Query 03

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course Mth 279

10/9 715pm

Section 2.4.*********************************************

Question: 1. How long will it take an investment of $1000 to reach $3000 if it is compounded annually at 4%?

F=p(1+i)^n

3000=1000(1+.04)^n

Log(3)/log(1.04)=n

N=28.01 years

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

3000=1000(1+(0.04/4))^(n*4)

3=(1.01)^(n*4)

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

n=27.602 years

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

F=P(e^(i*n))

@& Notation note: i is standard notation for sqrt(-1), and the exponential function e^(i x) is very important starting around Assignment 12 or so. Best to avoid using i for anything except sqrt(-1), especially in a context involving exponential functions.

No problem here, but I would suggest r as a better letter (standing for 'rate').*@

3000=1000(e^(0.04*n))

3=e^(0.04*n)

ln(3)=ln(e^(0.04*n))

1.09861=0.04n

N=27.485 years

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Your solution: in between marks above

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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:

F=P(1+i)^n

3000=1000(1+i)^15

3=(1+i)^15

1.07598=1+i

i=0.07598=7.6%

@& Ideally your solution would start with the differential equation y ' = k y, or if you prefer y ' = r y (the letter standing for the growth rate doesn't really matter, though per my previous note I would avoid 'i' in this situation.)*@

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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:

F=Pr^t

100,000=40,000(r)^72

2.5=r^72

R=1.0128

200,000=100,000(1.0128)^t

2=(1.0128)^t

Log(2)/log(1.0128)=t

t=54.49 more days

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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).

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

P(0)=Ce^(k(0))-M/k

P_0=C(1)-M/k

C=P_0+M/k

P(t)=(P_0+m/k)e^(k*t)-M/k

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In terms of k and M, determine the minimum population required to achieve long-term growth.

P(t) will increase as long as P_0+m/k is great than zero so if M is greater than -P_0*k long-term growth will occur

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What migration rate is required to achieve a constant population?

M=-P_0*k

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Your solution:

@& You haven't actually solved the equation; and you might (or might not) be quoting a formula for the solution of the equation.

On a test you will be expected to set up and solve the differential equation, as opposed to quoting a formula for the solution.

So do be sure you know how to set up and solve this equation.

*@

@& dP/dt = k P + M

is solve by separating variables to get

dP / (k P + M) = dt

with solution

1 / k ln | k P + M | = t + c.

This leads immediatly to

k P + M = A e^(k t)

or

P = A e^(k t) - M / k,

where A is an arbitrary constant (equivalent to your C).

Be sure you can solve the problem starting with the differential equation.*@

confidence rating #$&*:ut of 3

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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?

For one year the population increases from P_0 to P_0*e^kt where t=1

The difference between these two populations would be the migration rate

M=(P_0*e^k*1)-P_0

M=P_0*(e^k-1)

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How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question?

When compared to the migration rate for steady population: M=P_0*k the difference is [P_0*(e^k-1)]-[P_0*k]=P_0(e^k-1-k)

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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:

Constant decay rate is dQ/dt=-kQ

Q(t)=Ce^(-kt)

If the half life is 120 days then Q(120)=0.5C

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

0.5=e^(-k*120)

ln0.5=(-k*120)

-k=-0.00577

Q(t)=Ce^(-0.00577*t)

Q(t)=3e^(-0.00577*t)

Not sure where to go from here when we must determine the rate needed to add.

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Given Solution:

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@& Good. But do check my notes, and be sure you can solve equations of this nature showing all the steps, starting with the setup of the differential equation.*@