Problem 10 - Applied Calculus

Question

Let \(w\) be the number of worms (in millions) and \(r\) the number of robins (in thousands) living on an island. Suppose \(w\) and \(r\) satisfy the following differential equations, which correspond to the slope field in Figure \(10.42\). $$ \frac{d w}{d t}=w-w r, \quad \frac{d r}{d t}=-r+w r $$ Solve these differential equations in the two special cases when there are no robins and when there are no worms living on the island.

Step-by-Step Solution

Answer

With no robins, worms grow exponentially: \(w(t)=Ce^t\). With no worms, robins decay exponentially: \(r(t)=De^{-t}\).

Step 1: Analyze when there are no robins

Set the number of robins, \(r\), to zero. With \(r = 0\), the first differential equation becomes \(\frac{dw}{dt} = w - wr = w\). This simplifies to \(\frac{dw}{dt} = w\), a simple differential equation.

Step 2: Solve the worm equation when \(r=0\)

Solve \(\frac{dw}{dt} = w\). The general solution to this differential equation is \(w(t) = Ce^t\), where \(C\) is the constant of integration determined by initial conditions.

Step 3: Consider when there are no worms

Set the number of worms, \(w\), to zero. With \(w = 0\), the second differential equation becomes \(\frac{dr}{dt} = -r + wr = -r\). This simplifies to \(\frac{dr}{dt} = -r\).

Step 4: Solve the robin equation when \(w=0\)

Solve \(\frac{dr}{dt} = -r\). The general solution to this differential equation is \(r(t) = De^{-t}\), where \(D\) is the constant of integration determined by initial conditions.

Key Concepts

Slope Field

A slope field is an intuitive way to visualize a family of solutions to a differential equation. It consists of small line segments or arrows plotted at various points, each with a slope corresponding to the differential equation's solution at that point. For the given exercises involving worms and robins, the slope field represents how the population of each changes over time.
This can be especially helpful when analyzing different initial conditions without directly solving the differential equations. Each curve in a slope field trajectory depicts how one particular set of initial conditions evolves.

General Solution

When we talk about a general solution to a differential equation, we mean a solution that includes all possible solutions. For example, for \(\frac{dw}{dt} = w\), the general solution is \(w(t) = Ce^t\). Similarly, for \(\frac{dr}{dt} = -r\), the solution is \(r(t) = De^{-t}\).
Here, \(C\) and \(D\) are constants that can vary, allowing the formula to represent many possible solutions. To move from a general to a specific solution, we need to know initial conditions. Only then can \(C\) and \(D\) be determined.

Initial Conditions

Initial conditions are values given for the variables at the beginning of a scenario, allowing us to find a specific solution from a general one.
In our exercise, imagining when \(r=0\) or \(w=0\), these are like initial condition problems for the respective equations \(\frac{dw}{dt} = w\) and \(\frac{dr}{dt} = -r\).
These conditions help us pin down the exact path among many possible ones provided by the general solution. Without them, we would not be able to solve for unique constants like \(C\) and \(D\).

Integration Constant

In the context of solving differential equations, an integration constant appears when taking an indefinite integral. It's a way to account for all possible solutions to a differential equation. For example, with \(\frac{dw}{dt} = w\), integrating both sides yields \(w(t) = Ce^t\), where \(C\) is the integration constant.
The value of \(C\) isn't known until we apply an initial condition. Similarly, in \(\frac{dr}{dt} = -r\), integrating gives us \(r(t) = De^{-t}\). Here, \(D\) serves as the integration constant. Applying known values from the situation, such as initial population sizes, allows us to find specific values for these constants.