यदि `x = a (cost + logtan "" (t) / (2)) , y = a sin t`, तब `dy / dx` का मूल्यांकन बिंदु `t = (pi) / (3)` पर कीजिये |

Options

A. `8/a`
B. `8(sqrt3)/a`
C. `(sqrt3)/a`
D. `8(sqrt2)/a`

Answer

B

Explanation

यहाँ ` x = a ( cos t + log tan "" ( t ) / ( 2 ) ) `
दोनों पक्षों का t के सापेक्ष अवकलन करने पर,
` ( dx ) /(dt ) = a[ - sin t + ( 1 ) / (tan t//2 ) ( d ) / ( dt ) (tan "" ( t ) /( 2 )) ] `
` rArr ( dx ) / (dt ) = a [ - sin t + ( 1 ) /( tan t//2 ) * sec ^ 2 "" ( t ) / ( 2 ) * ( 1 ) /( 2 ) ] `
` rArr (dx ) / (dt ) = a [- sin t + ( 1 cos t//2 ) /( 2 sin t//2) * ( 1 )/ ( cos^ 2 t //2 ) ] `
` rArr ( dx ) / (dt ) = a [ - sin t + ( 1 ) /( 2sin t //2 * cos t//2 ) ] `
` rArr (dx ) / (dt ) = a [ - sin t + ( 1 ) /(sin t ) ], [ therefore sin 2 theta = 2 sin theta cos theta ] `
` rArr (dx ) / ( dt ) = a [ ( 1 - sin ^2 t ) /( sin t )] `
` rArr (dx ) /(dt ) = a * ( cos ^ 2 t ) / ( sin t ) `
और ` y = a sin t`
दोनों पक्षों का t के सापेक्ष पुनः अवकलन करने पर,
` (dy ) /(dt) = a cos t `
अब ` ( dy ) / ( dx ) = ( dy //dt ) /( dx //dt ) = ( a cos t ) /( a (( cos ^ 2 t ) /( sin t))) `
` rArr ( dy ) / ( dx ) = ( sin t ) / ( cos t ) = tan t `
दोनों पक्षों का x के सापेक्ष अवकलन करने पर,
` (d^ 2 y ) / ( dx^ 2 ) = ( d ) / (dx ) ( tan t ) `
` rArr ( d^ 2 y ) / ( dx ^ 2 ) = ( d ) /( dt ) (tan t ) (dt ) /(dx ) `,
` [ therefore (d ) /(dx ) (f (t )) = ( d ) / (dt ) f ( t ) * ( dt ) / (dx )] `
` rArr ( d^2 y ) / ( dx ^ 2 ) = sec ^2 t * ( sin t ) / ( a cos ^ 2 t )`,
` [ therefore (dx ) / (dt ) = ( a cos ^2 t ) / ( sin t ) rArr (dt ) /(dx ) = (sin x ) /( a cos ^2 t ) ] `
` rArr ( d^2 y ) /( dx ^ 2 ) = (sint sec^ 4 t ) /( a ) `
` t = ( pi ) /( 3 ) ` पर,
` [ ( d^2y ) /( dx ^ 2 ) ] _ ( t = pi//3 ) = ( 1 ) /(a) sin ""( pi ) /( 3 ) *sec ^( 4 ) "" (pi ) / ( 3 ) `
` rArr [ (d^2y ) /( dx ^2 ) ] _ ( t = pi //3 ) = ( 1 ) /( a ) xx ( sqrt 3 ) /( 2 ) xx ( 2 ) ^ 4 = ( 8 sqrt 3 ) /( a ) `