Question

A particle moves along the x-axis so that its position at any time t≥0 is given by t=arccot(t).

a. Does the particle always move to the right or to the left? Support your answer using calculus.
b. Find a(t). What do your results tell you about the acceleration of the particle?
c. When is the speed of the particle increasing or decreasing? Explain.
d. Find the time when the particle stops. Explain.

Answer 1

Oh lamp you always have long questions...
I am not sure if I have time now for this but we can start doing something with it.

Lets start with a.
Wait a sec, this seems wrong:
t=arccot(t)
should be x=arccot(t)
Can you check that please

Answer 2

yeah it says t= but it could be a typo

Answer 3

It must be otherwise the question does not make sense.

To answer a. we need to find the derivative of arccot(t), oh crap this is really gonna be long. Sorry I am leaving soon and cannot be asked for this

Answer 4

ok thanks anyway

Answer 5

quick help:
\[ d/dt arc \cot t = -\frac{ 1 }{ x^2+1 } \]

Answer 6

with t as variable.
When t is increasing from 0 this function will take smaller and smaller negative numbers. (increasing)
Think the rest through

Answer 7

ok

Answer 8

Andras is on precisely the right track.
Look at dx/dt: This is the expression for velocity. If the velocity is +, motion is to the right. If the velocity is (-), motion is to the left.
Your dx/dt is (-). What does that tell you (in response to the question of Part a)?

Answer 9

@mathmale its only moving left, so the answer is no?

Answer 10

Because the derivative is negative, we conclude that the particle is always moving to the left (in the negative x-direction). "Yes" or "no" would not be an appropriate answer; we have to chose from "left" or "right".