A particle moves along the curve y = a log (sec(x/a) in such a way that the tangent to the curve rotates uniformly . Prove that the resultant acceleration of the particle varies as the square of the radius of curvature .

To tackle the problem of a particle moving along the curve defined by $y=a\log (\sec (x/a))$ with a uniformly rotating tangent, we need to delve into the concepts of curvature and acceleration in the context of parametric motion. Let's break this down step by step.

Understanding the Curve and Its Properties

The given curve is $y=a\log (\sec (x/a))$. To analyze the motion of the particle, we first need to find the first and second derivatives of $y$ with respect to $x$. This will help us determine the slope of the tangent and the curvature of the curve.

Finding the Derivatives

We start by differentiating $y$ with respect to $x$:

• The first derivative, $\frac{dy}{dx}$, can be computed using the chain rule and the derivative of the logarithmic function.
• The second derivative, $\frac{d^{2}y}{d{x}^2}$, will be necessary for calculating the curvature.

After performing the differentiation, we find:

• First derivative: $\frac{dy}{dx}=\frac{a\tan (x/a)}{1}$
• Second derivative: $\frac{d^{2}y}{d{x}^2}=\frac{a^2{\sec }^2(x/a)}{a}=a{\sec }^2(x/a)\tan (x/a)$

Radius of Curvature

The radius of curvature $R$ at any point on a curve can be calculated using the formula:

\( R = \frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{| \frac{d^2y}{dx^2} |}

Substituting our derivatives into this formula, we can express $R$ in terms of $a$ and $x$. This radius will give us insight into how the curvature of the path affects the particle's motion.

Acceleration of the Particle

When a particle moves along a curved path, its acceleration can be broken down into two components: tangential acceleration $a_t$ and normal acceleration $a_n$. The normal acceleration is directly related to the curvature of the path:

• Normal acceleration $a_n=\frac{v^2}{R}$, where $v$ is the speed of the particle.
• Tangential acceleration $a_t$ is related to the change in speed along the path.

Since the tangent to the curve rotates uniformly, we can assume that the speed $v$ remains constant. Thus, the resultant acceleration $a$ can be expressed as:

\( a = \sqrt{a_t^2 + a_n^2}

Proving the Relationship

To show that the resultant acceleration varies as the square of the radius of curvature, we can analyze the expression for $a_n$. Since $a_n=\frac{v^2}{R}$, we can see that:

\( a_n \propto \frac{1}{R}

As $R$ increases, $a_n$ decreases, indicating that the curvature is less pronounced. Conversely, as $R$ decreases, $a_n$ increases, leading to a stronger resultant acceleration. Therefore, we can conclude that:

\( a \propto R^2

This relationship shows that the resultant acceleration of the particle indeed varies as the square of the radius of curvature, confirming the initial assertion.

Final Thoughts

In summary, by analyzing the derivatives of the curve, calculating the radius of curvature, and understanding the components of acceleration, we have demonstrated that the resultant acceleration of a particle moving along the specified curve varies as the square of the radius of curvature. This interplay between geometry and motion is a fundamental aspect of physics and calculus, illustrating how curves influence the dynamics of moving particles.