   Chapter 4.1, Problem 56E

Chapter
Section
Textbook Problem

Population Growth The rate of growth d P / d t of a population of bacteria is proportional to the square root of t, where P is the population size and t is the time in days ( 0 ≤ t ≤ 10 ) . That is, d P d t = k t The initial size of the population is 500. After I day, the population has grown to 600. Estimate the population after 7 days.

To determine

To calculate: The population of bacteria whose growth rate is dhdt=kt after 7 days.

Explanation

Given:

The rate of growth dPdt of a population of bacteria is proportional to the square root of

t, where P is the population size and t is the time in days 0t10. The initial size of the population is 500. After 1 day, the population has grown to 600.

Formula used:

Basic integration rule,

F(x)dx=F(x)+C

Calculation:

Consider the rate at which the population of bacteria grows,

dPdt=kt

Here P is the population of bacteria and t is the time taken by the bacteria to grow (in days).

Integrate the above growth rate to find the population of bacteria after t days,

(dPdt)dt=(kt)dtP=k(t)3232+C=23k(t)32+C

In the provide problem it is given that initial population of bacteria is 500. That is P=500 at t=0. Substitute these value in above equation,

500=23k(0)32+CC=500

Rewrite the equation for the population of bacteria after t days

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