Calculus Problem - Maximum Velocity, Derivatives, etc.

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[SOLVED] Calculus Problem - Maximum Velocity, Derivatives, etc.

Homework Statement


A particle moves along a line so that at any time t its position given by x(t)=2[tex]\Pi[/tex]t + cos2[tex]\Pi[/tex]t.

What is the maximum velocity?

Homework Equations


We found:
v(t) = 2[tex]\Pi[/tex] - sin2t[tex]\Pi[/tex](2[tex]\Pi[/tex])
a(t) = 2[tex]\Pi[/tex](-cos2t*[tex]\Pi[/tex])(2[tex]\Pi[/tex])
all values of t when particle's at rest in [0,3]: t=1/4, 5/4, 9/4


The Attempt at a Solution



We tried setting the acceleration to zero and got t = 1/4, and plugged that into the velocity and got v(t) = 0, which makes no sense because the max velocity is not when it is at rest.

Any help would be GREATLY appreciated ... I have been working at this for 6 hours and am afraid that I am slowly withering away
 
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a(t) is NOT zero for t=1/4. You made a mistake.

If the cosine of something is zero, what is the sine? (This is the quick way to do this problem ...)
 
Avodyne said:
a(t) is NOT zero for t=1/4. You made a mistake.

If the cosine of something is zero, what is the sine? (This is the quick way to do this problem ...)

How is a(1/4) not zero? You'd have:

[tex]a(1/4) = -4 \pi^2 cos(2(1/4) \pi)[/tex]

The cosine of pi over 2 is zero.

The thing is that he's missing values for when a(t) is zero. t should have values of 1/4, 3/4, 5/4, 7/4, and 9/4. Some of these values, when plugged into the velocity equation, do not amount to zero velocity.
 
thank you hotcommodity! i see that now! you are a lifesaver, truly.