Answer to Question #286743 in Differential Geometry | Topology for Riya

Question #286743

Find the curvature, the radius and the center of curvature at a point.


r=1+ cos theta ,theta=π/2



1
Expert's answer
2022-01-13T17:29:08-0500

"r=f(\\theta)=1+cos\\theta \\Rightarrow f(\\frac{\\pi}{2})=1\\\\\nf'(\\theta)=-sin\\theta\\Rightarrow f'(\\frac{\\pi}{2})=-1\\\\\nf''(\\theta)=-cos\\theta \\Rightarrow f''(\\frac{\\pi}{2})=0\\\\\n\\therefore \\text{Radius of curvature}\\\\= \\dfrac{(f(\\frac{\\pi}{2})^2+f'(\\frac{\\pi}{2})^2)^\\frac{3}{2}}\n{f(\\frac{\\pi}{2})^2+2f'(\\frac{\\pi}{2})^2-f(\\frac{\\pi}{2})f''(\\frac{\\pi}{2})}\\\\\n=\\dfrac{2^\\frac{3}{2}}{3}=\\dfrac{2\\sqrt 2}{3}\\\\\n\\text{Now, curvature $=$}\\dfrac{1}{\\text{Radius of Curvature}}=\\dfrac{3}{2\\sqrt 2}"



"\\text{And, Center of curvature = }((y_0+R),(x_0+f'(x_0))R)\\\\\n\\texttt{[Where $x_0=\\frac{\\pi}{2}, y_0=f(\\frac{\\pi}{2})=1$]}\\\\\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\n(1+ \\frac{2\\sqrt 2}{3}, (\\frac{\\pi}{2}-1)\\frac{2\\sqrt 2}{3})\\\\"



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