यदि `x= (2cos theta -cos 2theta ) ` व ` y= (2sin theta - sin 2theta )` तब ` theta =(pi)/(2 )` पर ` (d^(2)y)/(dx^(2))` का मान ज्ञात कीजिए |

Explanation

यहाँ ` " "x = (2cos theta -cos 2theta )`
`'theta'` के सापेक्ष अवकलन करने पर
` " "(dx)/(d theta ) =(-2sin theta + 2sin 2theta )`
अब ` " "y= (2sin theta -sin 2theta ) `
` rArr " "(dy)/(d theta ) =(2cos theta - 2cos 2 theta )`
हम जानते है की
` " "((dy)/(dx) )= ((dy)/(dx) xx(d theta )/(dx))=(2cos theta -2cos 2 theta )/((-2sin theta + 2sin 2theta ))`
` " "=((cos theta -cos 2theta ))/((sin 2theta -sin theta ))`
` =(2sin ((3theta)/2)sin ""((theta )/(2)))/(2cos ((3theta )/(2))sin ""((theta )/(2)))= tan ((3theta )/(2))`
अब ` (d^(2)y)/(dx^(2) )=(d)/(dx) [tan ((3theta )/(2))]`
` " "= (3)/(2) sec ^(2)"" (3theta )/(2)*(d theta)/(dx) `
` " "=(3)/(2) sec ^(2) ""(3theta )/(2)*(1)/(2(sin 2theta -sin theta ))`
व ` ((d^(2)y)/(dx^(2)))_(theta =(pi)/(2))=(3)/(2)*sec^(2)""((3pi)/(4))*(1)/(2(sin pi -sin (pi)/(2)))`
` " "=(3)/(4)sec ^(2)""(pi)/(4) =(3)/(4)xx (sqrt(2))^(2) =(3)/(2)`
जहाँ ` " "sec ""(3pi)/(4) =sec (pi- (pi)/(4) )=- sec ""(pi)/(4)`