MATH UPDATE: in order to keep up with modern design standards, we are rounding the vertex of f(x)=|x|. please adjust your calculations accordingly.
MATH UPDATE: in order to keep up with modern design standards, we are rounding the vertex of f(x)=|x|. please adjust your calculations accordingly. 2 comments
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@doskel
\{x \in (\frac{-r}{\sqrt{2}},\frac{+r}{\sqrt{2}}):r\sqrt{2}-\sqrt{r^2-x^2},|x|\}