Query 03

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course Mth 279

1/3/14 around 9 pm

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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P_1(t) = A0*(1+ r)^t

$3000 =$1000*(1+.04)^t

(1.04)^t = 3

ln(1.04)*t = ln(3)

t = ln(3)/ln(1.04) = around 28 years

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

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P_4(t) = A0*(1+ r/4)^4t

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

(1.01)^4t = 3

ln(1.01)*4t = ln(3)

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

t = (ln(3)/ln(1.01))/4 = around 27.6 year

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

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P (t) = A0*e^(r*t)

$3000 =$1000*e^.04*t

e^.04*t = 3

.04t = ln(3)

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

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Your solution: 

 answers above

 

 

confidence rating #$&*:8232; 

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Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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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^(r*15)

e^(r*15) = 3

r*15 = ln(3)

r = ln(3)/15 = around .0732 or 7.32 percent

 

 

confidence rating #$&*:8232; 

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Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3

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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: 

population —> dP/dt = kP, which give the general solution of P(t) = Ce^(kt)

P(0) = 40,000, —> C = 40,0000

P(3) = 100,000 t = 3, solving for k—>

100,000 = 40,000e^(k*3)

e^(3k) = 2.5

3k = ln(2.5)

k = ln(2.5)/3 = approx. .3054 = 30.54%

To find t needed for population of 200,000

200,000 = 40,000e^(.3054*t)

e^(.3054*t) = 5

.3054*t = ln(5)

t = ln(5)/.3054 = around 5.2699 to 5.25 days

confidence rating #$&*:8232; 

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Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:

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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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e^(-kt)P’ - ke^(-kt)P =e^(-kt)M, we can show (e^(-kt)P)’ =e^(-kt)P’-ke^(-kt)P

( e^(-kt)P)’ = using product rule

= e^(-kt )*P’ + (e^-kt)’*P

= e^(-kt) *P’ + (-ke^(-kt)*P) = e^(-kt)P’ - ke^(-kt)P

now we put into the form

(e^(-kt)P)’ = e^(-kt)M, so taking integral of both sides

e^(-kt)P = Int(e^(-kt)M dt) = M* Int(e^(-kt) dt) = M*(-e^(-kt)/k + c)

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

e^(-kt)*P = -M/k*e^(-kt) + C, we can divide out e^-kt

P = -M/k + C/e^(-kt) = Ce^(kt) - (M/k), so our final solution is

P(t) = Ce^(kt) - (M/k), for P(0) = P

C*e^(k*0) - (M/k) = C - (M/k) = P

C = (P + (M/k))

Im not sure if I am going in the right direction here

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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 = k P + M —>

P ' - k P = M

( P e^(-k t) ) ' = M e^(-k t).

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

—>

P = -M / k + c / e^(-k t) = - M / k + c e^(k t).

If P(0) = P_0 then at t = 0 —> e^(kt) = 1 and

P_0 = - M / k + c

—>

c = P_0 + M / k

Therefore

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

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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 #$&*:8232;

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Given Solution: 

 

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Self-critique (if necessary): Im not to sure on the first part

 

 

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Self-critique rating:2

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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: 

P(t) = (P + (M/k))e^(-kt) - (M/k), letting t = 1 we have and setting P(t) = P, which is equal to P(0) or P(t-1) where t = 1

P = (P + (M/k))e^(-kt) - (M/k)—>

(P + (M/k))e^(-kt) - (M/k)

= Pe^-k + (M/k)e^-k - (M/k)

= Pe^-k - (M/k)(-e^-k + 1), for plugging in.

P = Pe^-k - (M/k)(1 - e^-k),

P - Pe^-k = - (M/k)(1 - e^-k),factoring out p

P(1 - e^-k) = - (M/k)(1 - e^-k)

P(1 - e^-k) / (1 - e^-k) = -M/k

P = -M/k, —> final equation M = -kP,

 

confidence rating #$&*:8232; 

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Given Solution: 

 

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Self-critique (if necessary):I think I worked the steps in the right manor but I could be wrong

 

 

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Self-critique rating:3

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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 —> dQ/dt =-kQ

—> Q(t) = Q_0e^e(-kt)

1/2 the time to reach 1/2 the original amount

—> .5Q_0=Q_0 *e^(-k *120)

—> k=-ln(.5) / 120 =.006

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

—>Q(t) = 3 e^(-.006 t).

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

when r=.018 we are adding .018 grams of material every day

 

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Good.

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confidence rating #$&*:8232; 

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Given Solution: 

 

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Self-critique (if necessary):OK

 

 

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Self-critique rating:3"

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