Question

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(\theta)=(\theta-2 \sin \theta) \mathbf{i}+(1-2 \cos \theta) \mathbf{j} $$

Short answer

The curve is smooth on the open interval \((- \infty, +\infty)\).

Step-by-step solution

Compute the Derivatives

Begin by taking the derivative of each component of the vector-valued function. The derivative of \((\theta-2 \sin \theta) \mathbf{i}\) with respect to \(\theta\) is \(\mathbf{i} + 2 \cos(\theta) \mathbf{i}\). For the \(j\) component, the derivative of \((1-2 \cos \theta)\mathbf{j}\) with respect to \(\theta\) is \(2 \sin(\theta) \mathbf{j}\).

Examine the Derivatives

The derivatives \(\mathbf{i} + 2 \cos(\theta)\) and \(2 \sin(\theta)\) are defined for all real numbers. Therefore, these components do not introduce any non-smoothness in the curve.

Conclusion

Since the derivatives exist for all real numbers, the curve is smooth on the entire real line. In terms of interval notation, this is \((- \infty, +\infty)\).