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Velocity and acceleration on curve x=4cos t, y=t², z=2t

Short answer

Differentiate r = 4 cos t i + t² j + 2t k to get v = −4 sin t i + 2t j + 2k and a = −4 cos t i + 2 j. At t=0, v = 2k, a = −4i+2j. At t=π/2, v = −4i+πj+2k, a = 2j.

A particle moves along the curve x = 4 cos t, y = t², z = 2t. Find velocity and acceleration at time t = 0 and t = π/2.

Position: r = 4 cos t i + t² j + 2t k. Velocity v = dr/dt, acceleration f = d²r/dt².

Velocity and acceleration

v = −4 sin t i + 2t j + 2 k
f = −4 cos t i + 2 j

At t = 0

v = −4(0) i + 0 j + 2 k = 2k
f = −4(1) i + 2 j = −4i + 2j

At t = π/2

v = −4(1) i + π j + 2 k = −4i + πj + 2k
f = −4(0) i + 2 j = 2j

(Speed at t = π/2: √(16 + π² + 4) = √(20 + π²) ≈ 5.46.)

Answer:

  • t = 0: velocity = 2k, acceleration = −4i + 2j.
  • t = π/2: velocity = −4i + πj + 2k, acceleration = 2j.

✓ Answer checked by Claude Opus 5.5 (high effort)

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