Question

Find the velocity, acceleration, and speed of a particle with position function r(t) = (-2t sin(t), -2t cos(t), -6t²) v(t) = (-2 cos(t) + 2t sin(t), 2 sin(t) + 2t cos(t), -12t) a(t) = (4 sin(t) + 2t cos(t), 4 cos(t) - 2t sin(t), -12) |v(t)| = $\sqrt{8 cos^2(t) - 8t sin(t) cos(t) + 148t^2}$

Name: find the velocity acceleration and speed of a particle with position function rt 2t sint 2t cost 6t2 vt 2 cost 2t sint 2 sint 2t cost 12t at 4 sint 2t cost 4 cost 2t sint 12 vt sqrt8 cos2t 8 27634
Uploaded: 2023-12-15T11:31:33-08:00
Duration: 4 min 53 s
Channel: Adi S
Description: find the velocity acceleration and speed of a particle with position function rt 2t sint 2t cost 6t2 vt 2 cost 2t sint 2 sint 2t cost 12t at 4 sint 2t cost 4 cost 2t sint 12 vt sqrt8 cos2t 8 27634

          Find the velocity, acceleration, and speed of a particle with position function
r(t) = (-2t sin(t), -2t cos(t), -6t²)
v(t) = (-2 cos(t) + 2t sin(t), 2 sin(t) + 2t cos(t), -12t)
a(t) = (4 sin(t) + 2t cos(t), 4 cos(t) - 2t sin(t), -12)
|v(t)| = $\sqrt{8 cos^2(t) - 8t sin(t) cos(t) + 148t^2}$

find the velocity acceleration and speed of a particle with position function rt 2t sint 2t cost 6t2 vt 2 cost 2t sint 2 sint 2t cost 12t at 4 sint 2t cost 4 cost 2t sint 12 vt sqrt8 cos2t 8 27634

Added by Cristina M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

12/15/2023

Step 1

To find the velocity, we take the derivative of the position function with respect to time: v(t) = dr(t)/dt = (-2sin(t) - 2tcos(t), -2cos(t) + 2tsin(t), -12t) Show more…

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Find the velocity, acceleration, and speed of a particle with position function r(t) = (-2t sin(t), -2t cos(t), -6t²) v(t) = (-2 cos(t) + 2t sin(t), 2 sin(t) + 2t cos(t), -12t) a(t) = (4 sin(t) + 2t cos(t), 4 cos(t) - 2t sin(t), -12) |v(t)| = $\sqrt{8 cos^2(t) - 8t sin(t) cos(t) + 148t^2}$