(dx)/(dt)7x cos t=(1)/(2)sin2t

dx/dt + x cos t = 1/2 sin 2 t

Find (dy)/(dx):x=a{cos t+(1)/(2)log tan^(2)(t)/(2)} and y=a sin t

If x=a sin2t(1+cos2t) and y=b cos2t(1-cos2t), find the values of (dy)/(dx) at t=(pi)/(4) and t=(pi)/(3)

If x = 2 cos t - cos 2 t and y =2 sin t - sin 2t, then (dy)/(dx) at t = (pi)/(2) is

If x=sin t-t cos t and y = t sin t +cos t, then what is (dy)/(dx) at point t=(pi)/(2)?

Find (dy)/(dx) when : x=a(2t+sin2t), y=a(1-cos2t)

If x=2 cos t+cos 2t and y=2 sin t-sin2t , then the value of (dy)/(dx) at t=(pi)/(4) is -

If =f''(t)cos t+f'(t)sin t,y=-f''(t)sin t+f'(t)cos t, then int[((dx)/(dt))^(2)+((dy)/(dt))^(2)]^((1)/(2))dt is equal to