Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.4 Length Of A Plane Curve - Exercises Set 5.4 - Page 376: 27

$$L = \pi $$

$$\eqalign{ & x = \cos 2t,\,\,\,\,\,y = \sin 2t,\,\,\,\,\left( {0 \leqslant t \leqslant \pi /2} \right) \cr & {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\cos 2t} \right] = - 2\sin 2t \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\sin 2t} \right] = 2\cos 2t \cr & {\text{Use the arc length formula }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr & L = \int_0^{\pi /2} {\sqrt {{{\left( { - 2\sin 2t} \right)}^2} + {{\left( {2\cos 2t} \right)}^2}} dt} \cr & L = \int_0^{\pi /2} {\sqrt {4{{\sin }^2}2t + 4{{\cos }^2}2t} dt} \cr & L = \int_0^{\pi /2} {\sqrt {4\left( {{{\sin }^2}2t + {{\cos }^2}2t} \right)} dt} \cr & L = \int_0^{\pi /2} {2dt} \cr & {\text{Integrating}} \cr & L = 2\left[ t \right]_0^{\pi /2} \cr & L = 2\left( {\frac{\pi }{2} - 0} \right) \cr & L = \pi \cr} $$