Question

The function \(f(x)=\sin n x,\) where \(n\) is a positive real number, has a local maximum at \(x=\pi /(2 n)\) Compute the curvature \(\kappa\) of \(f\) at this point. How does \(\kappa\) vary (if at all) as \(n\) varies?

Short answer

Answer: The curvature of the function \(f(x)=\sin nx\) at \(x=\pi/(2n)\) is given by \(\kappa=n^2\). As \(n\) increases, the curvature \(\kappa\) increases, and as \(n\) decreases, the curvature \(\kappa\) decreases.

Step-by-step solution

Find the first derivative of \(f(x)\)

To compute \(f'(x)\), we need to differentiate \(f(x)=\sin nx\) with respect to \(x\). Applying the chain rule, we have: \(f'(x)=\frac{d(\sin nx)}{dx}=\cos nx \cdot n\)

Find the second derivative of \(f(x)\)

To compute \(f''(x)\), we need to differentiate \(f'(x)=n\cos nx\) with respect to \(x\). Again, applying the chain rule, we have: \(f''(x)=\frac{d(n\cos nx)}{dx}=-n^2\sin nx\)

Evaluate the first and second derivatives at \(x=\pi/(2n)\)

We're interested in finding the curvature at \(x=\pi/(2n)\), so let's evaluate the derivatives at this point: \(f'(\pi/(2n))=n\cos \left(n\left(\frac{\pi}{2n}\right) \right)=n\cos(\pi/2)=0\) \(f''(\pi/(2n))=-n^2\sin \left(n\left(\frac{\pi}{2n}\right)\right)=-n^2\sin(\pi/2)=-n^2\)

Compute the curvature \(\kappa\)

The formula for curvature \(\kappa\) for a function is given by: \(\kappa = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}\) Now, we'll plug in our values of \(f'(\pi/(2n))\) and \(f''(\pi/(2n))\): \(\kappa = \frac{|-n^2|}{(1+(0)^2)^{3/2}}=\frac{n^2}{1}=n^2\)

Analyze how \(\kappa\) changes as \(n\) varies

Since \(\kappa=n^2\), we can conclude that the curvature at \(x=\pi/(2n)\) varies as the square of \(n\). When \(n\) increases, the curvature \(\kappa\) increases, and when \(n\) decreases, the curvature \(\kappa\) decreases.

Key concepts

Derivatives

Derivatives are a fundamental concept in calculus, used to determine how a function changes at any given point. They represent the rate of change or the slope of the function. In the given exercise, the first derivative of the function \( f(x) = \sin nx \) is obtained using the chain rule. This first derivative, \( f'(x) \), provides the slope of the tangent line to the curve at any point \( x \). Derivatives help us find maxima, minima, and points of inflection in a function, which are crucial for understanding the shape and behavior of graphs.
To find the first derivative, \( f'(x) \), we use the fact that the derivative of \( \sin u \) is \( \cos u \). Applying this to \( f(x) \), we have \( f'(x) = \cos nx \cdot n \), which simplifies to \( n \cos nx \). This tells us how the sine function transforms as \( n \) changes.
Derivatives have numerous applications, including optimization problems and modeling the real-world processes where rates of change need to be monitored and controlled.

Chain Rule

The chain rule is a powerful derivative technique used when differentiating composite functions. It enables us to unravel these functions into simpler components. In the context of this exercise, it's crucial for differentiating \( f(x) = \sin nx \).
The chain rule states that if a function \( y = g(f(x)) \) is the composition of two functions, its derivative is given by \( y' = g'(f(x)) \cdot f'(x) \). For the exercise, we differentiate the inside function, \( nx \), first, and then the outside function, \( \sin(u) \), where \( u = nx \).
Here's how this works in practice: when differentiating \( \sin nx \) with respect to \( x \), we follow these steps:

• Differentiate \( nx \) with respect to \( x \), which gives \( n \).
• Differentiate \( \sin nx \) with respect to \( nx \) to get \( \cos nx \).
• Multiply the two results: \( n \cos nx \).

This step is essential for finding the curvature of the function at the specified point. Understanding the chain rule is key to working with complex functions productively.

Trigonometric Functions

Trigonometric functions like sine and cosine play a significant role in mathematics, especially calculus. These functions describe relationships within a right-angled triangle and the unit circle, lending them to modeling oscillations and waves in various fields.
In the problem at hand, \( f(x) = \sin nx \) is a trigonometric function that oscillates between -1 and 1. The frequency of oscillation is controlled by the parameter \( n \), affecting how often peaks and troughs occur over a given interval. This periodic nature of trigonometric functions makes them particularly useful in analyzing wave patterns and circular motion.
Using trigonometric identities simplifies the derivation process. For example, when evaluating the derivative at specific points, we rely on properties such as \( \cos(\pi/2) = 0 \) and \( \sin(\pi/2) = 1 \), which were useful in computing the curvature at \( x = \pi/(2n) \). These identities enable accurate computation and deep insights into mathematical models using trigonometric functions.

Curvature Formula

The curvature formula quantifies how sharply a curve bends and is given by\[\kappa = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}\].
Curvature measures the rate of change of the curve's direction. A higher curvature indicates a tighter bend, while a lower curvature suggests a more gradual turn.
In this problem, we are asked to evaluate the curvature at \( x = \pi/(2n) \). To do this, we need both the first and second derivatives of the function, \( f(x) = \sin nx \). We find:

• First derivative, \( f'(x) = n \cos nx \).
• Second derivative, \( f''(x) = -n^2 \sin nx \).

Upon evaluating these derivatives at \( x = \pi/(2n) \), we find that \( f'(\pi/(2n)) = 0 \) and \( f''(\pi/(2n)) = -n^2 \). Thus, the curvature simplifies to \( \kappa = n^2 \).
The formula showcases how the curvature changes with different values of \( n \). As \( n \) increases, the curvature \( \kappa \) becomes larger, illustrating a tighter bend in the function's graph.