Population: from table of values, estimate P'(2004), P'(2006

[whytehaze]

the population of a small town is given:

year (t)- 2003 2004 2005 2006
Population P(t) (thousands)- 11.2 11.8 12.2 12.5

a. estimate P'(2004)
b. estimate P'(2006)
c. what are the units of P'(t)?

 

[galactus]

Re: Populations problem

One way is to do a regression. I tried a quadratic regression and got a very accurate equation with an R^2 of over .999

That's mighty good. I used 2003 as 0 and went from there.

\(\displaystyle y=-.075x^{2}+.655x+11.205\)

Or in prettier fractional format:

\(\displaystyle y=\frac{-3}{40}x^{2}+\frac{131}{200}+\frac{2241}{200}\)

Now, find the derivative at 2004.

 

[skeeter]

Re: Populations problem

the population of a small town is given:

year (t)- 2003 2004 2005 2006
Population P(t) (thousands)- 11.2 11.8 12.2 12.5

a. estimate P'(2004)
b. estimate P'(2006)
c. what are the units of P'(t)?

you can also get a fair estimate of P'(t) using the slope of a secant line

 

[galactus]

Re: Populations problem

you can also get a fair estimate of P'(t) using the slope of a secant line

Yes, skeet, that is probably what they're looking, for more than likely.

I thought I would try the regression for kicks.

 

[skeeter]

Re: Populations problem

I thought I would try the regression for kicks.

... and the kick was good! give yourself credit for a field goal.

 

[DR._Glockman]

Re: Populations problem

Using a logistic equation and assuming the carrying capacity is 13 (in thousands), I got

P(t) = 13/[1+.1607e^(-.45t)], P'(t) = [5.85(.1607e^(-.45t)]/[1+.1607e^(-.45t)]^2

P(0) = 11.200137..., P'(0) = .697...
P(1) = 11.791737..., P'(1) = .493...
P(2) = 12.202725..., P'(2) = .336...
P(3) = 12.480080..., P'(3) = .224...
P(4) = 12.663609..., P'(4) = .147...
P(5) = 12.783477..., P'(5) = .095...
P(6) = 12.861101..., P'(6) = .061...

It should be obvious that the limiting factor is 13 and the derivative tapers off to zero as t increases.