Find T, N, and B for the given space curve.
-$$r ( t ) = \frac { 4 } { 3 } ( 1 + t ) ^ { 3 / 2 } \mathbf { i } + \frac { 4 } { 3 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + 1 \mathrm { tk }$$
A) $T = \frac { 2 } { 3 } \sqrt { 1 + t } \mathbf { i } + \frac { 2 } { 3 } \sqrt { 1 - t } j + \frac { 1 } { 3 } k ; N = \frac { 1 } { 2 } \sqrt { 2 - 2 } \mathrm { ti } + \frac { 1 } { 2 } \sqrt { 2 + 2 t j } ; B = \frac { 1 } { 6 } \sqrt { 2 + 2 t i } + \frac { 1 } { 6 } \sqrt { 2 - 2 } \mathbf { j } + \frac { 2 \sqrt { 2 } } { 3 } \mathbf { k }$
B) $\mathrm { T } = \frac { 2 } { 3 } \sqrt { 1 + \mathrm { t } } \mathrm { i } - \frac { 2 } { 3 } \sqrt { 1 - \mathrm { t } } \mathrm { j } + \frac { 1 } { 3 } \mathrm { k } ; \mathrm { N } = \frac { 1 } { 2 } \sqrt { 2 - 2 \mathrm { t } } + \frac { 1 } { 2 } \sqrt { 2 + 2 \mathrm { t } } ; \mathbf { B } = - \frac { 1 } { 6 } \sqrt { 2 + 2 \mathrm { t } } \mathrm { i } + \frac { 1 } { 6 } \sqrt { 2 - 2 \mathrm { t } } \mathbf { j } + \frac { 2 \sqrt { 2 } } { 3 } \mathrm { k }$
C) $\mathrm { T } = 2 \sqrt { 1 + \mathrm { t } } - 2 \sqrt { 1 - t } \mathbf { j } + 1 \mathbf { k } ; \mathbf { N } = \frac { 3 } { 2 } \sqrt { 2 - 2 \mathrm { t } } + \frac { 3 } { 2 } \sqrt { 2 + 2 t } \mathbf { j } ; \mathbf { B } = - \frac { 1 } { 2 } \sqrt { 2 + 2 t \mathrm { i } } + \frac { 1 } { 2 } \sqrt { 2 - 2 \mathrm { t } } + 2 \sqrt { 2 } \mathbf { k }$
D) $\mathrm { T } = \frac { 1 } { 3 } \sqrt { 1 + \mathrm { t } } - \frac { 1 } { 3 } \sqrt { 1 - \mathrm { t } } \mathrm { j } + \frac { 1 } { 3 } \mathrm { k } ; \mathrm { N } = \frac { 1 } { 2 } \sqrt { 2 - 2 \mathrm { t } } \mathrm { i } + \frac { 1 } { 2 } \sqrt { 2 + 2 \mathrm { t } } ; \mathbf { B } = - \frac { 1 } { 6 } \sqrt { 2 + 2 \mathrm { t } } \mathrm { i } + \frac { 1 } { 6 } \sqrt { 2 - 2 \mathrm { t } } + \frac { \sqrt { 2 } } { 3 } \mathbf { k }$

Question

Quizplus · Accepted Answer

To find the tangent vector $\mathbf{T}(t)$, we first take the derivative of the position vector $\mathbf{r}(t)$, which gives

$$\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=\frac{4}{3}\cdot\frac{3}{2}(1+t)^{1/2}\mathbf{i}-\frac{4}{3}\cdot\frac{3}{2}(1-t)^{1/2}\mathbf{j}+\mathbf{k}=\frac{2\sqrt{1+t}}{3}\mathbf{i}-\frac{2\sqrt{1-t}}{3}\mathbf{j}+\mathbf{k}$$

Then, we normalize this vector by dividing it by its magnitude:

$$\mathbf{T}(t)=\frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}=\frac{\frac{2\sqrt{1+t}}{3}\mathbf{i}-\frac{2\sqrt{1-t}}{3}\mathbf{j}+\mathbf{k}}{\sqrt{\left(\frac{2\sqrt{1+t}}{3}ight)^2+\left(-\frac{2\sqrt{1-t}}{3}ight)^2+1}}=\frac{2\sqrt{1+t}\mathbf{i}-2\sqrt{1-t}\mathbf{j}+\mathbf{k}}{\sqrt{2+2t}}$$

To find the normal vector $\mathbf{N}(t)$, we take the derivative of $\mathbf{T}(t)$, which gives

$$\mathbf{a}(t)=\frac{d\mathbf{T}}{dt}=\frac{d}{dt}\left(\frac{2\sqrt{1+t}\mathbf{i}-2\sqrt{1-t}\mathbf{j}+\mathbf{k}}{\sqrt{2+2t}}ight)=\frac{1}{\sqrt{2+2t}}\left(\frac{\sqrt{1+t}}{3}\mathbf{i}+\frac{\sqrt{1-t}}{3}\mathbf{j}ight)$$

Finally, we normalize this vector to obtain $\mathbf{N}(t)$:

$$\mathbf{N}(t)=\frac{\mathbf{a}(t)}{\|\mathbf{a}(t)\|}=\frac{\frac{\sqrt{1+t}}{3}\mathbf{i}+\frac{\sqrt{1-t}}{3}\mathbf{j}}{\sqrt{\left(\frac{\sqrt{1+t}}{3}ight)^2+\left(\frac{\sqrt{1-t}}{3}ight)^2}}=\frac{\sqrt{2-t}\mathbf{i}+\sqrt{2+t}\mathbf{j}}{\sqrt{2}}$$

Finally, we can find the binormal vector $\mathbf{B}(t)$ using the cross product:

$$\mathbf{B}(t)=\mathbf{T}(t)	imes\mathbf{N}(t)=\left(\frac{2\sqrt{1+t}}{3}\mathbf{i}-\frac{2\sqrt{1-t}}{3}\mathbf{j}+\mathbf{k}ight)	imes\left(\frac{\sqrt{2-t}}{\sqrt{2}}\mathbf{i}+\frac{\sqrt{2+t}}{\sqrt{2}}\mathbf{j}ight)=-\frac{\sqrt{2+2t}}{6}\mathbf{i}+\frac{\sqrt{2-2t}}{6}\mathbf{j}+\frac{2\sqrt{2}}{3}\mathbf{k}$$

Find T, N, and B for the Given Space Curve r(t)=43(1+t)3/2i+43(1−t)3/2j+1tkr ( t ) = \frac { 4 } { 3 } ( 1 + t ) ^ { 3 / 2 } \mathbf { i } + \frac { 4 } { 3 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + 1 \mathrm { tk }r(t)=34​(1+t)3/2i+34​(1−t)3/2j+1tk

Find T, N, and B for the Given Space Curve $r ( t ) = \frac { 4 } { 3 } ( 1 + t ) ^ { 3 / 2 } \mathbf { i } + \frac { 4 } { 3 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + 1 \mathrm { tk }$