यदि `x=a(theta+sin theta),y=a(1-cos theta)` तो `theta=(pi)/(2)` पर `(d^(2)y)/(dx^(2))` ज्ञात करें |

Explanation

दिया है, `x=a(theta+sin theta)implies(dx)/(d theta)=a(1+cos theta)" "...(1)`
तथा `y=a(1-cos theta)implies(dy)/(d theta)=a sin theta" "...(2)`
अब `" "(dy)/(dx)=(dy//d theta)/(dx//d theta)=(a sin theta)/(a(1+cos theta))=(2sin""(theta)/(2)cos""(theta)/(2))/(2 cos^(2)""(theta)/(2))=tan""(theta)/(2)`
`implies" "(d^(2)y)/(dx^(2))=(d)/(dx)(tan""(theta)/(2))=(d)/(d theta)(tan""(theta)/(2)).(d theta)/(dx)`
`=(1)/(2)sec^(2)""(theta)/(2)//(dx)/(d theta)=((1)/(2)sec^(2)""(theta)/(2))/(a(1+costheta))`
`theta=(pi)/(2)" पर "(d^(2)y)/(dx^(2))=(1)/(2)sec^(2)""(pi)/(4).(1)/(a(1+cos""(pi)/(2)))=(1)/(a)`