Question

One model of Earth's population growth is \[P(t)=\frac{ 64 }{ 1+11e^-.08t }\] , where t is measured in years since 1990, and P is measured in billions of people. Which of the following statements are true? Check all that apply.

A. The population of Earth is increasing by a steady rate of 8% per year.

B. The carrying capacity of Earth is 64 billion people.

C. The population of Earth will grow exponentially without bound.

D. In 1990, there were 5.33 billion people.

Answer 1

@DemolisionWolf can you help with this please?

Answer 2

do you have a particular one or all of them you want to look at together?

Answer 3

all of them if you dont mind, because it says to select all of the answer choices that are correct, so more than 1 can be right I just dont know how to tell which ones :o

Answer 4

alright ^_^

Answer 5

The population of Earth is increasing by a steady rate of 8% per year

so, in the equation there is an 0.08 in the exponent term, so that would lead me to think this is true, but lets work it out and check.

plug in t=1, and solve for p, then do it for t=2

Answer 6

I'll do t=3..

Answer 7

why t=3?

Answer 8

(3, 6.6)

Answer 9

because we want to look at 3 years to compare 2 increasing rates, just to make sure it is 8% each year.

Answer 10

we could pick any 3 years,
like t=22, 23, and 24

Answer 11

oh okay, so what do I plug t=2 into, like which equation?

Answer 12

plug it into the population equation the problem gives

Answer 13

I got 5.7 :o

Answer 14

good, so lets to t=2 now
(1, 5.7)
(2, ? )
(3, 6.6)

Answer 15

\[IncreasingSteadyRate=\frac{ NextYrPopulation-ThisYearPopulation }{ ThisYearPopultion}\]

Answer 16

@ali1029
this first question is the longest of them all, fyi

Answer 17

okay, so which numbers would I plug in to that new equation? im sorry im so confused still D;

Answer 18

plug t into this one:
\[P(t)=\frac {64}{1+11e^{−.08t}}\]
becomes
\[P(t)=\frac {64}{1+11e^{−.08(2)}}\]


[P(t)=\frac {64}{1+11e^{−.08t}}\]

Answer 19

just like you did to get 5.7 when you plugged in t=1

Answer 20

I got 3.08

Answer 21

mm.... it should be twice that number, 6.16 not 3.08,
but anyways, lets continue so we can be dun with this problem

so now we have the population for years 1991, 1992, and 1993.
can you find the percent increase from year to year? (i posted up the formula to figure out percent increase earlier)

Answer 22

oh i thought we were looking for year 2, let me do the equation

Answer 23

I got 1% but I dont think that sounds right...

Answer 24

@ali1029
t= 2 means 1992
t=1 means 1991
t=3 means 1993
we've done it

Answer 25

^_^
ok ok ok, lets back track a second.

so we were given the equation:
\[P(t)=\frac {64}{1+11e^{−.08t}}\]

and it said that t is the number of years after the year 1990. so If I were to make a table with this idea, it would look like this:
yr | t
1990 0
1991 1
1992 2
1993 3

so when we plug in t=2 into the equation, we are finding the population of people on the earth for the year 1992.
make sense?

Answer 26

oh okay!

Answer 27

^_^

Answer 28

so here is a table of what we have found:

t | p
1 5.7
2 6.2
3 6.6

so we want to find the percent increase from one year to the next to figure out if the part A is true or false. if the percent incrase we solve for is 8% then it is true, if it is not 8% then it is false.

Answer 29

I got .0645 so it would be false, but is my math correct?

Answer 30

\[\frac{ 6.2-5.7}{ 5.7 } = 0.087\]
[\frac{ 6.2-5.7}{ 5.7 } = 0.087\]

Answer 31

OH! so it would be true?!

Answer 32

\[\frac{ 6.6-6.2}{ 6.2 } = 0.0645\]
[\frac{ 6.6-6.2}{ 6.2 } = 0.0645\]

Answer 33

so we get 8.7% and 6.4% from one year to the next, so is it increasing at a steady rate each year? I think not

Answer 34

yeah no, I just realized that! so would C be the only correct answer? I'm thinking D also, but I'm not sure because we got 5.7 not 5.3 :O

Answer 35

so lets look at D
if t=1 refers to year 1991
then t=0 refers to year 1990,
so you gotta plug in t=0 into our equation again, so:
\[P(t)=\frac {64}{1+11e^{−.08t}}\]
\[P(t)=\frac {64}{1+11e^{−.08(0)}}\]

[P(t)=\frac {64}{1+11e^{−.08t}}\]

Answer 36

I got no solution ._.

Answer 37

:(
so if we plug in t=0, then
e^-.08*0 = e^0 = 1
so 1+11(1) = 12
then 64/12 = ?

Answer 38

ugh i keep messing up my math! >:( but i see that D would be true now

Answer 39

would C be another possible answer or is it just D?

Answer 40

so a recap-
A: false
B: ?
C: ?
D: True

as for B and C, only one can be true. and thus the other would have to be false. so lets check which is true.
lets plug in a really large number into t,
lets do 900 which would be year 2890, so

-.08*(900) = -72
e^-72 = 5.3x10^-32
1+11(5.3x10^-32) = 1
64/1 = 64
so like 900 years from now, it says the population will be 64.0
the .0 part tells me that it population growth stops at 64 billion people.

clear as mud? ^_^


most this problem is about plugging in numbers for t then seeing what they mean.

Answer 41

oh that does make sense, so thats saying that the max capacity is 64 mill?

Answer 42

& C is false?

Answer 43

B: is true
C: is false,
yep ^_^

Answer 44

thank you SO much for all of your help!!! you're awesome!!! :)

Answer 45

ur welcome
you seem to put up the weirdest/hardest questions each day, lol

Answer 46

haha! trust me, i know :p

Answer 47

It's a & b