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Chapter 3: Problem 6

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.

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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鈥檚 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.

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Most popular questions from this chapter

A large tank initially contains 200 gal of brine in which 15 lb of salt is dissolved. Starting at \(t=0\), brine containing 4 lb of salt per gallon flows into the tank at the rate of \(3.5 \mathrm{gal} / \mathrm{min} .\) The mixture is kept uniform by stirring and the well-stirred mixture leaves the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). (a) How much salt is in the tank at the end of one hour? (b) How much salt is in the tank when the tank contains only 50 gal of brine?

An object weighing \(32 \mathrm{lb}\) is released from rest \(50 \mathrm{ft}\) above the surface of a calm lake. Before the object reaches the surface of the lake, the air resistance (in pounds) is given by \(2 v\), where \(v\) is the velocity (in feet per second). After the object passes beneath the surface, the water resistance (in pounds) is given by \(6 v\). Further, the object is then buoyed up by a buoyancy force of \(8 \mathrm{lb}\). Find the velocity of the object 2 sec after it passes beneath the surface of the lake.

A given family of curves is said to be self-orthogonal if its family of orthogonal trajectories is the same as the given family. Show that the family of parabolas \(y^{2}=2 c x+c^{2}\) is self orthogonal.

A tank initially contains 100 gal of brine in which there is dissolved \(20 \mathrm{lb}\) of salt. Starting at time \(t=0\), brine containing \(3 \mathrm{lb}\) of dissolved salt per gallon flows into the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. (a) How much salt is in the tank at the end of \(10 \mathrm{~min}\) ? (b) When is there \(160 \mathrm{lb}\) of salt in the tank?

A case of canned milk weighing \(24 \mathrm{lb}\) is released from rest at the top of a plane metal slide which is \(30 \mathrm{ft}\) long and inclined \(45^{\circ}\) to the horizontal. Air resistance (in pounds) is numerically equal to one-third the velocity (in feet per second) and the coefficient of friction is \(0.4\). (a) What is the velocity of the moving case 1 sec after it is released? (b) What is the velocity when the case reaches the bottom of the slide?
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