Query 03

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course Mth 279

3/13 6

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 P(t) = A_not (1 + r) ^ t

where P(t) is 3,000,

A_not is 1,000

r is 4%

t is number of years.

solve for t by dividing 3,000 by 1,000 and taking the ln of both sides to get into

ln 3 = t (1.04)

t = 28.011 years

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

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Our formula becomes P(t) = (1 + r / 4) ^ 4t * A_not

after dividing by 1,000 and taking the ln of each side we are left with

ln3 = 4t ln(1.01)

t = 27.6 years

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

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this becomes a limiting problem. where lim as n approaches infinity = Ae^rt

to solve for t take the ln of both sides to get t = ln3 / .04 = 27.465 years

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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 our origional equation. 3,000 = 1,000(1+r)^15

solve for r take ln of both sides.

ln3 / 15 = ln(1+r)

ln3 / 15 = 0.073241

raise it by e.

e^(0.073241) = 1+r

r = 0.076 or 7.6% annual rate assuming compounded annually. but it makes very little difference

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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 the same equation as always.

100,000 = 40,000(1+r)^72t

2.5 = (1+r)^72t take ln of both sides

ln2.5 = 72ln(1+r)

r = 1.28%

now to find how long it takes that 100k to grow to 200k

200k = 100k(1+.0128)^t

like always, solve for t by taking ln of both sides to find simply that t = 54.92 hrs

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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(t)/dt / P(t) = K + M integrate both sides wrt t

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If you divide both sides of the equation by P you get

dP/dt / P = k + M / P,

not k + M.

The equation is of the form

P ' - k P = M.

This needs to be solved as a linear nonhomogeneous equation. You can't solve it by separating variables.

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log ( P(t) ) = Kt + Mt

raise by e,

P(t) = e^(K + M)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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to have long term growth, Your migration rate and growth rate must be positive, so solve for M and K to be positive, K and M can = 0, because e^0 = 1. but it would not experience long term growth. so at least e^1.

solve our other equation,

K + M = Log(P(t)) / t

I think....

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What migration rate is required to achieve a constant population?

P(t) = Ce^kt + M / k where Migration rate is larger than the growth rate

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

Just the opposite of the growth rate if it goes back to the same level.

I get the concept, but I'm not 100% sure my formula is correct.

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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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equal and opposite, where M is negative M and growth rate is the same just opposite.

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Here you just have to figure out how much the population increases in a year. That's how many migrate away.

The key is the comparison with the migration rate of the previous situation.

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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:

going by the example 2 in the book. solve for k knowing that P(t) = 3 and P(360) = 4

so set them equal to each other and solve for k.

4 = 3e^360k

4/3 = e^360k

ln(4/3) = 360k

k = ln(4/3) / 360 = roughly 7.99x10^-4 days

so our equation becomes Q(t) = 3e^( (7.99x10^-4 ) t )

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If r is the rate at which new material is being added then we have

dQ/dt = r - k P.

You have to determine k from the half-life.

Then solve the equation as first-order linear nonhomogeneous, with r remaining unknown.

You'll get an integration constant, so you'll have r and your integration constant as unknown.

These two quantities can be adjusted to fit the given conditions (initial Q is 3 grams, Q(360 days) = 4 grams).

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Self-critique (if necessary):

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Not too bad, but you're still trying to solve everything by separating variables.

That just doesn't work on any of these problems.

Check my notes.

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