X=a (cost + logtant/2), y= asint find d2y/dt2 and d2y/dx2
Answers
x' = a{ -sin(t) + ½ sec²(t/2)/tan(t/2) } = acos(t)cot(t) : y' = acos(t) : dy/dx = tan(t)
From dy/dx=tan(t) t is the angle ψ that the tangent makes with x-axis.
1+ (dy/dx)² = sec²(t) : d²y/dx² = sec²(t). dt/dx = (1/a)sec^4(t)sin(t)
Because of the log, 0<t<π. As t increases from 0 to ½π the curve moves in the upper half of the plane from -∞ to the point (0,a) on the y-axis. From ½π to π it reflects this move in the y-axis to +∞.
Always ψ>0, dx/dt>0 and dy/dt<0 and dy/dx<0 only in the 1st quadrant.
d²y/dx²>0 so the curves are always convex downwards.
ρ = asec³(t)/{sec^4(t)sin(t)} = acot(t)
∴ The point on the evolute has coordinates..
x = a{cos(t)+log(tan(t/2)} – ρsin(ψ) = alog(tan(t/2)) : y = asin(t)+ρcos(ψ) = acosec(t)
This is the parametric equation of the evolute.
For (x,y) equation we can eliminate t with sin(t) = 2tan(t/2)/{1+tan²(t/2)}...
a/y = 2exp(x/a)/{1+exp(2x/a)} = sech(x/a) or y = acosh(x/a)