Please help in simplifying X = (? – sin ?), dA = (3/2 – ...

[MAC-A-TAC]

Please help in simplifying X = (? – sin ?), dA = (3/2 – ...

I need help simplifying this equation before integrating.
Given X = (? – sin ?), dA = (3/2 – 2 cos ? + ½ cos 2?) d?

M[sub:23x82ehc]Y[/sub:23x82ehc]= ? x dA = (? – sin ?) (3/2 – 2 cos ? + ½ cos 2?)

M[sub:23x82ehc]Y[/sub:23x82ehc]= (3/2? - 2?cos ? + 1/2 ?cos 2? - 3/2 sin ? + 2sin ? cos ? -1/2 sin ? cos 2?) d? :?


Thank you.

 

[galactus]

Re: Please help in simplifying this equation.

Is that the full problem as it is written?. If not, it would be better if we saw it as is.

 

[MAC-A-TAC]

Re: Please help in simplifying this equation.

Problem statement:
Determine the x - coordinate of the centroid of the area under one arch of the cycloid described by the parametric equations
x = (? – sin ?) and y = (1 - cos ?).

Thank you.

 

[Deleted member 4993]

Re: Please help in simplifying this equation.

Hello.

I need help simplifying this equation before integrating.
Given X = (? – sin ?), dA = (3/2 – 2 cos ? + ½ cos 2?) d?


M[sub:1ymdfyo6]Y[/sub:1ymdfyo6]= ? x dA = (? – sin ?) (3/2 – 2 cos ? + ½ cos 2?)

M[sub:1ymdfyo6]Y[/sub:1ymdfyo6]= (3/2? - 2?cos ? + 1/2 ?cos 2? - 3/2 sin ? + 2sin ? cos ? -1/2 sin ? cos 2?) d? :?


Thank you.

These are straight forward integartions - where is the problem? The hardest typical integral in this bunch is:

\(\displaystyle \int \theta \cdot\cos(\theta) d\theta \, = \, \theta \cdot\sin(\theta) \, - \,\int\sin(\theta) d\theta \, = \, \, \theta \cdot\sin(\theta) \, + \, \cos(\theta) \, + \, C\)

 

[galactus]

Re: Please help in simplifying this equation.

This is a cycloid formed from a circle of radius 1. One thing to know is that the area of a cycloid is 3 times that of the circle that formed it.

\(\displaystyle x=t-sin(t), \;\ y=1-cos(t)\)

So, the area of this one would be \(\displaystyle 3\pi\)

Now, here is the cool part. The centroid coordinates of a cycloid are given by:

\(\displaystyle \overline{x}=\frac{\int_{a}^{b}x(t)y(t)x'(t)dt}{\int_{a}^{b}y(t)x'(t)dt}\)

\(\displaystyle \overline{y}=\frac{\frac{1}{2}\int_{a}^{b}y(t)^{2}x'(t)dt}{\int_{a}^{b}y(t)x'(t)dt}\)

You have all the info to calculate this using the formula. Of course, a=0 and b=\(\displaystyle 2\pi\)

You can just look at the graph and see what the x-coordinate is, but y is not so obvious.

An interesting bit of trivia. The cycloid was found to be the solution of the famous Brachistochrone problem.

It was posed by Johann Bernoulli. Newton found the cycloid was the shape that resulted in a bead going from point A to point B in the shortest amount of time.