यदि ` x=a sin 2t (1+cos 2t ) ` व ` y= b cos 2t ( 1-cos 2t ) ` , तब सिद्ध कीजिए की
`" "((dy)/(dx) )_(t=(pi)/(4 ) ("पर ") ) =(b)/(a) `

Explanation

यहाँ ` x=a sin 2t (1+cos 2t ) `
`'t'` के सापेक्ष अवकलन करने पर,
` (dx)/(dy)=a [sin 2t (-2sin 2t ) +(1+cos 2t )(2cos 2t) ]`
` " "= 2 a *[-sin ^(2) 2t +cos 2t +cos ^(2) 2t ]`
` " "=2a (cos 4t +cos 4t )`
जहाँ ` (cos^(2)2t -sin ^(2) 2t =cos 4t) `
अब ` y= bcos 2t (1-cos 2t)`
`'t'` के सापेक्ष अवकलन करने पर
` (dy)/(dx) =b[cos 2t(sin 2t )+ (1-cos 2t ) (-2sin 2t ) ]`
` " "= (2b) [sin 2t cos 2t -sin 2t +sin 2t cos 2t]`
` " "= (2b) [2sin 2t cos 2t -sin 2t) `
` " "= (2b )[sin 4t -sin 2t ]`
इसलिए `" "(dy)/(dx) =((dy)/(dx)xx(dt)/(dx))=((2b)(sin 4t -sin 2t))/((2a)(cos 4t +cos 2t))`
` rArr " ((dy)/(dx) ) _((pi)/(4))=(b)/(a) ({sin (4(pi)/(4) )- sin (2xx(pi)/(4)))}/({cos (4xx(pi)/(4) )-cos (2xx(pi)/(4))})`
` rArr " "(b)/(a) {(sin pi -sin ""(pi)/(2))/(cos pi+cos ""(pi)/(2))}= (b)/(a) ((0-1))/((-1+0) )=(b)/(a) `