Interpretation of the Derivative — Question 5

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Question 5

The population of a certain city (in thousands) P(t)P(t), as a function of time tt (in years since 2000), is modeled by the function: P(t)=500+40t−0.5t2P(t) = 500 + 40t - 0.5t^2

  • (a) Find the rate of change of the population with respect to time.

  • (b) What is the population growth rate in the year 2010?

  • (c) At what year is the population growth rate zero?

  • (d) Interpret your result from (c) in the context of the population.

Original worksheet page 1: question and worked solution for 3-2-005
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Question 5 - Solution

We are given: P(t)=500+40t−0.5t2P(t) = 500 + 40t - 0.5t^2

(a) Rate of Change of Population:

The rate of change of the population is given by the derivative: P′(t)=ddt(500+40t−0.5t2)=40−tP'(t) = \frac{d}{dt}(500 + 40t - 0.5t^2) = 40 - t

P′(t)=40−t\boxed{P'(t) = 40 - t}

(b) Growth Rate in 2010:

Since tt is years since 2000, 2010 corresponds to t=10t = 10.

P′(10)=40−10=30P'(10) = 40 - 10 = 30

Interpretation: In 2010, the population was increasing at a rate of 30 thousand people per year.

P′(10)=30(thousand people/year)\boxed{P'(10) = 30 \quad \text{(thousand people/year)}}

(c) When is the Population Growth Rate Zero?

Set P′(t)=0P'(t) = 0: 40−t=0⇒t=4040 - t = 0 \Rightarrow t = 40

t=40\boxed{t = 40}

(d) Interpretation:

Since t=40t = 40 corresponds to the year 2000+40=20402000 + 40 = 2040, the population stops growing in 2040.

Conclusion: In 2040, the population reaches its maximum and stops increasing. After that, it begins to decline.

The population stops growing in 2040.\boxed{\text{The population stops growing in 2040.}}

Original worksheet page 2: question and worked solution for 3-2-005

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