Query 03

#$&*

course Mth 279

6/20 12

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

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

annually at 4%?

****

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

3000=1000(1 + 0.04)^t

t = approx 28.0 years

#$&*

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

****

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

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

t = approx 27.6 years

#$&*

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

****

A=Pe^(kt)

A(t) = 1000e^(.04*t)

3000 = 1000e^(.04*t)

t = approx 27.5 years

#$&*

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

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

15 years?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

3000 = 1000(1 + r)^15

3 = (1 + r)^15

r = .0759 = approx 7.6%

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

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:

P(0) = 40,000

P(72)= 100,000

P(t) = Ae^(k*t)

Ae^0 = A = 40,000

Ae^(72*k) = 100,000

divide second equation by the first

(Ae^(72*k))/A = 100,000/40,000

e^(72*k) = 2.5

ln(e^(72*k) = ln(2.5)

72*k = ln(2.5)

k = ln(2.5)/72

k = 0.012726

A = 40 000

P(t) = Ae^(.012726*t)

200,000 = 40,000*e^(.012726*t)

t = 126.5

126.5 - 72 = approx 56.5 more hours

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

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^(kt) -(M/k)

#$&*

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

****

e^(kt) must be greater than M/k

at t(0), P(0) = C - (M/k)

therefore C (minimum population) must be greater than M/k

#$&*

What migration rate is required to achieve a constant population?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Migration rate must be equal to rate of population increase k*P

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

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?

****

The migration count must equal the population growth for the year

#$&*

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

the preceding question?

****

The migration rate must equal the population increase kP. The rate of increase must equal the rate of

decrease.

#$&*

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:

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

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:

Q(t + T) = 1/2*Q(t)

T = ln(2)/k

120 = ln(2)/k

k = 0.00578

This is the rate of decay of the first element.

General solution

Q'(t) = -Ce^(kt)

3 = Ce^0

C = 3

Q'(t) = 3e^(-.00578*t) + Me^(kt)

4 = 3e^(-.00578*360) + M

M = 3.62/360 = k2 = .01

@&

k is ln(.5) / 120, or if you prefer -ln(2) / 120.

During a short time interval `dt, the approximate amount decaying will be -ln(2) / 120 * Q * `dt, where Q is the amount at the time in question. The amount added will be r `dt, where r is the rate at which the decay of the second substance is adding material.

This the net change in the amount of material will therefore be approximately

`dQ = - ln(2) / 120 * Q * `dt + r `dt.

In the limit as `dt approaches zero we get

dQ/dt = - ln(2) / 120 * Q + r.

The task is to solve this equation with the given conditions.

*@

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique rating:"

&#Good work. Let me know if you have questions. &#