hello i really need help with my review for calculuslll. ok this is one of my questions. The position vector is r(t)= (8sin3t)i - (8cos3t)j + (7csc3t)k find the acceleration verctor at t=π/6.

Question

anonymous · Answer

just differentiate component-wise (twice)

eg V(t) = < 24cos(3t) i ....
      a(t) = < -72sin(3t) i ....

anonymous · Answer

I already derivated the r(t) to get v but im stuck with the derivate of v(t) to get a (t)

calculusfunctions · Answer

v(t)= (24cos3t)i + (24sin3t)j − 21(csc3t)(cot3t)k

a(t)= -72(sin3t)i + 72(cos3t)j − [−63(csc3t)(cot²3t) − 63(csc³3t)]k

a(π/6) = -72(sin(π/2))i + 72(cos(π/2)) + [63(csc(π/2))(cot²(π/2)) + 63(csc³(π/2))]k

a(π/6) = -72(1)i + 72(0)j + [63(1)(0)² + 63(1)³]k

a(π/6) = -72i + 63k

|a(π/6)| = √(72² + 63²)

|a(π/6)| ≈ 96 

I hope that helps.

anonymous · Answer

Perfect thank you so much!!! That's the answer!!!

calculusfunctions · Answer

You're Welcome Anytime!