Question

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=\langle t \sin t, t \cos t, t\rangle $$

Short answer

The derivative \(\mathbf{r}'(t)\) of the given vector function \(\mathbf{r}(t)\) is \(\langle \sin(t) + t\cos(t), \cos(t) - t\sin(t), 1 \rangle\).

Step-by-step solution

Identify the components

First, we identify the functions that form the vector \(\mathbf{r}(t)\). These are \(t \sin(t)\), \(t \cos(t)\), and \(t\). Each of these will be differentiated separately.

Differentiate each component

Taking the derivative for each component: \[(t \sin(t))'\] using product rule is \(\sin(t) + t\cos(t)\), \[(t \cos(t))'\] using product rule is \(\cos(t) - t\sin(t)\), and \[(t)'\] is \(1\).

Combine the derivatives

Finally, we combine these all together to form the derivative of the vector function \(\mathbf{r}(t)\), i.e., \(\mathbf{r}'(t) = \langle \sin(t) + t\cos(t), \cos(t) - t\sin(t), 1 \rangle\).