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Chapter 3: Problem 23

When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time \(t\) (hours) was \(b=10^{6}+10^{4} t-10^{3} t^{2} .\) Find the growth rates at a. \(t=0\) hours. b. \(t=5\) hours. c. \(t=10\) hours.

Short Answer

Expert verified
Growth rates: at \(t=0\) is \(10^4\), at \(t=5\) is \(0\), and at \(t=10\) is \(-10^4\).

Step by step solution

01

Understand the Problem

We are given a population function \(b = 10^6 + 10^4 t - 10^3 t^2\) representing the size of a bacterium population at time \(t\) hours. We need to find the growth rate at specific times by differentiating this function with respect to time \(t\).
02

Differentiate the Population Function

To find the growth rate, differentiate \(b = 10^6 + 10^4 t - 10^3 t^2\) with respect to \(t\). The derivative \(b'(t)\) will give us the rate of growth:\[b'(t) = \frac{d}{dt}(10^6 + 10^4 t - 10^3 t^2) = 10^4 - 2 \times 10^3 t.\]
03

Calculate the Growth Rate at t=0

Substitute \(t=0\) into the derivative to find the growth rate at \(t=0\):\[b'(0) = 10^4 - 2 \times 10^3 \times 0 = 10^4.\]Thus, the growth rate at \(t=0\) is \(10^4\).
04

Calculate the Growth Rate at t=5

Substitute \(t=5\) into the derivative to find the growth rate at \(t=5\):\[b'(5) = 10^4 - 2 \times 10^3 \times 5 = 10^4 - 10^4 = 0.\]This means the growth rate at \(t=5\) is \(0\), indicating no growth.
05

Calculate the Growth Rate at t=10

Substitute \(t=10\) into the derivative to find the growth rate at \(t=10\):\[b'(10) = 10^4 - 2 \times 10^3 \times 10 = 10^4 - 2 \times 10^4 = -10^4.\]This implies the growth rate at \(t=10\) is \(-10^4\), indicating decline.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation Techniques
Differentiation is a core concept in calculus that allows us to find the rate at which a function changes at any point. When dealing with functions representing physical phenomena, like population size, differentiation helps us understand how quickly these quantities are changing.
To differentiate a given function, we apply rules such as the power rule, which states that the derivative of a term like \(ax^n\) is \(nax^{n-1}\), where \(a\) and \(n\) are constants and \(x\) is the variable.
This technique is crucial in understanding changing rates in various contexts:
  • By differentiating the population function \(b = 10^6 + 10^4 t - 10^3 t^2\), we get its derivative, \(b'(t) = 10^4 - 2\times 10^3 t\).
  • This represents the growth rate or the rate of change of the bacterium population over time.
  • By knowing the derivative formula, we can substitute different values of \(t\) to find the precise growth rate at those points in time.
Growth Rate Calculations
Calculating growth rates involves determining how much a population or any quantity grows over time. In mathematical terms, this means finding the derivative of a function that represents the population.
For our example with bacteria growth:
  • The derivative, \(b'(t) = 10^4 - 2\times 10^3 t\), shows how the growth rate changes with time \(t\).
  • At \(t=0\), the growth rate \(b'(0) = 10^4\). This positive value indicates the population is increasing rapidly since initially, no influence of time is affecting the growth.
  • For \(t=5\), \(b'(5) = 0\). At this point, the growth rate is zero, suggesting the population has reached a temporary steady state.
  • When \(t=10\), \(b'(10) = -10^4\), a negative growth rate signals the population begins to decline rapidly due to the overwhelming effects reflected in the negative term \(-2\times 10^3 t\).
This illustrates how a function's derivative provides insights into the behavior of dynamic systems over time.
Population Modeling
Population modeling is an important application of calculus that helps us understand how populations change. Such models can be linear or non-linear depending on factors like birth rates, death rates, and other environmental influences.
In the context of the exercise, the function \(b = 10^6 + 10^4 t - 10^3 t^2\) incorporates both growth and decline, showing an initially growing population that eventually stops and declines due to additional factors represented by the term \(-10^3 t^2\).
  • Initially, the population grows as indicated by the \(10^4 t\) term, representing a simple linear growth.
  • However, as time progresses, the \(-10^3 t^2\) term introduces a quadratic effect that models decline due to limited resources or other negative factors.
  • Understanding these elements through derivatives and model equations is crucial for predicting future population sizes and answering 'what if' scenarios in population science.
These models help in planning and managing biological resources and understanding ecological dynamics.

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Most popular questions from this chapter

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