/*! 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 15 Dog Weight For the first 9 month... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 15

Dog Weight For the first 9 months of life, the average weight \(w,\) in pounds, of a certain breed of dog increases at a rate that is inversely proportional to time, \(t,\) in months. A 1 -month-old puppy weighs 6 pounds, and a 9 -monthold puppy weighs 80 pounds. a. Write a differential equation describing the rate of change of the weight of the puppy. b. Give the particular solution for this differential equation on the basis of the information given. c. Estimate the weight of the puppy at 3 months and at 6 months. d. Why does this differential equation describe weight gain for only 8 months instead of for the life span of the dog?

Short Answer

Expert verified
Differential equation: \( \frac{dw}{dt} = \frac{k}{t} \). Weight equation: \( w = \frac{74}{\ln(9)} \ln(t) + 6 \). Use this to find weight at given months. Only valid for 9 months due to growth changes.

Step by step solution

01

Define the Differential Equation

Since the rate of change of weight \( w \) is inversely proportional to time \( t \), we can write this as \( \frac{dw}{dt} = \frac{k}{t} \), where \( k \) is a constant. This is the form of the differential equation we need.
02

Solve the Differential Equation

To find the solution, integrate both sides of the equation. \[\int dw = \int \frac{k}{t} \, dt\]This gives us \( w = k \ln(t) + C \), where \( C \) is an integration constant.
03

Apply Initial Conditions

We know that at \( t = 1 \), \( w = 6 \). Plugging these values into the equation \( w = k \ln(t) + C \), we get:\[6 = k \ln(1) + C\]Which simplifies to \( 6 = C \) since \( \ln(1) = 0 \).
04

Determine the Value of k

Use the condition at \( t = 9 \), where \( w = 80 \):\[80 = k \ln(9) + 6\]Solving for \( k \), we find:\[k = \frac{80 - 6}{\ln(9)}\]
05

Write the Particular Solution

Plugging in \( C = 6 \) and \( k = \frac{74}{\ln(9)} \) back into the general solution, we get:\[w = \frac{74}{\ln(9)} \ln(t) + 6\]
06

Estimate the Weight at 3 Months

Substitute \( t = 3 \) into the equation:\[w = \frac{74}{\ln(9)} \ln(3) + 6\]
07

Estimate the Weight at 6 Months

Substitute \( t = 6 \) into the equation:\[w = \frac{74}{\ln(9)} \ln(6) + 6\]
08

Explain Time Limitation of Model

This differential equation only applies up to 9 months because the inverse relationship model does not reflect how growth rates naturally slow and eventually cease as the dog approaches its full adult size.

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.

Inverse Proportionality
In mathematics, inverse proportionality describes a relationship where one variable increases while the other decreases. In the context of this exercise, the weight ( w ) of the puppy grows at a rate inversely related to time ( t ).
  • Imagine two quantities: weight and time.
  • If weight increases proportionally to a decrease in time, each unit of time has a smaller effect.
This is expressed as:\[\frac{dw}{dt} = \frac{k}{t}\] where k is a constant. It's like saying, "As the clock ticks, the rate of gaining weight gradually decreases." Over time, this affects how rapidly or slowly the pup's weight changes.
Integration
Integration is a powerful mathematical tool used to find a function when its derivative is known. When we integrate, we essentially reverse the process of differentiation to find the original function.
  • The differential equation given was: \( \frac{dw}{dt} = \frac{k}{t} \).
  • To solve it, we integrate: \( \int dw = \int \frac{k}{t} \, dt \).
  • This yields: \( w = k \ln(t) + C \).
Here C is the integration constant, which we'll determine using provided values. The function we find describes how weight changes over time.
Initial Conditions
Initial conditions help us find specific values for constants in general solutions of differential equations. They anchor the equation to known real-world values at certain points.
  • At one month, the weight ( w ) is 6 pounds. At this point, \( t = 1 \).
  • By substituting these values into our weight equation, \( w = k \ln(t) + C \), we get \( 6 = C \) since \( \ln(1) = 0 \).
  • Another condition is at nine months, where the puppy weighs 80 pounds. Joining this info gives more insight into k : \( 80 = k \ln(9) + 6 \).
Solving for k adjusts our general solution to reflect these precise conditions, allowing a more accurate description of the dog's growth at varying times.
Weight Estimation
Weight estimation involves using an established model to predict outcomes at different times. With the differential equation, we can project the puppy's weight for any given month.
  • Substituting t = 3 provides an estimated weight at three months: \( w = \frac{74}{\ln(9)} \ln(3) + 6 \).
  • Similarly, for six months: \( w = \frac{74}{\ln(9)} \ln(6) + 6 \).
This step transforms theoretical concepts into practical predictions. However, this equation eventually loses applicability beyond nine months since it hinges on an oversimplified pattern of growth鈥攄ictated solely by time and not accounting for natural maturity limits.

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

Worldwide Cropland The amount of arable and permanent cropland worldwide increased at a slow but relatively steady rate of 0.0342 million square kilometers per year between 1970 and \(1990 .\) In 1980 there were 14.17 million square kilometers of cropland. (Source: Ronald Bailey, ed., The True State of the Planet, New York: The Free Press for the Competitive Enterprise Institute, 1995) a. Write a differential equation representing the growth of cropland. b. Write a general solution for the differential equation in part \(a\) c. Write the particular solution for the amount of cropland. d. Use the equations to estimate the rate of change of cropland in 1970 and in 1990 and the amount of cropland in those years.

Identify the differential equation as one that can be solved using only antiderivatives or as one for which separation of variables is required. Then find a general solution for the differential equation. \(\frac{d y}{d x}=\frac{k}{y}\)

A demand function and \(a\) supply function for the same commodity is given. a. Locate the shutdown point. Write a sentence of interpretation for this point. b. Locate the point of market equilibrium. Write a sentence of interpretation for this point. \(D(p)=50-2 p\) hundred units: \(S(p)=\left\\{\begin{array}{ll}0 & \text { for } p<10 \\ 0.1 p^{2} & \text { for } p \geq 10\end{array}\right.\) hundred units; \(p\) dollars per unit

Write a sentence of interpretation for the probability statement in the context of the given situation. \(P(a \geq 2)=0.25,\) where \(a\) is the age, in years, of a car rented from Hertz at the Los Angeles airport on \(12 / 28 / 2013 .\)

The demand for ceiling fans can be modeled as \(D(p)=25.92\left(0.996^{p}\right)\) thousand ceiling fans where \(p\) is the price (in dollars) of a ceiling fan. a. According to the model, is there a price above which consumers will no longer purchase fans? If so, what is it? b. Calculate the amount that consumers are willing and able to spend to purchase 18 thousand ceiling fans. c. How many fans will consumers purchase when the market price is \(\$ 100\) d. Calculate the consumers' surplus when the market price is \(\$ 100\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.