find

Asked by  | 24th May, 2009, 08:58: PM

Expert Answer:

x=a sin2θ(1+cos2θ)

dx/dθ=2acos2θ(1+cos2θ)-2asin22θ=2acos 2θ+2acos2 4θ

Similarily:

y=a cos2θ(1-cos2θ)

dy/dθ=-2asin2θ(1-cos2θ)+2asin2θcos2θ=2asin2θ

Dividing:

dy/dx=2acos 2θ+2acos2 4θ/2asin2θ=tan

x=a sin2θ(1+cos2θ)

dx/dθ=2acos2θ(1+cos2θ)-2asin22θ=2acos 2θ+2acos2 4θ

Similarily:

y=a cos2θ(1-cos2θ)

dy/dθ=-2asin2θ(1-cos2θ)+2asin2θcos2θ=2asin2θ

Dividing:

dy/dx=2acos 2θ+2acos2 4θ/2asin2θ
             =(cos2θ+cos2 4θ)/sin2θ

 

 

Answered by  | 15th Jun, 2009, 12:10: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.