Chapter 12 - Vector-Valued Functions - 12.4 Exercises - Page 848: 28

$$\eqalign{ & {\bf{T}}\left( \pi \right) = - {\bf{j}} \cr & {\bf{N}}\left( \pi \right) = {\bf{i}} \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = 36 \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = 4\cos 3t{\bf{i}} + 4\sin 3t{\bf{j}},{\text{ }}t = \pi \cr & {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}{\bf{a}}\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {4\cos 3t{\bf{i}} + 4\sin 3t{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = - 12\sin 3t{\bf{i}} + 12\cos 3t{\bf{j}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ { - 12\sin 3t{\bf{i}} + 12\cos 3t{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = - 36\cos 3t{\bf{i}} - 36\sin 3t{\bf{j}} \cr & {\bf{a}}\left( \pi \right) = - 36\cos \left( {3\pi } \right){\bf{i}} - 36\sin \left( {3\pi } \right){\bf{j}} \cr & {\bf{a}}\left( \pi \right) = 36{\bf{i}} \cr & {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr & {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left\| {{\bf{v}}\left( t \right)} \right\|}},{\text{ }}{\bf{v}}\left( t \right) \ne 0 \cr & {\bf{T}}\left( t \right) = \frac{{ - 12\sin 3t{\bf{i}} + 12\cos 3t{\bf{j}}}}{{\sqrt {{{\left( { - 12\sin 3t} \right)}^2} + {{\left( {12\cos 3t} \right)}^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - 12\sin 3t{\bf{i}} + 12\cos 3t{\bf{j}}}}{{\sqrt {{{\left( { - 12\sin 3t} \right)}^2} + {{\left( {12\cos 3t} \right)}^2}} }} \cr & {\bf{T}}\left( t \right) = \frac{{ - 12\sin 3t{\bf{i}} + 12\cos 3t{\bf{j}}}}{{12}} \cr & {\bf{T}}\left( t \right) = - \sin 3t{\bf{i}} + \cos 3t{\bf{j}} \cr & {\bf{T}}\left( \pi \right) = - \sin 3\pi {\bf{i}} + \cos 3\pi {\bf{j}} \cr & {\bf{T}}\left( \pi \right) = - {\bf{j}} \cr & {\text{Find }}{\bf{T}}'\left( t \right) \cr & {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ { - \sin 3t{\bf{i}} + \cos 3t{\bf{j}}} \right] \cr & {\bf{T}}'\left( t \right) = - 3\cos 3t{\bf{i}} - 3\sin 3t{\bf{j}} \cr & {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = \pi \cr & {\bf{T}}'\left( \pi \right) = - 3\cos 3\pi {\bf{i}} - 3\sin 3\pi {\bf{j}} \cr & {\bf{T}}'\left( \pi \right) = 3{\bf{i}} \cr & {\text{Finding the Principal Unit Normal Vector}} \cr & {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr & {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = \pi \cr & {\bf{N}}\left( \pi \right) = \frac{{{\bf{T}}'\left( \pi \right)}}{{\left\| {{\bf{T}}'\left( \pi \right)} \right\|}} = \frac{{3{\bf{i}}}}{{\left\| {3{\bf{i}}} \right\|}} \cr & {\bf{N}}\left( \pi \right) = {\bf{i}} \cr & {\text{Find }}{a_{\bf{T}}}{\text{ and }}{a_{\bf{N}}}{\text{ at }}t = \pi \cr & {a_{\bf{T}}} = {\bf{a}} \cdot {\bf{T}} = \left( {36{\bf{i}}} \right)\left( { - {\bf{j}}} \right) \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = {\bf{a}} \cdot {\bf{N}} = \left( {36{\bf{i}}} \right)\left( {\bf{i}} \right) \cr & {a_{\bf{N}}} = 36 \cr & \cr & {\text{Summary}} \cr & {\bf{T}}\left( \pi \right) = - {\bf{j}} \cr & {\bf{N}}\left( \pi \right) = {\bf{i}} \cr & {a_{\bf{T}}} = 0 \cr & {a_{\bf{N}}} = 36 \cr} $$