- #1
jofree87
- 38
- 0
y[1+(y')^2] = k
First solve for dx in terms of y and dy, an then use the substitution y = ksin2(θ) to obtain a parametric form of the solution. The curve turns out to be a cycloid.
My attempt:
(y')^2 = k/y-1
dy/dx = sqrt(k/y-1)
dx = dy/[sqrt(k/y-1)]
then substitute y = ksin^2(θ)
dx = dy/[sqrt(1/sin^2(θ)-1]
dx = dy/[sqrt(cos^2(θ)/sin^2(θ)]
dx = dy/cot(θ)
I don't know where to go from here, but the parametric form of y should equal k(1-cos(θ)).
Any help would be appreciated.
First solve for dx in terms of y and dy, an then use the substitution y = ksin2(θ) to obtain a parametric form of the solution. The curve turns out to be a cycloid.
My attempt:
(y')^2 = k/y-1
dy/dx = sqrt(k/y-1)
dx = dy/[sqrt(k/y-1)]
then substitute y = ksin^2(θ)
dx = dy/[sqrt(1/sin^2(θ)-1]
dx = dy/[sqrt(cos^2(θ)/sin^2(θ)]
dx = dy/cot(θ)
I don't know where to go from here, but the parametric form of y should equal k(1-cos(θ)).
Any help would be appreciated.