(" vi ")quad x=3cos theta-cos^(3)theta,y=3sin theta-sin^(3)theta

x=3costheta-2cos^(3)theta, y = 3 sin theta- 2 sin^(3) theta

Find (dy)/(dx) if x=3 cos theta- 2cos^(3)theta,y=3 sin theta -2sin^(3) theta.

Find (dy)/(dx) if x=3 cos theta- 2cos^(3)theta,y=3 sin theta -2sin^(3) theta.

Find (dy)/(dx) if x=3 cos theta- 2cos^(3)theta,y=3 sin theta -2sin^(3) theta.

The normal to the curve x= 3 cos theta - cos^(3) theta , y=3 sin theta - sin^(3) theta at theta = (pi)/(4) -

If x=3cos theta-2cos^(3)theta,y=3sin theta-2sin^(3)theta, then (dy)/(dx) is

Show that the equation of the normal at any point 'theta' on the curvex = 3 cos theta -cos^(3) theta, y=3 sin theta -sin^(3) theta is 4(y cos^(3)theta -x sin^(3) theta)=3 sin 4 theta.

If x=3cos theta-cos^(3)theta y=3sin theta-sin^(3)theta find (dy)/(dx)

Show that the equation of normal at any point on the curve x = 3 cos theta - cos^(3) theta, y = 3 sin theta - sin^(3) theta "is" 4 (y cos^(3) theta - x sin^(3) theta) = 3 sin 4 theta .

(3cos theta+cos3 theta)/(3sin theta-sin3 theta)=