Problem 322 - Calculus Volume 1

Question

For the following exercises, consider the construction of a pen to enclose an area. Two poles are connected by a wire that is also connected to the ground. The first pole is 20 ft tall and the second pole is 10 ft tall. There is a distance of 30 ft between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire needed?

Step-by-Step Solution

Answer

The wire should be anchored approximately 10 ft from the taller pole.

Step 1: Understand the Problem

We have two poles of different heights: 20 ft and 10 ft, with a distance of 30 ft between them. We need to find the optimal point on the ground where the wire should be anchored so that the amount of wire used is minimized.

Step 2: Define Variables

Let the distance from the base of the taller pole (20 ft) to the point on the ground where the wire is anchored be \( x \). Hence, the distance from the base of the shorter pole (10 ft) to this point will be \( 30 - x \).

Step 3: Express Wire Length as a Function of x

The length of the wire from the top of the first pole to the ground is \( \sqrt{x^2 + 20^2} \), and from the top of the second pole to the ground is \( \sqrt{(30-x)^2 + 10^2} \). Thus, the total wire length \( L(x) \) is: \[ L(x) = \sqrt{x^2 + 400} + \sqrt{(30-x)^2 + 100} \]

Step 4: Find the Derivative

Find the derivative of \( L(x) \) with respect to \( x \):\[ L'(x) = \frac{x}{\sqrt{x^2 + 400}} - \frac{30-x}{\sqrt{(30-x)^2 + 100}} \]

Step 5: Set Derivative to Zero and Solve

To find the critical points, we solve \( L'(x) = 0 \):\[ \frac{x}{\sqrt{x^2 + 400}} = \frac{30-x}{\sqrt{(30-x)^2 + 100}} \]Cross-multiply and simplify to solve for \( x \).

Step 6: Calculation and Solution

Through simplification and calculation, we find that \( x \approx 10 \) ft gives us the minimum wire length. This requires algebraic simplification or numerical methods to solve the equation derived in Step 5.

Key Concepts

Calculus

Calculus is a fundamental branch of mathematics, focused on understanding change. It's like the math of motion and growth. In this particular problem, we're using calculus to optimize a situation, which means finding the best or most efficient solution. It's essential in solving real-world problems where we want to determine maximum or minimum values.

In the context of the exercise, calculus is applied to minimize the length of wire needed between the two poles. By using calculus, we're able to find the precise point where a small change will lead to the least amount of material used. Thus saving resources and cost. Calculus encompasses many tools, but for optimization, derivatives are crucial.

• Analyze change dynamically.
• Solve for maximum or minimum values.
• Utilize derivatives to describe change.

Derivative

The derivative is a key concept in calculus, representing the rate of change or slope of a function. Imagine it like the speedometer in a car—it tells you how fast something is changing at any given moment. In mathematical terms, it's the instantaneous rate of change.

For the wire problem, the derivative helps us find where the function's slope is zero. This is important because it indicates points where the function has a local minimum or maximum, which is what we seek here. Calculating the derivative of the function representing the length of the wire (\( L(x) \)) shows us how changes in the grounding point's position affect the wire length.

• Derivatives gauge the immediate rate of change.
• Vital for solving optimization problems.
• Used to find critical points, revealing potential minima or maxima.

Critical Points

Critical points are where the magic happens in optimization. These points are found by setting the derivative of a function to zero. At these points, the function stands still—meaning neither increasing nor decreasing—often corresponding to minimum or maximum values.

In the wire problem, after deriving the function, we set the derivative, \( L'(x) \), to zero. This allows us to explore where these critical points occur. By solving the resulting equation, critical points were identified. They tell us potential positions on the ground to minimize wire usage. Essentially, critical points act as the crossroads in our optimization journey.

• Where the derivative is zero or undefined.
• Key to identifying local minima and maxima.
• Crucial for making accurate optimizations.

Distance Minimization

Distance minimization is about finding the shortest or most efficient path. It involves using mathematics, specifically calculus, to reduce the length of something, like the wire in our exercise. The goal is to use as little as possible while still achieving the desired task.

In the context of the problem, distance minimization ensures the wire's length is minimized without compromising on the structural requirement. By employing calculus, we determine the precise location to anchor the wire, ensuring minimal material usage. This process can apply to various fields, optimizing routes, layouts, and logistics operations alike.

• Aim to shorten material use.
• Use calculus to analyze lengths mathematically.
• Helps in reducing costs and improving efficiency.