Download Lab 6 - Math TAMU

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