At What Point Does the Curve $y = \tan(x)$ Exhibit Maximum Curvature?

Unlocking the Geometry of $y = \tan(x)$: Finding Maximum Curvature

In differential geometry, the curvature $\kappa$ of a twice-differentiable function $y = f(x)$ is defined by the rigorous formula:

$$\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}$$

To find where the curve $y = \tan(x)$ achieves its maximum curvature, we must analyze its first and second derivatives.

The Mathematical Derivation

1. First Derivative ($y'$):

$$y' = \sec^2(x)$$

1. Second Derivative ($y''$):

$$y'' = 2\sec^2(x)\sec(x)\tan(x) = 2\sec^2(x)\tan(x)$$

Substituting these into the curvature formula gives:

$$\kappa(x) = \frac{|2\sec^2(x)\tan(x)|}{(1 + \sec^4(x))^{3/2}}$$

Locating the Peak

Curvature is maximized where the numerator is large relative to the denominator. For $y = \tan(x)$ on the interval $(-\pi/2, \pi/2)$, as $x$ approaches the vertical asymptotes at $\pm\pi/2$, both $\sec(x)$ and $\tan(x)$ grow infinitely large. However, the denominator's $(1 + \sec^4(x))^{3/2}$ grows at a cubic rate of $\sec^6(x)$, while the numerator grows roughly at the rate of $\sec^3(x) \cdot \tan(x) \approx \sec^4(x)$.

Consequently, as $x \to \pm\pi/2$, the curvature actually approaches zero because the denominator outpaces the numerator. Conversely, at $x = 0$ (the inflection point), the second derivative is zero, meaning $\kappa(0) = 0$.

By evaluating the derivative of the curvature function $\kappa'(x) = 0$, we find that the local maxima of curvature occur symmetrically between the inflection point and the asymptotes—specifically where the rate of change of the slope peaks relative to the squared slope itself.