Problem 92 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

HORTICULTURE A sprinkler system sprays water onto a garden in such a way that \(11 e^{-r^{2} / 10}\) inches of water per hour are delivered at a distance of \(r\) feet from the sprinkler. What is the total amount of water laid down by the sprinkler within a 5 -foot radius during a 20 -minute watering period?

Step-by-Step Solution

Answer

The total amount of water delivered is approximately 330.88 cubic inches.

Step 1: Identify the integral to solve

The amount of water delivered per hour is given by the function \(11 e^{-r^{2} / 10}\). To find the total amount of water within a 5-foot radius, integrate this function over the area it covers.

Step 2: Set up the integral in polar coordinates

Convert to polar coordinates where the area element \(dA = 2\pi r \, dr\). The integral becomes: \[ W = \int_{0}^{5} 11 e^{-r^{2}/10} 2 \pi r \, dr \]

Step 3: Simplify the integral

Simplify the integral to make it easier to evaluate: \[ W = 22 \pi \int_{0}^{5} r e^{-r^{2}/10} \, dr \]

Step 4: Apply substitution

Let \(u = r^{2}/10\) so that \(du = (1/5)r \, dr\) or \(r \, dr = 5 du\). The limits change accordingly from \(r = 0\) to \(r = 5\), thus \(u = 0\) to \(u = 5^{2}/10 = 2.5\). The integral becomes: \[ W = 110\pi \int_{0}^{2.5} e^{-u} \, du \]

Step 5: Integrate and evaluate

The integral of \(e^{-u}\) is \(-e^{-u}\). So, evaluate the limits: \[ W = 110\pi [-e^{-u}]_{0}^{2.5} = 110\pi [-(e^{-2.5} - e^{0})] = 110\pi [1 - e^{-2.5}] \]

Step 6: Calculate the answer for a 20-minute period

Since 20 minutes is \(1/3\) of an hour, the total water delivered is: \[ T = \frac{1}{3} W = \frac{1}{3} \cdot 110\pi [1 - e^{-2.5}] \]

Step 7: Final calculation

Approximate the value with \(e^{-2.5} \approx 0.0821\): \[ T \approx \frac{1}{3} \cdot 110 \pi (1 - 0.0821) \approx \frac{1}{3} \cdot 110 \pi \cdot 0.9179 \approx 105.38 \pi \approx 330.88 \text{ cubic inches} \]

Key Concepts

Integration Techniques

Integration is a fundamental tool in calculus, used to find areas, volumes, central points, and many useful things. In this exercise, we use integration to determine the total amount of water distributed by the sprinkler. Specifically, we need to evaluate the integral \[ W = 22 \pi \int_{0}^{5} r e^{-r^{2}/10} \, dr \]. This can seem tricky because of the exponential function involved. To simplify, we use the substitution method and convert the integral into a more manageable form. The key here is to recognize patterns and use appropriate techniques, like substitution, which we will discuss in detail next.

Polar Coordinates

Polar coordinates are particularly useful for problems involving symmetry around a point, like in this sprinkler problem. Instead of working with coordinates \(x\text{ and }y\) on a plane, we use radius \(r\) and angle \(\theta\). By converting the area element to \(dA = 2\pi r \, dr\), we transform a potentially complicated double integral into a simpler single-variable integral. This method simplifies computation and can reveal insights that aren't obvious in Cartesian coordinates.

Substitution Method

The substitution method simplifies complicated integrals by making a change of variable. In our solution, we let \(u = r^{2}/10\) transforming the integral into \[ W = 110\pi \int_{0}^{2.5} e^{-u} \, du \]. Here, substitution helps us convert the variable \(r\) into \(u\), making the integral easier to solve. The limits of integration also change accordingly, allowing us to integrate more straightforwardly. This method is extremely powerful for handling integrals involving algebraic and exponential functions.

Exponential Functions

Exponential functions, like \(e^{x}\), are vital in calculus due to their unique properties. In our sprinkler problem, the water distribution is modeled by an exponential decay function \([11 e^{-r^{2} / 10}]\). Exponential decay describes quantities decreasing over time or distance, making it suitable for this context. The integral of \(e^{-u}\) is \(-e^{-u}\), which simplifies our calculations significantly. Understanding how to integrate and manipulate exponential functions is essential for solving a wide range of real-world problems.