Question

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(\theta)=2 \cos ^{3} \theta \mathbf{i}+3 \sin ^{3} \theta \mathbf{j} $$

Short answer

The curve given by the vector-valued function \( \mathbf{r}(\theta) \) is smooth on the open intervals \( (0, \pi) \) and \( (\pi, 2\pi) \).

Step-by-step solution

Define the Function in Component Form

The vector-valued function \( \mathbf{r}(\theta) \) is given as \(2 \cos ^{3} \theta \mathbf{i}+3 \sin ^{3} \theta \mathbf{j}\).

Differentiate the Function Components

Compute the derivative of the function by differentiating each component separately with respect to \( \theta \). The derivative of \(2 \cos ^{3} \theta \) is \( -6 \cos^{2}(\theta) \sin(\theta) \) and the derivative of \(3 \sin ^{3} \theta \) is \(9 \sin^{2}(\theta) \cos(\theta)\). So the derivative function \(\mathbf{r}'(\theta)\) is: \( -6 \cos^{2}(\theta) \sin(\theta) \mathbf{i} + 9 \sin^{2}(\theta) \cos(\theta) \mathbf{j} \).

Solve for When the Derivative Equals Zero

Solving for when the derivative is zero, involves setting both \(-6 \cos^{2}(\theta) \sin(\theta) = 0\) and \(9 \sin^{2}(\theta) \cos(\theta) = 0\).These two equations yield that the derivative is equal to zero when \( \theta \) equals \( 0, \pi, 2\pi \). Consequently, between these values the function is smooth.

Identifying the Interval of Smoothness

The derivative never equals zero in the open intervals \( (0, \pi) \) and \( (\pi, 2\pi) \). Therefore, these are the intervals on which the function \( \mathbf{r}(\theta) \) is smooth.