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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) = 6100e0.014t

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

 Oct 28, 2014

Best Answer 

 #2
avatar+130477 
+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

 

 Oct 28, 2014
 #1
avatar+33658 
+5

P(t) = 6100*e0.014t

dP(t)/dt = 0.014*6100*e0.014t

t = ln(P/6100)/0.014

 

P = 2.38*6100 

t=ln(2.38×61006100)0.014t=61.9357491202416662

t = 61.9357 years

dPdt=0.014×6100×e(0.014×61.9357)×106365.2422dPdt=556485.1493803975229587

dP/dt = 556485 individuals per day

 

(Edited to correct the expression for t)

.

 Oct 28, 2014
 #2
avatar+130477 
+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
 #3
avatar+33658 
+10

Thanks Chris, I forgot the 0.014 in my original calculation of t.  I've now corrected that.

.

 Oct 28, 2014

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