Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the point of the curve at which the curvature is a maximum. \(y=\sin x ;-\pi \leq x \leq \pi\)

Step-by-Step Solution

Answer

Maximum curvature occurs at \(x = \frac{\pi}{2}\) and \(x = -\frac{\pi}{2}\) with a curvature of 1.

Step 1: Compute the First Derivative

The first derivative of the function gives the slope of the tangent at any point on the curve. For the given function, \(y = \sin x\), we calculate the first derivative: \(y' = \cos x\). This derivative will be used to find the curvature later.

Step 2: Compute the Second Derivative

The second derivative gives the rate of change of the slope, which we will need to calculate curvature. For \(y = \sin x\), the second derivative is \(y'' = -\sin x\).

Step 3: Compute the Curvature Formula

The formula for curvature \(\kappa\) at a point \(x\) on the curve \(y = f(x)\) is \(\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\). Substitute \(y' = \cos x\) and \(y'' = -\sin x\) into this formula: \(\kappa = \frac{|-\sin x|}{(1 + \cos^2 x)^{3/2}}\).

Step 4: Simplify the Curvature Expression

Simplify the curvature formula: \(\kappa = \frac{\sin x}{(1 + \cos^2 x)^{3/2}}\). Since curvature is the absolute value, we focus on the expression \(|\sin x|\).

Step 5: Find Critical Points of Absolute Curvature

We need to find the values of \(x\) that maximize \(|\sin x|\). Since \(|\sin x|\) reaches its maximum values of 1 when \(x = \frac{\pi}{2}\) and \(x = -\frac{\pi}{2}\) within the interval \(-\pi \leq x \leq \pi\). These are the critical points to consider.

Step 6: Evaluate Curvature at Critical Points

Calculate the curvature at \(x = \frac{\pi}{2}\): \(\kappa = \frac{1}{(1 + 0)^{3/2}} = 1\).Calculate the curvature at \(x = -\frac{\pi}{2}\):\(\kappa = \frac{1}{(1 + 0)^{3/2}} = 1\).The curvature value of 1 is consistent at both critical points. Since both points yield the same maximum curvature, we confirm that it is indeed the maximum on the interval.

Step 7: Confirm the Maximum Curvature Points

Since both \(x = \frac{\pi}{2}\) and \(x = -\frac{\pi}{2}\) result in a maximum curvature of 1, these are the points where the curvature is maximum.

Key Concepts

Curvature

Curvature describes how much a curve bends at a given point. It's a key concept in calculus that helps us understand the nature of a curve. Curvature can change at different points on a curve, providing insights into the shape of the curve.
For our problem with the function \(y = \sin x\), we're asked to find where the curvature is maximized. The formula for curvature \(\kappa\) is given by: \[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \] This formula combines both the first and second derivatives. It essentially measures how much the curve is deviating from being a straight line at any point. The greater the curvature, the sharper the bend at that point. In this problem, we found the maximum curvature at two points on the interval \(-\pi \leq x \leq \pi\). These points provide a deeper understanding of the behavior of the sine wave within one period.

First Derivative

The first derivative represents the slope of the tangent line to the curve at any given point. For the function \(y = \sin x\), the first derivative is \(y' = \cos x\). This tells us how the function is changing at a particular moment.
Here's what the first derivative indicates:

• If \(y' > 0\), the function is increasing.
• If \(y' < 0\), the function is decreasing.
• If \(y' = 0\), the function could be at a local maximum, minimum, or experiencing an inflection point.

The first derivative is crucial in calculating curvature. It shows not only if the curve is ascending or descending but also the steepness, which affects the curvature formula. Knowing \(y' = \cos x\) helps determine how the slope influences the curve's overall shape.

Second Derivative

The second derivative shows the rate of change of the first derivative. For the function \(y = \sin x\), the second derivative is \(y'' = -\sin x\). This derivative can inform us about the concavity of the function, which is important in understanding how a graph bends.
Here's what the second derivative tells us:

• If \(y'' > 0\), the graph is concave up at that point.
• If \(y'' < 0\), the graph is concave down at that point.
• If \(y'' = 0\), the graph may have an inflection point, where concavity changes.

The second derivative is vital for finding where the curvature is maximized, since it directly appears in the numerator of the curvature formula. For \(y = \sin x\), exploring \(y''\) helps us identify maximum bending points where the graph strongly curves.

Critical Points

Critical points are where the first derivative is zero or undefined, indicating potential maxima, minima, or points of inflection. In the context of the given exercise, we are particularly interested in critical points of the curvature function itself.
For the curvature \(\kappa\) of \(y = \sin x\), we simplified it to depend on \(|\sin x|\). The critical points for \(|\sin x|\) are where \(|\sin x|\) achieves maximum values. On the interval \(-\pi \leq x \leq \pi\), this occurs at \(x = \frac{\pi}{2}\) and \(x = -\frac{\pi}{2}\). Both these locations give us very valuable information because they correspond to where \(|\sin x| = 1\). These critical points represent the instances where the curve bends the most sharply, and the curvature is greatest. A thorough understanding of critical points allows us to pinpoint the areas of maximum interest in mathematical problems like these.