Chapter 12 - Vector-Valued Functions - Review Exercises - Page 864: 37

$$\eqalign{ & \left( {\text{a}} \right) \cr & {\bf{v}}\left( t \right) = \left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t,3} \right\rangle \cr & {\text{speed}} = 3\sqrt {{{\cos }^2}t{{\sin }^2}t + 1} \cr & {\bf{a}}\left( t \right) = \left\langle {6\cos t{{\sin }^2}t - 3{{\cos }^3}t,6\sin t{{\cos }^2}t - 3{{\sin }^3}t,0} \right\rangle \cr & \left( {\text{b}} \right) \cr & {\bf{v}}\left( \pi \right) = \left\langle {0,0,3} \right\rangle \cr & {\bf{a}}\left( \pi \right) = \left\langle {3,0,0} \right\rangle \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {{{\cos }^3}t,{{\sin }^3}t,3t} \right\rangle ,{\text{ }}t = \pi \cr & \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {{{\cos }^3}t,{{\sin }^3}t,3t} \right\rangle } \right] \cr & {\bf{v}}\left( t \right) = \left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t,3} \right\rangle \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t,3} \right\rangle } \right\| \cr & {\text{speed}} = \sqrt {{{\left( { - 3{{\cos }^2}t\sin t} \right)}^2} + {{\left( {3{{\sin }^2}t\cos t} \right)}^2} + {{\left( 3 \right)}^2}} \cr & {\text{speed}} = \sqrt {9{{\cos }^4}t{{\sin }^2}t + 9{{\sin }^4}t{{\cos }^2}t + 9} \cr & {\text{speed}} = \sqrt {9{{\cos }^2}t{{\sin }^2}t\left( {{{\cos }^2}t + {{\sin }^2}t} \right) + 9} \cr & {\text{speed}} = \sqrt {9{{\cos }^2}t{{\sin }^2}t + 9} \cr & {\text{speed}} = 3\sqrt {{{\cos }^2}t{{\sin }^2}t + 1} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t,3} \right\rangle } \right] \cr & {\text{By using the product rule and simplifying}} \cr & {\bf{a}}\left( t \right) = \left\langle {6\cos t{{\sin }^2}t - 3{{\cos }^3}t,6\sin t{{\cos }^2}t - 3{{\sin }^3}t,0} \right\rangle \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right){\text{ and }}{\bf{a}}\left( t \right){\text{ at }}t = \pi \cr & {\bf{v}}\left( \pi \right) = \left\langle { - 3{{\cos }^2}\pi \sin \pi ,3{{\sin }^2}\pi \cos \pi ,3} \right\rangle \cr & {\bf{v}}\left( \pi \right) = \left\langle {0,0,3} \right\rangle \cr & {\bf{a}}\left( \pi \right) = \left\langle {6\cos \pi {{\sin }^2}\pi - 3{{\cos }^3}\pi ,6\sin \pi {{\cos }^2}\pi - 3{{\sin }^3}\pi ,0} \right\rangle \cr & {\bf{a}}\left( \pi \right) = \left\langle {3,0,0} \right\rangle \cr & \cr & {\text{Summary}} \cr & \left( {\text{a}} \right) \cr & {\bf{v}}\left( t \right) = \left\langle { - 3{{\cos }^2}t\sin t,3{{\sin }^2}t\cos t,3} \right\rangle \cr & {\text{speed}} = 3\sqrt {{{\cos }^2}t{{\sin }^2}t + 1} \cr & {\bf{a}}\left( t \right) = \left\langle {6\cos t{{\sin }^2}t - 3{{\cos }^3}t,6\sin t{{\cos }^2}t - 3{{\sin }^3}t,0} \right\rangle \cr & \left( {\text{b}} \right) \cr & {\bf{v}}\left( \pi \right) = \left\langle {0,0,3} \right\rangle \cr & {\bf{a}}\left( \pi \right) = \left\langle {3,0,0} \right\rangle \cr} $$