If x = a (cos t + t sin t), y = a (sin t - t cos t),
then at t = `pi//4,(dy)/(dx)`

x = a (cos t + sin t), y = a (sin t-cos t)

x=a(cos t + log tan t/2), y =a sin t

If x=a(cos t+t sin t) and y=a(sin t-t cos t)

7. x=a(cos t+t sin t),y=a(sin t-t cos t) find dy/dx

x=a cos t, y=b sin t

If x=a ( tcos t- sin t ) ,y =a ( tsin t +cos t ),then (dy)/(dx) =

The distance, from the origin, of the normal to the curve, x = 2 cos t + 2t sin t, y = 2 sin t - 2t "cos t at t" = (pi)/(4) , is :

If x=a(cos t+t sin t) and y=a(sin t-t cos t), then find the value of (d^(2)y)/(dx^(2)) at t=(pi)/(4)

If x=a(cos t+t sin t) and y=a(sin t-t cos t), then find the value of (d^(2)y)/(dx^(2)) at t=(pi)/(4)

If x=a(cos t+t sin t) and y=a(sin t-t cos t),then find the value of (d^(2)y)/(dx^(2)) at t=(pi)/(4)