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Triple product for r = a cos t i + a sin t j + at tanα k

Short answer

Compute derivatives: r' = (-a sin t, a cos t, a tan α), r'' = (-a cos t, -a sin t, 0). r'×r'' = a² tan α sin t i - a² tan α cos t j + a² k. Then r·(r'×r'') = a³ t tan α.

If r = a cos t i + a sin t j + at tanα k, find [r dr/dt d²r/dt²].

[Reading: r = a cos t i + a sin t j + a t tan α k, with a and α constants.]

We need [r, dr/dt, d²r/dt²] = r · (dr/dt × d²r/dt²).

Derivatives

r   = (a cos t,  a sin t,  a t tan α)
r′  = (−a sin t, a cos t,  a tan α)
r″  = (−a cos t, −a sin t, 0)

r′ × r″

r′ × r″ = ( a cos t·0 − a tan α·(−a sin t),
            a tan α·(−a cos t) − (−a sin t)·0,
            (−a sin t)(−a sin t) − (a cos t)(−a cos t) )
        = ( a² tan α sin t, −a² tan α cos t, a² )

Dot with r

r · (r′ × r″) = a cos t · a² tan α sin t − a sin t · a² tan α cos t + a t tan α · a²
              = a³ tan α sin t cos t − a³ tan α sin t cos t + a³ t tan α
              = a³ t tan α

Answer: [r r′ r″] = a³ t tan α.

✓ Answer checked by Claude Opus 5.5 (high effort)

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