`x=(t-sin t), y=(1-cos t):then(d^(2)y)/(dx^(2)) at` t = `pi/2`

If x=e^(t)sin t,y=e^(t)cos t then (d^(2)y)/(dx^(2)) at x=pi is

If x = e ^(t ) sin t, y = e ^(t) cos t, then (d ^(2) y )/(dx ^(2)) at t = pi is

If x=a(t-sin t) and y=a(1+cos t) then (dy)/(dx)=

If x= sin t and y= sin^(3)t , then (d^(2)y)/(dx^(2)) at t=pi/2 is

x=t cos t,y=t+sin t. Then (d^(2)x)/(dy^(2)) at t=(pi)/(2) is

x=a cos t,y=a sin t then (d^(2)(y))/(dx^(2)) at t=(pi)/(3),(pi)/(6),(pi)/(2),(pi)/(4)

x=t cos t,y=t+sin t. Then (d^(2)tau)/(dy^(2)) at t=(pi)/(2) is (pi+4)/(2)(b)-(pi+4)/(2)-2(d) none of these

If x=t^(3)+t+5 and y=sin t then (d^(2)y)/(dx^(2))

If y = sin t and x = 2 t then (d^(2)y)/(dx^(2)) =