Question

Find the principal unit normal vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=\sqrt{2} t \mathbf{i}+e^{t} \mathbf{j}+e^{-t} \mathbf{k}, \quad t=0 $$

Short answer

The principal unit normal vector to the curve at \(t = 0\) is \([0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\).

Step-by-step solution

Compute the first derivative

\(\frac{d\mathbf{r}}{dt} = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k}\). For \(t = 0\), the derivative becomes \(\sqrt{2} \mathbf{i} + 1\mathbf{j} - 1\mathbf{k}\).

Normalize the first derivative

The magnitude of \(\frac{d\mathbf{r}}{dt}\) at \(t = 0\) is \(\sqrt{(\sqrt{2})^2 + 1^2 + (-1)^2} = 2\). The tangent unit vector \(\mathbf{T}\) at \(t = 0\) is \([\frac{\sqrt{2}}{2}, \frac{1}{2}, -\frac{1}{2}]\).

Compute the second derivative

The second derivative \(\frac{d^2\mathbf{r}}{dt^2} = e^{t} \mathbf{j} + e^{-t} \mathbf{k}\). For \(t = 0\), it results in \(1\mathbf{j} + 1\mathbf{k}\).

Normalize the second derivative

The magnitude of \(\frac{d^2\mathbf{r}}{dt^2}\) at \(t = 0\) is \(\sqrt{1^2 +1^2} = \sqrt{2}\). Hence, the derivative of the tangent unit vector \(\mathbf{T'}\) at \(t = 0\) is \([\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\).

Find the principal unit normal vector

The principal unit normal vector \(\mathbf{N}\) at \(t = 0\) can be found using the formula \(\mathbf{N} = \frac{\mathbf{T'}(t)}{||\mathbf{T'}(t)||}\). Substituting the values, we get \(\mathbf{N} = \frac{[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]}{||[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]||}\). The corresponding vector \(\mathbf{N}\) is \([0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\).