Find the curvature κ(t) of the parabola x(t) = (t, t^2) for -∞ < t<∞ I've found d2x/dt2 to be 2 and dx/dt to be 2t using κ(t) = d2dt2/(1 + (dx/dt)^2)^3/2 is there any way I can easily simplify this?
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Originally Posted by Playthious is there any way I can easily simplify this? No, there isn't. Regards. Fernando Revilla
Originally Posted by Playthious Find the curvature κ(t) of the parabola x(t) = (t, t^2) for -∞ < t<∞ I've found d2x/dt2 to be 2 and dx/dt to be 2t using κ(t) = d2dt2/(1 + (dx/dt)^2)^3/2 I assume you mean (d2x/dt2)/(1+ (dx/dt)^2)^2/3 in your notation. is there any way I can easily simplify this? Well, you could put in the values you got for the first and second derivatives! $\displaystyle \kappa(t)= \frac{2}{(1+ 4t^2)^{3/2}}$
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