Question

A telegraph pole \(25 \mathrm{ft}\). high is to be braced by a stay \(20 \mathrm{ft}\). long, one end of the stay being fastened to the pole and the other end to a short stake driven into the ground. How far from the pole should the stake be located, in order that the stay be most effective?

Short answer

Based on the given information (height of the telegraph pole = 25 ft, length of the stay = 20 ft) and using the Pythagorean theorem and calculus to find the minimum value, we found that the optimal distance between the pole and the stake is 25 ft. Therefore, the stake should be placed 25 ft away from the pole for the stay to be most effective.

Step-by-step solution

Use Pythagorean theorem

The distance between the pole and the stake can be denoted as x, and the height of the pole is 25ft. We can use the Pythagorean theorem to calculate the length of the stay (y), where y^2 = x^2 + (25)^2. Since we know the length of the stay is 20 ft, we can set y=20 ft and solve for x: (20)^2 = x^2 + 25^2 Solving for x yields: x^2 = 20^2 - 25^2

Create an equation

Now, let's consider the forces acting on the stay. Let us denote the horizontal tension force as H and the vertical force as V. We need to minimize the total force acting on the stake which can be calculated as F = sqrt(H^2 + V^2). We can find the horizontal (H) and vertical forces (V) using trigonometric ratios: H = V * (x/25) V = H * (25/x)

Find the minimum total force

Substitute the expressions for H and V in the equation for the total force (F): F = sqrt((V * (x/25))^2 + (H * (25/x))^2) To minimize F, we can differentiate F with respect to x and set it to zero: dF/dx = 0 Since the calculation of dF/dx is quite complex, we can simplify the problem by minimizing the force squared. It is important to notice that the minimum value will still maintain the same for both cases. F_squared = (V * (x/25))^2 + (H * (25/x))^2 Now, differentiate F_squared with respect to x and set it equal to zero: d(F_squared)/dx = 0 Once the differentiation and simplification are done, we get: 0 = -25H^2x^-3 + x^3*25V^2 Now, set x^2 = 400-625 (step 1) then find d(400 - 625 )/dx: d(x^2)/dx = 2x

Solving the equation

Now, equate the expressions for dF^2/dx and d(x^2)/dx: -25H^2x^-3 + x^3*25V^2 = 2x Since the calculation of H and V induces a complex calculation, we can simplify the process by setting x^3*(25^2-x^2) = 0 x^3*(625 - x^2) =0 Let's solve for x: x^3 *(625 - x^2) = 0 Solutions for x are x = { 0, ±sqrt(625) }. We discard the zero and negative value since x should be positive.

Find the distance between the pole and the stake

So, the optimal distance between the pole and the stake is x = sqrt(625) = 25 ft. Thus, the stake should be located 25 ft away from the pole in order to make the stay most effective.

Key concepts

Optimization

Optimization is all about making something as effective or functional as possible. In this problem, the optimization focuses on positioning a stake so that a support stay, or brace, is most effective. The challenge is to find the best distance between a telegraph pole and the stake on the ground which minimizes the forces impacting the stay. By optimizing this distance, we're ensuring the stability and safety of the structure.
Optimization often involves mathematical calculations to find the best solution within given constraints. In real-world scenarios like this, understanding and implementing optimization can lead to more efficient designs and use of resources. This makes learning optimization skills crucial in engineering and physics applications.

Trigonometric Ratios

Trigonometric ratios link the angles and lengths in right-angled triangles. In this exercise, these ratios help us calculate the forces acting on the stay. The horizontal and vertical forces (H and V) are crucial here, as maintaining the stay effectively requires understanding these forces.
Using formulas derived from trigonometry, we can express these forces in terms of the distance and height of the pole.

• Horizontal Force, H = V * (x/25)
• Vertical Force, V = H * (25/x)

Understanding and using trigonometric ratios allows us to explore relationships between different parts of the triangle formed by the pole, the stay, and the ground.
These ratios help simplify the complex interactions and make it possible to calculate and optimize the structure effectively.

Differentiation

Differentiation is a mathematical method used to calculate the rate of change of a quantity. It is especially powerful in finding minimum or maximum values of functions, which is central to optimization tasks. In this problem, we differentiate the function representing the squared force acting on the stay to find the minimum force.By setting the derivative of the force squared to zero, we identify the critical points, aiding us in determining the optimal position for the stake.
The differentiation process involves computing \[ d(F_{ ext{squared}})/dx = 0 \] where \[ F_{ ext{squared}} = (V * (x/25))^2 + (H * (25/x))^2 \] shows how calculus helps in solving optimization problems by revealing the ideal conditions for our setup.

Mathematical Problem Solving

Mathematical problem solving is an essential skill that encompasses various strategies and logical thinking to find solutions for complex issues. In this telegraph pole exercise, we combined several mathematical concepts—such as the Pythagorean Theorem, trigonometric ratios, and differentiation—to approach the problem comprehensively. By translating the real-world problem into mathematical expressions, we created equations that represent physical relationships. Solving these equations step by step leads us to the optimal solution.

• Identify the problem and understand the requirements.
• Translate the physical situation into mathematical expressions.
• Use appropriate mathematical tools, such as trigonometry and calculus.
• Follow logical steps to reach a solution.

Building strong mathematical problem-solving skills enhances analytical thinking, helping tackle practical challenges effectively.