• Sep 2nd 2009, 03:22 PM
asweet1
My calculus book gives me the formula for a human population as 906e^0.008(t-1800) I am supose to find out the population for 2010.
here is the chart

year 1940 1950 1960 1970 1980 1990 2000
People (millions) 2300 2520 3020 3700 4450 5300 6090

I tried to plug these numbers in but did not get anything close to these numbers I am lost.

Some additional info: The table is longer it actually gives 20 centuries but the ones above are the last few. t=time and g(t) is population. PLEASE HELP . The book says the answr is 4.86 billion people but I don't understand.

The actual question is what does the approximate population function p given predict for the year 2010?
p(t)=906e^0.008(t-1800)
• Sep 2nd 2009, 07:48 PM
mr fantastic
Quote:

Originally Posted by asweet1
My calculus book gives me the formula for a human population as 906e^0.008(t-1800) I am supose to find out the population for 2010.
here is the chart

year 1940 1950 1960 1970 1980 1990 2000
People (millions) 2300 2520 3020 3700 4450 5300 6090

I tried to plug these numbers in but did not get anything close to these numbers I am lost.

Some additional info: The table is longer it actually gives 20 centuries but the ones above are the last few. t=time and g(t) is population. PLEASE HELP . The book says the answr is 4.86 billion people but I don't understand.

The actual question is what does the approximate population function p given predict for the year 2010?
p(t)=906e^0.008(t-1800)

Substitute the value of t corresponding to 2010 into p(t)=906e^0.008(t-1800) and calculate the value of p.
• Sep 2nd 2009, 11:46 PM
pacman
p(t) = 906e^0.008(t-1800),

for the year 2000, the population was 6,090 millions based on the table, plugging it back and solve for t, we have

6,090 = 906e^[0.008(t - 1800)

ln(6090/906) = 0.008(t - 1800) ln e, (ln e = 1]

t = 1800 + [ln 6090/906]/0.008 = 2038 a.d.

it should be 2000 a.d., the discrepancy is due to the fact that the logarithmic function is not THAT perfect, an approximation only

p(t) = 906e^0.008(t-1800), using this outdated model, we have

for the year 2010 a.d.

asweet, this is what you want to solve: for t = 2010,

p(2010) = 906e^[0.008(2010-1800)] = 4,862 millions = 4.862 billions [ population explosion]

asweet1, you may screwed this up, p(t) = 906e^[0.008(t-1800)];

the expression "0.008(t-1800)" is the exponent of e. I hope it helps YOU.
• Sep 3rd 2009, 07:44 AM
asweet1
The question also asks for the prediction of the population in the year 1000 B.C.E. I used the equation do a.d but I got 1805.i and the book says the answer is 0.17 people what am I doing wrong?
• Sep 3rd 2009, 08:06 AM
pacman
p(t) = 906e^0.008(t-1800),

at t = 1000 a.d.

P(1000) = (906)e^0.008(1000 - 1800) = (906)e^[0.008(-800)]

P(1000) = (906)e^[0.008(1000 - 1800)] =

P(1000) = (906)e^(-6.4)

P(1000) = (906)/[(2.71828)^6.4]

P(1000) = 906/601.842

P(1000) = 1.5054 million

see the graph.

The exponential equation is a way off, it predicts too far from the true value, 800 years is too big to consider.

Even at to time of Jesus, the earth population was around 100 million to 300 million.

But, mathematically, it helps us to appreciate the beauty of MATH.
• Sep 3rd 2009, 08:13 AM
asweet1
how does the book get the answer 0.17 people? I tried using negatives but it doesnt work
• Sep 3rd 2009, 04:18 PM
pacman
t = 1800 + [ln P(t)/906]/0.008

= 1800 + [ln(0.17/906)]/0.008

= 1800 - 1073

= 727 a. d.

if we use t = 0.17 million = 170,000 people

the year would be 727 a.d.

which is below 1000 a.d.

Notice that the exponential function is only good when the prediction is between 1940 to 2000, any farther than that is way off tangent. . . . but it serves its purpose, to predict the population...