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Chapter 14: Problem 186

Consider \(f(x)=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), \quad x \in\left(0, \frac{\pi}{2}\right)\). A normal to \(y=f(x)\) at \(x=\frac{\pi}{6}\) also passes through the point: (A) \(\left(\frac{\pi}{4}, 0\right)\) (B) \((0,0)\) (C) \(\left(0, \frac{2 \pi}{3}\right)\) (D) \(\left(\frac{\pi}{6}, 0\right)\)

Short Answer

Expert verified
(D) \(\left(\frac{\pi}{6}, 0\right)\) is the point through which the normal passes.

Step by step solution

01

Find the function f(x) at x = 蟺/6

Evaluate the function \( f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) \) at \( x = \frac{\pi}{6} \). First, calculate \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \). Then substitute this value into the function:\[\sqrt{\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}} = \sqrt{\frac{\frac{3}{2}}{\frac{1}{2}}} = \sqrt{3}\]Thus, \( f\left(\frac{\pi}{6}\right) = \tan^{-1}\left(\sqrt{3}\right) = \frac{\pi}{3} \).
02

Calculate the derivative of f(x)

To find the equation of the normal, we must first find the derivative of \( f(x) \). The function is \( f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) \). Let \( u(x) = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \).The derivative of \( f(x) \) is given by:\[f'(x) = \frac{d}{dx}\left(\tan^{-1}(u(x))\right) = \frac{1}{1 + u(x)^2} \cdot \frac{du}{dx}\]We will calculate \( \frac{du}{dx} \) separately.
03

Derive u(x) and find its derivative

Given \( u(x) = \sqrt{\frac{1+\sin x}{1-\sin x}} \), use the quotient rule and chain rule:\[\frac{d}{dx}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) = \frac{1}{2\sqrt{\frac{1+\sin x}{1-\sin x}}} \cdot \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)\]Find the derivative of \( \frac{1+\sin x}{1-\sin x} \) using the quotient rule:\[= \frac{(\cos x)(1-\sin x) + (1+\sin x)(-\cos x)}{(1-\sin x)^2} = \frac{2\cos x}{(1-\sin x)^2}\]Therefore,\[\frac{du}{dx} = \frac{\cos x}{\sqrt{\frac{1+\sin x}{1-\sin x}}(1-\sin x)^2}\]
04

Calculate the slope of the tangent at x = 蟺/6

Substitute \( x = \frac{\pi}{6} \) into the derivative expression.Since \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), and \( u \left(\frac{\pi}{6}\right) = \sqrt{3} \), substitute these values:\[u(u^2 + 1) = 1 + 3 = 4\]\[f'\left(\frac{\pi}{6}\right) = \frac{1}{4} \cdot \frac{\frac{\sqrt{3}}{2}}{\sqrt{3} \times \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}\]Hence, the slope of the tangent is \( \frac{\sqrt{3}}{2} \) at \( x = \frac{\pi}{6} \).
05

Determine the equation of the normal line

The slope of the normal is the negative reciprocal of the tangent slope. Hence,\[\text{normal slope} = -\frac{2}{\sqrt{3}}\]Use the point-slope form for the equation of the line, using point \( (\frac{\pi}{6}, \frac{\pi}{3}) \):\[y - \frac{\pi}{3} = -\frac{2}{\sqrt{3}} (x - \frac{\pi}{6})\]This simplifies to:\[y = -\frac{2}{\sqrt{3}}x + \frac{\pi}{3} + \frac{\pi}{3\sqrt{3}}\]
06

Check if given points lie on the normal line

Verify which of the given points satisfy the normal line's equation:- For \((0,0)\): \(0 = -\frac{2}{\sqrt{3}} \times 0 + \frac{\pi}{3} + \frac{\pi}{3\sqrt{3}} \), which does not hold.- For \(\left(\frac{\pi}{4}, 0\right)\): \(0 = -\frac{2}{\sqrt{3}} \times \frac{\pi}{4} + \frac{\pi}{3} + \frac{\pi}{3\sqrt{3}} \), which also does not hold.- For \(\left(\frac{\pi}{6}, 0\right)\): Substituting, \(-\frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} + \frac{\pi}{3} + \frac{\pi}{3\sqrt{3}} \), which holds.- For \((0, \frac{2\pi}{3})\): \(\frac{2\pi}{3} eq -\frac{2}{\sqrt{3}} \times 0 + \frac{\pi}{3} + \frac{\pi}{3\sqrt{3}} \), which does not hold.Hence, the point \(\left(\frac{\pi}{6}, 0\right)\) lies on the normal line.

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Key Concepts

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

Arctan function
The arctan function, also known as the inverse tangent function, is an essential concept in calculus. It is denoted as \( \tan^{-1}(x) \) or \( \text{arctan}(x) \) and represents the angle whose tangent is \( x \). This function is crucial when working with trigonometric equations and when finding angles based on given tangent values.
  • The range of the arctan function is \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), which means it outputs values only within those limits.
  • The function is continuous and smooth, and it is increasing since its derivative is always positive.
In the provided exercise, the arctan function helps determine \( f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)\). Understanding this allows us to accurately compute the needed angle or value of \( x \) through known trigonometric relationships.
Derivative of inverse functions
Understanding the derivative of inverse functions is a fundamental topic in calculus. For the arctan function, the derivative is derived using the chain rule and the rules for inverse functions.
  • For \( y = \tan^{-1}(x) \), the derivative is \( \frac{dy}{dx} = \frac{1}{1+x^2} \).
  • When dealing with functions nested within the arctan, like \( f(x) = \tan^{-1}(u(x)) \), the derivative is computed as \( f'(x) = \frac{1}{1+u(x)^2} \cdot \frac{du}{dx} \).
In our example, finding the derivative of \( f(x) \) involves first finding \( \frac{du}{dx} \) for the inside function \( u(x) = \sqrt{\frac{1+\sin x}{1-\sin x}} \) and then combining this with the derivative expression for the inverse tangent. Grasping this concept is key to analyzing changes and behaviors in functions involving inverse trigonometric components.
Normal line to a curve
A normal line to a curve at a given point is a line perpendicular to the tangent at that point. To find the equation of the normal line, one must first calculate the slope of the tangent line and then find the negative reciprocal to get the normal slope.
  • The slope of the tangent is derived from the function's derivative. In this exercise, we calculated the tangent slope as \( \frac{\sqrt{3}}{2} \) at \( x = \frac{\pi}{6} \).
  • The slope of the normal line is therefore \( -\frac{2}{\sqrt{3}} \), found by taking the negative reciprocal.
Using the point-slope formula with the slope of the normal and the point \( \left( \frac{\pi}{6}, \frac{\pi}{3} \right) \), we established the equation of the normal line. Understanding normal lines enhances our ability to study orthogonal relationships and distances relative to curves.
Quotient rule in differentiation
The quotient rule is a technique used in calculus for finding the derivative of a function that is the ratio of two differentiable functions. It is particularly helpful for problems involving ratios, such as \( \frac{1+\sin x}{1-\sin x} \) in this exercise.
  • The quotient rule states that for \( h(x) = \frac{f(x)}{g(x)} \), the derivative \( h'(x) \) is \( \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \).
  • In our solution, this rule was used to find \( \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) \), resulting in a derivative that helps in further calculations.
The effective use of the quotient rule simplifies complex differentiations and ensures precise handling of functions expressed as fractions.

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Most popular questions from this chapter

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