Problem 20 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A miser spends money at a rate proportional to the amount he has. Suppose that right now he has \(\$ 100,000\) stowed under his mattress; he does not pay any taxes and does not earn any return on his money. Assume that this is all the money he has and that he has no other source of income. At the moment he is spending the money at a rate of \(\$ 10,000\) per year. (a) At what rate will he be spending money when he has \(\$ 50,000 ?\) (b) At what time will the amount of money be down to \(\$ 10,000\) ?

Step-by-Step Solution

Answer

The miser will be spending money at a rate of \$50,000 per year when he has \$50,000 left, and the money will be down to \$10,000 at \(t = \ln(10)\) years.

Step 1: Set Up Differential Equation

Setting up the differential equation, we know that the rate of spending is given by \(dM/dt = -kM\). Here, \(M\) represents the amount of money the miser has at time \(t\), and \(k\) is the constant of proportionality. The negative sign indicates that the amount of money is decreasing over time. We are given that when \(M = 10,000\), \(dM/dt = -10,000\). Inserting these values into the equation we get \(k = 1\).

Step 2: Solve Differential Equation

The general solution to the differential equation \(\frac{dM}{dt} = -kM\) is \(M(t) = M_0 e^{-kt}\), where \(M_0\) is the initial amount of money. In this case, \(M_0 = 100,000\) and \(k = 1\). Therefore, the solution is \(M(t) = 100,000 e^{-t}\).

Step 3: Compute Rate at $50,000

For part (a), we need to determine the rate of change when the amount of money is \$50,000. This means solving \(\frac{dM}{dt} = -kM\), which is \(\frac{dM}{dt} = -50,000\). Therefore, he will be spending money at a rate of \$50,000 per year when he has \$50,000 left.

Step 4: Determine Time for $10,000

For part (b), we need to find the time it takes for the scholar's money to decrease to \$10,000. This means solving equation \(M(t) = 100,000 e^{-t}\) for \(t\) when \(M = 10,000\). Express the equation like this: \(e^{-t} = 10,000 / 100,000\). Solve for \(t\) we then get \(t = \ln(10)\).

Key Concepts

Rate of Change

In the context of differential equations and calculus, the 'rate of change' is a measure of how a quantity, such as money in a bank account or population size, changes with respect to another variable—often time. The rate can be constant, like a car traveling at a steady speed, or it can be variable, as in the case of our miser whose spending changes according to the money he has left.

For our miser, the rate of change of his money over time is proportional to the amount he has, which is mathematically represented by the differential equation \( \frac{dM}{dt} = -kM \) where \( M \) is the amount of money, \( t \) is time, and \( k \) is the proportionality constant. Differential equations like this allow us to model and predict the behavior of dynamic systems over time.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This is a common pattern in real-world phenomena like radioactivity, population decline, or, in our case, a miser spending his savings.

In our exercise, the miser's money \( M(t) \) decreases over time following the equation \( M(t) = 100,000 e^{-kt} \) where \( e \) is the base of the natural logarithm and \( t \) is the time in years. Exponential functions are crucial in understanding how systems evolve over time, especially when the rate of change is itself changing.

Proportional Relationships

Proportional relationships occur when two quantities change at the same rate. If one quantity doubles, the other also doubles; if one is cut in half, the other responds accordingly. This is the foundation for the concept of direct variation used in many mathematical models, including our miser's spending habits.

The equation \( \frac{dM}{dt} = -kM \) represents a proportional relationship between the rate of spending money (\(\frac{dM}{dt}\)) and the amount of money he currently has (\(M\)). When the amount of money decreases, the rate at which he spends also decreases, both by the factor determined by the constant \(k\).

Solving Differential Equations

Solving differential equations involves finding a function or formula that describes how one variable evolves as another changes. It's a crucial part of understanding dynamic systems in physics, biology, economics, and more. The solutions to these equations can take many forms, but in the case of our exercise, an exponential function describes the miser's spending over time.

To solve the given differential equation \( \frac{dM}{dt} = -kM \) with initial conditions, we found the general solution \( M(t) = M_0 e^{-kt} \) where \( M_0 \) is the initial amount of money and substituted the given values to model the miser's financial situation. It's this type of structured approach to problem-solving that helps demystify complex dynamic behaviors.