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MATH 148 Lab Key #6 3/7/2017 (1) Suppose that a population is divided into three age classes and that 75% of the females of age zero and 30% of the females of age one survive until the end of the next breeding season. Moreover, assume that females of age zero have an average of 1 female offspring, females of age one have an average of 4 female offspring, and females of age two have an average of 2 female offspring. The initial population consists of 100 females of age zero, 200 females of age one, and 100 females of age two. (a) Find the Leslie matrix and initial age distribution. (b) Find the age distributions at times t = 1 and t = 2. Solution: (a) 1 L = 0.75 0 4 2 0 0 . 0.3 0 The initial age distribution is given by 100 N (0) = 200 . 100 (b) 2 100 100 + 800 + 200 1100 = 75 . 0 200 = 75 0 100 60 60 2 1100 1100 + 300 + 120 1460 = 825 . 0 75 = 825 0 60 23 23 1 4 N (1) = L · N (0) = 0.75 0 0 0.3 1 4 N (2) = L · N (1) = 0.75 0 0 0.3 1 (2) Consider the Leslie matrix 2 0.4 L= 0 0 5 0 0.9 0 8 1 0 0 0 0 0.8 0 (a) Determine the number of age classes in the population. (b) Determine the fraction of two-year-olds that survive until the end of the next breeding season. (c) Determine the average number of female offspring of a 1-year-old female. Solution: (a) There are 4 age classes in the population. (b) 80% or 45 of the age 2 population survives until the end of the next breeding season. (c) On average, an age 1 female gives birth to 5 female offspring. (3) Let A = (1, 2, 3) and B = (−1, 3, −6) . Find a unit vector in the direction −→ of AB. Solution: −→ AB = (−1 − 1, 3 − 2, −6 − 3) = (−2, 1 − 9) . −→ Therefore, a unit vector in the direction of AB is given by 1 2 1 9 (−2, 1, −9) = √ · (−2, 1, −9) = − √ , √ , − √ . |(−2, 1, −9)| 4 + 1 + 81 86 86 86 − → → (4) Let − a = h−3, −1i and b = h1, 2i . − → − → → → → (a) Find − a + 2 b . Graph the position vectors − a and 2 b and illustrate − a + − → 2 b using the Triangle or Parallelogram Law. − → − → → → → (b) Find − a − b . Graph the position vectors − a and − b and illustrate − a − − → b using the Triangle or Parallelogram Law. Solution: − → → (a) − a + 2 b = h−3, −1i + h2, 4i = h−1, 3i . [Graph using either the parallelogram or triangle rules.] − → → (b) − a − b = h−3, −1i − h1, 2i = h−4, −3i . [Graph using either the parallelogram or triangle rules.] 2 (5) A triangle has vertices A = (2, 1, 5) , B = (−1, −3, 7) , C = (2, −4, 1) . Find the angle at the vertex B. Solution: → −→ − → −→ − BA · BC = BA BC cos θ. −→ − → BA · BC ∴ cos θ = −→ − → BA BC −→ − → where θ is the angle at B. Now, BA = (3, 4,−2) and BC = (3, −1, −6) . There√ √ −→ − → − → fore, BA · BC = 9 − 4 + 12 = 17. Also, BA = 9 + 16 + 4 = 29 and − √ → √ BC = 9 + 1 + 36 = 46. 17 √ . 29 · 46 17 ∴ θ = cos−1 √ . 29 · 46 ∴ cos θ = √ → (6) Find the components of the vector − x that has magnitude 5 and forms an ◦ angle of 240 measured counterclockwise from the x-axis. Solution: − → x = (5 cos 240◦ , 5 sin 240◦ ) . Now, 240◦ is in Quadrant III and has a reference angle of 60◦ . Therefore, cos 240◦ = √ − cos 60◦ = − 21 and sin 240◦ = − 3 2 . → ∴− x = √ ! 5 5 3 − ,− . 2 2 3