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MATH 1030-
Homework #
Instructions: Do the following problems on a separate sheet of paper.
- The population of zombies in Hurricane, Utah is doubling every 6 days! When Mildred counted this morning, there were 245 of them. We will use the doubling time formula with Tdouble = 6 and Q 0 = 245.
Q = 245(2)t/^6
(a) How many zombies will there be in 3 weeks? 3 weeks in days is 21 days, so we’ll use t = 21.
Q = 245(2)^21 /^6 = 2, 771. 859
so in 3 weeks there will be 2,772 zombies.
(b) When was there only one zombie?
We’ll use Q = 1 and solve for t.
1 = 245(2)t/^6 1 245
= 2t/^6
log
= log 2t/^6
log
t 6
log 2
log 2451 log 2
= t
− 47 .62 = t
So there was only 1 zombie 47.62 days ago.
(c) When will the entire town (population 8,250) become zombies? We’ll use Q = 8, 250 and solve for t.
8 , 250 = 245(2)t/^6 8250 245
= 2t/^6
log
t 6
log 2
log (^8250245) log 2
= t
30 .44 = t
So there will be 8,250 zombies in Hurricane 30.44 days from now.
- Carbon-14 has a half-life of 5730 years. We’ll use the half-life formula with Thalf = 5730.
Q = Q 0 (.5)t/^5730
(a) How long does it take for the amount of Carbon-14 to decrease by 10%? To see how long it takes to decrease by 10%, let’s start with Q 0 = 100 and Q = 90, which is 10% less and solve for t.
90 = 100(.5)t/^5730 .9 =. 5 t/^5730
log .9 =
t 5730
log. 5
5730 log. 9 log. 5
= t
870 .98 = t
So it will take 870.98 years to decrease by 10%.
(b) What percentage of its original Carbon-14 does a fossil have that is 200,000 years old?
A fossil starts with 100% of its Carbon-14, so we’ll use Q 0 = 100 and t = 200, 000.
Q = 100(.5)^200 ,^000 /^5730 =. 000000003
so there is only a tiny percentage of the original Carbon-14, i.e., .000000003% of it is left.
(c) How old is a fossil that has 30% of its original amount of Carbon-14? We’ll use Q 0 = 100 again and put Q = 30 to solve for t.
30 = 100(.5)t/^5730 .3 =. 5 t/^5730
log .3 = t 5730
log. 5
log. 3 log. 5
= t
9 , 952 .81 = t
So if a fossil has 30% of its Carbon-14, it has been 9,952.81 years since it died.
(c) If there were 18 monkey attacks this year, in what year will only 5 monkey attacks occur? Again we’ll use Q 0 = 18 with now as the starting time. We want to know when there are 5 attacks, so we’ll set Q = 5 and solve for t.
5 = 18(.938)t 5 18
=. 938 t
log
= t log. 938
20 .01 = t
So there will only be 5 attacks in 20 years, i.e., in 2028.
- Suppose that Alfred has an ant colony who’s population is doubling every 5 months. We’ll use the doubling time formula with Tdouble = 5 and t in months.
Q = Q 0 (2)t/^5
(a) How often does the population of the ant colony triple? To find how often it triples, we’ll set Q 0 = 1, Q = 3, and solve for t.
3 = 2t/^5
log 3 = t 5
log 2
log 3 log 2
= t
7 .925 = t
So the ant colony triples in size every 7.925 months.
(b) If there were 450 ants one year ago, how many are there now?
We’ll use Q 0 = 450 as the starting value with one year ago being the starting time. Right now is 12 months after the starting time, so set t = 12.
Q = 450(2)^12 /^5 = 2, 375. 11
So there are 2,375 ants right now.
(c) If there were 450 ants one year ago, when will the population reach 10,000? We’ll use Q 0 = 450 as the starting value again with one year ago being the starting time. We want to know when there are 10,000 ants, so we set Q = 10, 000 and solve for t.
10 , 000 = 450(2)t/^5
= 2t/^5
log
t 5
log 2
log 10450 ,^000 log 2
= t
22 .37 = t
So there will be 10,000 ants 22.37 months after the starting time, which was 12 months ago. So there will be 10,000 ants 10.37 months from now.
- Suppose the number of alien visits is cut in half every 200 years. Further suppose that there were 860 alien visits in 1950. We’ll use the half-life formula with Thalf = 200, t in years, and Q 0 = 860 the starting value where 1950 is the starting time. Q = 860(.5)t/^200
(a) How many alien visits were there in 2000 BCE? To find the number of alien visits in 2000 BCE, we notice that 2000 BCE is 2000 + 1950 = 3950 years before the starting time. So we’ll set t = −3950.
Q = 860(.5)−^3950 /^200 = 758, 299 , 667. 6
So there were about 758 million alien visits in 2000 BCE.
(b) In what year will there be only 100 alien visits?
To find when there will be 100 visits, we’ll use Q = 100 and solve for t.
100 = 860(.5)t/^200 100 860
=. 5 t/^200
log
t 200
log. 5
log (^100860) log. 5
= t
620 .87 = t
So there will be 100 visits 620.87 years after the starting time of 1950, i.e., towards the end of the year
(c) What is the percentage rate of decay per year? (hint: use t = 1) To find the decay rate per year, let’s change Q 0 to be 100 and we’ll investigate what happens to 100 in 1 year; that is, we’ll set t = 1. Q = 100(.5)^1 /^200 = 99. 65 so after one year, the value of 100 decreases to 99.65. That means that it decreased by 0.35. So the decay rate is 0.35% per year.
