Question

The population in the world in 2006 was 6.53 billion people and was growing at a rate of 1.14% per year. Assuming that this growth rate continues, the model represents the population P (in billions of people) in year t. Answer the following questions.


[A] What will be the anticipated world population in 2020?

[B] For what value of t will the world population be 12 billion? What date corresponds to this value of t?

Answer 1

nobody know the answer to this one

Answer 2

this is a question on differential equations...:) are you sure it is there for you?

Answer 3

what do you mean?

Answer 4

yes we can do this

Answer 5

@akshay did you get your probability answer?

Answer 6

@treyhud the formula is
\[P(t)=6.53e^{.0114t}\] where t is years after 2006

Answer 7

2020 is 14 years after 2006 to replace t by 14 to get answer to first part

Answer 8

yes.. satellite... but its good you r helping.. i am not in touch with it for 2 years.. thanks:)
i was searching for the format.

Answer 9

\[.0114\times 14=.1596\] and \[6.53e^{.1596}=7.64\] rounded

Answer 10

@akshay look at the very end of that problem
gives and easy way to do it without using a solid. very simple actually.

Answer 11

we can work through it if you like

Answer 12

P(t)=6.53(1.8377)^r-2006=12 so the value of t is 1.8377 ? i dont know if that is right

Answer 13

@treyhud answer to second question put
\[12=6.53e^{.0114t}\] and solved for t

Answer 14

t is number of years after 2006

Answer 15

@satellite i am not getting you??

Answer 16

for this problem or probability one?

Answer 17

probability one..

Answer 18

ok let me finish this one and then start another thread

Answer 19

ok sure.

Answer 20

@treyhud is it clear what is going on here? we do not work with the numbers 2006,2020 etc. instead you put
2006 = 0 that is when we start counting
2020=14 that is 14 years later

Answer 21

ok yeah that makes since with the years part. I am just trying to solve for t for the second part. What are you getting cause my number seems a little off

Answer 22

formula for continuous grown it
\[P=P_0e^{rt}\] where
\[P_0\] is initial population, r is rate as a decimal and t is time

Answer 23

in the last problem you have P=12 so put
\[12=6.53e^{.0114t}\] and solve for t

Answer 24

you do this via
\[\frac{12}{6.53}=e^{.0114t}\]
\[\ln(\frac{12}{6.53})=.0114t\]
\[t=\ln(\frac{12}{6.53})\div .0014\]

Answer 25

so it would look like this \[t=6.53e^.0114/12\]

Answer 26

oh no. you have to get the variable out of the exponent by taking the log

Answer 27

ok thats my problem then

Answer 28

first step is divide both sides by 6.53
second step is take the natural log of both sides

Answer 29

then divide the result by .0014

Answer 30

you get
\[t=\ln(\frac{12}{6.53})\div .0014\]

Answer 31

434.64?

Answer 32

i made a typo sorry. you need to divide by .0114, not .0014

Answer 33

oh

Answer 34

that is the rate of growth, .0114

Answer 35

should have written
\[t=\ln(\frac{12}{6.53})\div .0114\]

Answer 36

my mistake. i copied wrong. you get around 53

Answer 37

so 53 years after 2006 is 2059

Answer 38

gotcha. thanks alot. u taught me better on this chat deal than my professor has all semester

Answer 39

i will take that as a compliment. yw