Question

The function \(f(x)=\sin n x,\) where \(n\) is a positive real number, has a local maximum at \(x=\frac{\pi}{2 n}\). 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 at \(x = \frac{\pi}{2n}\) is equal to \(n^2\). As \(n\) increases, the curvature increases quadratically, making the function more curvy. Conversely, as \(n\) decreases, the curvature decreases, making the function less curvy.

Step-by-step solution

Find the first and second derivatives of \(f(x)\)

To find the curvature, we need to compute the second derivative of the function. As a starting point, let's first find the first and second derivatives of the function \(f(x) = \sin(nx)\). First derivative: $$f'(x) = \frac{d}{dx}[\sin(nx)] = n\cos(nx)$$ Second derivative: $$f''(x) = \frac{d^2}{dx^2}[\sin(nx)] = -n^2\sin(nx)$$

Compute the curvature formula and evaluate it at \(x = \frac{\pi}{2n}\)

The curvature, \(\kappa\), can be computed using the formula: $$\kappa(x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}$$ Now, we will evaluate the curvature at the point \(x = \frac{\pi}{2n}\) using the first and second derivatives computed in the previous step. First, evaluate the first and second derivatives at \(x = \frac{\pi}{2n}\): $$f'\left(\frac{\pi}{2n}\right) = n\cos\left(n\frac{\pi}{2n}\right) = n\cos\left(\frac{\pi}{2}\right) = 0$$ $$f''\left(\frac{\pi}{2n}\right) = -n^2\sin\left(n\frac{\pi}{2n}\right) = -n^2\sin\left(\frac{\pi}{2}\right) = -n^2$$ Now, substitute the evaluated values into the curvature formula: $$\kappa\left(\frac{\pi}{2n}\right) = \frac{|-n^2|}{(1 + (0)^2)^{3/2}} = \frac{n^2}{1} = n^2$$ So, the curvature \(\kappa\) at \(x = \frac{\pi}{2n}\) is equal to \(n^2\).

Analyze the effect of varying \(n\) on the curvature \(\kappa\)

Now that we have the curvature \(\kappa = n^2\) at \(x = \frac{\pi}{2n}\), we can analyze how the curvature varies as the value of \(n\) changes. As the value of \(n\) increases, the curvature \(\kappa\) increases quadratically. This means that for larger values of \(n\), the curvature of \(f(x)\) at \(x = \frac{\pi}{2n}\) is greater, indicating that the function becomes more "curvy" or bends more sharply at the given point. On the other hand, as \(n\) decreases, the curvature \(\kappa\) also decreases. This implies that for smaller values of \(n\), the curvature of \(f(x)\) at \(x = \frac{\pi}{2n}\) is less pronounced, and the function appears less curvy or bends less sharply at the specified point. In conclusion, the curvature of the function \(f(x) = \sin(nx)\) at \(x = \frac{\pi}{2n}\) varies quadratically with the value of \(n\). As \(n\) increases, the curvature increases, making the function more curvy, whereas, as \(n\) decreases, the curvature decreases, making the function less curvy.

Key concepts

Calculus and Curvature

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. A crucial application of calculus is in understanding the shapes and behaviors of curves, specifically through the concept of curvature. The curvature of a function gives us a quantitative measure of how much a curve deviates from being a straight line (flat curve) at a particular point. It’s a way to gauge the 'curviness' of the function.

In the context of our exercise, where the function given is \(f(x)=\sin(nx)\), calculus drives the calculation of its curvature \(\kappa\) at a specific point. This is done by leveraging derivatives, which are foundational tools in calculus for analyzing the rate at which something changes.

Derivatives and Their Role

Derivatives represent the rate at which a function is changing at any point and are fundamental in calculus. The first derivative of a function gives the slope of the tangent line at any point on the function's curve, which indicates the direction and rate of change of the function. For the function \(f(x)=\sin(nx)\), the first derivative \(f'(x)=n\cos(nx)\) tells us how \(f(x)\) is changing with respect to \(x\).

The second derivative provides information about the curvature and concavity of the function’s graph. It tells us how the rate of change itself is changing. For our function, the second derivative \(f''(x)=-n^2\sin(nx)\) is key to calculating curvature, which involves evaluating it at the point of interest—here, \(x=\frac{\pi}{2n}\). When the second derivative is evaluated at this point, it influences the curvature formula leading to a calculation that reflects how 'curvy' the graph is at that specific point.

Trigonometric Functions

Trigonometric functions—like sine, cosine, and tangent—are periodic functions that are pivotal in describing cyclical phenomena, including waves and oscillations. The function in our exercise, \(\sin(nx)\), is a trigonometric function with a frequency that is modified by the parameter \(n\). As \(n\) increases, the waves of the sine function become more tightly packed.

This attribute of \(\sin(nx)\), influenced by changes in \(n\), directly affects the curvature. At the local maximum \(x=\frac{\pi}{2n}\), the sine function has a particular curvature based on \(n\). Given that \(\sin\) of \(\frac{\pi}{2}\) is 1, the curvature is related to the square of \(n\) at this point. The analysis of trigonometric functions in calculus often involves such evaluations at specific points to determine properties like maxima, minima, and points of inflection, which convey essential insights about the function’s geometry and behavior.