Find T, N, and B for the given space curve.
-$$\mathbf { r } ( \mathrm { t } ) = \left( \mathrm { t } ^ { 2 } - 9 \right) \mathbf { i } + ( 2 \mathrm { t } - 9 ) \mathbf { j } + 4 \mathbf { k }$$
A) $T = \frac { t } { t ^ { 2 } + 1 } \mathbf { i } + \frac { 1 } { t ^ { 2 } + 1 } j ; N = \frac { 1 } { t ^ { 2 } + 1 } \mathbf { i } - \frac { t } { t ^ { 2 } + 1 } j ; B = - k$
B) $T = \frac { t } { \sqrt { t ^ { 2 } + 1 } } \mathrm { i } + \frac { 1 } { \sqrt { t ^ { 2 } + 1 } } \mathrm { j } ; \mathbf { N } = \frac { 1 } { \sqrt { t ^ { 2 } + 1 } } \mathrm { i } - \frac { \mathrm { t } } { \sqrt { \mathrm { t } ^ { 2 } + 1 } } \mathrm { j } ; \mathbf { B } = - \mathbf { k }$
C) $T = \frac { t } { t ^ { 2 } + 1 } \mathbf { i } + \frac { 1 } { t ^ { 2 } + 1 } j ; N = \frac { 1 } { t ^ { 2 } + 1 } \mathbf { i } - \frac { t } { t ^ { 2 } + 1 } j ; B = k$
D) $T = \frac { t } { t ^ { 2 } + 1 } \mathrm { i } - \frac { 1 } { t ^ { 2 } + 1 } j ; N = - \frac { 1 } { t ^ { 2 } + 1 } i - \frac { t } { t ^ { 2 } + 1 } j ; B = - k$

Question

Quizplus · Accepted Answer

We start by finding the first derivative, which is 
$$\mathbf { r } ^ { ' } ( \mathrm { t } ) = 2 \mathrm { t } \mathbf { i } + 2 \mathbf { j }$$
The magnitude of $\mathbf { r } ^ { ' } ( \mathrm { t } )$ is 
$$\left\| \mathbf { r } ^ { ' } ( \mathrm { t } ) \right\| = \sqrt { 4 \mathrm { t } ^ { 2 } + 4 } = 2 \sqrt { \mathrm { t } ^ { 2 } + 1 }$$
So the unit tangent vector is 
$$\mathbf { T } = \frac { \mathbf { r } ^ { ' } ( \mathrm { t } ) } { \left\| \mathbf { r } ^ { ' } ( \mathrm { t } ) \right\| } = \frac { 2 \mathrm { t } \mathrm { i } + 2 \mathrm { j } } { 2 \sqrt { \mathrm { t } ^ { 2 } + 1 } } = \frac { \mathrm { t } } { \sqrt { \mathrm { t } ^ { 2 } + 1 } } \mathrm { i } + \frac { 1 } { \sqrt { \mathrm { t } ^ { 2 } + 1 } } \mathrm { j }$$
The second derivative is 
$$\mathbf { r } ^ { '' } ( \mathrm { t } ) = 2 \mathbf { i }$$
The magnitude of $\mathbf { r } ^ { '' } ( \mathrm { t } )$ is constant: $\left\| 2 \mathbf { i } \right\| = 2$. So the curvature is 
$$\kappa = \frac { \left\| \mathbf { r } ^ { '' } ( \mathrm { t } ) \right\| } { \left\| \mathbf { r } ^ { ' } ( \mathrm { t } ) \right\| ^ { 3 } } = \frac { 2 } { ( 2 \sqrt { \mathrm { t } ^ { 2 } + 1 } ) ^ { 3 } } = \frac { 1 } { ( \mathrm { t } ^ { 2 } + 1 ) ^ { 3 / 2 } }$$
The normal vector points in the direction of the centripetal acceleration, which is 
$$\mathbf { a } _ { c } = \kappa \left\| \mathbf { r } ^ { ' } ( \mathrm { t } ) \right\| ^ { 2 } \mathbf { N }$$
Taking the cross product of $\mathbf { r } ^ { ' } ( \mathrm { t } )$ and $\mathbf { a } _ { c }$ gives the binormal vector, $\mathbf { B } = \mathbf { T } \times \mathbf { N }$.

To find $\mathbf { N }$, we solve for $\mathbf { a } _ { c }$. Since $|\mathbf{r}'(t)| = 2\sqrt{t^2+1}$, we get 
$$\mathbf{a}_c = \kappa|\mathbf{r}'(t)|^2 |\mathbf{T}| $$
Substituting in the expressions we obtained, we get 
$$
\mathbf{a}_c = \frac{4t}{(t^2+1)^{3/2}} \begin{pmatrix}
t/\sqrt{t^2+1} \
1/\sqrt{t^2+1}
\end{pmatrix}
= \frac{4t^2}{(t^2+1)^{5/2}} \begin{pmatrix}
t \
1
\end{pmatrix}
$$
Hence, the magnitude of $\mathbf{a}_c$ is
$$
|\mathbf{a}_c| = \frac{4t^2}{(t^2+1)^{5/2}}\sqrt{t^2+1} = \frac{4t^2}{(t^2+1)^{3/2}}
$$
Since this is positive for all $t$, we get 
$$
\mathbf{N} = \frac{\mathbf{a}_c}{|\mathbf{a}_c|} = \frac{t}{\sqrt{t^2+1}} \mathbf{i} - \frac{1}{\sqrt{t^2+1}}\mathbf{j}
$$
Finally, we calculate the binormal vector:
$$
\mathbf{B} = \mathbf{T} \times \mathbf{N} = \begin{pmatrix}
t/\sqrt{t^2+1}\
1/\sqrt{t^2+1}\
0
\end{pmatrix} \times \begin{pmatrix}
t/\sqrt{t^2+1}\
-1/\sqrt{t^2+1}\
0
\end{pmatrix} = \begin{pmatrix}
0\
0\
-1
\end{pmatrix} = -\mathbf{k}
$$
Hence, the correct answer is B.

Find T, N, and B for the Given Space Curve r(t)=(t2−9)i+(2t−9)j+4k\mathbf { r } ( \mathrm { t } ) = \left( \mathrm { t } ^ { 2 } - 9 \right) \mathbf { i } + ( 2 \mathrm { t } - 9 ) \mathbf { j } + 4 \mathbf { k }r(t)=(t2−9)i+(2t−9)j+4k

Find T, N, and B for the Given Space Curve $\mathbf { r } ( \mathrm { t } ) = \left( \mathrm { t } ^ { 2 } - 9 \right) \mathbf { i } + ( 2 \mathrm { t } - 9 ) \mathbf { j } + 4 \mathbf { k }$