#$&* course M 279 2-19 1 Section 2.4.*********************************************
.............................................
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: $3000 = $1000e^(r*15) e^(r*15) = 3 r*15 = ln(3) r = ln(3)/15 = approx. .0732 or 7.32% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 4. A population experiences growth rate k and migration rate M, meaning that when the population is P the rate at which new members are added is k P, but the rate at they enter or leave the population is M (positive M implies migration into the population, negative M implies migration out of the population). This results in the differential equation dP/dt = k P + M. Given initial condition P = P_0, solve this equation for the population function P(t). where p(t) = -k and equation is P’ + p(t)P = M we mult by integrating constant, e^(Int( p(t) dt) = e^(-kt) e^(-kt)P’ - ke^(-kt)P =e^(-kt)M, we can show (e^(-kt)P)’ =e^(-kt)P’-ke^(-kt)P ( e^(-kt)P)’ = using product rule = e^(-kt )*P’ + (e^-kt)’*P = e^(-kt) *P’ + (-ke^(-kt)*P) = e^(-kt)P’ - ke^(-kt)P now we put into the form (e^(-kt)P)’ = e^(-kt)M, so taking integral of both sides e^(-kt)P = Int(e^(-kt)M dt) = M* Int(e^(-kt) dt) = M*(-e^(-kt)/k + c) = -Me^(-kt)/k + C e^(-kt)*P = -M/k*e^(-kt) + C, we can divide out e^-kt P = -M/k + C/e^(-kt) = Ce^(kt) - (M/k), so our final solution is P(t) = Ce^(kt) - (M/k), for P(0) = P C*e^(k*0) - (M/k) = C - (M/k) = P C = (P + (M/k)) P(t) = (P + (M/k))e^(-kt) - (M/k), final solution imposing given init cond. ????This one stumped me for some reason so it’s a good chance I messed it up somewhere. I spent close to an hour battling this problem?????????? #$&* In terms of k and M, determine the minimum population required to achieve long-term growth. **** #$&* What migration rate is required to achieve a constant population? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For P(t) = (P + (M/k))e^(-kt) - (M/k) As t infintity P(t) (P + (M/k))*1 - (M/k) For (P + (M/k))*1 - (M/k) P + (M/k) - (M/k) , Population in terms of M/k is P = -((M/k) - (M/k)) = 0 I believe that for population greater than 0 Population will grow long term, but this doesn’t seem right to me?? ??????I’ve worked this out several ways and keep getting zero, So does that mean I messed up my differential equation somewhere or am I approaching it wrong????? Even using this approach…. (P + (M/k))e^(-kt) - (M/k), we can start by rearranging Pe^(-kt) + (M/k)e^(-kt) - (M/k) = Pe^(-kt) + (M/k)(e^(-kt) - 1), as tinfin, e^(-kt) 1 P(t)_tinfin = P*1 + (M/k)*(1-1) = P Constant Population means dP/dt = 0 so using dP/dt = kP + M, setting dP/dt equal to zero 0 = kP + M M = -kP, which is is easy enough to understand for population to stay the same the amount of mirgration( or number leaving herd) would have to be equal but opposite to the rate at which the herd is growing.
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 5. Suppose that the migration in the preceding occurs all at once, annually, in such a way that at the end of the year, the population returns to the same level as that of the previous year. How many individuals migrate away each year? **** #$&* How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question? **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(t) = (P + (M/k))e^(-kt) - (M/k), letting t = 1 we have and setting P(t) = P, which is equal to P(0) or P(t-1) where t = 1 P = (P + (M/k))e^(-kt) - (M/k), working on the right side (P + (M/k))e^(-kt) - (M/k) = Pe^-k + (M/k)e^-k - (M/k) = Pe^-k - (M/k)(-e^-k + 1), this is the form I want plugging back into original equ. P = Pe^-k - (M/k)(1 - e^-k), subtracting Pe^-k from right side P - Pe^-k = - (M/k)(1 - e^-k), where we can factor P from P - Pe^-k P(1 - e^-k) = - (M/k)(1 - e^-k), dividing both sides by (1 - e^-k) P(1 - e^-k) / (1 - e^-k) = -M/k, which reduces to P = -M/k, mult thru bu -k and we finally get M = -kP, which is the same answer we got previously. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 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)
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: