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Cartesian vector to cylindrical

Short answer

At (1,2,3), ρ=√5, φ=63.43°. A_x=1, A_y=2, A_z=1. A_ρ = A_x cosφ + A_y sinφ = 2.236, A_φ = -A_x sinφ + A_y cosφ = 0, A_z=1. So A_cy = 2.236 a_ρ + a_z.

Transform A_c = x a_x + xy a_y + a_z at the point (1, 2, 3) in Cartesian co-ordinate system to A_cy in cylindrical co-ordinate system.

Given: A_c = x a_x + xy a_y + a_z at the point (1, 2, 3). Find A_cy (cylindrical).

Step 1: Cylindrical coordinates of the point

ρ = √(1² + 2²) = √5 = 2.236
φ = tan⁻¹(2/1) = 63.43°        cosφ = 1/√5 = 0.4472,  sinφ = 2/√5 = 0.8944
z = 3

Step 2: Cartesian components at the point

A_x = x = 1,   A_y = xy = 2,   A_z = 1

Step 3: Transform

A_ρ = A_x cosφ + A_y sinφ = 1(0.4472) + 2(0.8944) = 0.4472 + 1.7889 = 2.236
A_φ = −A_x sinφ + A_y cosφ = −1(0.8944) + 2(0.4472) = −0.8944 + 0.8944 = 0
A_z = 1

Check: |A|² = 1 + 4 + 1 = 6, and 2.236² + 0 + 1² = 5 + 1 = 6 ✓

Answer: A_cy = 2.236 a_ρ + a_z (that is, √5 a_ρ + a_z) at (1, 2, 3).

General form: A_ρ = ρ cos²φ + ρ² cosφ sin²φ, A_φ = −ρ sinφ cosφ + ρ² sinφ cos²φ, A_z = 1. At ρ = √5 these give the same values as above.

✓ Answer checked by Claude Opus 5.5 (high effort)

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