Problem 26 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

It is April and, with the arrival of warm weather a pile of snow has turned into a puddle and the puddle is drying up. At time \(t=0\) there are 4 gallons of water in the puddle. The rate of change of water in the puddle is given by \(f(t)=-e^{0.2 t}\) gallons per hour. How much water is in the puddle \(t\) hours from now? When will the puddle dry up?

Step-by-Step Solution

Answer

The amount of water in the puddle at any given time \(t\) is given by \(V(t) = 9 - 5e^{0.2t}\) gallons. The puddle will dry up when \(t = \frac{ln 1.8}{0.2}\) hours (approximately 4.42 hours).

Step 1: Find the Volume of Water at Time 't'

Since the volume of water is changing at the rate of \(f(t) = -e^{0.2t}\) gallons/hour, the volume of water at time 't' would be the initial volume plus the change in volume, which is the integral of the rate function from 0 to 't'. This amounts to implementing the formula: \[V(t) = V(0) + \int_0^t f(u) \, du\] with \(V(0) = 4\) gallons and \(f(u) = -e^{0.2u}\) gallons/hour.

Step 2: Integral Calculation

We need to calculate integral \(\int_0^t -e^{0.2u} du\). Using the exponential rule for integration (\(\int e^{au} du = \frac{1}{a}e^{au} + C\), where C is a constant of integration), and substituting \(a = 0.2\) and the limits of integration, we get: \[-\frac{1}{0.2} [e^{0.2t} - e^0] = -5 e^{0.2t} + 5\]

Step 3: Compute Volume Function

Substituting this value in the expression derived in Step 1, we get the volume at time 't': \(V(t) = 4 -5 e^{0.2t} + 5 = 9 - 5e^{0.2t}\) gallons.

Step 4: Find the Time When Puddle Dries Up

The puddle will dry up when the volume of the water becomes zero. Equate \(V(t)\) to 0 and solve for 't'. The equation becomes: \(0 = 9 - 5e^{0.2t}\). Upon solving, we get \(e^{0.2t} = 1.8\) and thus, \(t = \frac{ln 1.8}{0.2}\).

Key Concepts

Exponential functions

Exponential functions are an essential part of calculus and many real-world applications. They are functions of the form \( f(t) = a \cdot e^{bt} \), where \( a \) and \( b \) are constants and \( e \) is the base of the natural logarithm (approximately 2.718). These functions describe processes that grow or decay at rates proportional to their current value.
In this exercise, the function \( f(t) = -e^{0.2t} \) represents the rate of change of water volume in the puddle. Since the exponential has a negative coefficient, it describes a decay process, meaning the amount of water decreases over time.

• Exponential functions are powerful for modeling natural phenomena such as population growth, radioactive decay, and in this case, the drying of a puddle.
• The rate can change rapidly or slowly, depending on the power \( b \); here, \( b = 0.2 \), which indicates a moderate rate of change.
• This function serves as our stepping stone to understanding how to find integral solutions related to changes over time.

Rate of change

The concept of rate of change is fundamental in calculus, especially in understanding how a quantity changes over time. In mathematics, it's typically denoted by a function such as \( f(t) \), indicating change over time \( t \).
For this puddle exercise, the rate at which water evaporates is given by \( f(t) = -e^{0.2t} \). This implies:

• At any point of time \( t \), the amount of water in the puddle decreases at an exponential rate.
• The negative sign indicates a reduction in the quantity, which means your puddle is drying up.
• This rate function helps us calculate the integral, a key step in finding the total change over a specific time period.

Understanding the rate of change is crucial because it translates complex processes into manageable mathematical expressions, paving the way for integration.

Integration

Integration is a core concept in calculus, used to find the accumulation of quantities—the reverse process of differentiation. When you integrate a rate of change function, you can determine the overall change in quantity.
In our example of the puddle:

• The integral: \( \int_0^t -e^{0.2u} \, du \) accumulates the total change in water volume over time from 0 to \( t \).
• Using exponential integration rules yields \( -\frac{1}{0.2} (e^{0.2t} - e^0) \), simplifying the way we find the water volume at any future time \( t \).
• This integration helps us create a function, \( V(t) = 9 - 5e^{0.2t} \), that represents the water volume at any time \( t \).

Integration turns the rate of change into a comprehensive picture of how much water remains in the puddle continuously over time.

Integral calculus

Integral calculus involves techniques to calculate integrals, allowing us to solve real-world problems like the drying puddle scenario.
With integral calculus, we take a step-by-step approach to solving the exercise.

• Firstly, we treat the problem statement as a rate of change scenario given by \( f(t) \).
• Applying definite integral concepts, we found the change in water volume over a time interval \( [0, t] \).
• Combining these with the initial condition \( V(0) = 4 \) gallons, the integration results give us \( V(t) = 9 - 5e^{0.2t} \).
• To find when the puddle dries up, we solve \( 0 = 9 - 5e^{0.2t} \), which leads us to find \( t \) using logarithmic calculations as \( t = \frac{\ln 1.8}{0.2} \).

This process demonstrates how integral calculus provides us with tools to solve practical, dynamic systems by converting rates into total quantities, predicting outcomes in the study of changes.