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, \(\kappa\), at the point \(x = \pi/(2n)\) is equal to \(n^2\). As \(n\) increases, the curvature also increases, and as \(n\) decreases, the curvature decreases.

Step-by-step solution

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

To find the first derivative, we use the chain rule: $$\frac{d}{dx}(\sin(nx)) = n\cos(nx).$$ So \(f'(x) = n\cos(nx)\).

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

Using the chain rule, we can find the second derivative: $$\frac{d^2}{dx^2}(\sin(nx)) = -n^2\sin(nx).$$ So \(f''(x) = -n^2\sin(nx)\).

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

At \(x=\pi/(2n)\), we have: $$f'(\pi/(2n)) = n\cos(n(\pi/(2n))) = n\cos(\pi/2) = 0,$$ and $$f''(\pi/(2n)) = -n^2\sin(n(\pi/(2n))) = -n^2\sin(\pi/2) = -n^2.$$

Compute the curvature \(\kappa\)

Plugging the values of the derivatives into the formula for curvature, we have: $$\kappa = \frac{\lvert -n^2 \rvert}{(1 + (0)^2)^{3/2}} = n^2.$$

Analyze the variation of \(\kappa\) with respect to \(n\)

From the expression for \(\kappa\), we see that it is equal to \(n^2\). Since \(n\) is a positive real number, as \(n\) increases, the curvature \(\kappa = n^2\) also increases. Similarly, if \(n\) decreases, the curvature also decreases. Thus, as \(n\) varies, the curvature \(\kappa = n^2\) also varies. The curvature increases with increasing \(n\) and decreases with decreasing \(n\).