The position of a particle in motion in the plane at time t is
X(t)=−3tI+sin(−8t)J
At time any t, determine the following:
(a) the speed of the particle is: ______
(b) the unit tangent vector to X(t) is: ____ I+ _____ J

Question

The position of a particle in motion in the plane at time t is 
X(t)=−3tI+sin(−8t)J 
At time any t, determine the following: 
(a) the speed of the particle is: ______ 
(b) the unit tangent vector to X(t) is: ____   I+ _____  J

anonymous · Answer

attachment

anonymous · Answer

@e.mccormick  ?

anonymous · Answer

For speed you did $\|\mathbf X'(t)\|$

anonymous · Answer

What was I supposed to do? I thought that was it

anonymous · Answer

How are you getting negatives if you square things?

anonymous · Answer

$$
[(-3t)']^2=[-3]^2=9
$$

anonymous · Answer

$$
[(\sin(-8t))']^2=[-8\cos(-8t)]^2=64\cos^2(-8t)
$$

anonymous · Answer

hmm  it doesnt like that either though

anonymous · Answer

did you add them and take the square root?

anonymous · Answer

oh duh....

anonymous · Answer

forgot square root

anonymous · Answer

so the unit tangent vector would be what I had before divided by what we just put in?

anonymous · Answer

Well, each component is divided by the speed, but each component is the derivative of the X components.

anonymous · Answer

I dont understand. What are the x components?

anonymous · Answer

-3t  and sin(-8t)

anonymous · Answer

so divide those by the derivative of x(t) correct?

anonymous · Answer

-3 / (-64cos(64)-9) ?

anonymous · Answer

nvm -3 / sqrt(64cos(-8t)**2 + 9)

agent0smith · Answer

It looks like you solved this one...?