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Chapter 3: Problem 26

A right triangle is formed using the points \((-1,0)\), a point \((\cos \theta, \sin \theta)\) on the first quadrant of the unit circle, and the point \((\cos \theta, 0)\), as shown in the figure. What is the maximum possible area of the triangle?

Short Answer

Expert verified
The maximum possible area of the triangle is \(\frac{3}{4}\) square units.

Step by step solution

01

Identify Triangle Vertices

The vertices of the triangle are given as: \(A(-1, 0)\), \(B(\text{cos} \theta, \text{sin} \theta)\), and \(C(\text{cos} \theta, 0)\).
02

Determine Base and Height

The base of the triangle is the horizontal distance between points \(A\) and \(C\), and the height is the vertical distance between point \(B\) and \(C\).
03

Calculate Base of the Triangle

The base length \(b\) can be calculated as \(b = \text{cos} \theta - (-1) = \text{cos} \theta + 1\).
04

Calculate Height of the Triangle

The height \(h\) is the \(y\)-coordinate of point \(B\), which is \(h = \text{sin} \theta\).
05

Write Area Formula

The area \(A\) of the right triangle is given by: \(A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (\text{cos} \theta + 1) \times \text{sin} \theta\).
06

Use Double-Angle Formula

Recall that \text{sin}(2\theta) = 2 \text{sin} \theta \text{cos} \theta\(. Thus, the area can be rewritten as: \)A = \frac{1}{2} \times (\text{cos} \theta + 1) \times \text{sin} \theta = \frac{1}{2} \text{sin} \theta \text{cos} \theta + \frac{1}{2} \text{sin} \theta\( = \frac{1}{4} \text{sin} (2 \theta) + \frac{1}{2} \text{sin} \theta\).
07

Find Maximum Value

The maximum value of \text{sin}(2\theta) is \(1\), occurring when \(2\theta = \frac{\text{pi}}{2}\) or \(\theta = \frac{\text{pi}}{4}\). Thus, the maximum area is: \(A = \frac{1}{4} \times 1 + \frac{1}{2} \times 1 \rightarrow A = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\). The maximum area is thus \(\frac{3}{4}\) square units.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are equations that hold true for all values of the given variables. They are foundational in the study of trigonometric functions and are particularly useful when solving problems like maximizing the area of a triangle. In this exercise, we used the double-angle formula: \( \text{sin}(2\theta) = 2 \text{sin} \theta \text{cos} \theta \). This identity simplifies the product of sine and cosine into a single trigonometric function. Such identities help to transform complex expressions into simpler forms, making calculations easier. Remember to familiarize yourself with other key identities such as Pythagorean identities, sum and difference formulas, and product-to-sum formulas, as they are incredibly useful in various mathematical problems.
Unit Circle
The unit circle is a circle with a radius of one centered at the origin of the coordinate plane. It is a fundamental tool in trigonometry. Every point \( (\text{cos} \theta, \text{sin} \theta) \) on the circumference represents the cosine and sine of an angle \( \theta \). In the context of the given problem, understanding the unit circle helps to locate and visualize points \( \text{cos} \theta \) and \( \text{sin} \theta \) in the first quadrant. By plotting these points, you can better grasp how the triangle鈥檚 dimensions change with different angles \( \theta \). The key takeaway is that the unit circle translates abstract trigonometric functions into concrete geometric points, essential for problems involving angles and distances.
Calculus
Calculus allows us to find maximum and minimum values of functions, which is critical in optimization problems. To maximize the area of the right triangle, you need to determine the values of \( \theta \) that yield the highest area. By using the derivative of the area function and finding where it equals zero, you locate the angle that maximizes the area. Although this problem solves for the maximum through substitution, calculus concepts underpin the methodology. In general, knowing how to take the derivative and interpret it to find critical points is essential. Calculus provides the tools needed to handle changes and optimize outcomes, making it a powerful framework for solving a wide range of mathematical problems.

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Most popular questions from this chapter

Differentiate implicity to find \(d y / d x\) and \(d^{2} y / d x^{2}\). $$ x^{2}-y^{2}=5 $$

Certain chemotherapy dosages depend on a patient's surface area. According to the Mosteller model, \({ }^{17}\) $$ S=\frac{\sqrt{h w}}{60} $$ where \(h\) is the person's height in centimeters, \(w\) is the person's weight in kilograms, and \(S\) is the approximation to the person's surface area in \(\mathrm{m}^{2}\). Use this formula. Assume that a female's height is a constant \(160 \mathrm{~cm}\), but she is on a diet. If she loses \(3 \mathrm{~kg}\) per month, how fast is her surface area decreasing at the instant she weighs \(60 \mathrm{~kg}\) ?

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Certain chemotherapy dosages depend on a patient's surface area. According to the Mosteller model, \({ }^{17}\) $$ S=\frac{\sqrt{h w}}{60} $$ where \(h\) is the person's height in centimeters, \(w\) is the person's weight in kilograms, and \(S\) is the approximation to the person's surface area in \(\mathrm{m}^{2}\). Use this formula in Exercises 37 and 38 . Assume that a male's height is a constant \(180 \mathrm{~cm}\), but he is on a diet. If he loses \(4 \mathrm{~kg}\) per month, how fast is his surface area decreasing at the instant he weighs \(85 \mathrm{~kg}\) ?

Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. When no interval is specified, use the real line \((-\infty, \infty)\). $$ f(x)=\frac{1}{3} x^{3}-x+\frac{2}{3} $$
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