बिंदु 't' पर वक्र `x=cos^3t : y=bsin^3t` के स्पर्शी का समीकरण ज्ञात कीजिए |

Find dy/dx, x=3 cos t-2cos^3t , y=3sint-2sin^3t .

x= sin^3t/sqrt x cos2t , y= cos^3t/ sqrt x cos2t .

Find (dy)/(dx) , if x=a log t and y=bsin t

x=3sin t-2 sin ^3t, y=3 cos t -2cos^3t

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

Show that the equation of normal at any point 't' on the curve: x = 3 cos t - cos^3 t and y = 3 sint - sin^3 t is: 4(ycos^3t - x sin^3t)=3 sin 4t .

If x=a cos ^(3) t,y =a sin ^(3)t,then " at" t = (pi)/(3) ,(dy)/(dx) =

Find dy/dx x=cos^(3)t/sqrtcos2t , y = sin^(3)t/sqrt cos2t