# Brachistochrone homework

1. Sep 17, 2011

### Liquidxlax

1. The problem statement, all variables and given/known data

consider a sigle loop of the cycloid with a fixed value of a. A car is released at a point P0 from rest anywhere on the track between the origin and the lowest point P, that is P0 has a parameter 0<theta0 < pi. show that the time for the cart to roll from P0 to P is given by the integral

time( P0]/sub] -> P) = $\sqrt{\frac{a}{g}}\int \sqrt{\frac{1 - cos\vartheta}{cos\vartheta_{0} - cos\vartheta}}d\vartheta$

integral is from theta naught to pi

and prove the integral equals $\pi\sqrt{\frac{a}{g}}$ the integral may be tricky and you can use theta = pi -2(alpha)
2. Relevant equations

$\frac{df}{dx}=\frac{d}{dy}\frac{df/dx'}$

3. The attempt at a solution

1/2 mv2 = mg(y-y1

v = $\sqrt{2g(y-y1}$

dt = ds/v

T = $\int\frac{\sqrt{1+(x')2}}{\sqrt{2g(y-y1}}$ dy

test the Euler formula to get

y = a(1 - cos(theta)) y' = asin(theta)

x = a (theta - sin(theta)) x' = a - acos(theta)

This is now the part I am having a problem with

Now i substitute my y and x' in

T = $\int\frac{\sqrt{1 + (a( 1 - cos(\vartheta))2}}{\sqrt{2ga(cos(\vartheta) - cos(\vartheta0)}}(asin(\vartheta)$

and somehow that equals the top equation. Then i need to integrate it, which i get no where near close to what it wants no matter what the substitution and trig identities

not sure why all the itex isn't working

2. Sep 18, 2011

### Spinnor

Re: Brachistochrone

Do this help?

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3. Sep 18, 2011

### Liquidxlax

Re: Brachistochrone

it sure did sometimes i'm such a retard, x and y do not depend on each other they depend on theta...

because i was doing sqrt ( 1 y'^2) dx

thank you very much

4. Sep 18, 2011

### Spinnor

Re: Brachistochrone

x and y do depend on each other but the answer had to be in terms of theta so that was a more useful parameter?