query 3

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course M 279

2-19 1

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

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

Using P_1(t) = A0*(1+ r)^t, where r = .04 & A0 = $1000

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

(1.04)^t = 3

ln(1.04)*t = ln(3)

t = ln(3)/ln(1.04) = approx. 28 years

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

Using P_4(t) = A0*(1+ r/4)^4t, where r = .04 & A0 = $1000

$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 = approx. 27.6 years

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

Using P (t) = A0*e^(r*t), where r = .04 & A0 = $1000

$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:

?????I put my answers within the normal marks?????????

@& no problem*@

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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 = approx. .0732 or 7.32%

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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:

For population we have dP/dt = kP, so general solution is P(t) = Ce^(kt)

P(0) = 40,000, so C = 40,0000

P(3) = 100,000 where t = 3, so we can solve 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 = approx. .5.2699 or approx. 5.25 days

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Given Solution:

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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).

where p(t) = -k and equation is P’ + p(t)P = M

we mult by integrating constant, e^(Int( p(t) dt) = e^(-kt)

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))

P(t) = (P + (M/k))e^(-kt) - (M/k), final solution imposing given init cond.

????This one stumped me for some reason so it’s a good chance I messed it up somewhere. I spent close to an hour battling this problem??????????

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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:

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

As t infintity P(t) (P + (M/k))*1 - (M/k)

For (P + (M/k))*1 - (M/k)

P + (M/k) - (M/k) , Population in terms of M/k is

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

I believe that for population greater than 0 Population will grow long term, but this doesn’t seem right to me??

??????I’ve worked this out several ways and keep getting zero, So does that mean I messed up my differential equation somewhere or am I approaching it wrong?????

Even using this approach….

(P + (M/k))e^(-kt) - (M/k), we can start by rearranging

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

= Pe^(-kt) + (M/k)(e^(-kt) - 1), as tinfin, e^(-kt) 1

P(t)_tinfin = P*1 + (M/k)*(1-1) = P

Constant Population means dP/dt = 0 so using

dP/dt = kP + M, setting dP/dt equal to zero

0 = kP + M

M = -kP, which is is easy enough to understand for population to stay the same the amount of mirgration( or number leaving herd) would have to be equal but opposite to the rate at which the herd is growing.

@& Good.

dP/dt = k P + M can be written

P ' - k P = M,

a first-order linear nonhomogeneous equation.

Using integrating factor e^(-k t) the equation becomes

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

Integrating both sides

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

so that

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

If P(0) = P_0 then at t = 0 we have e^(kt) = 1 and

P_0 = - M / k + c

so

c = P_0 + M / k

and our solution is

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

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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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?

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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), working on the right side

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

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

= Pe^-k - (M/k)(-e^-k + 1), this is the form I want plugging back into original equ.

P = Pe^-k - (M/k)(1 - e^-k), subtracting Pe^-k from right side

P - Pe^-k = - (M/k)(1 - e^-k), where we can factor P from P - Pe^-k

P(1 - e^-k) = - (M/k)(1 - e^-k), dividing both sides by (1 - e^-k)

P(1 - e^-k) / (1 - e^-k) = -M/k, which reduces to

P = -M/k, mult thru bu -k and we finally get

M = -kP, which is the same answer we got previously.

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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 the do contribute.

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Given Solution:

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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:

Knowing dQ/dt = -kQ, where Q is quanitity present of material easy to solve ..

dQ/dt = -kQ

dQ/Q = -k dt, taking integrals of both sides we get

ln|Q| = -kt + C

Q = e^(-kt + C) = e^(C)*e^(-kt) = Ce^(-kt)

Knowing that equation for half-life is defined as

Q(t) = Ce^(-kt), where Q(t)/C is always going to equal 2( wrt half-life). Now we have

2 = e^(-kt)

@& for the half-life we the factor is .5, not 2.*@

ln(2) = -kt, so where t is referring to half-life

t = ln(2)/-k, we can’t solve t but we don’t have to it was given so we can solve for k.

k = ln(2)/-t = ln(2)/-120 = -.006

Now we have dQ/dt = -.0006Q + R, where R is constant quanity being added from 2nd substance

dQ/dt = -.0006Q + R, rearranging

dQ/dt + .0006Q = R

Now of form dQ/dt + p(t)Q = g(t), Finding integration constant

e^(Int(.0006 dt)) = e^(.0006t), mult thru we get

e^(.0006t)Q’ + e^(.0006t)Q =e^(.0006t)R, we can show

( e^(.0006t)P)’ = using product rule

= e^(.0006t)*Q’ + (e^.0006t)’*Q

= e^(.0006t) *Q’ + (.0006e^(.0006t)*Q)

= e^(.0006t)Q’ + .0006e^(.0006t)Q

now we put into the form

(e^(.0006t)Q)’ = e^(.0006t)R, so taking integral of both sides

e^(.0006t)Q = Int(e^(.0006t)R dt) = R* Int(e^(.0006t) dt) = R*(e^(.0006t)/.0006 + c)

= Re^(.0006t)/.0006 + C

e^(.0006t)*Q = Re^(.0006t)/.0006 + C, we can divide out e^.0006t

Q = R/.0006 + C /e^(.0006t) = R/.0006 + Ce^(-.0006t) , so our final solution is

Q(t) = Ce^(-.0006t) + R/.0006, for Q(0) = 3

C*e^(-.0006*0) + R/.0006 = C + R/.0006 = 3

C = (3 - R/.0006)

Q(t) = (3 - R/.0006)e^(-.0006t) + R/.0006, setting t = 360 and Q(360) = 4 we have

4 = (3 - R/.0006)e^(-.0006*360) + R/.0006, rearranging right hand side

(3 - R/.0006)e^(-.0006*360) + R/.0006

= 3e^(-.0006*360) - (R/.0006)e^(-.0006*360) + R/.0006

= 3*.806 - (R/.0006)*(.806 + 1)

= 2.417 - (R/.0006)*1.806, Now we can solve for R

4 = 2.417 - (R/.0006)*1.806

-(R/.0006)*1.806 = 1.583

R*(-3010) = 1.583

R = approx. -.000525

So the rate of decay for the second substance must be at least .000525

????I’m not sure this is correct, seems like when I calculated C = (3 - R/.0006) that this may not have been correct???????

@& 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.

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Given Solution:

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@& Good work. Check my notes and let me know if you have questions.*@