/*! 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 54 The figure shows an angle \(\the... [FREE SOLUTION] | 91Ó°ÊÓ

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The figure shows an angle \(\theta\) between 0 and \(\pi / 2 .\) Let \(T(\theta)\) be the area of triangle \(A B C\), and let \(S(\theta)\) be the arca of the segment of the circle cut by the chord \(A B\). Find \(\lim _{\theta \rightarrow \infty} T(\theta) / S(\theta)\)

Short Answer

Expert verified
The short answer is: \(\lim_{\theta \rightarrow \infty} \frac{T(\theta)}{S(\theta)} = 0\)

Step by step solution

01

Write \(T(\theta)\) as a function of \(\theta\)

Since \(A\), \(B\), \(O\) lie on the circle, we know that \(OA = OB = r\) (the circle has a constant radius). We also know the angle \(AOB = \theta\). So, we can use the following formula to find the area of triangle \(A B C\): Area of triangle \(A B C\) = \(T(\theta) = \frac{1}{2}ab\sin(\theta)\), where \(a\) and \(b\) are the lengths of the two sides adjacent to angle \(\theta\). In our case, \(a = b = r\), so we can write the area of the triangle as: \(T(\theta) = \frac{1}{2} r^2 \sin(\theta)\).
02

Write \(S(\theta)\) as a function of \(\theta\)

Next, let's compute the area of the circular segment \(S(\theta)\). The circular segment is obtained by subtracting the area of the triangle \(ABC\) from the area of the sector \(AOB\): \(S(\theta) = A_{sector} - T(\theta)\). The area of a sector can be expressed as: \(A_{sector} = \frac{1}{2}r^2\theta\), where \(r\) is the radius and \(\theta\) is the central angle. Therefore, we can write \(S(\theta)\) as: \(S(\theta) = \frac{1}{2} r^2\theta - \frac{1}{2} r^2 \sin(\theta)\).
03

Compute \(\lim_{\theta \rightarrow \infty} \frac{T(\theta)}{S(\theta)}\)

Now, let's compute the limit of the ratio \(T(\theta) / S(\theta)\) as \(\theta\) approaches infinity: \(\lim_{\theta \rightarrow \infty} \frac{T(\theta)}{S(\theta)} = \lim_{\theta \rightarrow \infty} \frac{\frac{1}{2} r^2 \sin(\theta)}{\frac{1}{2} r^2\theta - \frac{1}{2} r^2 \sin(\theta)}\) To make the limit computation easier, we can cancel out the \(r^2\) term from both the numerator and the denominator: \(\lim_{\theta \rightarrow \infty} \frac{\sin(\theta)}{\theta - \sin(\theta)}\) Since \(\lim_{\theta \rightarrow \infty} \sin(\theta) = 0\), the limit becomes: \(\lim_{\theta \rightarrow \infty} \frac{0}{\theta} = 0\) So, the limit of the ratio \(T(\theta) / S(\theta)\) as \(\theta\) approaches infinity is 0.

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Key Concepts

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

Limit of a Function
Understanding the concept of a limit is fundamental to calculus and analysis. Limits help us understand the behavior of functions as they approach a certain value or infinity. Consider the mathematical expression \( \lim_{x \to a} f(x) \), which represents the limit of the function \( f(x) \) as \( x \) approaches \( a \) . If the function \( f(x) \) becomes arbitrarily close to a specific value \( L \) as \( x \) approaches \( a \), then we say that the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \). This concept is crucial when dealing with continuous functions and helps to understand behavior in situations that are not easily evaluated directly, such as when dealing with infinity or undefined expressions.

In our exercise, we are asked to find the limit of the ratio \( T(\theta) / S(\theta) \) as \( \theta \) approaches infinity. Limit computations require careful consideration of the terms involved, especially when the independent variable reaches toward infinity. The result we obtain tells us how the relationship between the area of a triangle and the area of a sector changes as the angle \( \theta \) grows without bound.
Trigonometric Functions
Trigonometric functions are the backbone of trigonometry and are essential in various mathematical computations, particularly those involving right triangles and circles. Basic trigonometric functions include sine, cosine, and tangent. In the context of this exercise, the sine function \( sin(\theta) \) describes the ratio between the length of the opposite side to the angle \( \theta \) and the hypotenuse in a right-angled triangle.

These functions are also periodic, meaning they repeat their values in regular intervals. For example, \( sin(\theta) \) has a period of \( 2\pi \) radians, meaning that it repeats every \( 2\pi \) radians. Knowing this periodic nature is crucial when examining the behavior of \( sin(\theta) \) as \( \theta \) approaches infinity, as seen in the exercise. In this case, since \( sin(\theta) \) oscillates between -1 and 1, its limit does not exist as \( \theta \) approaches infinity. However, in the context of the ratio provided in the problem, the effect of the \( sin(\theta) \) term vanishes, leading to the limit being zero.
Area of a Sector
The area of a sector of a circle is a measure of the enclosed space by a circle's arc and two radii. To visualize it, think of a pizza slice where the crust is part of the circle's circumference, and the pointy end is at the circle's center. This area can be calculated using the formula \( A_{sector} = \frac{1}{2}r^2\theta \) where \( r \) is the radius of the circle, and \( \theta \) is the central angle in radians. The formula is derived from the proportion of the circle's entire area \( \pi r^2 \) that the sector covers, which is \( \theta \) out of the total angle in a circle, \( 2\pi \) radians.

In our exercise, understanding the area of a sector is key to finding \( S(\theta) \) , which represents the difference between the area of the sector and the area of the triangle within it. By subtracting \( T(\theta) \) from the area of the sector, we identify the area specific to the circular segment alone.
Area of a Triangle
For any triangle, the area can be calculated if the lengths of the base and height are known, using the simple formula \( Area = \frac{1}{2}base \times height \). However, when we are dealing with triangles within a circle (inscribed triangles), another formula often provides a more convenient method of calculation: \( Area = \frac{1}{2}ab \sin(\theta) \) , where \( a \) and \( b \) are the lengths of any two sides, and \( \theta \) is the angle between those sides. This is particularly useful when dealing with non-right triangles where the height is not readily available.

In the specific exercise we are discussing, the triangle is defined by the radii and chord of a circle, leading to the sides \( a \) and \( b \) being equal to the radius \( r \) , and \( \theta \) being the central angle. Hence, the area of triangle \( ABC \) is expressed as \( T(\theta) = \frac{1}{2} r^2 \sin(\theta) \) . This relationship between the radius, angle, and area is crucial in solving problems that involve a sector and its inscribed or central triangle.

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Most popular questions from this chapter

The set \(S\) of rational numbers \(x\) with \(x^{2}<2\) has rational upper bounds but no least rational upper bound. The argument goes like this. Suppose that \(S\) has a least rational upper bound and call it \(x_{0} .\) Then either $$x_{0}^{2}=2, \quad \text { or } \quad x_{0}^{2}>2, \quad \text { or } \quad x_{0}^{2} < 2.$$ (a) Show that \(x_{0}^{2}=2\) is impossible by showing that if \(x_{0}^{2}=\) 2, then \(x_{0}\) is not rational. (b) Show that \(x_{0}^{2}>2\) is impossible by showing that if \(x_{0}^{2}>2,\) then there is a positive integer \(n\) for which \(\left(x_{0}-\frac{1}{n}\right)^{2}>2,\) which makes \(x_{0}-\frac{1}{n}\) a rational upper bound for \(S\) that is less than the least rational upper bound \(x_{0}\). (c) Show that \(x_{0}^{2}<2\) is impossible by showing that if \(x_{0}^{2}<2,\) then there is a positive integer \(n\) for which \(\left(x_{0}+\frac{1}{n}\right)^{2}<2 .\) This places \(x_{0}+\frac{1}{n}\) in \(S\) and show's that \(x_{0}\) cann0ot be an upper bound for \(S\).

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