/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Lawn Sprinkler A lawn sprinkler ... [FREE SOLUTION] | 91Ó°ÊÓ

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Lawn Sprinkler A lawn sprinkler is constructed in such a way that \(d \theta / d t\) is constant, where \(\theta\) ranges between \(45^{\circ}\) and \(135^{\circ}(\) see figure). The distance the water travels horizontally is $$x=\frac{v^{2} \sin 2 \theta}{32}, \quad 45^{\circ} \leq \theta \leq 135^{\circ}$$ where \(v\) is the speed of the water. Find \(d x / d t\) and explain why this lawn sprinkler does not water evenly. What part of the lawn receives the most water? FOR FURTHER INFORMATION For more information on the "calculus of lawn sprinklers," see the article "Design of an Oscillating Sprinkler" by Bart Braden in Mathematics Magazine. To view this article, go to MathArticles.com.

Short Answer

Expert verified
The water's horizontal distance changes at a rate of \(dx/dt = \frac{v^2}{32} \cdot 2 \cos(2\theta) \cdot \text{const}\). The sprinkler does not water evenly since \(dx/dt\) varies with \(\theta\). The largest amount of water is delivered to the part of the lawn where the sprinkler's angle is \(90^{\circ}\).

Step by step solution

01

Find the Derivative of the Horizontal Distance Function

Given \(x = \frac{v^2 \sin(2\theta)}{32}\), the derivative with respect to time \(t\) would be: \(dx/dt = \frac{d}{dt} [\frac{v^2 \sin(2\theta)}{32}]\). Apply the chain rule, taking the derivative of the outside function and then multiplying it by the derivative of the inside function. This gives: \(dx/dt = \frac{v^2}{32} \cdot 2 \cos(2\theta) \cdot \frac{d\theta}{dt}\).
02

Substitute the Constant Rate of Change of the Angle

Substitute \(d\theta/dt\) with the given constant value into the equation to solve for \(dx/dt\). Therefore, \(dx/dt = \frac{v^2}{32} \cdot 2 \cos(2\theta) \cdot \text{const}\).
03

Discuss the Sprinkler's Performance

Because \(dx/dt\) is not a constant value (due to its dependency on the variable \(\theta\)), the water does not cover the lawn evenly, meaning different areas will receive different amounts of water. More specifically, areas where the angle \(\theta\) makes the cosine function \(\cos(2\theta)\) maximized (resulting in larger values for \(dx/dt\)) will receive more water.
04

Identify the Lawn Part that Receives More Water

The cosine function achieves its maximum value of 1 at \(2\theta = 0\). Hence, the angle \(\theta\) at which sprinkler waters the most is at \(\theta = 0\). But since construction of this sprinkler only allows \(\theta\) to range from \(45^{\circ}\) to \(135^{\circ}\), the part of the lawn that receives the most water will be where the value of \(\theta\) is centered in this interval, i.e., at \(\theta = 90^{\circ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. It helps us find the rate of change of a function that is dependent on another function.

In our sprinkler problem, we have a composite function involving trigonometric functions: \( x = \frac{v^2 \sin(2\theta)}{32} \).
To apply the chain rule, we differentiate the outer function, which is the division by 32 here, and multiply by the derivative of the inner function, which involves \(\sin(2\theta)\).

The chain rule simplifies problems involving functions nested within others. This makes it a powerful tool for tackling complex problems like those involving rates of change in physics and engineering.

  • Differentiate the outer function first.
  • Find the derivative of the inner function.
  • Multiply the derivatives together to get the final result.
Rate of Change
Understanding the rate of change is crucial in describing how one quantity changes concerning another.

In this exercise, we are interested in how the horizontal distance covered by the water spray \(x\) changes over time, expressed as \(dx/dt\).

The sprinkler problem showed that \(dx/dt = \frac{v^2}{32} \cdot 2 \cos(2\theta) \cdot \frac{d\theta}{dt}\).
Here, \(\frac{d\theta}{dt}\) is a constant, but because \(\theta\) varies, \(dx/dt\) changes too.

The concept of rate of change is widely applicable beyond this problem, explaining everything from speed in physics to changes in economic indicators.

  • Identify the quantities you want to explore.
  • Use calculus techniques to find their interdependence.
  • Determine how changes in one aspect affect the other over time.
Trigonometric Functions
Trigonometric functions like sine and cosine are vital in analyzing scenarios involving angles and periodic motions.

For the sprinkler, the distance \(x\) depends on \(\sin(2\theta)\) and \(\cos(2\theta)\).
These functions describe how the angle \(\theta\) affects the horizontal reach of the water.

Trigonometric functions oscillate between -1 and 1, and this behavior directly influences the rate at which \(dx/dt\) changes.

Understanding these functions is essential in fields like engineering and physics, where systems often operate in cycles.

  • Recognize the role of angles in your problem.
  • Understand how sine or cosine affects the result.
  • Consider the periodic nature and range of these functions.
Derivative
The derivative is a foundational concept in calculus that signifies how a function changes when the input changes.

In our sprinkler example, finding \(dx/dt\) involves differentiating the function \(x = \frac{v^2 \sin(2\theta)}{32}\) with respect to time.

The derivative tells us the instantaneous rate of change — how quickly or slowly the water's reach changes over time.

Understanding derivatives allows us to model and make predictions about real-world scenarios.
It's used in various fields, including physics, economics, and any discipline that involves dynamic change.

  • Identify the variable with respect to which you need to differentiate.
  • Apply appropriate differentiation techniques.
  • Interpret the physical meaning of the derivative in the context of the problem.

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