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Minima and maxima of a cycloid

  • Thread starter Bipolarity
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The cycloid is defined by the parametric equations
[itex] x = a(t-sin(t)) [/itex] and [itex] y = a(1-cos(t)) [/itex]

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

I differentiated first to get

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

Then, by the chain rule:

[itex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{sin(t)}{1-cos(t)} [/itex]

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

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

Where is my mistake?

BiP
 

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  • #2
LCKurtz
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The cycloid is defined by the parametric equations
[itex] x = a(t-sin(t)) [/itex] and [itex] y = a(1-cos(t)) [/itex]

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

I differentiated first to get

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

Then, by the chain rule:

[itex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{sin(t)}{1-cos(t)} [/itex]

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

But according to the graph on my textbook,
the minima occur when [itex] t = aπN [/itex], 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##.
 

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