Question

The radius of curvature of curve y=f(x ) at any point on curve is

Answer 1

Answer:

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Answer 2

Answer:

The radius of curvature of a curve y= f (x) at a point is

$\dfrac{(1+(\dfrac{dy}{dx})^{2})^{3/2}}{|\dfrac{d^{2}y}{dx^{2}} |}$.

Explanation:

The distance from the vertex to the center of curvature is known as the radius of curvature (represented by R).

- In polar coordinates

$r=r(\Theta) , \\the radius of curvature \rho=\frac{1}{K} \frac{\left[r^{2}+\left(\frac{d r}{d \theta}\right)^{2}\right]^{3 / 2}}{\left|r^{2}+2\left(\frac{d r}{d \theta}\right)^{2}-r \frac{d^{2} r}{d \theta^{2}}\right|}$

$R=1 / K , where R is the radius of curvature and K is the curvature.\\ R=\frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{3 / 2}}{\left|\frac{d^{2} y}{d x^{2}}\right|} where K is the curvature of the curve, K=d T / d s , (Tangent vector function) \mathrm{R} the radius of curvature$