# How can you calculate the rate at which a population is growing? ?

If t is in years since 1790 and P is in billions, we use the following logistic function to model india's population between 1790 and 1930

194/1+36e^(-0.00314545t)What is the population when it is growing fastest?At what time is the population growing fastest?

Update:

the equation is p(t)= 194/1+36e^(-0.0314545t)

Relevance
• 1 month ago

Find the inflection point

Now, this function is from a special place in hell.  We're going to simplify things a bit

p(t) = 194 / (1 + 36 * e^(-0.0314545t))

p(t) = 194 * (1 + 36 * e^(-0.0314545t))^(-1)

-0.0314545 = a

p(t) = 194 * (1 + 36 * e^(at))^(-1)

p'(t) = 194 * (-1) * (1 + 36 * e^(at))^(-2) * 36 * e^(at) * a

p'(t) = -194 * 36 * a * e^(at) / (1 + 36 * e^(at))^2

-194 * 36 * a = b

p'(t) = b * e^(at) / (1 + 36 * e^(at))^2

u = e^(at)

u' = a * e^(at)

v = (1 + 36 * e^(at))^2

v' = 2 * (1 + 36 * e^(at)) * 36 * a * e^(at) = 72a * e^(at) * (1 + 36 * e^(at))

p''(t) = b * (v * u' - u * v') / v^2

b is just a constant, no need to worry about that

v^2 is always positive.  No need to worry about that.

Let v * u' - u * v' = 0

(1 + 36 * e^(at))^2 * a * e^(at) - e^(at) * 72a * e^(at) * (1 + 36 * e^(at)) = 0

a * e^(at) * (1 + 36 * e^(at))^2 - 72a * e^(2at) * (1 + 36 * e^(at)) = 0

Again, a is a constant, e^(at) will never be 0 and 1 + 36 * e^(at) will never be 0.  Divide through by all 3

1 + 36 * e^(at) - 72 * e^(at) = 0

1 - 36 * e^(at) = 0

1 = 36 * e^(at)

1/36 = e^(at)

ln(1/36) = at

-2 * ln(6) = at

-0.0314545 = a

-2 * ln(6) = -0.03124545 * t

2 * ln(6) / 0.03124545 = t

This is when it is changing at its fastest

p(t) = 194 / (1 + 36 * e^(-0.0314545t))

194 / (1 + 36 * e^(-0.0314545 * 2 * ln(6) / 0.03124545)) =>

194 / (1 + 36 * e^(-2 * ln(6)) =>

194 / (1 + 36 * e^ln(1/36)) =>

194 / (1 + 36 * (1/36)) =>

194 / (1 + 1) =>

194 / 2 =>

97