#$&* course MTH 279 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 $3000 in 15 years? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This time we are searching for the rate instead of the time. Time is given to be 15 years. 3000 = 1000e^(r * 15) Dividing by 1000 we get: 3 = e^(r*15) Taking the natural log of both sides will get rid of e. ln(e^(r*15)) = ln(3) r * 15 = ln(3) Reducing again we solve for r to get r = ln(3) / 15 = 7.32 % confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Population growth is modeled by the diff. equation dP/dt = kP. The general solution of P' = kP is P(t) = C(e^kt). We are told that the population starts at 40,000, which is at time t = 0. We are also told that in 72 hours, or 3 days, the population grows to 100000, or P(3) = 100000. Therefore we can set up the equation: 100000 = 40000(e^k*t). So plugging in, 100000 = 40000(e^k*3). Now we can solve for k. If we simplfy the equation we get: 2.5 = e^3k. Taking the ln of both sides we get 3k = ln(2.5). Dividing by 3 we get k = ln(2.5)/3 = 30.54%. To get the population to 200000, we set up the equation: 200000 = 40000(e^0.3054*t). Simplifying we get 5 = e^(0.3054t). Taking the ln of both sides gives us ln(5) = 0.3054t. Solving for t by dividing we get t = ln(5)/0.3054 = 5.2699 days. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 4. A population experiences growth rate k and migration rate M, meaning that when the population is P the rate at which new members are added is k P, but the rate at they enter or leave the population is M (positive M implies migration into the population, negative M implies migration out of the population). This results in the differential equation dP/dt = k P + M. Given initial condition P = P_0, solve this equation for the population function P(t). **** First I will rewrite the equation dP/dt = k P + M to be in standard form, P' - kP = M. Integrating using integrating factor e^-kt, gives me (Pe^-kt)' = Me^-kt. Then I will integrate to get Pe^-kt = -M/ke^-kt + C. Since P = P_0, at t = 0, e^-kt = 1, we get that P_0 = -M/k + C. Solving for C, we get C = P_0 + M/k So P(t) = -M/k + (P_0 + M/k)e^kt. #$&* In terms of k and M, determine the minimum population required to achieve long-term growth. **** We know that e^-kt = 1 (positive), P(t) will increase as long as C is a positive number. #$&* What migration rate is required to achieve a constant population? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To achieve a constant population, as long as M is greater than -k * P_0, the population will increase. If not, it will decrease. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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? **** #$&* How does this compare to the migration rate required to achieve a steady population, as determined in the preceding question? **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each year the population increases multiplied by e^k. The number of individuals who migrate away each year can be calculated by P_0(e^k) - P_0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Not sure if I lost myself here....kindof confused. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: 6. A radioactive element decays with a half-life of 120 days. Another substance decays with a very long half-life producing the first element at what we can regard as a constant rate. We begin with 3 grams of the element, and wish to increase the amount present to 4 grams over a period of 360 days. At what constant rate must the decay of the second substance add the first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know the formula dQ/dt = -kQ. From this we can get Q(t) = Q_0 * e^-kt. Since we are looking for a half-life of 120 days, we know that t = 120. So we can simplify to have 0.5* Q_0 = Q_0 * e^(-k*120). We can divide, and then take the ln of both sides and solve for k. We end up with k = -ln(0.5) / 120 = 0.006. Now we can plug in our numbers, Q(t) = 3*(e^(-0.006 * t)), which is the rate of decay. The rate the material is lost at is dQ/dt = -0.018 * e^(-0.006*t). At the start, we lose 0.018 grams per day, so as long as we add as much as we lose, we won't have a change. So we know we need to add 0.018 grams each day. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:"