Minima and maxima of a cycloid

[Bipolarity]

The cycloid is defined by the parametric equations
x = a(t-sin(t)) and y = a(1-cos(t))

I am trying to find the set of points of relative extrema of a cycloid.

I differentiated first to get

\frac{dx}{dt} = a(1-cos(t)) and \frac{dy}{dt} = a*sin(t)

Then, by the chain rule:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{sin(t)}{1-cos(t)}

This fraction is 0 whenever t = πN where N is nonnegative integer.
Thus, the derivative is 0 whenever t = πN where N is nonnegative integer.

But according to the graph on my textbook,
the minima occur when t = aπN, where N is nonnegative integer.

Where is my mistake?

BiP

 

[LCKurtz]

The cycloid is defined by the parametric equations
x = a(t-sin(t)) and y = a(1-cos(t))

I am trying to find the set of points of relative extrema of a cycloid.

I differentiated first to get

\frac{dx}{dt} = a(1-cos(t)) and \frac{dy}{dt} = a*sin(t)

Then, by the chain rule:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{sin(t)}{1-cos(t)}

This fraction is 0 whenever t = πN where N is nonnegative integer.
Thus, the derivative is 0 whenever t = πN where N is nonnegative integer.

But according to the graph on my textbook,
the minima occur when t = aπN, where N is nonnegative integer.

Where is my mistake?

BiP

You are correct the $\sin t = 0$ when $t = n\pi$. $n$ can be any integer. But if you examine your graph, you will see you get max points for those values of $t$. For what values of $t$ does the derivative $\frac{dy}{dx}$ fail to exist? You might look there for the mins. And I don't think you will find that as $t = n\pi$.