Find the parameter $c$ such that the parabola $y = cx^2$ has curvature $40$ at the origin.
Any thoughts on where to start? I currently have $r(t) = \lbrace cx^2, y\rbrace $, $r'(t) = \lbrace 2cx,1\rbrace$ , $r''(t) = \lbrace 2c,0\rbrace $. Is my thinking right?
In rectangular coordinates if the curve is an explicit function of $y=y(x)$ curvature is directly
$$\kappa =\dfrac {|y''|}{\left(1+y'^{2}\right)^{\frac {3}{2}}} $$
so if $y=cx^2$ you have $y'=2cx;\;y''=2c$
substitute in the formula
$$\kappa(x)=\dfrac {|\,2c\,|}{\sqrt{\left(1+4c^2x^2\right)^3}} $$
and $\kappa(0)=|\,2c\,|$
$\kappa(0)=40\to |\,2c\,|=40\to c =\pm 20$
hope this helps