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Radioactive decay: The amount remaining \(A\), in grams, of a radioactive substance is a function of time \(t\), measured in days since the experiment began. The equation of change for \(A\) is $$ \frac{d A}{d t}=-0.05 A $$ a. What is the exponential growth rate for \(A\) ? b. If initially there are 3 grams of the substance, find a formula for \(A\). c. What is the half-life of this radioactive substance?

Short Answer

Expert verified
a. The rate is -0.05. b. \( A = 3e^{-0.05t} \). c. Half-life is approximately 13.86 days.

Step by step solution

01

Identifying the exponential growth rate

The given equation \( \frac{dA}{dt} = -0.05A \) represents an exponential decay equation. The rate of decay is represented by the coefficient of \( A \), which is \(-0.05\). Therefore, the exponential growth rate \( r \) is \(-0.05\), indicating decay rather than growth.
02

Solving the differential equation

The differential equation \( \frac{dA}{dt} = -0.05A \) is separable. We can rewrite it as \( \frac{dA}{A} = -0.05 dt \). Integrating both sides gives \( \int \frac{1}{A} \, dA = \int -0.05 \, dt \). The integrations yield \( \ln |A| = -0.05t + C \), where \( C \) is a constant of integration.
03

Finding the initial condition

We use the given initial condition, \( A(0) = 3 \), to find the constant \( C \). Substituting \( t = 0 \) and \( A = 3 \) into the integrated equation gives \( \ln 3 = C \). Thus, \( C = \ln 3 \).
04

Expressing the solution in terms of exponential function

Having \( \ln |A| = -0.05t + \ln 3 \), we exponentiate both sides to solve for \( A \), resulting in \( |A| = e^{-0.05t + \ln 3} \). This simplifies to \( A = Ce^{-0.05t} \) where \( C = 3 \). Therefore, the equation for \( A \) is \( A = 3e^{-0.05t} \).
05

Calculating the half-life

The half-life \( T_{1/2} \) is the time at which the quantity \( A(t) \) is half of the original quantity. Thus, setting \( A = \frac{3}{2} \) in \( 3e^{-0.05t} = \frac{3}{2} \), we can solve for \( t \). Dividing by 3, \( e^{-0.05t} = \frac{1}{2} \). Taking the natural logarithm of both sides gives \( -0.05t = \ln \frac{1}{2} \). Solving for \( t \) gives \[ t = \frac{\ln 0.5}{-0.05} = \frac{-0.6931}{-0.05} \approx 13.86 \text{ days}. \]

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Key Concepts

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

Exponential Functions
Exponential functions are a type of mathematical function used to model scenarios where quantities grow or decay at a proportional rate over time. Radioactive decay is a perfect example of exponential decay. It follows a predictable pattern where the quantity of the substance decreases exponentially as time passes.
Simply put, the general form of an exponential function related to decay is described by the equation:
\[ A(t) = A_0 e^{rt} \]
Where:
  • \( A(t) \) is the amount of substance remaining at time \( t \)
  • \( A_0 \) is the initial quantity of the substance
  • \( r \) is the decay rate (a negative number for decay situations)
  • \( e \) is the base of the natural logarithm, approximately equal to 2.718
This form shows how exponential decay allows us to easily predict how much of a material remains after a certain period of time given a constant decay rate. It's also used in other contexts such as population growth, or interest calculations, but the basis remains the same - change at a steady percentage rate.
Differential Equations
Differential equations provide a way to mathematically describe how a quantity changes over time, which makes them essential in modeling natural phenomena such as radioactive decay.
In our given exercise, we encounter the differential equation \( \frac{dA}{dt} = -0.05A \). This is a first-order separable differential equation. Here, \( \frac{dA}{dt} \) stands for the rate of change of the amount with respect to time, and \( -0.05A \) indicates that the rate of change is proportional to the current amount of radioactive substance.
To solve this, we rearrange the equation to allow integration: \[ \frac{dA}{A} = -0.05 \, dt \]By integrating both sides, we can express this relationship through natural logarithms, leading to a solution of the form \( \ln |A| = -0.05t + C \), where \( C \) is an integration constant defined by initial conditions.
Understanding and solving differential equations can help solve problems related to decay, growth, and other dynamic systems.
Half-Life Calculation
Calculating half-life allows us to determine how long it takes for half of a given quantity of a substance to decay. This concept is particularly important in fields like radioactive decay, pharmacokinetics, and physics.
In the case of the radioactive substance from the exercise, we use the exponential decay formula to determine the half-life, \( T_{1/2} \).
We set the equation: \[ A(t) = \frac{A_0}{2} \]
Substituting our solution for \( A(t) = 3e^{-0.05t} \) gives:
  • \( 3e^{-0.05t} = \frac{3}{2} \)
  • Simplifying, \( e^{-0.05t} = \frac{1}{2} \)
  • Taking natural logarithms, we solve: \( -0.05t = \ln \frac{1}{2} \)
  • Finally, \( t = \frac{\ln 0.5}{-0.05} \)
This yields \( t \approx 13.86 \) days, meaning in roughly 13.86 days, the substance will reduce to half of its initial amount. Mastering half-life calculations enables scientists and professionals to predict how substances diminish over time, crucial for safety and effectiveness in applications such as medical treatment and nuclear energy.

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