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o Tools Window Help Screen Shot 2019-04-25 at 8.15.42 PM Q Search 3. Complete the solution ofthe brachistochrone problem begun in Section 2-2 and show that the desired curve is a cycloid with a cusp at the initial point at which the particle is released. Show also that if the particle is projected with an initial kinetic energy Imug that the brachistochrone is still a cycloid passing through the two points with a cusp at a height z above the initial point given by uo-2gz. Screen Shot Screen Shot 2019-04.. .5.42 PM Screen S 201 08 PM

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Functional we want to extremize is arclength 、dr-+av-divided by velocity /2g 2gy This leads to the integral, 2一20(1 + x*) x--(1 + x2 ) → 2ga Hereon, a-> k . Use the change of variables y-k sinでの㈠ φ = 2 sin sin () J2-dyJkksin (4)

-) -Xe, e i Half-angle formula-sine) 4 (2 cos φ 2), making the integral [13] into, 2e ) ( k(1-cos Even if we reverse the change of variables (usin inr) Im(e2)cos'-sin21-2sin,you get a t 2+ which you can't invert to get y(x). You must settle fora transcendental equation x(y)-k(2sin parametric solution. The parameter is φ as in the change of variables, and you get,
Here $x(\phi ) \ and \ y(\phi )$ represents a cycloid

Given initial kinetic energy ==> KE =

PE = mgz

PE = KE