Question

A particle moves along the trajectory of parabola $y=a{x}^2$. Assuming speed to be constant, find the radius of curvature at origin.

• A. $\frac{1}{2a}$
• B. $\frac{1}{a}$
• C. $\frac{1}{2ax}$
• D. $\frac{2}{a}$

Answer 1

The correct option is A $\frac{1}{2a}$

By formula, radius of curvature for the curve in the form of $y=f\left(x\right)$ is given as $R=\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}{\left|\frac{d^{2}y}{{dx}^2}\right|}$
Here, $y=a{x}^2$, $\frac{dy}{dx}=2ax$, $\frac{d^{2}y}{{dx}^2}=2a$
Thus, $R=\frac{{[1+{\left(2ax\right)}^2]}^{3/2}}{\left|2a\right|}$
For $x=0,R=\frac{1}{2a}$