/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Growth of Bacteria The growth of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Growth of Bacteria The growth of bacteria makes it necessary to time-date some food products so that they will be sold and consumed before the bacteria count is too high. Suppose for a certain product the number of bacteria present is given by $$ f(t)=500 e^{0.1 t} $$ where \(t\) is time in days and the value of \(f(t)\) is in millions. Find the number of bacteria present at each time. (a) 2 days (b) 4 days (c) 1 week

Short Answer

Expert verified
After 2 days: 610.7 million, 4 days: 745.9 million, 1 week: 1006.9 million.

Step by step solution

01

- Identify the formula

The formula given for the number of bacteria is: \[ f(t) = 500 e^{0.1 t} \]where \( t \) is the time in days.
02

- Evaluate for 2 days

Substitute \( t = 2 \) in the formula to find the number of bacteria after 2 days:\[ f(2) = 500 e^{0.1 \times 2} \]Calculate the exponent and the expression:\[ f(2) = 500 e^{0.2} \]Using a calculator, find \( e^{0.2} \approx 1.2214 \):\[ f(2) = 500 \times 1.2214 = 610.7 \]Therefore, the number of bacteria after 2 days is approximately 610.7 million.
03

- Evaluate for 4 days

Substitute \( t = 4 \) in the formula to find the number of bacteria after 4 days:\[ f(4) = 500 e^{0.1 \times 4} \]Calculate the exponent and the expression:\[ f(4) = 500 e^{0.4} \]Using a calculator, find \( e^{0.4} \approx 1.4918 \):\[ f(4) = 500 \times 1.4918 = 745.9 \]Therefore, the number of bacteria after 4 days is approximately 745.9 million.
04

- Evaluate for 1 week

Substitute \( t = 7 \) in the formula to find the number of bacteria after a week (7 days):\[ f(7) = 500 e^{0.1 \times 7} \]Calculate the exponent and the expression:\[ f(7) = 500 e^{0.7} \]Using a calculator, find \( e^{0.7} \approx 2.0138 \):\[ f(7) = 500 \times 2.0138 = 1006.9 \]Therefore, the number of bacteria after 7 days (1 week) is approximately 1006.9 million.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

exponential functions
An exponential function is a type of function where a constant base is raised to a variable exponent. It’s generally written in the form \( f(x) = a \cdot b^{x} \) where \( a \) is a constant, \( b \) is the base and \( x \) is the variable exponent. These functions are characterized by their rapid increase or decrease. In the context of bacteria growth, the base of the exponential function is the constant \( e \) (Euler's number, approximately 2.718), and it models how the bacteria population increases over time. Exponential functions frequently appear in natural processes, like population growth, radioactive decay, and even compound interest. They are crucial for understanding how quantities can grow or decline quickly over time.
time-dependent growth
Time-dependent growth refers to how the quantity of something (like bacteria) changes over time. In our exercise, the function \( f(t) = 500 e^{0.1 t} \) models the growth of bacteria over days. Here are the key aspects to understand:
  • \( t \) represents time, measured in days.
  • The expression \( e^{0.1 t} \) describes how the growth rate influences the bacteria count as time passes.
  • \( f(t) \) gives us the number of bacteria (in millions).
For example, after 2 days, 4 days, or 1 week, you can plug in the values of \( t \) into the formula to see how the bacteria count changes. Therefore, understanding time-dependent growth helps us predict and manage processes that change over time, ensuring safety and efficiency in various fields like food preservation and medicine.
natural exponential function
The natural exponential function involves the constant \( e \), which is approximately equal to 2.718. It is fundamental in mathematics and widely used in modeling natural processes. The general form of a natural exponential function is \( f(t) = a e^{kt} \), where:
  • \( a \) is a constant that represents the initial quantity.
  • \( e \) is Euler's number.
  • \( k \) is the growth rate.
  • \( t \) is time.
In the bacteria growth exercise, the function \( f(t) = 500 e^{0.1 t} \) uses \( k = 0.1 \), indicating the growth rate of the bacteria population per day. The term \( 500 \) represents the initial bacteria count in millions. The natural exponential function provides a precise and efficient way to model real-life growth phenomena, making it indispensable in various scientific and engineering disciplines.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At the World Championship races held at Rome's Olympic Stadium in \(1987,\) American sprinter Carl Lewis ran the 100 -m race in 9.86 sec. His speed in meters per second after \(t\) seconds is closely modeled by the function $$f(t)=11.65\left(1-e^{-t / 1.27}\right)$$ (Source: Banks, Robert B., Towing Icebergs, Falling Dominoes, and Other Adventures in Applied Mathematics, Princeton University Press.) (a) How fast was he running as he crossed the finish line? (b) After how many seconds was he running at the rate of \(10 \mathrm{m}\) per sec?

Decay of Radium Find the half-life of radium- 226 , which decays according to the function \(A(t)=A_{0} e^{-0.00043 t},\) where \(t\) is time in years.

(a) Explain why a polynomial function of even degree cannot have an inverse. (b) Explain why a polynomial function of odd degree may not be one-to-one.

Population Decline A midwestern city finds its residents moving to the suburbs. Its population is declining according to the function defined by $$ P(t)=P_{0} e^{-0.04 t} $$ where \(t\) is time measured in years and \(P_{0}\) is the population at time \(t=0 .\) Assume that \(P_{0}=1,000,000\) (a) Find the population at time \(t=1\) (b) Estimate the time it will take for the population to decline to \(750,000\). (c) How long will it take for the population to decline to half the initial number?

(Modeling) Solve each problem. See Example 11 . Employee Training A person learning certain skills involving repetition tends to learn quickly at first. Then learning tapers off and skill acquisition approaches some upper limit. Suppose the number of symbols per minute that a person using a keyboard can type is given by $$ f(t)=250-120(2.8)^{-0.5 t} $$ where \(t\) is the number of months the operator has been in training. Find each value. (a) \(f(2)\) (b) \(f(4)\) (c) \(f(10)\) (d) What happens to the number of symbols per minute after several months of training?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.