Question

Find the unit tangent vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=6 \cos t \mathbf{i}+2 \sin t \mathbf{j}, \quad t=\frac{\pi}{3} $$

Short answer

The unit tangent vector to the curve \( \mathbf{r}(t)=6 \cos t \mathbf{i}+2 \sin t \mathbf{j} \) at \( t=\frac{\pi}{3} \) is \( \mathbf{T}\left(\frac{\pi}{3}\right) = -\frac{3}{2} \mathbf{i}+ \frac{1}{2\sqrt{3}} \mathbf{j} \).

Step-by-step solution

Calculate the Derivative of the curve

To find the tangent vector, the first step is to compute the derivative of the curve. The derivative gives us a vector that points in the direction of the curve's motion at any point, which is the tangent vector. The derivative of \( \mathbf{r}(t) \) is:\[ \mathbf{r}'(t) = -6 \sin t \mathbf{i}+2 \cos t \mathbf{j}\]

Evaluate the Derivative at the given point

Now we substitute \( t=\frac{\pi}{3} \) into the derivative:\[ \mathbf{r}'\left(\frac{\pi}{3}\right) = -6 \sin \left(\frac{\pi}{3}\right) \mathbf{i}+2 \cos \left(\frac{\pi}{3}\right) \mathbf{j} = -6 \times \frac{\sqrt{3}}{2} \mathbf{i}+2 \times \frac{1}{2} \mathbf{j} = -3\sqrt{3} \mathbf{i}+ \mathbf{j}\]

Normalize the vector obtained

To make this tangent vector a unit vector, we normalize it by dividing it by its magnitude. The magnitude of vector \( \mathbf{r}'\left(\frac{\pi}{3}\right) \) is:\[ ||\mathbf{r}'\left(\frac{\pi}{3}\right)|| = \sqrt{(-3\sqrt{3})^2 + 1^2} = 2\sqrt{3}\]So the unit tangent vector \( \mathbf{T}(t) \) at \( t=\frac{\pi}{3} \) is:\[ \mathbf{T}\left(\frac{\pi}{3}\right) = \frac{\mathbf{r}'\left(\frac{\pi}{3}\right)}{||\mathbf{r}'\left(\frac{\pi}{3}\right)||} = \frac{-3\sqrt{3} \mathbf{i}+ \mathbf{j}}{2\sqrt{3}} = -\frac{3}{2} \mathbf{i}+ \frac{1}{2\sqrt{3}} \mathbf{j}\]