Question

Find the principal unit normal vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=3 \cos t \mathbf{i}+3 \sin t \mathbf{j}, \quad t=\frac{\pi}{4} $$

Short answer

The principal unit normal vector to the curve at \(t=\frac{\pi}{4}\) is \(\mathbf{N}(\frac{\pi}{4}) = \frac{-1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\)

Step-by-step solution

Find the derivative of the position vector

The derivative of the position vector, which will give us the tangent vector, can be found using the standard rules for differentiation. The derivative is found as follows: \(\mathbf{r}'(t)=-3 \sin t \mathbf{i} + 3 \cos t \mathbf{j}\)

Normalize the tangent vector

Before we take the derivative of \(\mathbf{r}'(t)\) to find the normal vector, we need to normalize \(\mathbf{r}'(t)\). Normalizing means to make the vector unit length. This is achieved by dividing the vector with its magnitude. Let's call this normalized vector \(\mathbf{T}(t)\), where \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{| \mathbf{r}'(t) |}\).

Find the derivative of the unit tangent vector

Now that we have \(\mathbf{T}(t)\), the next step is to differentiate it to find the normal vector. So, \(\mathbf{T}'(t) = -\mathbf{i} - \mathbf{j}\).

Normalize the derivative of the unit tangent vector

The derivative \(\mathbf{T}'(t)\) is not necessarily a unit vector, so we need to normalize it again. After normalizing we get \(\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}\).

Find the principal unit normal vector at \(t=\frac{\pi}{4}\)

Substituting \(t=\frac{\pi}{4}\) into \(\mathbf{N}(t)\) will give us the principal unit normal vector at the specified value of the parameter, which is what the problem is asking for. So, \(\mathbf{N}(\frac{\pi}{4}) = \frac{-1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{j}\).