Chapter 12 - Vector-Valued Functions - 12.3 Change Of Parameter; Arc Length - Exercises Set 12.3 - Page 866: 12

$$L = \frac{{\sqrt 5 {\pi ^2}}}{2}$$

$$\eqalign{ & {\bf{r}}\left( t \right) = {t^2}{\bf{i}} + \left( {\cos t + t\sin t} \right){\bf{j}} + \left( {\sin t - t\cos t} \right){\bf{k}};\,\,\,\,\,\,\,\,\,\,\,0 \leqslant t \leqslant \pi \cr & \cr & {\text{Calculate }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}{\bf{i}} + \left( {\cos t + t\sin t} \right){\bf{j}} + \left( {\sin t - t\cos t} \right){\bf{k}}} \right] \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{t^2}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\left( {\cos t + t\sin t} \right)} \right]{\bf{j}} + \frac{d}{{dt}}\left[ {\left( {\sin t - t\cos t} \right)} \right]{\bf{k}} \cr & {\bf{r}}'\left( t \right) = 2t{\bf{i}} + \left( { - \sin t + \sin t + t\cos t} \right){\bf{j}} + \left( {\cos t - \cos t + t\sin t} \right){\bf{k}} \cr & {\bf{r}}'\left( t \right) = 2t{\bf{i}} + t\cos t{\bf{j}} + t\sin t{\bf{k}} \cr & \cr & \left\| {{\bf{r}}'\left( t \right)} \right\| = \left\| {2t{\bf{i}} + t\cos t{\bf{j}} + t\sin t{\bf{k}}} \right\| \cr & \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( {t\cos t} \right)}^2} + {{\left( {t\sin t} \right)}^2}} \cr & \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2} + {t^2}{{\cos }^2}t + {t^2}{{\sin }^2}t} \cr & \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2} + {t^2}\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \cr & \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {5{t^2}} \cr & \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt 5 t \cr & \cr & {\text{Find the arc length using }}L = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr & {\text{for the interval }}\,0 \leqslant t \leqslant \pi ,\,\,{\text{ }}a = 0{\text{ and }}b = \pi . \cr & L = \int_0^\pi {\sqrt 5 t} dt \cr & \cr & {\text{Integrate and evaluate}} \cr & L = \sqrt 5 \left[ {\frac{{{t^2}}}{2}} \right]_0^\pi \cr & L = \sqrt 5 \left[ {\frac{{{\pi ^2}}}{2} - \frac{{{0^2}}}{2}} \right] \cr & L = \frac{{\sqrt 5 {\pi ^2}}}{2} \cr} $$