what is radius of curvature?.......
Answers
Answer:
In differential geometry, the radius of curvature, R, is the reciprocal of the curvature. For a curve, it equals the radius of the circular arc which best approximates the curve at that point. For surfaces, the radius of curvature is the radius of a circle that best fits a normal section or combinations thereof.
i hope this helps you
If γ : ℝ → ℝn is a parametrized curve in ℝn then the radius of curvature at each point of the curve, ρ : ℝ → ℝ, is given by[3]
{\displaystyle \rho ={\frac {\left|{\boldsymbol {\gamma }}'\right|^{3}}{\sqrt {\left|{\boldsymbol {\gamma }}'\right|^{2}\,\left|{\boldsymbol {\gamma }}''\right|^{2}-\left({\boldsymbol {\gamma }}'\cdot {\boldsymbol {\gamma }}''\right)^{2}}}}}{\displaystyle \rho ={\frac {\left|{\boldsymbol {\gamma }}'\right|^{3}}{\sqrt {\left|{\boldsymbol {\gamma }}'\right|^{2}\,\left|{\boldsymbol {\gamma }}''\right|^{2}-\left({\boldsymbol {\gamma }}'\cdot {\boldsymbol {\gamma }}''\right)^{2}}}}}.
As a special case, if f(t) is a function from ℝ to ℝ, then the radius of curvature of its graph, γ(t) = (t, f(t)), is
{\displaystyle \rho (t)={\frac {\left|1+f'^{\,2}(t)\right|^{\frac {3}{2}}}{\left|f''(t)\right|}}.}{\displaystyle \rho (t)={\frac {\left|1+f'^{\,2}(t)\right|^{\frac {3}{2}}}{\left|f''(t)\right|}}.}