/*! 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 76 A population of minnows in a lak... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 76

A population of minnows in a lake is estimated to be 100,000 at the beginning of the year 2010 . Suppose that \(t\) years after the beginning of 2010 the rate of growth of the population \(p(t)\) (in thousands) is given by \(p^{\prime}(t)=(3+0.12 t)^{3 / 2} .\) Estimate the projected population at the beginning of the year 2015

Short Answer

Expert verified
The projected population at the beginning of 2015 is approximately 106,560 minnows.

Step by step solution

01

Calculate Time Interval

The problem asks for the population at the beginning of the year 2015. Since the reference year is 2010, we need to determine the number of years from 2010 to 2015. This is a time interval of 5 years.
02

Set Up the Initial Problem

The initial population at the beginning of 2010 is given as 100,000 minnows, which can be expressed as 100 in thousands (since the rate is given in thousands). You want to find the population at the beginning of 2015, or at time \( t = 5 \).
03

Integrate the Rate of Growth Function

To find the population at time \( t = 5 \), integrate the rate of growth function \( p'(t) = (3 + 0.12t)^{3/2} \) from \( t = 0 \) to \( t = 5 \). This will give you the change in population (in thousands) over this time period.
04

Perform the Integration

The indefinite integral of \((3 + 0.12t)^{3/2}\) can be found using a substitution method, letting \( u = 3 + 0.12t \), which simplifies the integral to \( \frac{1}{0.12} \int u^{3/2} \, du = \frac{1}{0.12} \cdot \frac{2}{5} u^{5/2} + C \). After computing, this gives \( \frac{10}{3} \left(3+0.12t\right)^{5/2} + C \).
05

Evaluate Definite Integral

Evaluate the definite integral from \( t = 0 \) to \( t = 5 \):\[ \left. \frac{10}{3} (3 + 0.12t)^{5/2} \right|_{0}^{5} = \frac{10}{3} (3.6)^{5/2} - \frac{10}{3} (3)^{5/2} \].Calculate this to find the change in population.
06

Calculate the Change in Population

Compute the values:- \( (3 + 0.12 \times 5)^{5/2} = (3.6)^{5/2} \approx 7.164 \)- \( (3)^{5/2} \approx 5.196 \)The change is \( \frac{10}{3} \times (7.164 - 5.196) \approx \frac{10}{3} \times 1.968 \approx 6.56 \)Thus, the population change is approximately 6.56 thousand minnows.
07

Determine the Projected Population

Add the change in population to the initial population:\( 100 + 6.56 \approx 106.56 \)The projected population at the beginning of 2015 is approximately 106,560 minnows.

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 91影视!

Key Concepts

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

Definite Integral
The definite integral is a fundamental concept in calculus. It helps us find the accumulated change over a specified interval. In this case, we are interested in the population of minnows over a period of five years from 2010 to 2015.

When you have a function that describes the rate of growth, such as the rate of growth function for the minnow population, integrating this function over time gives you the total change in population.

To carry out a definite integral, you need to specify the start and end points. Here, these are from time \( t = 0 \) to \( t = 5 \). This interval is used to measure how the population grows over those years. The result of a definite integral is a numerical value that represents this total change, which then can be added to the initial population to determine the population at a specific future time.
Rate of Growth Function
A rate of growth function describes how a population or quantity changes over time. In our problem, the rate of growth function is given by \( p'(t) = (3 + 0.12t)^{3/2} \).

This function is crucial because it models the growth behavior of the minnow population in the lake. The expression \((3 + 0.12t)^{3/2}\) reflects how growth changes as time \( t \) passes. The use of an exponent suggests that the growth accelerates over time, meaning it doesn't just grow at a steady pace, but likely increases more rapidly as time progresses.

To utilize this function for calculating the total change in population, you need to integrate it over the desired time interval. This will give the total number of thousands of minnows added to the population from the start of 2010 to the start of 2015.
Change in Population
The change in population over a given period can be calculated by integrating the rate of growth function between specific time intervals.

In the example provided, the difference between the population in 2015 and the population in 2010 is determined by evaluating the definite integral of the growth rate function from \( t = 0 \) to \( t = 5 \). This involves calculating the values of \((3.6)^{5/2}\) and \((3)^{5/2}\), and finding their difference. The result is then multiplied by the factor \( \frac{10}{3} \) to convert it to thousands of minnows, as detailed in the step-by-step method provided in the solution.

Using this approach, we estimate a change of 6.56 thousand minnows, meaning the projected population at the beginning of 2015 is 106,560 minnows. By understanding this process, you can see the advantage of using calculus to make accurate predictions based on current growth trends.

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

$$ \int x^{2} e^{-2 x^{3}} d x $$

(a) Evaluate the integral \(\int(5 x-1)^{2} d x\) by two methods: first square and integrate, then let \(u=5 x-1\). (b) Explain why the two apparently different answers obtained in part (a) are really equivalent.

(a) Let $$ I=\int_{0}^{a} \frac{f(x)}{f(x)+f(a-x)} d x $$ Show that \(I=a / 2\). [Hint: Let \(u=a-x\), and then note the difference between the resulting integrand and \(1 .]\) (b) Use the result of part (a) to find $$ \int_{0}^{3} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} d x $$ (c) Use the result of part (a) to find $$ \int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x $$

$$ \int \frac{\sec ^{2} x d x}{\sqrt{1-\tan ^{2} x}} $$

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$ V=V_{p} \sin (2 \pi f t) $$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts \((\mathrm{V})\), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). \((1 \mathrm{~Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the \(\boldsymbol{r m s}\) or root-mean-square value of \(V\). By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$ V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}} $$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f\), and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral.] (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{~V}\) at a frequency of \(60 \mathrm{~Hz}\). What is the peak voltage at such an outlet?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.