1. **State the problem:** Convert the vector $\mathbf{A} = 3\mathbf{a}_x + 4\mathbf{a}_y$ into cylindrical components at $\theta = 53.1^\circ$ and find the radial component $A_r$. 2. **Formula used:** The cylindrical components $A_r$ and $A_\theta$ relate to Cartesian components $A_x$ and $A_y$ by: $$A_r = A_x \cos\theta + A_y \sin\theta$$ $$A_\theta = -A_x \sin\theta + A_y \cos\theta$$ 3. **Given values:** $$A_x = 3, \quad A_y = 4, \quad \theta = 53.1^\circ$$ 4. **Calculate $A_r$:** $$A_r = 3 \cos 53.1^\circ + 4 \sin 53.1^\circ$$ 5. **Evaluate trigonometric functions:** $$\cos 53.1^\circ \approx 0.6, \quad \sin 53.1^\circ \approx 0.8$$ 6. **Substitute and simplify:** $$A_r = 3 \times 0.6 + 4 \times 0.8 = 1.8 + 3.2 = 5.0$$ **Final answer:** $$\boxed{A_r = 5.0}$$