Particle motion problem

[vileoxidation1]

I am working on a problem involving the chain rule. The problem is as follows: The position of an object as a function of time is s(t) = sin(3t) – 2t + 4. Determine where in the interval [0, 3] the object is moving to the right and to the left.

So, I know the basic process for solving this problem. I know that I need to find the derivative, and then where it is positive, the object is moving right, and where it is negative the object is moving left. I have found the derivative, which is s’(t) = 3cos(3t) – 2, with a period of (2*pi)/3. So now I need to find the zeros, and then determine if the function is positive or negative in the various intervals.

I found two of the zeros, by setting the function equal to zero and solving for t:
3cos(3t) = 2
cos(3t) = 2/3
3t = 0.841
t = 0.28 +/- [(2*pi)/3]

So this process gives me two of the zeros, which are 0.28 and 2.37. However, I know there is a third zero in between these two, since I graphed s’(t) and saw that the third zero is at 1.81. But, I do not know how to find this third zero (1.81) algebraically, which is how this problem is supposed to be done. Can someone help me figure out how to find this third zero by hand? Thanks in advance!

 

[stapel]

First, please let me say "Thank you!" for showing your work and reasoning so nicely!

3cos(3t) = 2
cos(3t) = 2/3
3t = 0.841

In the last step above, you used the inverse-cosine function (I'll bet) on your calculator. But remember that the inverse-cosine function gives only one of the (infinitely-many) solutions to the original cosine equation. The value returned by your calculator is the first-quadrant answer. Isn't there another one in the second quadrant? Given by something like 3t = 2*pi - arccos(2/3)?

 

[vileoxidation1]

Ah ha! Thank you. I knew it was something relatively simple like that, I just couldn't wrap my mind around it!

 

[stapel]

Ah ha! Thank you. I knew it was something relatively simple like that...

I think we've all missed that, at least once!