# Minima and maxima of a cycloid

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

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LCKurtz
Homework Helper
Gold Member
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 $t = n\pi$.