Question

Find the principal unit normal vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=6 \cos t \mathbf{i}+6 \sin t \mathbf{j}+\mathbf{k}, \quad t=\frac{3 \pi}{4} $$

Short answer

The principal unit normal vector to the curve at the given point is \( \frac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) \).

Step-by-step solution

Compute the Derivative \( r'(t) \)

Finding the derivative of the given vector \( r(t) = 6 \cos t \mathbf{i}+6 \sin t \mathbf{j}+\mathbf{k} \), we have \( r'(t) = -6 \sin t \mathbf{i} + 6 \cos t \mathbf{j} \).

Compute the Magnitude of \( r'(t) \)

The magnitude of \( r'(t) \) can be found by the formula \[ |r'(t)| = \sqrt{(-6 \sin t)^2 + (6 \cos t)^2} = \sqrt{36(\sin^2 t + \cos^2 t)} = 6 \].

Compute the Unit Tangent Vector \( T(t) \)

The unit tangent vector to the curve at any point \( t \) is given by the formula \( T(t) = \frac{r'(t)}{|r'(t)|} \). Substitute \( r'(t) \) and \( |r'(t)| \) from step 1 and step 2 respectively to obtain \( T(t) = \frac{-\sin t \mathbf{i} + \cos t \mathbf{j}}{6} \).

Compute the Derivative \( T'(t) \)

Finding the derivative of the vector \( T(t) \), we have \( T'(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \).

Compute the Magnitude of \( T'(t) \)

The magnitude of \( T'(t) \) can be found by the formula \[ |T'(t)| = \sqrt{(-\cos t)^2 + (- \sin t)^2} = \sqrt{\cos^2 t + \sin^2 t} = 1 \].

Compute the Principal Unit Normal Vector \( N(t) \)

Substitute \( T'(t) \) and \( |T'(t)| \) from step 4 and step 5 respectively into the formula \( N(t) = \frac{T'(t)}{|T'(t)|} \) to obtain \( N(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \).

Evaluate at the given Parameter Value

Finally, substitute \( t = \frac{3 \pi}{4} \) into \( N(t) \) to find the principal unit normal vector at the given point. \[ N \left( \frac{3 \pi}{4} \right) = -\cos \left( \frac{3 \pi}{4} \right) \mathbf{i} - \sin \left( \frac{3 \pi}{4} \right) \mathbf{j} \] After simplification, we get \[ N(t) = \frac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) \].