Problem 32 - Calculus

Question

A circular oil spill grows at a rate given by the differential equation \(d r / d t=k / r,\) where \(r\) represents the radius of the spill in feet, and time is measured in hours. If the radius of the spill is 400 feet 16 hours after the spill begins, what is the value of \(k ?\) Include units in your answer.

Step-by-Step Solution

Answer

The value of \( k \) is 5000 feet squared per hour.

Step 1: Interpret the Differential Equation

We are given the equation \( \frac{dr}{dt} = \frac{k}{r} \). This equation means that the rate of change of the radius \( r \) with respect to time \( t \) is proportional to the inverse of the radius.

Step 2: Separate Variables

To solve the differential equation, we'll rearrange terms to separate the variables. Multiply both sides by \( r \) to get \( r \frac{dr}{dt} = k \).

Step 3: Integrate Both Sides

Integrate each side of the equation with respect to \( t \). This gives \( \int r \, dr = \int k \, dt \). The left side becomes \( \frac{1}{2} r^2 + C_1 \) and the right side becomes \( kt + C_2 \), where \( C_1 \) and \( C_2 \) are constants of integration.

Step 4: Simplify the Integrated Equation

Equate the two sides to get \( \frac{1}{2} r^2 = kt + C \), where \( C = C_2 - C_1 \) is a constant.

Step 5: Utilize Initial Conditions

We need to find the value of \( C \). Use the condition that \( r = 400 \) at \( t = 16 \). Substitute these values into the equation giving \( \frac{1}{2} (400)^2 = k(16) + C \). Simplifying, we have \( 80000 = 16k + C \).

Step 6: Derive a Secondary Condition and Solve for k

To find another condition, consider when the spill begins, \( r = 0 \) at \( t = 0 \). However, realistically the spill cannot have been 0, so assume \( C = 0 \) to find \( k \). Thus, \( 16k = 80000 \).

Step 7: Calculate the Value of k

Solve \( 16k = 80000 \) for \( k \). Divide both sides by 16 to find \( k = \frac{80000}{16} = 5000 \).

Step 8: Add Units to the Result

Since \( k \) is derived from the equation \( \frac{dr}{dt} \), where \( dr \) has units of feet and \( dt \) has units of hours, \( k \) will have units of feet squared per hour.

Key Concepts

Separation of Variables

Separation of variables is a common technique used to solve differential equations. It's very handy when you have an equation where one side can depend entirely on one variable (usually, this is your unknown function), and the other side depends on the other variable.

Here's what you do:

• You rearrange the equation so that all terms with one variable are on one side, and all terms with the other variable are on the other. In the oil spill problem, this involved making the terms involving the radius, \(r\), and the time, \(t\), go to opposite sides of the equation.
• This involved multiplying both sides by \(r\) to cancel out the \(r\) in the denominator, resulting in \(r \frac{dr}{dt} = k\).

Once you successfully separate the variables, the next steps are to integrate each side, tackling the differential equation piece by piece. This sets the stage for finding the solution.

Initial Conditions

Initial conditions are crucial when solving differential equations because they allow us to find specific solutions. A differential equation typically has a family of solutions. Initial conditions help narrow down to a single specific solution.

In our example, the initial condition given was that the radius of the spill, \(r\), is 400 feet when the clock hits 16 hours. When you plug these initial conditions into your integrated equation, it provides you with a means to determine the integration constant, \(C\), in the equation \( \frac{1}{2} r^2 = kt + C \).

• This means substituting \(r = 400\) and \(t = 16\) into the general solution to solve for \(C\).
• The computation leads you to set up the equation \(80000 = 16k + C\) to find the relation between \(C\) and \(k\).

By allowing us to link these constants with real-life constraints, initial conditions ensure the differential equation solution becomes a precise model instead of just a rough estimate.

Integration Constants

Integration constants always appear when you integrate one side of a differential equation. These constants emerged in our equations post-integration, reflecting the fact that when you integrate, it's not just one function you potentially reach, but rather an entire family of functions, all related by these constant terms.

Let’s consider the process again:

• After separation of variables in the oil spill problem, we integrated both sides to get \( \frac{1}{2} r^2 + C_1 = kt + C_2 \).
• Combining the constants into \(C = C_2 - C_1\) simplifies our equation to \( \frac{1}{2} r^2 = kt + C \).

Integration constants are particularly useful because they allow the general solution to a differential equation to be adjusted to meet specific conditions or restraints, as highlighted by initial conditions. Once you solve for these constants using the provided data in the problem, you convert a general potential prediction about future states into one that exactly aligns with past observed states.