1. **Problem statement:** Given the vector function $$\mathbf{r}(t) = (t\sin t + \cos t)\mathbf{i} + (\sin t - t\cos t)\mathbf{j}$$ for $$t > 0$$, find the acceleration vector $$\mathbf{a}$$ expressed in the form $$\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}$$ without explicitly finding the unit tangent vector $$\mathbf{T}$$ and the unit normal vector $$\mathbf{N}$$. 2. **Recall the formula:** The acceleration vector can be decomposed as $$\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}$$ where - $$a_T = \frac{d|\mathbf{v}|}{dt}$$ is the tangential component (rate of change of speed), - $$a_N = \kappa |\mathbf{v}|^2$$ is the normal component (related to curvature $$\kappa$$ and speed), - $$\mathbf{v} = \mathbf{r}'(t)$$ is the velocity vector. 3. **Step 1: Compute velocity $$\mathbf{v}(t) = \mathbf{r}'(t)$$** $$\mathbf{v}(t) = \frac{d}{dt}[(t\sin t + \cos t)\mathbf{i} + (\sin t - t\cos t)\mathbf{j}]$$ Calculate each component: - $$\frac{d}{dt}(t\sin t + \cos t) = \sin t + t\cos t - \sin t = t\cos t$$ - $$\frac{d}{dt}(\sin t - t\cos t) = \cos t - (\cos t - t\sin t) = t\sin t$$ So, $$\mathbf{v}(t) = (t\cos t)\mathbf{i} + (t\sin t)\mathbf{j}$$ 4. **Step 2: Compute acceleration $$\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$$** $$\mathbf{a}(t) = \frac{d}{dt}[(t\cos t)\mathbf{i} + (t\sin t)\mathbf{j}]$$ Calculate each component: - $$\frac{d}{dt}(t\cos t) = \cos t - t\sin t$$ - $$\frac{d}{dt}(t\sin t) = \sin t + t\cos t$$ So, $$\mathbf{a}(t) = (\cos t - t\sin t)\mathbf{i} + (\sin t + t\cos t)\mathbf{j}$$ 5. **Step 3: Compute speed $$|\mathbf{v}(t)|$$** $$|\mathbf{v}(t)| = \sqrt{(t\cos t)^2 + (t\sin t)^2} = \sqrt{t^2(\cos^2 t + \sin^2 t)} = \sqrt{t^2} = t$$ (since $$t > 0$$) 6. **Step 4: Compute tangential acceleration $$a_T = \frac{d}{dt}|\mathbf{v}(t)| = \frac{d}{dt} t = 1$$ 7. **Step 5: Compute curvature $$\kappa$$ using formula $$\kappa = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^3}$$** In 2D, the cross product magnitude is: $$|\mathbf{v} \times \mathbf{a}| = |v_x a_y - v_y a_x|$$ Calculate: - $$v_x = t\cos t$$ - $$v_y = t\sin t$$ - $$a_x = \cos t - t\sin t$$ - $$a_y = \sin t + t\cos t$$ So, $$v_x a_y - v_y a_x = (t\cos t)(\sin t + t\cos t) - (t\sin t)(\cos t - t\sin t)$$ Expand: $$= t\cos t \sin t + t^2 \cos^2 t - t\sin t \cos t + t^2 \sin^2 t$$ Simplify terms: $$t\cos t \sin t - t\sin t \cos t = 0$$ Remaining terms: $$t^2 (\cos^2 t + \sin^2 t) = t^2$$ Therefore, $$|\mathbf{v} \times \mathbf{a}| = t^2$$ 8. **Step 6: Compute curvature $$\kappa$$** $$\kappa = \frac{t^2}{t^3} = \frac{1}{t}$$ 9. **Step 7: Compute normal acceleration $$a_N = \kappa |\mathbf{v}|^2 = \frac{1}{t} \times t^2 = t$$ 10. **Step 8: Write acceleration in $$\mathbf{T}$$ and $$\mathbf{N}$$ form:** $$\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N} = 1 \mathbf{T} + t \mathbf{N}$$ **Final answer:** $$\boxed{\mathbf{a} = \mathbf{T} + t \mathbf{N}}$$