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When sugar is dissolved in water, the amount \(A\) that remains undissolved after \(t\) minutes satisfies the differential equation \(d A / d t=-k A(k>0) .\) If \(25 \%\) of the sugar dissolves after 1 min, how long does it take for half of the sugar to dissolve?

Short Answer

Expert verified
It takes approximately 2.415 minutes for half of the sugar to dissolve.

Step by step solution

01

Understand the differential equation

The equation \( \frac{dA}{dt} = -kA \) describes exponential decay. The rate at which the sugar dissolves is proportional to the amount of sugar remaining undissolved.
02

Solve the differential equation

The equation \( \frac{dA}{dt} = -kA \) is separable. Integrating both sides with respect to time, we get \( \int \frac{1}{A} \, dA = \int -k \, dt \), which leads to \( \ln|A| = -kt + C \). Exponentiating both sides, we have \( A(t) = A_0 e^{-kt} \), where \( A_0 \) is the initial amount of sugar.
03

Calculate the constant of proportionality

We use the condition that 25% of the sugar dissolves after 1 minute. This means \( A(1) = 0.75A_0 \). Plugging into the equation, we get \( 0.75A_0 = A_0 e^{-k} \), simplifying gives \( 0.75 = e^{-k} \). Taking the natural logarithm of both sides, we find \( k = -\ln(0.75) \).
04

Find the time for 50% dissolution

We want \( A(t) = 0.5A_0 \). Using the formula from Step 2, we have \( 0.5A_0 = A_0 e^{-kt} \), which simplifies to \( 0.5 = e^{-kt} \). Taking the natural logarithm of both sides, \( \ln(0.5) = -kt \). Solve for \( t \) using \( k \) from Step 3: \( t = \frac{-\ln(0.5)}{-\ln(0.75)} \).
05

Simplify the expression for time

Compute \( t = \frac{\ln(0.5)}{\ln(0.75)} \) using a calculator. The natural logarithm of \( 0.5 \) is approximately \(-0.693\) and of \( 0.75 \) is approximately \(-0.287\). Thus, \( t \approx \frac{-0.693}{-0.287} \approx 2.415 \).

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Key Concepts

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

Exponential Decay
Exponential decay is a process where a quantity decreases over time at a rate proportional to its current value. In simple terms, the larger the quantity, the faster it decreases.
For example, when sugar dissolves in water, it follows an exponential decay model. The differential equation representing this is \( \frac{dA}{dt} = -kA \), where \( A \) is the amount of sugar undissolved, \( t \) is time, and \( k \) is a positive constant that represents the decay rate.
  • If large amounts of sugar are present, it will dissolve quickly at first.
  • As the amount decreases, the rate slows down proportionately.
Exponential decay is ubiquitous in nature, seen in processes like radioactive decay, cooling of objects, and more.
This concept illustrates how rapid changes can slow as quantities get smaller, making exponential decay vital to understanding many real-world phenomena.
Separable Differential Equations
A separable differential equation is a form of differential equation that can be separated into two separate integrals. This makes it easier to solve mathematically.
The equation \( \frac{dA}{dt} = -kA \) is an example. By manipulating this equation, we move all the \( A \)-related terms to one side and the \( t \)-related terms to the other:
  • Rearrange: \( \frac{1}{A} \, dA = -k \, dt \)
  • Now, each side can be integrated separately.
By doing this, we essentially 'peel apart' the variables, allowing us to solve for \( A \) independently of \( t \).
The ability to separate variables is a powerful tool in solving differential equations, allowing us to handle a broad spectrum of problems with simplicity.
Integrating Factor
An integrating factor is a mathematical tool used to solve linear differential equations, especially when an equation isn't easily separable.
While our example equation \( \frac{dA}{dt} = -kA \) was separable, other equations are not so straightforward. In these cases, an integrating factor allows for simplification.
The integrating factor, denoted typically as \( \mu(t) \), is chosen based on the differential equation form and is used to transform it into an easily integrable form. This method allows us to:
  • Multiply the entire differential equation by \( \mu(t) \) to make the left side an exact derivative.
  • Solve directly through integration.
Using an integrating factor provides a structured approach to tackling more complex differential equations beyond simple separable ones.
Natural Logarithm
A natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is approximately 2.718.
It is an essential function in mathematics, especially in calculus and complex situations involving growth and decay.
In the example, the natural logarithm appears when solving the differential equation. After integrating both sides, we get:
  • \( \ln |A| = -kt + C \)
  • Using properties of logarithms, we simplify to determine other variables as needed.
The natural logarithm helps streamline solving differential equations because it simplifies expressions involving exponential functions.
Its properties are vital in calculus, enabling techniques to solve equations involving growth, decay, and compounding processes.

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