Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A mold grows at a rate that is proportional to the amount present. Initially there is 3 oz of this mold, and 10 hours later there is 5 oz. (a) How much mold is there at the end of 1 day? (b) When is there 10 oz of the mold?

Short Answer

Expert verified
(a) There will be approximately 8.25 oz of mold at the end of 1 day. (b) There will be 10 oz of the mold after approximately 23.18 hours.

Step by step solution

01

Exponential growth formula

The exponential growth formula is given by: \(A(t) = A_0 e^{kt}\) Where \(A(t)\) is the amount present at time \(t\), \(A_0\) is the initial amount of mold, \(k\) is the constant of proportionality, and \(t\) is the time in hours. In this problem, \(A_0 = 3\) oz and at \(t = 10\) hours, we have \(A(10) = 5\) oz.
02

Find the constant of proportionality (k)

To find the value of \(k\), we will first plug in the given values into the exponential growth formula: \(5 = 3 e^{10k}\) Now, we will solve for \(k\): \(\frac{5}{3} = e^{10k}\) \(10k = \ln{\left(\frac{5}{3}\right)}\) \(k = \frac{\ln{\left(\frac{5}{3}\right)}}{10}\)
03

(a) How much mold is there at the end of 1 day?

Since there are 24 hours in a day, we need to find the amount of mold at \(t = 24\) hours. Plug the value of \(k\) we found in Step 2 into the exponential growth formula and replace \(t\) with \(24\): \(A(24) = 3 e^{\frac{24\ln\left(\frac{5}{3}\right)}{10}}\) Evaluate the expression: \(A(24) \approx 8.25\) oz There will be approximately 8.25 oz of mold at the end of 1 day.
04

(b) When is there 10 oz of the mold?

To find out when there will be 10 oz of mold, plug the value of \(k\) we found in Step 2 into the exponential growth formula and set \(A(t) = 10\): \(10 = 3 e^{\frac{t\ln\left(\frac{5}{3}\right)}{10}}\) Now, we will solve for \(t\): \(\frac{10}{3} = e^{\frac{t\ln\left(\frac{5}{3}\right)}{10}}\) \(t\ln\left(\frac{5}{3}\right)= 10\ln{\left(\frac{10}{3}\right)}\) \(t = \frac{10\ln{\left(\frac{10}{3}\right)}}{\ln{\left(\frac{5}{3}\right)}}\) Evaluate the expression: \(t \approx 23.18\) hours There will be 10 oz of the mold after approximately 23.18 hours.

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 Vaia!

Key Concepts

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

Ordinary Differential Equations
Understanding ordinary differential equations (ODEs) is essential when dealing with exponential growth problems. In the context of mold growth, we describe how the quantity of mold changes over time using these equations. The exponential growth formula is a classic ODE:

\[A(t) = A_0 e^{kt}\]

Here, the equation models how the mold increases as a function of time, taking into account the initial amount and the growth rate. Solving ODEs involves finding a function that satisfies this equation, typically requiring knowledge of calculus. ODEs are fundamental in modeling real-world situations where change is involved, allowing us to predict future states based on current conditions.
Proportional Relationships
A key concept in solving exponential growth problems is understanding proportional relationships. This means that the rate of change of a quantity is directly proportional to the current amount of that quantity. In simpler terms, the bigger the quantity, the faster it grows.
  • This relationship is captured by the constant of proportionality, \(k\), in the exponential growth equation.
  • For mold growth, the more mold exists, the more rapidly it increases, up until it reaches the limits of its environment or resources.
Proportional relationships underlie many natural processes, making this concept incredibly useful in fields such as biology, chemistry, and physics. By understanding this relationship, students can better tackle complex growth problems.
Initial Conditions
Initial conditions refer to the starting values of a problem, which are crucial in determining the particular solution to an ordinary differential equation. In our case, the initial amount of mold, \(A_0 = 3\) oz, acts as a pivotal piece of information.
  • These conditions allow us to solve for other variables in the equation, such as the growth rate \(k\).
  • Knowing the initial amount means we can predict future growth scenarios with accuracy.
Initial conditions set the stage for the entire growth process. Whether it’s mold growing on bread or a population expanding, without them, we cannot have tailored predictions. They provide the "starting line" for any growth-related calculation.
Rate of Change
The rate of change answers the question: how quickly does the mold grow over time? Mathematically represented by \(k\) in the exponential formula, it reflects how proportionality impacts growth.
  • To find \(k\), we used known amounts at different times, allowing us to describe this growth precisely.
  • In our solution, by solving \(\ln\left(\frac{5}{3}\right)\), we determine how fast the mold grows from 3 oz to 5 oz in 10 hours.
Understanding the rate of change helps in comparing different growth scenarios. It gives insight into how an entity evolves, making it indispensable in applications ranging from forecasting economic trends to analyzing population dynamics.

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

An object weighing \(12 \mathrm{lb}\) is placed beneath the surface of a calm lake. The buoyancy of the object is \(30 \mathrm{lb}\); because of this the object begins to rise. If the resistance of the water (in pounds) is numerically equal to the square of the velocity (in feet per second) and the object surfaces in \(5 \mathrm{sec}\), find the velocity of the object at the instant when it reaches the surface.

At 10 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant the temperature of the coffee was \(180^{\circ} \mathrm{F}\), and 10 minutes later it was \(160^{\circ} \mathrm{F}\). Assume the constant temperature of the kitchen was \(70^{\circ} \mathrm{F}\). (a) What was the temperature of the coffee at \(10: 15\) A.m.? (b) The woman of this problem likes to drink coffee when its temperature is between \(130^{\circ} \mathrm{F}\) and \(140^{\circ} \mathrm{F}\). Between what times should she have drunk the coffee of this problem?

A man is pushing a loaded sled across a level field of ice at the constant speed of \(10 \mathrm{ft} / \mathrm{sec}\). When the man is halfway across the ice field, he stops pushing and lets the loaded sled continue on. The combined weight of the sled and its load is \(80 \mathrm{lb}\); the air resistance (in pounds) is numerically equal to \(\frac{3}{4} v\), where \(v\) is the velocity of the sled (in feet per second); and the coefficient of friction of the runners on the ice is \(0.04\). How far will the sled continue to move after the man stops pushing?

The air in a room whose volume is 10,000 cu ft tests \(0.15 \%\) carbon dioxide. Starting at \(t=0\), outside air testing \(0.05 \%\) carbon dioxide is admitted at the rate of 5000 cu \(\mathrm{ft} / \mathrm{min}\). (a) What is the percentage of carbon dioxide in the air in the room after \(3 \mathrm{~min} ?\) (b) When does the air in the room test \(0.1 \%\) carbon dioxide.

Assume that the population of a certain city increases at a rate proportional to the number of inhabitants at any time. If the population doubles in 40 years, in how many years will it triple?

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.

Sign-up for free