Question

the population of toledo, ohio, in the year 2000 was approximately 510,000. assume the population is increasing at a rate of 4.6 % per year.
a. write the exponential function that relates the total population, ( p(t) ), as a function of ( t ), the number of years since 2000.
( p(t) =)

b. use part a. to determine the rate at which the population is increasing in ( t ) years. use exact expressions.
( p(t) =) people per year

c. use part b. to determine the rate at which the population is increasing in the year 2006. round to the nearest person per year
( p(6) =) people per year

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Explanation:

Step1: Write the exponential function

The general form of an exponential growth function is \(P(t)=P_0(1 + r)^t\), where \(P_0\) is the initial population and \(r\) is the growth rate.
Given \(P_0 = 510000\) and \(r=0.046\), so \(P(t)=510000(1 + 0.046)^t=510000(1.046)^t\).

Step2: Differentiate the function

Using the formula \(\frac{d}{dt}(a^t)=a^t\ln(a)\) (where \(a = 1.046\) and \(y = P(t)\)), \(P^{\prime}(t)=\frac{d}{dt}(510000(1.046)^t)\).
By the constant - multiple rule \((cf(x))^\prime=c\cdot f^\prime(x)\) (\(c = 510000\), \(f(t)=(1.046)^t\)), we have \(P^{\prime}(t)=510000\times(1.046)^t\ln(1.046)\).

Step3: Calculate the value for \(t = 6\)

Substitute \(t = 6\) into \(P^{\prime}(t)\).
\(P^{\prime}(6)=510000\times(1.046)^6\ln(1.046)\).
First, \((1.046)^6\approx1.300\) (using a calculator for \((1.046)^6=1.046\times1.046\times1.046\times1.046\times1.046\times1.046\)), \(\ln(1.046)\approx0.045\).
\(P^{\prime}(6)=510000\times1.300\times0.045\).
\(P^{\prime}(6)=510000\times0.0585 = 29835\).

Answer:

a. \(P(t)=510000(1.046)^t\)
b. \(P^{\prime}(t)=510000(1.046)^t\ln(1.046)\)
c. \(P^{\prime}(6)\approx29835\)