# doubling time

• Aug 16th 2009, 08:06 AM
skeske1234
doubling time
The population of the world is doubling every 35 a. In 1987 the total population reached 5 billion. find the projected world population for the year 2001.
USE THE EXPONENTIAL GROWTH FORMULA.
y=y_o(e)^(kt)
35a=5e^(1987t)
ln7a/1987=k

y=5e^[(ln7a/1987)(2001)]
=35.6
this is wrong, the answer is 6.6 billion
• Aug 16th 2009, 08:13 AM
skeeter
Quote:

Originally Posted by skeske1234
The population of the world is doubling every 35 a. In 1987 the total population reached 5 billion. find the projected world population for the year 2001.
USE THE EXPONENTIAL GROWTH FORMULA.
y=y_o(e)^(kt)
35a=5e^(1987t)
ln7a/1987=k

doubling every 35 years?

let 1987 be t = 0

$y = 5e^{kt}$

$10 = 5e^{35k}$

solve for k, then use the original equation to evaluate y when t = 14 (2001)
• Aug 16th 2009, 08:15 AM
skeske1234
Quote:

Originally Posted by skeeter
doubling every 35 years?

let 1987 be t = 0

$y = 5e^{kt}$

$10 = 5e^{35k}$

solve for k, then use the original equation to evaluate y when t = 14 (2001)

no, doubling every 35 a.

There is an a there.. and no that is not a typo because there are other questions in the textbook that read a following a number too.. I am not too sure about this eg.

Uranium-28 has a half life of 4.5x10^9 a.
Find the mass that remains from this sample afer 1000 a. Find the rate of decrease of the mass after 10000 a.

So there is an a? :S
• Aug 16th 2009, 08:20 AM
skeeter
fine ... every $35a$

let 1987 be t = 0

$y = 5e^{kt}$

$10 = 5e^{(35a)k}$

solve for $k$ in terms of $a$, then use the original equation to evaluate y when t = 14 (2001)