.04 ^ n

ln (3) = ln (1.04) * n

n = ln (3) / ln (1.04)

= about 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)

ln (3) = ln (1.01) * 4n

4n = (ln(3) / ln(1.01))

n = ((ln(3) / ln(1.01)) / 4

= about 27.6 years

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How long will it take if compounded continuously at the same annual rate?

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using A(t) = A0 * e ^ (rt)

3000 = 1000 e ^ (0.04t)

3 = e ^ (0.04t)

ln(3) = 0.04t

t = ln(3) / 0.04

= about 27.5 years

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Your solution:

Solutions were done below each respective question.

Confidence rating: 3

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Given Solution:

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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

3 = (1 + i) ^ 15

15th root of 3 = 1 + i

1.076 = 1 + i

= 0.076 = 7.6%

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Given Solution:

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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 x(t) = x e^(kt) and 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

now solve for t

200,000 = 100,000 e ^ (0.013t)

2 = e^(0.013t)

ln(2) = 0.013t

t = ln(2) / 0.013

= about 53.31 hours

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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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dP/dt = k P + M has the general solution of P(t) = c e^(kt) - (M/k)

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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) - so, the minimum population must be greater than (M/k)

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What migration rate is required to achieve a constant population?

The migration rate M, must equal the rate of population increase k P.

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Your solution:

Written below each respective question.

Confidence rating: 2

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Given Solution:

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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 individuals migrating away each year would equal the number that migrates in, keeping the population at the same level.

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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 (P) would have to equal the rate of population increase (k P). The equal rates would in turn keep the population steady.

@&

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 they do contribute.

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Your solution:

Solutions were done below each respective question.

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Given Solution:

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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:

dQ/dt = -kQ

Q(t) = Q0 e^(-kt)

0.5 Q0 = Q0 * e ^ (-k * 120)

K = -ln(0.5) / 120 = 0.006

--so--

Q(t) = Q0 e^(-0.006 t)

Q(t) = 3 e ^ (-0.006 t)

4 = 3 e^(-0.006 * 360) + M

4 - M = 3 e^(-0.006 * 120)

M = 4 - 3 e^-0.006 * 120) = 3.65

--so--

dQ/dt = -0.18 e ^ (-0.006t) WRT material added by constant rate of decay

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During a short time interval `dt, the approximate amount decaying will be -ln(2) / 120 * Q * `dt, where Q is the amount at the time in question. The amount added will be r `dt, where r is the rate at which the decay of the second substance is adding material.

This the net change in the amount of material will therefore be approximately

`dQ = - ln(2) / 120 * Q * `dt + r `dt.

In the limit as `dt approaches zero we get

dQ/dt = - ln(2) / 120 * Q + r.

The task is to solve this equation with the given conditions.

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Given Solution:

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Self-critique (if necessary): Not exactly sure about the run-down on this one. I think I have a general frame work, but think I’m a bit off on narrowing that down into an exact execution as far as the start-to-end process.

Self-critique rating: OK

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Self-critique (if necessary):

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