Originally Posted by

**HallsofIvy** What, exactly, is your difficulty? The derivatives should be very easy. From r= a(1- cos t), r'= a sin t and r"= a cos t.

The numerator is $\displaystyle (a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$$\displaystyle = (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$$\displaystyle = (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

The denominator is $\displaystyle a^2(1- cos t)^2+ 2a^2 sin^2 t- a^2(1- cos t)(cos t)$$\displaystyle = a^2 - 2a^2 cos t+ a^2 cos^2 t+ 2a^2 sin^2 t- a^2 cos t+ a^2 cos^2 t$$\displaystyle = 3a^2- 3a^2 cos t= 3a^2(1- cos t)$

So the fraction is $\displaystyle \frac{2^{3/2}a^3 (1- cos t)^{3/2}}{3 a^2 (1- cos 6)}= \frac{2^{3/2}}{3}a (1- cos t)^{1/2}= \frac{2^{3/2}}{3}a \sqrt{1- cos t}$

Now use a "half-angle" formula: $\displaystyle sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}$