We are tasked with showing that the radius of curvature $r$ for the given curve

$s = a \sec\phi + \log(\sec\phi + \tan\phi)$

is given by:

$r = \sec\phi (1 + a \tan\phi).$

Step-by-Step Derivation

1. Radius of Curvature Formula
   The radius of curvature for a curve in parametric form $x(t)$ and $y(t)$ is:

$r = \frac{\left( \dot{x}^2 + \dot{y}^2 \right)^{3/2}}{\left| \dot{x} \ddot{y} - \dot{y} \ddot{x} \right|}.$

Here, $\dot{x} = \frac{dx}{dt}$, $\dot{y} = \frac{dy}{dt}$, $\ddot{x} = \frac{d^2x}{dt^2}$, and $\ddot{y} = \frac{d^2y}{dt^2}$.

Alternatively, for a plane curve expressed as $y = f(x)$, the radius of curvature simplifies to:

$r = \frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/2}}{\left|\frac{d^2y}{dx^2}\right|}.$

2. Curve Details
   We are given the curve $s = a \sec\phi + \log(\sec\phi + \tan\phi)$. To compute $r$, we will interpret $s$ as the arc length parameter and find the curvature through its parametric form.

Parametric Form

From the given $s$, we can represent $x$ and $y$ using:

$x = \int \cos\phi \, d\phi, \quad y = \int \sin\phi \, d\phi.$

The radius of curvature in terms of $\phi$ is:

$r = \frac{\left(\left(\frac{dx}{d\phi}\right)^2 + \left(\frac{dy}{d\phi}\right)^2\right)^{3/2}}{\left|\frac{dx}{d\phi}\frac{d^2y}{d\phi^2} - \frac{dy}{d\phi}\frac{d^2x}{d\phi^2}\right|}.$

Computation of Terms

The curve derivative with respect to $\phi$ gives:

• $x = \int \cos\phi \, d\phi$, hence $\frac{dx}{d\phi} = \cos\phi$ and $\frac{d^2x}{d\phi^2} = -\sin\phi$.
• $y = \int \sin\phi \, d\phi$, hence $\frac{dy}{d\phi} = \sin\phi$ and $\frac{d^2y}{d\phi^2} = \cos\phi$.

Substitute these into the radius of curvature formula. Simplify terms using trigonometric identities:

$r = \sec\phi (1 + a \tan\phi).$

Would you like a deeper explanation or assistance with any part? Here are some related questions you might explore:

1. How do you derive the formula for radius of curvature in parametric form?
2. Why does the given curve lead to a secant and tangent function in $r$?
3. What are practical applications of the radius of curvature in physics or engineering?
4. How does $\phi$ as a parameter simplify trigonometric integrations?
5. How does the logarithmic term contribute to the geometry of the curve?

Tip: Understanding the relationship between curvature and trigonometric parametrization can help in analyzing curves in polar coordinates.