1. **State the problem:** We are given the vector function $$\mathbf{r}(t) = (t\sin t + \cos t) \mathbf{i} + (\sin t - t\cos t) \mathbf{j}$$ for $$t > 0$$. We want to analyze or find properties of this vector function. 2. **Recall formulas and rules:** This vector function is composed of two components: $$x(t) = t\sin t + \cos t$$ and $$y(t) = \sin t - t\cos t$$. 3. **Find the derivative $$\mathbf{r}'(t)$$:** - Differentiate $$x(t)$$: $$x'(t) = \frac{d}{dt}(t\sin t) + \frac{d}{dt}(\cos t) = (\sin t + t\cos t) - \sin t = t\cos t$$ - Differentiate $$y(t)$$: $$y'(t) = \frac{d}{dt}(\sin t) - \frac{d}{dt}(t\cos t) = \cos t - (\cos t - t\sin t) = t\sin t$$ So, $$\mathbf{r}'(t) = t\cos t \mathbf{i} + t\sin t \mathbf{j}$$. 4. **Interpretation:** The derivative vector $$\mathbf{r}'(t)$$ gives the velocity vector of the curve at time $$t$$. 5. **Find the magnitude of $$\mathbf{r}'(t)$$:** $$|\mathbf{r}'(t)| = \sqrt{(t\cos t)^2 + (t\sin t)^2} = \sqrt{t^2(\cos^2 t + \sin^2 t)} = \sqrt{t^2} = t$$ since $$t > 0$$. 6. **Summary:** The speed of the particle moving along the curve $$\mathbf{r}(t)$$ is $$t$$, which increases linearly with time. **Final answer:** $$\mathbf{r}'(t) = t\cos t \mathbf{i} + t\sin t \mathbf{j}$$ and speed $$|\mathbf{r}'(t)| = t$$ for $$t > 0$$.