It is a well-known fact that curvature of a circle with radius $R$ is equal to $\frac{1}{R}$. But if I write parameterization like $\gamma=(R\cos\frac{\tau}{R},R\sin\frac{\tau}{R})$ (because it must be a natural parametrization with $\vert \frac{d\gamma}{d\tau}\vert=1$) and then differentiate it twice I obtain $\frac{d^2\gamma}{d\tau^2}=(-\frac{1}{R}\cos\frac{\tau}{R},-\frac{1}{R}\sin\frac{\tau}{R})$ which gives $k=\vert \frac{d^2\gamma}{d\tau^2} \vert = \frac{\sqrt2}{R}$. What am I missing?..
Very stupid question about curvature of a circle
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calculus
differential-geometry
curves
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0Where did $\sqrt 2$ came? – 2017-02-11
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2$\cos^2{a}+\sin^2{a}=1$, not 2. – 2017-02-11
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0They say there are no stupid questions, only stupid answers. – 2017-02-11
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0Oh, sorry!.. Surely yes! I did a very stupid mistake: $\vert \frac{d^2\gamma}{d\tau^2} \vert= \sqrt{\frac{1}{R^2}(\cos^2\frac{\tau}{R}+\sin^2\frac{\tau}{R})} = \frac{1}{R}$, so everything is fine. – 2017-02-11
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3I've seen the same mistake made many times by students rushing through computations ... – 2017-02-11
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1The best part of this question was stopping to think at the end and knowing you'd made a mistake somewhere. You can answer your own question and highlight your error for others to learn from. – 2017-02-11