#$&* course Mth 279 resent 3/4/13 Section 2.4.*********************************************
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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 $3000 in 15 years? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using f = P(1 + i)^n 3000 = 1000 (1 + i )^15 and solving for i, 3 = (1+i)^15 15th root of 3 = 1 + i 1.076 = 1 + i 0.076 = i or i = 7.6% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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: Using the formula for exponential growth, x(t) = x e^(kt) solving for k, 100,000 = 40,000 e^(k * 72) 2.5 = e^(k * 72) ln(2.5) = k * 72 k = ln(2.5) / 72 = 0.013 approx. solving for t, 200,000 = 100,000 e^(0.013t) 2 = e^(0.013t) ln(2) = 0.013t t = ln(2) / 0.013 = approximatly 54.5 more hours confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 4. A population experiences growth rate k and migration rate M, meaning that when the population is P the rate at which new members are added is k P, but the rate at they enter or leave the population is M (positive M implies migration into the population, negative M implies migration out of the population). This results in the differential equation dP/dt = k P + M. Given initial condition P = P_0, solve this equation for the population function P(t). **** dP/dt = k P + M has the general solution of P(t) = c e^(kt) - (M/k) #$&* In terms of k and M, determine the minimum population required to achieve long-term growth. **** the increase of the population e^(kt) must be greater than (M/k) at t = 0, P(t) = c - (M/k) which means that the minimum population must be greater than (M/k) #$&* What migration rate is required to achieve a constant population? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The migration rate M, must equal the rate of population increase k P. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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 number of individuls that migrate away each year equals the number that migreates in each year. which would keep the population the same as the previous year. #$&* How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question? **** The migration rate M, must equal the rate of population increase k P. They are the same, bc to keep a steady or constant population, these rates must be equal. #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
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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: first the half life of the first substance is 120, 120 = ln(2) / k and solving for k, k = 0.006 using a form of the equation, Q(t) = C e^(-kt) 4 = 3e^(-0.006 * 120) + M 4 - M = 3e^(-0.006 * 120) M = 4 - 3e^(-0.006 * 120) = 2.54
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