Query_03 

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course Mth 279

2/24 20

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:

The compound interest formula is A = P ( 1 + r / n ) ^ ( n t )

Plugging in the given information we have

3000 = 1000 ( 1 + 0.04)^t

3 = 1.04 ^ t

We can use logarithims to solve for t.

log(3) = log(1.04^t)

log(3) = t log(1.04)

log(3) / log(1.04) = t

t = 28 years if compounded annually.

If compounded quarterly, we introduce an n value of 4.

3000 = 1000 ( 1 + 0.04 / 4)^4t

3 = (1 + 0.01)^(4 t )

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

log(3) = 4 t log(1.01)

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

t = 27. 6 years

The continuous compound interest formula is A = P e ^ (r t)

3000 = 1000 e ^ (0.04 t)

3 = e^(0.04 t)

log(3) = 0.04 t

log(3) / 0.04 = t

t = 27.47

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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:

A = P (1 + r)^ 15

3000 = 1000 (1 + r)^15

3 = (1 + r )^15

3 = 1 + r^15

2 = r^15

15th root 2 = r

r = 1.047

4.7 percent annual return.

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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:

A = P e^( r t)

100000 = 40000 ^ (72 r)

100 = 40 e ^ (72 r)

2.5 = e^(72 r)

log(2.5) = 72 r

r = 0.0127 is the constant growth rate.

Now we plug in 200 000 as the A value.

200 000 = 100 000 e ^ (0.0127 t)

log (2) = 0.0127 t

t = 54.6 hours.

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Good, but you should start with the differential equation, not its general solution.

The rate of population growth is proportional to the population, so

dP/dt = kP

which can be written

P ' - k P = 0,

a first-order linear homogeneous equation which we solve in the usual manner, obtaining general solution

P(t) = Ae^(kt), A > 0.

P(0) = 40,000 and P(3) = 100,000, giving us the equations

A e^(k * 0) = 40 000

A e^(k * 3) = 100 000.

We solve for A and k, obtaining A = 40 000 and k = .3, approx.., so that

P(t) = 40 000 e^(.3 t).

To find t when population reaches 200,000 we solve

P(t) = 200 000

40 000 e^(.3 t) = 200 000

e^(.3 t) = 5

.3 t = ln(5)

t = ln(5) / .3 = 5, very approximately.

The time require is about 5 days.

This is about 2 days after the population reaches 100 000 (again this is very approximate).

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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:

P’ = kP + M

This is a separable equation.

P’ / (kP + M ) = 1

int P / (kP + M ) = int 1

1 / k ln (k P + M) = t + c

k P + M = e^(kc) e^(kt) = C e^(kt)

The general solution is

P(t) = (c / k) e^(kt) - M / k

When t = 0 , P(0) = P_0

P_0 = (c / k) e^ 0 - M / k

P_0 = c / k - M / k

c = P_0 + M / k

P(t) = [(k P_0) + M) / k ] e^(kt) - M / k

P(t) = (P_0 - m / K) e^(kt) - M / k

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P(t) = (P_0 + M / K) e^(kt) - M / k

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Constant population will be achieved when

P_0 - M / k = 0

M = k P_0

P’ must remain positive to ensure long term growth.

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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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Self-critique (if necessary):

I do not understand how to being or setup this problem.

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To start:

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 ).

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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 (tau)

dQ / dt = -kQ

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

Q(t + tau) = 1 / 2 Q(t)

Since Q(t) = C e^(-kt)

e^(-kt) = 1 / 2

Tau = (ln 2) / k

??? Does the decay (what is leaving the substance) from the second substance join the first substance while the first substance is still decaying?

??? Does this decay being added to the first substance also decay in time?

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As soon as the second substance decays into the first, it starts decaying at the rate of the first.

If the rate at which the second substance adds to the first is r, then the equation for the amount of the first would be

dQ/dt = -k Q + r.

This equation should be solved for the given conditions.

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