Question

Find the principal unit normal vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=t \mathbf{i}+\frac{1}{2} t^{2} \mathbf{j}, \quad t=2 $$

Short answer

The principal unit normal vector to the curve at t=2 is \( \mathbf{N}(2)= \frac{2\mathbf{i} - \mathbf{j}}{\sqrt{5}} \).

Step-by-step solution

Find the tangent vector

First, we will take the derivative of \( \mathbf{r}(t)\) to obtain the tangent vector \( \mathbf{T}(t)\). Given \( \mathbf{r}(t)=t \mathbf{i}+\frac{1}{2} t^{2} \mathbf{j}\), we differentiate with respect to t to find \( \mathbf{T}(t)= \mathbf{r}'(t)= \mathbf{i}+t\mathbf{j}\).

Normalize the tangent vector

Next we take the magnitude of \( \mathbf{T}(t)\), and divide \( \mathbf{T}(t)\) by its magnitude to get the Unit Tangent vector \( \mathbf{T}(t)\). The magnitude of \( \mathbf{T}(t)\) is: \(\|\mathbf{T}(t)\| = \sqrt{1^2 + t^2}\) and so the unit tangent vector is \(\mathbf{T}(t)= \frac{\mathbf{T}(t)}{\|\mathbf{T}(t)\|} = \frac{\mathbf{i}+t\mathbf{j}}{\sqrt{1^2 + t^2}}= \frac{\mathbf{i}+t\mathbf{j}}{\sqrt{1+t^2}}\) .

Step 3:Compute the derivative of the unit tangent vector

The normal vector is the derivative of unit tangent vector. Differentiating \( \mathbf{T}(t)\) with respect to t yields \(\mathbf{T}'(t)= \frac{t\mathbf{i} - \mathbf{j}}{(1+t^2)^\frac{3}{2}}\).

Normalize to find the principal unit normal vector

Lastly, we normalize this vector to find the principal unit normal vector. The magnitude of \( \mathbf{T}'(t)\) is: \( \|\mathbf{T}'(t)\| = \sqrt{t^2 + 1^2}\). So the principal unit normal vector \(\mathbf{N}(t)\) at t=2 is \(\mathbf{N}(2)= \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} = \frac{t\mathbf{i} - \mathbf{j}}{\sqrt{t^2 + 1}} = \frac{2\mathbf{i} - \mathbf{j}}{\sqrt{5}} \).