Chapter 12 - Vector-Valued Functions - Review Exercises - Page 864: 52

\[\begin{align} & \mathbf{T}\left( \frac{\pi }{6} \right)=-\frac{\sqrt{3}}{4}\mathbf{i}+\frac{1}{4}\mathbf{j} \\ & \mathbf{N}\left( \frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}\mathbf{i}-\frac{1}{2}\mathbf{j} \\ & {{a}_{\mathbf{T}}}=0 \\ & {{a}_{\mathbf{N}}}=0 \\ \end{align}\]

\[\begin{align} & \mathbf{r}\left( t \right)=3\cos 2t\mathbf{i}+3\sin 2t\mathbf{j},\text{ }t=\frac{\pi }{6} \\ & \text{Calculate }\mathbf{v}\left( t \right)\text{ and }\mathbf{a}\left( t \right) \\ & \mathbf{v}\left( t \right)=\mathbf{r}'\left( t \right) \\ & \mathbf{v}\left( t \right)=\frac{d}{dt}\left[ 3\cos 2t\mathbf{i}+3\sin 2t\mathbf{j} \right] \\ & \mathbf{v}\left( t \right)=-6\sin 2t\mathbf{i}+6\cos 2t\mathbf{j} \\ & \mathbf{a}\left( t \right)=\mathbf{v}'\left( t \right) \\ & \mathbf{a}\left( t \right)=\frac{d}{dt}\left[ -6\sin 2t\mathbf{i}+6\cos 2t\mathbf{j} \right] \\ & \mathbf{a}\left( t \right)=-12\cos 2t\mathbf{i}-12\sin 2t\mathbf{j} \\ & \mathbf{a}\left( \frac{\pi }{6} \right)=-12\cos \left( \frac{\pi }{3} \right)\mathbf{i}-12\sin \left( \frac{\pi }{3} \right)\mathbf{j} \\ & \mathbf{a}\left( \frac{\pi }{6} \right)=-6\mathbf{i}-6\sqrt{3}\mathbf{j} \\ & \text{Find the unit Tangent Vector }\mathbf{T}\left( t \right) \\ & \mathbf{T}\left( t \right)=\frac{\mathbf{v}\left( t \right)}{\left\| \mathbf{v}\left( t \right) \right\|},\text{ }\mathbf{v}\left( t \right)\ne 0 \\ & \mathbf{T}\left( t \right)=\frac{-12\cos 2t\mathbf{i}-12\sin 2t\mathbf{j}}{\left\| -12\cos 2t\mathbf{i}-12\sin 2t\mathbf{j} \right\|}=\frac{-6\sin 2t\mathbf{i}+6\cos 2t\mathbf{j}}{12} \\ & \mathbf{T}\left( t \right)=-\frac{1}{2}\sin 2t\mathbf{i}+\frac{1}{2}\cos 2t\mathbf{j} \\ & \mathbf{T}\left( \frac{\pi }{6} \right)=-\frac{1}{2}\sin 2\left( \frac{\pi }{6} \right)\mathbf{i}+\frac{1}{2}\cos 2\left( \frac{\pi }{6} \right)\mathbf{j} \\ & \mathbf{T}\left( \frac{\pi }{6} \right)=-\frac{\sqrt{3}}{4}\mathbf{i}+\frac{1}{4}\mathbf{j} \\ & \text{Find }\mathbf{T}'\left( t \right) \\ & \mathbf{T}'\left( t \right)=\frac{d}{dt}\left[ -\frac{1}{2}\sin 2t\mathbf{i}+\frac{1}{2}\cos 2t\mathbf{j} \right] \\ & \mathbf{T}'\left( t \right)=-\cos 2t\mathbf{i}-\sin 2t\mathbf{j} \\ & \text{Evaluate }\mathbf{T}'\left( t \right)\text{ at }t=\frac{\pi }{6} \\ & \mathbf{T}'\left( \frac{\pi }{6} \right)=-\cos 2\left( \frac{\pi }{6} \right)\mathbf{i}-\sin 2\left( \frac{\pi }{6} \right)\mathbf{j} \\ & \mathbf{T}'\left( \frac{\pi }{6} \right)=-\frac{1}{2}\mathbf{i}-\frac{\sqrt{3}}{2}\mathbf{j} \\ & \text{Finding the Principal Unit Normal Vector} \\ & \mathbf{N}\left( t \right)=\frac{\mathbf{T}'\left( t \right)}{\left\| \mathbf{T}'\left( t \right) \right\|} \\ & \text{Evaluate }\mathbf{N}'\left( t \right)\text{ at }t=\frac{\pi }{6} \\ & \mathbf{N}\left( \frac{\pi }{6} \right)=\frac{\mathbf{T}'\left( \pi /6 \right)}{\left\| \mathbf{T}'\left( \pi /6 \right) \right\|}=\frac{-\frac{1}{2}\mathbf{i}-\frac{\sqrt{3}}{2}\mathbf{j}}{\left\| -\frac{1}{2}\mathbf{i}-\frac{\sqrt{3}}{2}\mathbf{j} \right\|}=\frac{\sqrt{3}}{2}\mathbf{i}-\frac{1}{2}\mathbf{j} \\ & \text{Find }{{a}_{\mathbf{T}}}\text{ and }{{a}_{\mathbf{N}}} \\ & {{a}_{\mathbf{T}}}=\mathbf{a}\cdot \mathbf{T}=\left( -6\mathbf{i}-6\sqrt{3}\mathbf{j} \right)\left( -\frac{\sqrt{3}}{4}\mathbf{i}+\frac{1}{4}\mathbf{j} \right) \\ & {{a}_{\mathbf{T}}}=0 \\ & {{a}_{\mathbf{N}}}=\mathbf{a}\cdot \mathbf{N}=\left( 6\mathbf{i}-6\sqrt{3}\mathbf{j} \right)\left( \frac{\sqrt{3}}{2}\mathbf{i}-\frac{1}{2}\mathbf{j} \right) \\ & {{a}_{\mathbf{N}}}=0 \\ & \\ & \text{Summary} \\ & \mathbf{T}\left( \frac{\pi }{6} \right)=-\frac{\sqrt{3}}{4}\mathbf{i}+\frac{1}{4}\mathbf{j} \\ & \mathbf{N}\left( \frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}\mathbf{i}-\frac{1}{2}\mathbf{j} \\ & {{a}_{\mathbf{T}}}=0 \\ & {{a}_{\mathbf{N}}}=0 \\ \end{align}\]