Question

The depth (in feet) of water distributed by a rotating lawn sprinkler in an hour is \(k e^{-r / 10}, 0 \leq r \leq 10\), where \(r\) is the distance from the sprinkler and \(k\) is a constant. Determine \(k\) if 100 cubic feet of water is distributed in 1 hour.

Short answer

The value of \(k\) is approximately 0.00465.

Step-by-step solution

Understanding the Problem

We want to find the constant \(k\) in the equation for the depth of water distributed by the sprinkler. We know that the total water distributed over the area is 100 cubic feet.

Recognizing the Formula

The given formula for the depth of water is \(d(r) = k e^{-r / 10}\), where \(d(r)\) represents the depth at distance \(r\) from the sprinkler. This function is only valid for \(0 \leq r \leq 10\).

Setting up the Integral

The volume of water distributed, \(V\), can be calculated by integrating the depth function over the entire area it covers. Since the area is a circular disk, we use cylindrical coordinates and set up the integral: \[ V = \int_{0}^{2\pi} \int_{0}^{10} k e^{-r / 10} \cdot r \ dr \ d\theta \], where \(r\) is the radius and \(\theta\) is the angle in radians.

Solving the Inner Integral

First, solve the integral with respect to \(r\): \[ \int_{0}^{10} k e^{-r / 10} r \ dr = k \int_{0}^{10} r e^{-r / 10} dr \].To solve \(\int r e^{-r/10} dr\), use integration by parts with \(u = r\) and \(dv = e^{-r/10} dr\).

Integration by Parts

Perform integration by parts:- Let \(u = r\), thus \(du = dr\).- Let \(dv = e^{-r/10} dr\), thus \(v = -10 e^{-r/10}\).The integration gives us:\[ \int r e^{-r/10} dr = -10re^{-r/10} + \int 10e^{-r/10} dr = -10re^{-r/10} - 100e^{-r/10} + C \],evaluated from \(0\) to \(10\).

Evaluate Definite Integrals

Evaluate the expression from \(r=0\) to \(r=10\): \[ \left[-10re^{-r/10} - 100e^{-r/10}\right]_{0}^{10} = \left[-10(10)e^{-10/10} - 100e^{-10/10}\right] + \left[10(0)e^{0} + 100e^{0}\right] \] \[ = [-100e^{-1} - 100e^{-1}] + [0 + 100] \] \[ = -200e^{-1} + 100 \].

Integrating Over θ

Now integrate over \(\theta\):\[ \int_{0}^{2\pi} (-200e^{-1} + 100) k \ d\theta = (-200e^{-1} + 100)k \int_{0}^{2\pi} d\theta = (200\pi)(100 - 200/e)k \].

Solve for k

Set the integral equal to 100 cubic feet:\[ (200\pi)(100 - 200/e)k = 100 \]. Solve for \(k\) by dividing both sides by \((200\pi)\):\[ k = \frac{100}{200\pi(100 - 200/e)} \].

Calculating k

Compute the value of \(k\) using the expression found in Step 8. Using a calculator, find the numerical approximation:\[ k \approx \frac{100}{200\pi(100 - 73.2051)} \approx 0.00465 \].

Key concepts

Cylindrical Coordinates

Understanding cylindrical coordinates is crucial when dealing with problems involving rotation and circular symmetry, such as sprinkler systems or other circular setups. Cylindrical coordinates consist of three variables:

• Distance from a fixed origin (\(r\))
• Angular displacement from a reference direction (\(\theta\))
• Height along the vertical direction (\(z\))

In our specific problem, imagine a rotating sprinkler distributing water over a circular area. This situation perfectly fits the cylindrical coordinate system, as it naturally describes phenomena with a central axis (like the sprinkler).
By setting up our integrals in cylindrical coordinates, we can easily account for all water distributed in the entire area: a key step in solving the problem.
Effectively using this coordinate system allows us to simplify problems by aligning our mathematical view with the symmetry of the physical situation, providing more flexibility and ease when integrating, particularly in problems involving rotation.

Volume of Revolution

When approaching problems like how much water is distributed by a sprinkler, the concept of volume of revolution comes into play. This involves taking a shape and "revolving" it around an axis to create a 3D object.
In our problem, the water from the sprinkler creates a "revolution" of depth – a distribution that fans out around the sprinkler to form a kind of circular disk.
By integrating the depth function across this circular shape using cylindrical coordinates, we can determine the total volume of water distributed. This gives us a beautiful and effective method to calculate the total effect of a rotating system over a given time.

• This concept is mainly applicable when the object or function in question rotates about an axis.
• It is a fundamental idea in determining volume in many physical and engineering applications.

Hence, through the volume of revolution, we can get a clearer picture of how quantities like volume or area behave under rotation, making it essential to a range of real-world scenarios.

Definite Integrals

Definite integrals are a fundamental part of calculus that help calculate the total value over a specific interval. They are particularly beneficial in solving problems involving area and volume, as they provide a precise accumulated value over a designated range.
In the problem of the sprinkler, the integral is used to calculate how much water is distributed over a circular area. We set it up as a double integral, first integrating over the radius \(r\) to find the effect of depth at various distances, and then over the angle \(\theta\) to account for the complete circular coverage.

• When solving for \(k\), a constant in our equation, we use the solution of these integrals to ensure that the total water distributed equals a specific value: in this case, 100 cubic feet.
• This helps in applying conditions and constraints inherent in real-world problems to our mathematical models.

Definite integrals, thus, are vital tools to transform physical problems into solvable mathematical forms, providing an exact measure to otherwise intuitive or estimated queries.