   Chapter 11.5, Problem 102E Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

Epidemics Another epidemic follows the curve P = 200 1 + 20 , 000 e − 0.549 t  million people, where P is the number of people infected and t is in years. How fast is the epidemic growing after 10 years? After 20 years? After 30 years? (Round your answers to two significant digits.) [HINT: See Example 3.]

To determine

To calculate: The rate at which the epidemic is growing after 10 years, 20 years and 30 years if the epidemic is approximated with the equation P=2001+20,000e0.549tmillion people where P is the number of people infected and t is the number of years after the start of the epidemic.

Explanation

Given Information:

The flu epidemic is approximated with the equation P=2001+20,000e0.549tmillion people where P is the number of people infected and t is the number of years after the start of the epidemic and the time after which the rate of growth of the epidemic is calculated is 10 years, 20 years and 30 years respectively.

Formula used:

The following derivative formula is used:

ddx(ex)=ex

Calculation:

Consider that the flu epidemic is approximated with the equation P=2001+20,000e0.549tmillion people where P is the number of people infected and t is the number of years after the start of the epidemic.

Re-write P=2001+20,000e0.549tmillion people as P=200(1+20,000e0.549t)1.

The rate of change of P(t) is equal to the rate at which the epidemic grows.

dPdt=200(1+20,000e0.549t)11[20000(0.549e0.549t)]=200(1+20,000e0.549t)2[20000(0.549e0.549t)]=2196000e0.549t(1+20,000e0.549t)2

Substitute t=10 in the equation to calculate the rate at which the epidemic grows after 10 years.

P(10)=2196000e0.549(10)(1+20,000e0

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