/*! 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 7 Suppose that mold grows at a rat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 7

Suppose that mold grows at a rate proportional to the amount present. If there are initially \(500 \mathrm{~g}\) of mold and \(6 \mathrm{~h}\) later there are \(600 \mathrm{~g}\), determine the amount of mold present after one day. When is the amount of mold \(1000 \mathrm{~g}\) ?

Short Answer

Expert verified
After one day, the amount of mold is approximately 1036.8 g. The mold reaches 1000 g in approximately 22.52 hours.

Step by step solution

01

Identify constants and initial conditions

Let the initial amount of mold be denoted as \(M_0 = 500 \, \text{g}\). Let the amount of mold at time \( t \) be denoted as \(M(t)\). Given \(M(6) = 600 \, \text{g}\). The mold grows at a rate proportional to the amount present, so we use the equation \( \frac{dM}{dt} = kM \) where \( k \) is the constant of proportionality.
02

Solve the differential equation

The equation \( \frac{dM}{dt} = kM \) is separable: \( \frac{1}{M} dM = k \, dt \). Integrating both sides, we get \( \ln|M| = kt + C \) where \( C \) is the constant of integration. Exponentiating both sides, we have \( M = e^{kt+C} = e^C e^{kt} = M_0 e^{kt} \). Therefore, the solution to the differential equation is \( M(t) = M_0 e^{kt} \).
03

Determine the value of k

Using the given conditions \( M(0) = 500 \, \text{g} \) and \( M(6) = 600 \, \text{g} \): \( 600 = 500 e^{6k} \). Solving for \( k \): \( \frac{600}{500} = e^{6k} \Rightarrow \frac{6}{5} = e^{6k} \Rightarrow \ln\left(\frac{6}{5}\right) = 6k \Rightarrow k = \frac{1}{6} \ln\left(\frac{6}{5}\right) \).
04

Calculate the amount of mold after one day

Substitute \( k = \frac{1}{6} \ln\left(\frac{6}{5}\right) \) and \( t = 24 \, \text{h} \) into \( M(t) = M_0 e^{kt} \): \( M(24) = 500 e^{24 \left( \frac{1}{6} \ln\left( \frac{6}{5} \right) \right) } \). Simplify this expression to find \( M(24) = 500 e^{4 \ln\left( \frac{6}{5} \right)} = 500 \left( \frac{6}{5} \right)^4 \). Calculate \( \left( \frac{6}{5} \right)^4 \) to get \( M(24) \).
05

Determine when the amount of mold is 1000 g

To find the time when the mold is 1000 g, let \( M(t) = 1000 \) and substitute into \( M(t) = 500 e^{kt} \): \( 1000 = 500 e^{kt} \Rightarrow 2 = e^{kt} \Rightarrow \ln(2) = kt \Rightarrow t = \frac{\ln(2)}{k} \). Using \( k = \frac{1}{6} \ln\left( \frac{6}{5} \right) \), calculate \( t = \frac{\ln(2)}{ \frac{1}{6} \ln\left( \frac{6}{5} \right) } = 6 \frac{ \ln(2)}{\ln(6/5)} \) to determine the time.

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.

Proportional Growth
Proportional growth refers to a process where the growth rate of a quantity is directly proportional to its current size. In simple terms, the larger the amount, the faster it grows. This concept is very common in natural phenomena such as population growth, radioactive decay, and in this case, mold growth. The mathematical representation of this concept involves a constant of proportionality, denoted as \(k\), leading to the equation \(\frac{dM}{dt} = kM\). This means that the rate of change of the mold's mass at any time is proportional to its current mass. Therefore, if you double the mass of the mold, its growth rate will also double.
Separable Differential Equations
Separable differential equations are a class of differential equations that can be simplified by separating variables. In the mold growth problem, we start with the equation \(\frac{dM}{dt} = kM\). To solve this, we separate the variables so that all terms involving \(M\) are on one side and all terms involving \(t\) are on the other. This gives us \(\frac{1}{M} dM = k \, dt\). By integrating both sides, we obtain the natural logarithm of \(M\) on one side and a linear function of \(t\) on the other, leading to \(\ln|M| = kt + C\). This step transforms the equation into a form that can be solved with basic integration techniques.
Exponential Functions
Exponential functions are essential in describing processes with constant proportional growth. In solving the mold growth problem, after integration, we get \(\ln|M| = kt + C\). By exponentiating both sides, we convert the logarithmic form back to the exponential form: \(M = e^{kt + C} = e^C \, e^{kt}\). Setting \(M_0 = e^C\) simplifies this to \(M(t) = M_0e^{kt}\). This function shows that the amount of mold at any time \(t\) is determined by the initial amount \(M_0\) and grows exponentially over time with a growth rate of \(k\).
Initial Conditions
Initial conditions are the values of the function and its derivatives at the start of the problem. These conditions help determine the specific solution to the differential equation. In our exercise, the initial condition is that the mold mass at \(t = 0\) is \(500 \, \mathrm{g}\), written mathematically as \(M(0) = 500\). This allows us to solve for constants in our exponential function. Additionally, we are given that \(M(6) = 600 \, \mathrm{g}\), which helps in calculating the constant of proportionality \(k\). These initial conditions ensure the solution fits the specific situation.
Constant of Proportionality
The constant of proportionality \(k\) is fundamental in determining the rate at which mold grows over time. To find \(k\), we use the information provided: initially \(500 \, \mathrm{g}\) of mold grows to \(600 \, \mathrm{g}\) after 6 hours. Inserting these values into the equation \(600 = 500e^{6k}\) and solving for \(k\) gives \(k = \frac{1}{6} \ln\left(\frac{6}{5}\right)\). This value of \(k\) is then used to make predictions about future growth, such as finding the amount of mold after one day or determining when it will reach \(1000 \, \mathrm{g}\).

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

Suppose that an object of mass \(1 \mathrm{~kg}\) is thrown with a downward initial velocity of \(5 \mathrm{~m} / \mathrm{s}\) and is subjected to an air resistance equivalent to the instantaneous velocity. (a) Find the velocity of the object and the distance fallen at time \(t\). (b) How far does the object drop after \(5 \mathrm{~s}\) ?

(Tumor and Organism Growth) Ludwig von Bertalanffy (1901-1972) made valuable contributions to the study of organism growth, including the growth of tumors, based on the relationship between body size of the organism and the metabolic rate. He theorized that weight is directly proportional to volume and that the metabolic rate is proportional to surface area. This indicates that surface area is proportional to \(V^{2 / 3}\) because \(V\) is measured in cubic units and \(S\) in square units. Therefore, Bertalanffy studied the IVP \(d V / d t=a V^{2 / 3}-b V, V(0)=V_{0}\). (a) Solve this initial value problem to show that volume is given by \(V(t)=\) \(\left[\frac{a}{b}-\left(\frac{a}{b}-V_{0}^{1 / 3}\right) e^{-b t / 3}\right]^{3}\). (b) Find \(\lim _{t \rightarrow \infty} V(t)\) and explain what the limit represents. (c) This initial value problem is similar to that involving the Logistic equation \(d V / d t=a V^{2}-b V, V(0)=V_{0}\). Compare the limiting volume between the Logistic equation and the model proposed by Bertalanffy in the cases when \(a / b>1\) and when \(a / b<1\).

After 10 days, \(800 \mathrm{~g}\) of a radioactive element remain, and after 15 days, \(560 \mathrm{~g}\) remain. What is the half-life of this element?

Find the amount of salt \(y(t)\) in a tank with initial volume \(V_{0}\) gallons of liquid and \(y_{0}\) pounds of salt using the given conditions. (a) \(R_{1}=4 \mathrm{gal} / \mathrm{min}, R_{2}=4 \mathrm{gal} / \mathrm{min}, S_{1}=2\) \(\mathrm{lb} / \mathrm{gal}, V_{0}=400, y_{0}=0\). (b) \(R_{1}=4 \mathrm{gal} / \mathrm{min}, R_{2}=2 \mathrm{gal} / \mathrm{min}, S_{1}=1\) \(\mathrm{lb} / \mathrm{gal}, V_{0}=500, y_{0}=20\). (c) \(R_{1}=4 \mathrm{gal} / \mathrm{min}, R_{2}=6 \mathrm{gal} / \mathrm{min}, S_{1}=1\) \(\mathrm{lb} / \mathrm{gal}, V_{0}=600, y_{0}=100\). Describe how the values of \(R_{1}\) and \(R_{2}\) affect the volume of liquid in the tank.

An object that weighs \(48 \mathrm{lb}\) is released from rest at the top of a plane metal slide that is inclined \(30^{\circ}\) to the horizontal. Air resistance \((\mathrm{lb})\) is numerically equal to one-half the velocity (ft/s), and the coefficient of friction is \(\mu=1 / 4\). Using Newton's second law of motion by summing the forces along the surface of the slide, we find the following forces: (a) the component of the weight parallel to the slide: \(F_{1}=48 \sin 30^{\circ}=24\); (b) the component of the weight perpendicular to the slide: \(N=48 \cos 30^{\circ}=\) \(24 \sqrt{3}\); (c) the frictional force (against the motion of the object): \(F_{2}=-\mu N=-\frac{1}{4} \times\) \(24 \sqrt{3}=-6 \sqrt{3}\); and (d) the force due to air resistance (against the motion of the object): \(F_{3}=-\frac{1}{2} v\). Because the mass of the object is \(m=\) \(48 / 32=3 / 2\), we find that the velocity of the object satisfies the initial value problem \(m d v / d t=F_{1}+F_{2}+F_{3}\) or $$ \frac{3}{2} \frac{d v}{d t}=24-6 \sqrt{3}-\frac{1}{2} v, \quad v(0)=0 . $$ Solve this problem for \(v(t)\). Determine the distance traveled by the object at time \(t\), \(x(t)\), if \(x(0)=0\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

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.