Problem 65 - Calculus Early Transcendentals

Question

a. Two poles of heights \(m\) and \(n\) are separated by a horizontal distance \(d .\) A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when \(\theta_{1}=\theta_{2}\) (see figure). b. Fermat's Principle states that when light travels between two points in the same medium (at a constant speed), it travels on the path that minimizes the travel time. Show that when light from a source \(A\) reflects off a surface and is received at point \(B\), the angle of incidence equals the angle of reflection, or \(\theta_{1}=\theta_{2}\) (see figure).

Step-by-Step Solution

Answer

In summary, we were able to show that the configuration requiring the least amount of rope occurs when the angles are equal (\(\theta_1 = \theta_2\)). By applying the same principle to Fermat's Principle for light reflection, we demonstrated that the angle of incidence equals the angle of reflection when light reflects off a surface.

Step 1: Calculate the rope length

Let \(x\) be the horizontal distance from the base of the pole with height \(m\) to the point where the rope touches the ground. Then, the horizontal distance from the base of the pole with height \(n\) to the point where the rope touches the ground is \(d-x\). Furthermore, we can label the length of the rope from the top of the first pole to the ground as \(l_1\) and the length from the ground to the top of the other pole as \(l_2\). Using the right triangles formed by the poles, we can write the rope lengths in terms of the pole heights, horizontal distances, and trigonometric functions of the angles: \(l_1 = \frac{m}{\sin \theta_1}\), \(l_2 = \frac{n}{\sin \theta_2}\). The total rope length \(L\) is then: \(L=l_1+l_2 = \frac{m}{\sin \theta_1} + \frac{n}{\sin \theta_2}\).

Step 2: Find the derivatives

To find the configuration that requires the least amount of rope, we need to minimize \(L\). To do this, we can differentiate \(L\) with respect to \(x\) and set the derivative equal to zero. Using the Chain Rule, we first find the partial derivatives of \(L\) with respect to \(\theta_1\) and \(\theta_2\), and then use the trigonometric identities to relate the angles \(\theta_1\) and \(\theta_2\) to \(x\). Differentiate \(L\) with respect to \(\theta_1\) and \(\theta_2\): \(\frac{\partial L}{\partial \theta_1} = -\frac{m}{\sin^2 \theta_1} \cos \theta_1\), \(\frac{\partial L}{\partial \theta_2} = -\frac{n}{\sin^2 \theta_2} \cos \theta_2\). Now we need to relate the angles \(\theta_1\) and \(\theta_2\) to \(x\). Using trigonometry, we can find the following relationships: \(x = m \cot \theta_1\), \(d-x = n \cot \theta_2\). Differentiate both equations with respect to \(x\): \(\frac{dx}{d\theta_1} = -m \csc^2 \theta_1\), \(\frac{dx}{d\theta_2} = n \csc^2 \theta_2\). Now differentiate \(L\) with respect to \(x\) using the Chain Rule: \(\frac{dL}{dx} = \frac{\partial L}{\partial \theta_1} \frac{dx}{d\theta_1} + \frac{\partial L}{\partial \theta_2} \frac{dx}{d\theta_2} = 0\). By combining these equations and simplifying, we obtain the equation: \(\frac{m}{\sin^3 \theta_1} = \frac{n}{\sin^3 \theta_2}\).

Step 3: Prove that the minimal rope length configuration occurs when \(\theta_1 = \theta_2\)

To prove this, we will use the equation we derived in Step 2: \(\frac{m}{\sin^3 \theta_1} = \frac{n}{\sin^3 \theta_2}\). We will first see that this equation implies that \(m \sin^2 \theta_2 = n \sin^2 \theta_1\). Then, we will use the properties of right triangles to show that \(\theta_1 = \theta_2\) is the minimal configuration. Divide both sides of the equation by \(\sin \theta_1 \sin \theta_2\): \(\frac{m}{\sin \theta_1 \sin^2 \theta_2} = \frac{n}{\sin^2 \theta_1 \sin \theta_2}\). Simplifying, we find that: \(m \sin^2 \theta_2 = n \sin^2 \theta_1\). Now consider two right triangles with angles \(\theta_1\) and \(\theta_2\). We can label the altitude of the triangle with angle \(\theta_1\) as \(h_1\) and the base as \(b_1\), and the altitude of the triangle with angle \(\theta_2\) as \(h_2\) and the base as \(b_2\). Using the sine function, we find: \(h_1 = m \sin \theta_1\), \(h_2 = n \sin \theta_2\), \(b_1 = m \cos \theta_1\), \(b_2 = n \cos \theta_2\). Using the equation \(m \sin^2 \theta_2 = n \sin^2 \theta_1\), we find: \(h_1^2 = h_2^2\). Since the triangles are right triangles, we have: \(h_1^2 + b_1^2 = h_2^2 + b_2^2\). Substituting and simplifying, we obtain: \(m^2 \cos^2 \theta_1 = n^2 \cos^2 \theta_2\). Taking the square root and dividing by \(m \cos \theta_1\) on both sides, we find: \(\frac{n}{m} \frac{\cos \theta_2}{\cos \theta_1} = 1\). This equation implies that \(\theta_1 = \theta_2\), which means that the minimal rope length configuration occurs when \(\theta_1 = \theta_2\). For part b, we can use the result we derived for part a in the context of light reflection. Since the travel time of light between points \(A\) and \(B\) is directly proportional to the length of the path that light takes, we can apply the principle of minimal rope length to Fermat's Principle. By demonstrating that the minimal configuration for rope length occurs when \(\theta_1 = \theta_2\), we have simultaneously shown that the angle of incidence equals the angle of reflection (\(\theta_1 = \theta_2\)) when light reflects off a surface.

Key Concepts

Minimization Problems in Calculus

Minimization problems are fundamental aspects of calculus, where the goal is to find the lowest value of a function under given constraints. This type of problem is crucial across various fields, from engineering to economics, whenever efficiency and cost reduction are essential. In these problems, the function representing the quantity to be minimized is often related to physical quantities like distance, area, or, as in our exercise, the length of a rope.

To solve a minimization problem, we generally need to calculate the derivative of the function and find where this derivative is zero – these points are the potential minima. However, we also must ensure that these points are indeed minima by checking the second derivative or using other methods like the First Derivative Test. In the exercise, we aim to minimize the total length of the rope connecting two poles, which directly applies these principles.

Fermat's Principle

Fermat's Principle is a concept from the realm of optics that states that the path taken by light between two points will be the one that requires the least time when light speed is constant. This principle can be thought of as the 'economic route' light takes, optimizing its path much like a minimization problem in calculus. When applied to the reflection of light, Fermat's Principle leads to the law of reflection.

In our exercise, we analogically apply this principle to demonstrate that the angles of incidence and reflection are equal to minimize the path of the rope, not just light. This also elegantly links geometric properties of light paths to optimization problems solved with calculus, showcasing the interdisciplinary nature of these concepts.

Derivatives in Calculus

Derivatives stand at the core of calculus, offering a way to describe change and rate of change. The central role of derivatives in optimization problems, including the one in our exercise, lies in their ability to determine the slope of the tangent line to the function's graph at any given point. When the derivative is zero, the slope is flat, and the function could either have a maximum, a minimum, or neither at that point.

In practice, finding the derivative of a function with respect to a variable—known as differentiation—can reveal critical information about the function's behavior. For optimization, derivatives allow us to locate potential minima and is aptly displayed in our exercise by differentiating the total rope length to find the minimum. Understanding how to take derivatives, especially in complex relations involving trigonometric functions, is crucial in solving many real-world problems.

Trigonometric Functions

Trigonometric functions—sine, cosine, tangent, and their reciprocals—describe relationships between the angles and sides of a triangle. These functions have vast applications in fields such as physics, engineering, and of course, calculus. In the context of our exercise, trigonometric functions are used to express the lengths of the rope in terms of the poles' heights and the angles at which the rope meets the ground.

Understanding the properties and graphs of these functions is essential when dealing with oscillatory motion, waves, and, as highlighted by this problem, optimizing lengths and distances. The use of sine in our problem emphasizes how these functions aid in representing and solving real-life scenarios mathematically, and subtleties like the reciprocal relationship between sine and the resultant rope lengths illustrate the complexity that can arise in such optimizations.