Question

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(t)=\frac{2 t}{8+t^{3}} \mathbf{i}+\frac{2 t^{2}}{8+t^{3}} \mathbf{j} $$

Short answer

The curve defined by the vector-valued function \(\mathbf{r}(t)\) is smooth on the intervals \((-∞, -2)\) and \((-2, ∞)\)

Step-by-step solution

Identify the Denominator

Firstly, look at the denominators of the i and j components of the vector-valued function. The denominator in both components is \(8 + t^{3}\).

Equate the Denominator to Zero

To identify the values of t at which these components might not be defined, set the denominator equal to zero and solve for t: \(8 + t^{3} = 0\). That implies \(t^{3} = -8\) and thus \(t = -2\).

Identify the Intervals

The curve defined by \(\mathbf{r}(t)\) is not differentiable (and therefore not smooth) at \(t = -2\). With this, it can be concluded that the curve is smooth on the open intervals \((-∞, -2)\) and \((-2, ∞)\). These are the open intervals on which the curve is smooth.