Saturation Math Example 1

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Example 1

medium

The logistic function

P(t) = \dfrac{1000}{1+e^{-0.5(t-6)}}

models population growth. Find

P(0)

P(6)

, and

\lim_{t\to\infty}P(t)

Solution

1
$P(0) = \frac{1000}{1+e^{-0.5(0-6)}} = \frac{1000}{1+e^{3}} = \frac{1000}{1+20.09} \approx \frac{1000}{21.09} \approx 47.4$ .
2
$P(6) = \frac{1000}{1+e^{0}} = \frac{1000}{1+1} = 500$ . At $t=6$ , population is at half the maximum.
3
$\lim_{t\to\infty}P(t)$ : as $t\to\infty$ , $e^{-0.5(t-6)}\to0$ , so $P\to\frac{1000}{1+0}=1000$ . The carrying capacity (saturation level) is $L=1000$ .

Answer

P(0)\approx47.4

;

P(6)=500

;

\lim_{t\to\infty}P(t)=1000

The logistic function starts slow, grows rapidly in the middle, then saturates at

L

(the carrying capacity). The inflection point occurs at

t=x_0

where

P=L/2

; here at

t=6

P=500

About Saturation

Saturation is the phenomenon where a growing quantity approaches a limiting value asymptotically, with the rate of growth decreasing as the limit is approached.

Learn more about Saturation →

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