Find the principle unit normal vector to the curve given below at the specified point. $$\mathbf { r } ( t ) = t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } , \quad t = 3$$
A) $N ( 3 ) = \frac { - 24 } { \sqrt { 145 } } \mathbf { i } - \frac { 1 } { \sqrt { 145 } } \mathbf { j }$
B) $N ( 3 ) = \frac { - 24 } { \sqrt { 145 } } \mathbf { i } + \frac { 1 } { \sqrt { 145 } } \mathbf { j }$
C) $N ( 3 ) = \frac { - 24 } { \sqrt { 577 } } \mathbf { i } - \frac { 1 } { \sqrt { 577 } } \mathbf { j }$
D) $N ( 3 ) = \frac { - 24 } { \sqrt { 577 } } \mathbf { i } + \frac { 1 } { \sqrt { 577 } } \mathbf { j }$
E) $N ( 3 ) = \frac { 24 } { \sqrt { 577 } } \mathbf { i } - \frac { 1 } { \sqrt { 577 } } \mathbf { j }$

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The Answer of Find the principle unit normal vector to the curve given below at the specified point. $$\mathbf { r } ( t ) = t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } , \quad t = 3$$
A) $N ( 3 ) = \frac { - 24 } { \sqrt { 145 } } \mathbf { i } - \frac { 1 } { \sqrt { 145 } } \mathbf { j }$
B) $N ( 3 ) = \frac { - 24 } { \sqrt { 145 } } \mathbf { i } + \frac { 1 } { \sqrt { 145 } } \mathbf { j }$
C) $N ( 3 ) = \frac { - 24 } { \sqrt { 577 } } \mathbf { i } - \frac { 1 } { \sqrt { 577 } } \mathbf { j }$
D) $N ( 3 ) = \frac { - 24 } { \sqrt { 577 } } \mathbf { i } + \frac { 1 } { \sqrt { 577 } } \mathbf { j }$
E) $N ( 3 ) = \frac { 24 } { \sqrt { 577 } } \mathbf { i } - \frac { 1 } { \sqrt { 577 } } \mathbf { j }$

Find the Principle Unit Normal Vector to the Curve Given r(t)=ti+4t2j,t=3\mathbf { r } ( t ) = t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } , \quad t = 3r(t)=ti+4t2j,t=3

Find the Principle Unit Normal Vector to the Curve Given $\mathbf { r } ( t ) = t \mathbf { i } + 4 t ^ { 2 } \mathbf { j } , \quad t = 3$