Question

In Exercises \(27-34,\) find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t^{3} \mathbf{j} $$

Short answer

The vector-valued function \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} \) is smooth over the open interval \( (-\infty, +\infty) \).

Step-by-step solution

Compute the Derivatives

First, differentiate each component of the vector-valued function \( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} \). The derivative of \( t^2 \) with respect to \( t \) is \( 2t \), and the derivative of \( t^3 \) with respect to \( t \) is \( 3t^2 \). So, \( \mathbf{r}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} \).

Identify the Domain

For this function, the function is defined for all real numbers. Therefore, the domain of \( \mathbf{r}'(t) \) is \( (-\infty, +\infty) \).

Check for Continuity

In order for the function \( \mathbf{r}(t) \) to be smooth, its derivative \( \mathbf{r}'(t) \) must be continuous. Since the derivative is composed of polynomials, it is a continuous function over \( (-\infty, +\infty) \).