Wanting to derive formula for curvature I stumbled upon problem with calculation. The formula I wanted to derive is as follows: $$\kappa=\frac{\lvert\frac{d\vec r}{dt}\times\frac{d^2\vec r}{dt^2}\rvert}{\lvert\frac{d\vec r}{dt}\rvert^3}$$
One of the definitions of curvature is rate of change of unit tangent vector with respect to arc length. i.e. $$\lvert\frac{d\hat T}{ds}\rvert=\lvert\frac{d\hat T}{dt}\cdot\frac{dt}{ds}\rvert=\lvert\frac{d\hat T}{dt}\rvert\cdot\frac{1}{\lvert ds/dt\rvert}$$ Therefore: $$\kappa=\lvert\frac{d}{dt}\Biggl(\frac{d\vec r/dt}{\lvert d\vec r/dt\rvert}\Biggr)\rvert\cdot\frac{1}{\lvert d\vec r/dt\rvert}$$ Calculating derivative by quotient rule and taking into account absolute value I am left with: $$\lvert\frac{d^2\vec r}{dt^2}\cdot\lvert d\vec r/dt\rvert-\frac{[d\vec r/dt]^2}{\lvert d\vec r/dt\rvert}\frac {d^2\vec r}{dt^2}\rvert\cdot\frac{1}{\lvert d\vec r/dt\rvert^3}$$since $$[d\vec r/dt]^2 = \lvert d\vec r/dt\rvert^2$$whole expression with absolute value reduces to zero...
HINT: To differentiate $|d\vec r/dt|$, you need to differentiate $$\left|\frac{d\vec r}{dt}\right|^2 = \frac{d\vec r}{dt}\cdot\frac{d\vec r}{dt}$$ by the product rule and then deduce what you want by the chain rule.