Questions: Given r(t) = t i + t^2 j + t k, find the binormal vector B(0). Write your answer using standard unit vector notation.

Transcript text: Given $\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t \mathbf{k}$, find the binormal vector $\mathbf{B}(0)$. Write your answer using standard unit vector notation.

Solution

Solution Steps

To find the binormal vector $\mathbf{B}(0)$, we need to follow these steps:

Compute the tangent vector $\mathbf{T}(t)$ by differentiating $\mathbf{r}(t)$ with respect to $t$ and then normalizing it.
Compute the normal vector $\mathbf{N}(t)$ by differentiating $\mathbf{T}(t)$ with respect to $t$ and then normalizing it.
The binormal vector $\mathbf{B}(t)$ is the cross product of $\mathbf{T}(t)$ and $\mathbf{N}(t)$.
Evaluate $\mathbf{B}(t)$ at $t = 0$.

Step 1: Compute the Tangent Vector $\mathbf{T}(t)$

The tangent vector $\mathbf{T}(t)$ is obtained by differentiating $\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}$ with respect to $t$: \[ \mathbf{r}'(t) = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + \mathbf{k} \] Normalize $\mathbf{r}'(t)$ to get $\mathbf{T}(t)$: \[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{\mathbf{i} + 2t \mathbf{j} + \mathbf{k}}{\sqrt{1 + (2t)^2 + 1}} = \frac{\mathbf{i} + 2t \mathbf{j} + \mathbf{k}}{\sqrt{2 + 4t^2}} \]

Step 2: Compute the Normal Vector $\mathbf{N}(t)$

Differentiate $\mathbf{T}(t)$ with respect to $t$ to find $\mathbf{N}(t)$: \[ \mathbf{T}'(t) = \frac{d}{dt}\left(\frac{\mathbf{i} + 2t \mathbf{j} + \mathbf{k}}{\sqrt{2 + 4t^2}}\right) \] At $t = 0$, $\mathbf{T}(0) = \frac{\mathbf{i} + \mathbf{k}}{\sqrt{2}}$. The derivative simplifies to: \[ \mathbf{T}'(0) = \frac{2 \mathbf{j}}{\sqrt{2}} = \sqrt{2} \mathbf{j} \] Normalize $\mathbf{T}'(0)$ to get $\mathbf{N}(0)$: \[ \mathbf{N}(0) = \frac{\sqrt{2} \mathbf{j}}{\sqrt{2}} = \mathbf{j} \]

Step 3: Compute the Binormal Vector $\mathbf{B}(0)$

The binormal vector $\mathbf{B}(t)$ is the cross product of $\mathbf{T}(t)$ and $\mathbf{N}(t)$: \[ \mathbf{B}(0) = \mathbf{T}(0) \times \mathbf{N}(0) = \left(\frac{\mathbf{i} + \mathbf{k}}{\sqrt{2}}\right) \times \mathbf{j} \] Calculate the cross product: \[ \mathbf{B}(0) = \frac{1}{\sqrt{2}} (\mathbf{i} \times \mathbf{j} + \mathbf{k} \times \mathbf{j}) = \frac{1}{\sqrt{2}} (\mathbf{k} - \mathbf{i}) \] Simplify: \[ \mathbf{B}(0) = \frac{1}{\sqrt{2}} \mathbf{k} - \frac{1}{\sqrt{2}} \mathbf{i} = -\frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{k} \]