   Chapter 3.7, Problem 25E

Chapter
Section
Textbook Problem

In Example 6 we considered a bacteria population that doubles every hour. Suppose that another population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number n of bacteria after t hours and use it to estimate the rate of growth of the bacteria population after 2.5 hours.EXAMPLE 6  To determine

To find: the expression for the number of bacteria after t, and estimate the rate of growth of the bacteria population after 2.5 hours.

Explanation

Given:

The initial population n0 is 400.

Population of the bacteria triples in one hour as per the rate expression given below.

n(t)=n0×3t (1)

Calculation:

Expression for n bacteria in t hours.

Substitute the 400 for n0 in (2).

n(t)=400×3t (2)

Determine the rate of growth of population after 2.5 hours.

Differentiate equation (2) with respect to t.

dndt=400×3t(ln3)=400×3t(1

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