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Exponential Growth and Decay Models

Key Questions

What is the general formula for exponential growth of a population?

Answer:

Population $[P] = C e^{k t}$ $[P]= Ce^[kt$

Explanation:

If the rate of growth $P$ $P$ is proportional to itself, then with respect to time $t$ $t$ ,

$\frac{d P}{d t} = k P$ $[dP]/dt=kP$ , ....inverting both sides, ..... $\frac{d t}{d P} = \frac{1}{k P}$ $dt/[dP]=[1]/[kP$ and so integrating both sides

$\int d t = \int \frac{d P}{k P}$ $intdt=int[dP]/[kP$ , thus,..... $t = \frac{1}{k} ln P +$ $t=1/klnP +$ a constant............ $[1]$ $[1]$

Suppose $P$ $P$ is some value $C$ $C$ when $t = 0$ $t=0$ , substituting

$0 = \frac{1}{k} ln C +$ $0=1/klnC+$ constant, therefore the constant $= - \frac{1}{k} ln C$ $= -1/klnC$ and so substituting this value for the constant in ... $[1]$ $[1]$ we have ,

$t = \frac{1}{k} [ln P - ln C]$ $t= 1/k[ln P-lnC]$ = $\frac{1}{k} ln [\frac{P}{C}]$ $1/k ln[P/C]$ , therefore , $k t = ln [\frac{p}{C}]$ $kt=ln[p/C]$ [ theory of logs] and so

$e^{k t} = \frac{P}{C}$ $e^[kt]=P/C$ ......giving $P = C e^{k t}$ $P=Ce^[kt$ . The constant $k$ $k$ will represent the excess of births over deaths or vice versa for a decreasing rate.

Barry H. · 1 · Mar 2 2018

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Applications of Definite Integrals

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