   Chapter 9.5, Problem 50E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Endangered species population It is determined that a wildlife refuge can support a group of up to 120 of a certain endangered species. If 75 are introduced onto the refuge and their population after t years is given by p ( t ) = 75 ( 1 + 4 t t 2 + 16 ) find the rate of population growth after t years. Find the rate after each of the first 7 years.

To determine

To calculate: The rate of population growth of a certain endangered species after t years and after each of the first 7 years, if the population after t years is given as p(t)=75(1+4tt2+16) when 75 endangered species introduced onto the refuge.

Explanation

Given Information:

The expression is p(t)=75(1+4tt2+16)=(75+300tt2+16)

Formula Used:

The quotient rule for the derivative of the two function, ddx(fg)=gdfdxfdgdxg2.

The sum and difference rule of derivate of functions, ddx[u(x)±v(x)]=ddxu(x)±ddxv(x).

The simple power rule of derivative ddx(xn)=nxn1.

Calculation:

Consider the provided population growth of a certain endangered species after t years is p(t)=75(1+4tt2+16).

Differentiate the provided population growth function,

dp(t)dt=ddt[75(1+4tt2+16)]=ddt[75+300tt2+16]

Use the sum and difference rule of derivative, ddx[u(x)±v(x)]=ddxu(x)±ddxv(x).

dp(t)dt=ddt[75(1+4tt2+16)]=ddt(75)+ddt(300tt2+16)=ddt(300tt2+16)

Use the quotient rule for the derivative of the two function f(x) and g(x) is, ddx(fg)=gdfdxfdgdxg2.

dp(t)dt=(t2+16)ddt(300t)(300t)ddt(t2+16)(t2+16)2

Use the simple power rule of derivative ddx(xn)=nxn1.

dp(t)dt=(t2+16)(300)(300t)(2t+0)(t2+16)2=(300t2+4800)600t2(t2+16)2=300t2+4800(t2+16)2

Therefore, the rate of population growth after t years is p'(t)=300t2+4800(t2+16)2.

Now, the rate of change population growth for each year is given as,

Substitute 1 for t in the rate function p'(t)=300t2+4800(t2+16)2.

p(1)=(300(1)2+4800((1)2+16)2)=(300(1)+4800(17)2)15

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