query 03

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course Mth 279

6/15 10:10 pmI feel like the assignments for this week got progressively easier

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You've been working conscientiously and asking good questions. That tends to make things get progressively easier.

Also, of course, I started out with things that shouldn't, but do, typically cause problems for students during a differential equations course. Better to get them out of the way early so they don't get in the way when you need them to learn other things.

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Section 2.4.

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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 (1+ r/n) ^(nt)

annually:

3000 = 1000 (1+.04/1)^t

3 = 1.04^t

log(3) = t log(1.04)

t = log(3)/log(1.04) = 28.011 approx.

 quarterly:

3000 = 1000 (1+.04/4)^(4t)

3 = (1.01)^4t

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

t = log(3)/(4log(1.01)) = 27.602 approx.

continuously:

A = Pe^(rt)

3000 = 1000 e^(.04t)

3 = e^(.04t)

ln3 = .04t

t = ln(3)/.04 = 27.465 approx.

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 Do I need to have these equations memorized???

 

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You should have encountered compound interest in high school, and seen how the limiting values of these expressions are related to the definition of the exponential function.

I suggest searching under 'Exponential Functions and Compound Interest'.

To answer your question, you should simply know this.

These ideas will be reinforced by this section.

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Self-critique 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: 

 3000 = 1000e^(15r)

ln(3) = 15r

ln(3)/15 = r

r = .07324 approx.

7.32% approximately 

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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: 

 

 first we must find k

P(t) = Ce^(kt)

100000 = 40000 e^(72k)

2.5 = e^72k

ln2.5 = 72k

k = ln(2.5)/72

200000 = 100000e^(kt)

2 = e^(ln(2.5)/72 t)

ln2 = ln(2.5)/72 t

t = ln(2)/ln(2.5)*72 = 54.465 hours approx.

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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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: 

 

 dP/dt -kP = M

p(t) = -k

P(t) = -kt

u = e^(-kt)

e^(-kt) dP/dt = ke^(-kt) P = e^(-kt)M

integrate both sides

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

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

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

Any value of P_0 will allow the population to grow as long as k is positive

P_0e^(kt) = -M/k

M = -P_0*k e^(kt)

 

 

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Since e^(k t) is positive and increasing, P(t) will therefore be increasing as long as P_0 + M / k > 0.

Similarly P(t) will be decreasing if P_0 + M / k < 0.

So the threshold migration rate occurs when P_0 + M / k = 0, giving us M = - k P_0.

If M > - k P_0, population increases.

If M < - k P_0, population decreases

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confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 I'm not sure about the second question

 

 

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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: 

 

 I know I'm supposed to combine the compounding annually equation with the migration one somehow.

I'm not really sure how to do that.

But this question is different from the last one because in the last one, it was asking about continuous migration. This question is presuming that the migration happens all at once. Although the average migration for the year will be the same, their individual graphs will be different. This is similar to a previous assignment when we were estimating averages by assuming the small business made the same every month

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary):

 

 I can't quite figure out how to answer the first part. Although I'm pretty sure I have to combine the annual compounding one with the population one

 

 

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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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Self-critique rating:

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You're making very good progress.

Check my notes.

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