Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the point of the curve at which the curvature is a maximum. \(y=\ln \cos x\) for \(-\pi / 2

Step-by-Step Solution

Answer

The curvature is maximum at the point \((0, 0)\).

Step 1: Find the First Derivative

First, find the first derivative of the function. Given that \( y = \ln \cos x \), apply the chain rule: \( \frac{dy}{dx} = \frac{1}{\cos x} \cdot ( -\sin x) = -\tan x \).

Step 2: Find the Second Derivative

Now, find the second derivative of \( y \). Since \( \frac{dy}{dx} = -\tan x \), the second derivative is found by differentiating \( -\tan x \): \( \frac{d^2y}{dx^2} = -\sec^2 x \).

Step 3: Use the Curvature Formula

Use the formula for curvature \( \kappa \) of the curve: \[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \] Substitute \( y' = -\tan x \) and \( y'' = -\sec^2 x \) to get: \[ \kappa = \frac{|\sec^2 x|}{(1 + \tan^2 x)^{3/2}} \]

Step 4: Simplify the Curvature Formula

Recognize that \( 1 + \tan^2 x = \sec^2 x \). This simplifies the curvature \( \kappa \) to: \[ \kappa = \frac{1}{\sec^3 x} = \cos^3 x \] since the absolute value is removed due to the context of \(-\pi/2 < x < \pi/2\).

Step 5: Find the Derivative of the Curvature

Differentiate \( \cos^3 x \) to find \( \kappa' \) and identify critical points. \( \frac{d}{dx}(\cos^3 x) = 3\cos^2 x(-\sin x) = -3\cos^2 x\sin x \).

Step 6: Solve \( \kappa' = 0 \) for Critical Points

Set \( -3\cos^2 x \sin x = 0 \). This equation holds when \( \sin x = 0 \) or \( \cos x = 0 \) (but \( \cos x = 0 \) is outside the interval). Thus, \( \sin x = 0 \) gives \( x = 0 \).

Step 7: Verify Maximum Curvature at \( x = 0 \)

Check values around \( x = 0 \) to ensure it is a maximum. Since the curvature formula \( \cos^3 x \) decreases as \( x \) moves away from 0 (because \( \cos x \) decreases), \( x = 0 \) is a maximum.

Key Concepts

First Derivative

The first derivative is a powerful tool in calculus, as it helps us understand the rate of change of a function. In this exercise, we start with the function given by \[y = \ln \cos x\quad \text{for} \quad -\pi/2 < x < \pi/2\]To find the first time derivative, we apply the chain rule. This rule allows us to differentiate complex expressions by breaking them into simpler parts. For our function, the derivative of \[y = \ln \cos x\]is:\[\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x\]This shows that the rate of change of the function with respect to x is \(-\tan x\),indicating how the function slopes down or up. Knowing the first derivative tells us how the function behaves, at least locally, and prepares us for further exploration like finding critical points and analyzing concavity.

Second Derivative

The second derivative of a function gives us insight into the curvature of the function, which is essentially how "bendy" it is. In our exercise, after finding the first derivative \(-\tan x,\)we proceed to find the second derivative by differentiating it once more. This results in:\[\frac{d^2y}{dx^2} = -\sec^2 x\]The second derivative helps us understand how the rate of change (the first derivative) itself is changing. If the second derivative is positive, it means the function is curving upwards, like a smile. If it's negative, the function is curving downwards, like a frown.In our context, \(-\sec^2 x\)means the curve is always bending downwards, indicating it has a concave form throughout the interval. Knowing this sets the stage for applying the curvature formula to delve deeper into the behavior of the curve.

Curvature Formula

Curvature is a measure of how quickly a curve departs from being a straight line. For the curve defined by \(y = \ln \cos x\),we use the curvature formula:\[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\]where \(y' = -\tan x\)and \(y'' = -\sec^2 x.\)By substituting these values, we simplify the expression:\[\kappa = \frac{|\sec^2 x|}{(1 + \tan^2 x)^{3/2}}\]Recognizing that \(1 + \tan^2 x = \sec^2 x,\)we further simplify to:\[\kappa = \frac{1}{\sec^3 x} = \cos^3 x\]within the range of \(-\pi/2 < x < \pi/2.\)Finally, we find where the curvature reaches its maximum by investigating the critical points through differentiation. In this exercise, the maximum occurs at \(x = 0,\)where the curve is perfectly flat and \(\kappa = 1,\)indicating maximum curvature directly at this point. Using curvature analysis, we find that our curve is most curved at this specific point.