NBHM 2020 PART A Question 3 Solution

A point is moving along the curve $y = x^2$ with unit speed. What is the magnitude of its acceleration at the point $(1/2, 1/4)$?
Solution:
Fact: If a particle is in motion along a curved path at a constant speed then its acceleration at any point of its path is inversely proportional to the radius of curvature of the path at that point.  First, we find the curvature $K$ of $y=x^2$ at the point $(1/2, 1/4)$. Now, $$K = \frac{y{''}(x)}{((1+y^{'}(x))^2)^{\frac{3}{2}}} = \frac{2}{(1+4x+4x^2)^{\frac{3}{2}}}.$$Substituting the given point $x=\frac{1}{2}$ gives $K = \frac{1}{\sqrt{2}}$. Now, the radius of curvature = $\frac{1}{\text{curvature}} = \sqrt{2}$.
The required acceleration = radius of curvature = $\frac{1}{\text{curvature}} = \frac{1}{\sqrt{2}}$.

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