If `x=asin2t(1+cos2t)` and `y=bcos2t(1-cos2t)` , show that at `t=pi/4` , `(dy)/(dx)=b/a` .

Answer

N/a

Explanation

`:' x = a sin2t(1+cos2t)` nd ` y = b cos2t(1-cos2t)`
`:. (dx)/(dt) = a[sin2t.(d)/(dt)(1+cos2t)+(1+cos2t).(d)/(dt)sin2t]`
`=a[sin2t.(-sin2t).(d)/(dt)2t+(1+cos2t).cos2t.(d)/(dt)2t]`
`=-2asin^(2)2t+2acos2t(1+cos2t)`
`rArr (dx)/(dt) = -2a[sin^(2)2t-cos2t(1+cos2t)]`
and `(dy)/(dt) = b [cos2t.(d)/(dt) (1-cos2t)+1(1-cos2t).(d)/(dt)cos 2t]`
`= b[cos2t.(sin2t)'(d)/(dt)2t+(1-cos2t)(-sin2t).(d)/(dt)2t]`
` = b [2sin2t.cos2t+2(1-cos2t)-(-sin2t)]`
`=2b[sin2t.cos2t-(1-cos2t)sin2t]`
`:.(dy)/(dx)=(dy//dt)/(dx//dt) = (-2b[-sin2t.cos2t+(1-cos2t)sin2t])/(-2a[sin^(2)2t-cos2t(1+cos2t)])`
`rArr (dy/dx)_(t=pi//4) = (b)/(a)([-sin '(pi)/(2)cos'(pi)/(2)+(1-cos'(pi)/(2))sin'(pi)/(2)])/([sin^(2)'(pi)/(2)-cos'(pi)/(2)(1+cos'(pi)/(2))])`
`=b/a.((0+1))/((1-0)), [:' sin' (pi)/(2)=1 "and" cos'(pi)/(2) = 0]`
`= b/a`