#$&* course Mth 279 1/3/14 around 9 pm Section 2.4.*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* 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 = $1000e^(r*15) e^(r*15) = 3 r*15 = ln(3) r = ln(3)/15 = around .0732 or 7.32 percent confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3 ********************************************* 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: 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 #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* Question: 4. A population experiences growth rate k and migration rate M, meaning that when the population is P the rate at which new members are added is k P, but the rate at they enter or leave the population is M (positive M implies migration into the population, negative M implies migration out of the population). This results in the differential equation dP/dt = k P + M. Given initial condition P = P_0, solve this equation for the population function P(t). **** e^(-kt)P’ - ke^(-kt)P =e^(-kt)M, we can show (e^(-kt)P)’ =e^(-kt)P’-ke^(-kt)P ( e^(-kt)P)’ = using product rule = e^(-kt )*P’ + (e^-kt)’*P = e^(-kt) *P’ + (-ke^(-kt)*P) = e^(-kt)P’ - ke^(-kt)P now we put into the form (e^(-kt)P)’ = e^(-kt)M, so taking integral of both sides e^(-kt)P = Int(e^(-kt)M dt) = M* Int(e^(-kt) dt) = M*(-e^(-kt)/k + c) = -Me^(-kt)/k + C e^(-kt)*P = -M/k*e^(-kt) + C, we can divide out e^-kt P = -M/k + C/e^(-kt) = Ce^(kt) - (M/k), so our final solution is P(t) = Ce^(kt) - (M/k), for P(0) = P C*e^(k*0) - (M/k) = C - (M/k) = P C = (P + (M/k)) Im not sure if I am going in the right direction here #$&* In terms of k and M, determine the minimum population required to achieve long-term growth. **** #$&* What migration rate is required to achieve a constant population? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dP/dt = k P + M —> P ' - k P = M ( P e^(-k t) ) ' = M e^(-k t). P e^(-k t) = -M / k e^(-k t) + c —> P = -M / k + c / e^(-k t) = - M / k + c e^(k t). If P(0) = P_0 then at t = 0 —> e^(kt) = 1 and P_0 = - M / k + c —> c = P_0 + M / k Therefore P(t) = -M / k + (P_0 + M / k) e^(k t).
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Im not to sure on the first part ------------------------------------------------ Self-critique rating:2 ********************************************* Question: 5. Suppose that the migration in the preceding occurs all at once, annually, in such a way that at the end of the year, the population returns to the same level as that of the previous year. How many individuals migrate away each year? **** #$&* How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question? **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(t) = (P + (M/k))e^(-kt) - (M/k), letting t = 1 we have and setting P(t) = P, which is equal to P(0) or P(t-1) where t = 1 P = (P + (M/k))e^(-kt) - (M/k)—> (P + (M/k))e^(-kt) - (M/k) = Pe^-k + (M/k)e^-k - (M/k) = Pe^-k - (M/k)(-e^-k + 1), for plugging in. P = Pe^-k - (M/k)(1 - e^-k), P - Pe^-k = - (M/k)(1 - e^-k),factoring out p P(1 - e^-k) = - (M/k)(1 - e^-k) P(1 - e^-k) / (1 - e^-k) = -M/k P = -M/k, —> final equation M = -kP, confidence rating #$&*:8232; ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I think I worked the steps in the right manor but I could be wrong ------------------------------------------------ Self-critique rating:3 ********************************************* 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: 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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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:3"