Question

Find the open interval(s) on which the curve given by the vector-valued function is smooth. $$ \mathbf{r}(t)=\frac{1}{t-1} \mathbf{i}+3 t \mathbf{j} $$

Short answer

The curve is smooth on the open intervals \( (-\infty, 1) \) and \( (1, \infty) \).

Step-by-step solution

Differentiate

First, differentiate the vector function \( \mathbf{r}(t) \). The derivative of \( \mathbf{r}(t) \) is given by \( \mathbf{r}'(t) \). This gives \( \mathbf{r}'(t)= \left( \frac{d}{dt} \frac{1}{t-1}, \frac{d}{dt} 3t \right) = \left( -\frac{1}{(t-1)^2}, 3 \right) \).

Find Where the Derivative is Undefined

Now we need to find when this derivative is undefined. A curve is not differentiable where its derivative is undefined. By looking at the function \( \mathbf{r}'(t)= \left( -\frac{1}{(t-1)^2}, 3 \right) \), we can see that the derivative is undefined when \( t=1 \), because division by zero is undefined.

Identify the Open Intervals

Since the only point where the function is not differentiable is \( t=1 \), we can break the function's domain into two open intervals that do not include \( t=1 \). These intervals are all the values of t less than 1 and greater than 1. In other words, the function is smooth on the open intervals \( (-\infty, 1) \) and \( (1, \infty) \).