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Chapter 10: Problem 31

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

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

Step by step solution

01

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 \(2 \sin(\theta) \mathbf{j}\).
02

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.
03

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 \((- \infty, +\infty)\).

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Most popular questions from this chapter

True or False? In Exercises 67-70, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If a particle moves along a sphere centered at the origin, then its derivative vector is always tangent to the sphere.

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

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The \(z\) -component of the derivative of the vector-valued function \(\mathbf{u}\) is 0 for \(t\) in the domain of the function. What does this information imply about the graph of \(\mathbf{u}\) ?
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