LS 30B Homework 4 Solutions Feb 9, 2018
Solutions:
5.1.2: Choose an initial value for X and use values of r that are less than, greater than, and equal
to one to test the above statements by computing two values of X for each value of r.
Solution: Choose X0 = 1,
r = 1:
X1 = 1 ∗ 1 = 1, X2 = 1 ∗ X1 = 1 ∗ 1 = 1.
r = 2:
X1 = 2 ∗ 1 = 2, X2 = 2 ∗ 2 = 4.
r = 1/2:
1 1 11 1
X1 = ∗ 1 = , X2 = = .
2 2 22 4
5.1.3: An inedible alga is growing on a pond in a city park. Only a small part of the pond is now
covered by the algae, but the area covered is doubling each day. The city decides to remove the
algae once it covers half the pond. If the pond will be completely overgrown in thirty days, on
what day will it be half covered? (Hint: Try working backward.)
Solution: The pond will be completely overgrown in thirty days. Since the algae doubles each day
the pond will be half overgrown with algae on the twenty ninth day. On the twenty eighth day it
will be less than half covered. So the city will remove the algae on the twenty ninth day.
5.1.4: A rabbit population is growing at 10% a year. If there are 10 rabbits this year and time is
discrete, how many will there be in 10 years? Use a loop in SageMath to check your answer.
Solution: In this case r = 1.1 and X0 = 10. So
X10 = (1.1)10 ∗ 10 = 25.937424601.
(Round one way or the other to get the actual rabbit population.)
Sage code:
init = 10
r = 1.1
for i in range(0,10):
x = r*init
init = x
init
1
, LS 30B Homework 4 Solutions Feb 9, 2018
5.1.5: While we have been working with r > 1, representing growth, r can be less than 1, repre-
senting a quantity that decreases over time. The half-life of a radioactive element is the amount of
time needed for half the element to decay. What fraction of the initial amount of such an element
will remain after ten half-lives?
1 1 1
Solution: Each half life we lose 2 of the element. So after ten iterations we get 210
= 1024 .
5.1.6: When money in a bank account accrues compound interest, the interest earned in one time
period is added to the principal, and then the sum is used as the base for the next time period.
(a) If you start off with $1000 and earn 2% interest that is compounded annually, how much
money will you have in 5 years? In 10 years? In 20 years?
(b) How long will it take you to accumulate $10,000?
Solution:
(a) In this case r = 1.02 and X0 = 1000. So
X5 = (1.02)5 ∗ 1000 = 1104.0808032
X10 = (1.02)10 ∗ 1000 = 1218.99441999,
X20 = (1.02)20 ∗ 1000 = 1485.94739598
(b) To get to 10000 we solve, 10000 = (1.02)n ∗ 1000. Simplifying, we get 10 = (1.02)n . Solving
this we that after 117 years we will have 10000.
5.1.7: If r = 1.2 and X0 = 0.42, what is X1 ? X2 ?
Solution: X1 = 1.2(0.42)(1−0.42) = 0.29232 and X2 = 1.2(0.29232)(1−0.29232) = 0.24824282112.
6.2.16: Similarly, for another function g : R2 → R3 , choose vectors for g(e1 ), g(e2 ) and work
through the reasoning above to find the matrix representation of g. What are the dimensions of
this matrix?
Solution:
a
g(e1 ) = b
c
d
g(e2 ) = e
f
2
Solutions:
5.1.2: Choose an initial value for X and use values of r that are less than, greater than, and equal
to one to test the above statements by computing two values of X for each value of r.
Solution: Choose X0 = 1,
r = 1:
X1 = 1 ∗ 1 = 1, X2 = 1 ∗ X1 = 1 ∗ 1 = 1.
r = 2:
X1 = 2 ∗ 1 = 2, X2 = 2 ∗ 2 = 4.
r = 1/2:
1 1 11 1
X1 = ∗ 1 = , X2 = = .
2 2 22 4
5.1.3: An inedible alga is growing on a pond in a city park. Only a small part of the pond is now
covered by the algae, but the area covered is doubling each day. The city decides to remove the
algae once it covers half the pond. If the pond will be completely overgrown in thirty days, on
what day will it be half covered? (Hint: Try working backward.)
Solution: The pond will be completely overgrown in thirty days. Since the algae doubles each day
the pond will be half overgrown with algae on the twenty ninth day. On the twenty eighth day it
will be less than half covered. So the city will remove the algae on the twenty ninth day.
5.1.4: A rabbit population is growing at 10% a year. If there are 10 rabbits this year and time is
discrete, how many will there be in 10 years? Use a loop in SageMath to check your answer.
Solution: In this case r = 1.1 and X0 = 10. So
X10 = (1.1)10 ∗ 10 = 25.937424601.
(Round one way or the other to get the actual rabbit population.)
Sage code:
init = 10
r = 1.1
for i in range(0,10):
x = r*init
init = x
init
1
, LS 30B Homework 4 Solutions Feb 9, 2018
5.1.5: While we have been working with r > 1, representing growth, r can be less than 1, repre-
senting a quantity that decreases over time. The half-life of a radioactive element is the amount of
time needed for half the element to decay. What fraction of the initial amount of such an element
will remain after ten half-lives?
1 1 1
Solution: Each half life we lose 2 of the element. So after ten iterations we get 210
= 1024 .
5.1.6: When money in a bank account accrues compound interest, the interest earned in one time
period is added to the principal, and then the sum is used as the base for the next time period.
(a) If you start off with $1000 and earn 2% interest that is compounded annually, how much
money will you have in 5 years? In 10 years? In 20 years?
(b) How long will it take you to accumulate $10,000?
Solution:
(a) In this case r = 1.02 and X0 = 1000. So
X5 = (1.02)5 ∗ 1000 = 1104.0808032
X10 = (1.02)10 ∗ 1000 = 1218.99441999,
X20 = (1.02)20 ∗ 1000 = 1485.94739598
(b) To get to 10000 we solve, 10000 = (1.02)n ∗ 1000. Simplifying, we get 10 = (1.02)n . Solving
this we that after 117 years we will have 10000.
5.1.7: If r = 1.2 and X0 = 0.42, what is X1 ? X2 ?
Solution: X1 = 1.2(0.42)(1−0.42) = 0.29232 and X2 = 1.2(0.29232)(1−0.29232) = 0.24824282112.
6.2.16: Similarly, for another function g : R2 → R3 , choose vectors for g(e1 ), g(e2 ) and work
through the reasoning above to find the matrix representation of g. What are the dimensions of
this matrix?
Solution:
a
g(e1 ) = b
c
d
g(e2 ) = e
f
2
