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The population of the human race in 2000 was 6100 million, with an annual growth rate of 1.4%. The population (in millions) as a function of time (years) is described reasonably well by the equation

**P _{(t)} = 6100e^{0.014t}**

Determine:

- When the population was/will be 2.38 times that at the census date (in years, relative to the census date, with 4 decimal places)
- The population at that time (in whole millions, as an integer)
- The rate of growth of the population at that time (in number of individuals per day, as an integer: 1 year has 365.2422 days)
Years = years

Pop = million

Rate = individuals per day

Oli96 Oct 28, 2014

#2**+13 **

I get something a little different from Alan, here

The beginning population number doesn't matter, so the first part is.....

2.38P = Pe^{.014t} divide by P and take the ln of both sides

ln(2.38) = lne^{.014t} and by a log property, we have (remember, lne = 1)

ln(2.38) =.014t divide both sides by .014

ln(2.38)/.014 = t = 61.9357 years

CPhill Oct 28, 2014

#1**+5 **

P(t) = 6100*e^{0.014t}

dP(t)/dt = 0.014*6100*e^{0.014t}

t = ln(P/6100)/0.014

P = 2.38*6100

$${\mathtt{t}} = {\frac{{ln}{\left({\frac{{\mathtt{2.38}}{\mathtt{\,\times\,}}{\mathtt{6\,100}}}{{\mathtt{6\,100}}}}\right)}}{{\mathtt{0.014}}}} \Rightarrow {\mathtt{t}} = {\mathtt{61.935\: \!749\: \!120\: \!241\: \!666\: \!2}}$$

t = 61.9357 years

$${\mathtt{dPdt}} = {\frac{{\mathtt{0.014}}{\mathtt{\,\times\,}}{\mathtt{6\,100}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\mathtt{0.014}}{\mathtt{\,\times\,}}{\mathtt{61.935\: \!7}}\right)}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{6}}}}{{\mathtt{365.242\: \!2}}}} \Rightarrow {\mathtt{dPdt}} = {\mathtt{556\,485.149\: \!380\: \!397\: \!522\: \!958\: \!7}}$$

dP/dt = 556485 individuals per day

(Edited to correct the expression for t)

.

Alan Oct 28, 2014

#2**+13 **

Best Answer

I get something a little different from Alan, here

The beginning population number doesn't matter, so the first part is.....

2.38P = Pe^{.014t} divide by P and take the ln of both sides

ln(2.38) = lne^{.014t} and by a log property, we have (remember, lne = 1)

ln(2.38) =.014t divide both sides by .014

ln(2.38)/.014 = t = 61.9357 years

CPhill Oct 28, 2014