WebWork 12 1. The storage shed shown below is the shape of a halfcylinder of radius and length . (a) What is the volume of the shed? volume = (b) The shed is filled with sawdust whose density (mass/unit volume) at any point is proportional to the distance of that point from the floor. The constant of proportionality is . Calculate the total mass of sawdust in the shed. mass = 2. A lamp has two bulbs, each of a type with an average lifetime of 4 hours. The probability density function for the lifetime of a bulb is . What is the probability that both of the bulbs will fail within 4 hours? Hint: Note the two bulbs operate independently which means the probability they both fail is the product of the probabilities that each fails. 3. (1 pt) In 1950 an experiment was done observing the time gaps between successive cars on the Arroyo Seco Freeway. The data show that the density function of these time gaps was given approximately by where is the time in seconds and is a constant. a. Find the constant . b. Find the cumulative distribution function graph of density function . You should also sketch a rough and the cumulative distribution function . (You cannot submit this on webwork but try it for practice.) c. Find the mean and median time gap. Mean = Median = 4. While taking a walk along the road where you live, you accidentally drop your glove, but you don't know where. The probability density for having dropped the glove kilometers from home (along the road) is a. What is the probability that you dropped it within 1 kilometer of home? b. At what distance from home is the probability that you dropped it with km of home equal to ? 5. Consider the function shown below. 6. What type of function is shown in this graph? A. density function (pdf) B. cumulative distribution function (cdf) (Be sure that you can give reasons for your answer.) What value must have? 6. Solve the following initial value problem. (Note this problem may use techniques from previous sections.) 7. Find the solution to satisfying 8. a. Paramecia in a pond sample are growing according to the principle of Malthusian growth (with no food or space limitations). Initially, there are 1600 Paramecia. Four hours later, the population has 1900 individuals. Find the population of Paramecia as a function of time Find at what time the population doubles. b. A large population of 11000 is transferred to where a limited diet affects their growth dynamics. The Paramecia now satisfy the population dynamics of the differential equation: Solve this differential equation (using 11000 as the starting population at Find what happens to the population as Limiting Population = 9. A mothball with volume has an initial volume mothball slowly evaporates according to the differential equation After 6 months, the volume of the mothball has decreased to 0.125 cm , i.e., a. Solve this differential equation and find cm . The where is a function of only. b. Find how long it takes for the mothball to disappear. months. You should sketch a graph of 10. a. A population of yeast is growing according to a Malthusian growth model. Suppose that it satisfies the initial value problem where is in hours. Solve this differential equation and determine how long it takes for this population to double. Doubling time = hr. b. Because of competition from another organism in the broth, the yeast has dwindling supplies of food for growth. An approximate model with a time varying growth rate from this competition is given by the following: Solve this differential equation. c. Find the maximum of this population and when this occurs. Maximum population = Time of Max Population = hr. Also, determine when the population returns to 1000. Time when population is 1000 again = hr. You should sketch a graph for this population. 11. Suppose that a population develops according to the logistic equation where is measured in weeks. (a) What is the carriying capacity? (b) Is the solution increasing or decreasing when capacity? is between and the carriying (c) Is the solution increasing or decreasing when is greater than the carriying capacity? 12. In many population growth problems, there is an upper limit beyond which the population cannot grow. Many scientists agree that the earth will not support a population of more than 16 billion. There were 2 billion people on earth at the start of 1925 and 4 billion at the beginning of 1975. If is the population, measured in BILLIONS, years after 1925, an appropriate model is the differential equation (a) Solve this differential equation. (b) The population in 2015 will be (c) The population will be 9 billion some time in the year billion. Note that the data in this problem are out of date, so the numerical answers you'll obtain will not be consistent with current population figures. Hint: (a) Separate variables and use the given information to solve for . (b) Evaluate . (c) Solve for the appropriate time. Sol: (1) (a) (b) Sol: (2) 1 2 r l 2 2 Lkr 3 3 0.39958 Sol: (3) (a) (b) (c) a 0.119 P t 1 e 0.119t mean 5.824, Sol: (4) (a) (b) median 8.403 1 e 7 0.999088 y 0.428 km Sol: (5) Cumulative distribution function (cdf) Since the function is increasing, thus, the function is a cdf. c 0.667 Sol: (6) Sol: (7) y t 1 6t 2 y t 9e 2 x 7 Sol: (8) (a) P t 1600e 19 ln 16 t 4 t d 0.029 hour (b) P t 6000e 0.1t 5000 Limiting population 5000 Sol: (9) (a) (b) k 0.6 , V t 1.7 0.2t t 8.5 months 3 Sol: (10) (a) Y t 1000e0.02t Doubling time 34.66 hr (b) (c) Y t 1000e 0.020.001t t Maximum population 1000 Time of Max Population 20 hr Unanswered: Also, determine when the population returns to 1000. Time when population is 1000 again = You should sketch a graph for this population. Sol: (11) (a) carriying capacity 1000 hr. (b) (c) increasing decreasing Sol: (12) (a) y t (b) (c) 16 1 7e -1.694610-2 t 6.34 billion In 2055 the population will be 9 billion. \fSol: (1) (a) (b) Sol: (2) 1 2 r l 2 2 Lkr 3 3 0.39958 Sol: (3) (a) (b) (c) a 0.119 P t 1 e 0.119t mean 5.824, Sol: (4) (a) (b) median 8.403 1 e 7 0.999088 y 0.428 km Sol: (5) Cumulative distribution function (cdf) Since the function is increasing, thus, the function is a cdf. c 0.667 Sol: (6) Sol: (7) y t 1 6t 2 y t 9e 2 x 7 Sol: (8) (a) P t 1600e 19 ln 16 t 4 t d 0.029 hour (b) P t 6000e 0.1t 5000 Limiting population 5000 Sol: (9) (a) (b) k 0.6 , V t 1.7 0.2t t 8.5 months 3 Sol: (10) (a) Y t 1000e0.02t Doubling time 34.66 hr (b) (c) Y t 1000e 0.020.001t t Maximum population 1000 Time of Max Population 20 hr Unanswered: Also, determine when the population returns to 1000. Time when population is 1000 again = You should sketch a graph for this population. Sol: (11) (a) carriying capacity 1000 hr. (b) (c) increasing decreasing Sol: (12) (a) y t (b) (c) 16 1 7e -1.694610-2 t 6.34 billion In 2055 the population will be 9 billion