Query 03

#$&*

course MTH 279

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

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

****

#$&*

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

****

#$&*

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

****

#$&*

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A = P(1+r/n)^(nt)

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

3 = 1.04^t

ln(3) = ln(1.04)t

t = ln(3)/ln(1.04) ≈ 28.01 years if compounded annually at 4% annually.

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

3 = (1.01)^(4t)

ln(3) = 4tln(1.01)

t = ln(3)/(4ln(1.01)) ≈ 27.6 years if compounded quarterly at 4% annually.

3000 = 1000e^(0.04t)

3 = e^(0.04t)

ln(3) = 0.04t

t = ln(3)/0.04 ≈ 27.46 years if compounded continuously at 4% annually.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

Question: 2. What annual rate of return is required if an investment of $1000 is to reach $3000 in 15 years?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3000 = 1000(1 + r)^(15)

3 = (1+r)^(15)

3^(1/15) = 1+r

r = 3^(1/15)-1 ≈ 0.0759 The annual rate of return required would be 7.59%.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Rate of change of Bacteria = growth rate * # of bacteria

B’(t) = kB(t)

dB/dt = kB

ʃdB/B = ʃkdt

ln(B) + c_1 = kt + c_2

ln(B) = kt + C

B(t) = Ce^(kt)

Initial Conditions: B(0) = 40000 and B(72) = 100000.

40000 = Ce^(k*0) >> C = 40000

100000 = 40000e^(72k)

5/2 = e^(72k)

ln(5)-ln(2) = 72k

[ln(5)-ln(2)]/72 = k ≈ 0.0127

200000 = 100000e^({[ln(5) - ln(2)]t}/72)

2 = e^({[ln(5) - ln(2)]t}/72)

ln(2) = {[ln(5) - ln(2)]t}/72

72ln(2) = (ln(5) - ln(2))t

[72ln(2)]/[ln(5)-ln(2)] = t ≈ 54.4658

It would take approximately 54 hours and 28 minutes to reach 200,000 bacteria.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

Self-critique rating: OK

*********************************************

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).

****

#$&*

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

****

#$&*

What migration rate is required to achieve a constant population?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

dP/dt - kP = M

µ = e^(ʃ-kdt) = e^(-kt)

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

d/dt(Pe^(-kt)) = Me^(-kt)

ʃd/dt(Pe^(-kt)) dt = ʃMe^(-kt)dt

Pe^(-kt) + c_1 = -Me^(-kt)/k + c_2

Pe^(-kt) = -Me^(-kt)/k + C

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

To have long-term growth, the population growth [Ce^(kt)] must be greater than the migration [-M/k]. If we say that the population is at its minimum when t = 0, then we have P(t) = C - M/k, with C being the initial population. We then see that the minimum population required to achieve net-growth would be ≥ M/k. To achieve a constant population, we must have a migration rate (M) that is equal to the population growth rate. M = kP

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

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?

****

#$&*

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

****

#$&*

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The number of individuals that migrate out would be equal to the population growth. Imagine that there are initially 2 individuals in a population. The population grows by 2 in that year, to make a population of 4. The migration out must be 2, so that the population returns to the initial population point of 2.

This is the same as the migration rate required to achieve a steady population. In the example I gave, k would be 1. However, it can be any value, as long as the rate for the population growth is equal to the rate for the migration.

@&

See if you agree with the following:

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 the do contribute.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First element:

Q(t) = Ce^(-kt)

100 = 50e^(-120k)

½ = e^(-120k)

ln(1/2) = -120k

k = -ln(1/2)/120 ≈ 0.0058 = The decay rate of the first element.

We now know that the amount of the first element is Q(t) = 3e^(-0.006t) at time t. We know that to keep the amount of element at equilibrium the rate that material must be added is equal to the rate that the element decays, Q’(t) = -0.018e^(-0.006t). This means that the second element must decay at constant rate of -0.018 to maintain a constant amount of the first element. The second element must contribute at least 0.018g of new element 1 per day to increase to 4g.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

This problem took me awhile, and I wasn’t sure exactly which direction to go at it. I think I got the answer, but I’m not entirely sure.

@&

I agree with your result.

Here's my reasoning, just for comparison:

The decay rate is constant so

dQ/dt = -kQ,

where Q is quantity present.

This is easily solved. We obtain

Q(t) = Q_0 e^(-kt)

The half-life is the time required to reach half the original quantity. Since this requires 120 days we have Q(120) = .5 Q_0 so that

.5 Q_0 = Q_0 * e^(-k * 120)

which we solve for k. The result is

k = -ln(.5) / 120 = .006

so our model is

Q(t) = Q_0 e^(-.006 t).

Left to itself, our original 3 grams will therefore decay in such a way that

Q(t) = 3 e^(-.006 t).

If, however, we add material at an appropriate constant rate r, the quantity of material present may be kept constant.

The rate at which the material is lost is

dQ/dt = -.018 e^(-.006 t)

At the initial instant t = 0, this rate is -.018, meaning that material is being lost at the rate of .018 grams / day.

If we add material continuously at this constant rate, then none will be lost.

Thus r = .018, meaning that we must continuously add .018 grams of new material per day.

It's worth noting that if rather than adding material continuously we just add a chunk consisting of .018 grams of the material at the end of each day, the amount of material will increase over time, with the beginning-of-day amount increasing toward a limiting value which is in excess of the original 3 grams. It would be interesting to calculate this limiting value.

*@

------------------------------------------------

Self-critique rating:

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