If `x=a(cos 2 theta+2 theta sin 2 theta) " and" y=a(sin 2 theta - 2 theta cos 2 theta)`, find `(d^(2)y)/(dx^(2)) " at" theta =(pi)/(8)`.

Explanation

`x= a ( cos 2theta + 2theta sin 2 theta)`
`" " rArr (dx)/(dtheta) =a (-2sin 2theta + 2sin 2theta + 4theta cos 2 theta)`
`rArr (dx)/(dtheta ) = a ( 4theta cos 2theta ) ................ (1)`
`y= a(sin 2 theta - 2theta cos 2theta)`
`" " rArr (dy)/(dtheta) = a( 2cos 2theta + 4 theta sin 2theta - 2cos 2theta )`
`rArr (dy)/(dtheta) = a(4theta sin 2theta).............. (2)`
using (1) and (2)
`" " rArr (dy)/(dx) = (a(4theta sin 2 theta))/( a ( 4theta cos 2 theta))`
`" " rArr (dy)/(dx) = ( sin 2 theta )/( cos 2 theta) = tan 2 theta `
Differentiating again with respect to x, we get
`" " rArr (d^(2)y)/(dx^(2)) = 2 sec^(2) 2 theta. (dtheta )/(dx)`
`" " rArr (d^(2)y)/( dx^(2)) = 2 sec^(2) 2 theta . (1)/( a( 4theta cos 2theta))`
`" "(d^(2)y)/( dx^(2)) _(theta = (pi)/(8)) = 2 sec^(2)""(pi)/(4). (1)/(a(4(pi)/(8)cos""(pi)/(4)))`
`= ( 8 sqrt2)/( pi a)`