If a 0, x=a(t+ t) and y=a(1- t) , then d^2 y d x^2 at t= 2 3 is

Step-by-step solution

Given, Differentiate Eq. (i) w.r.t. t , we get d x d t =a(1+ t) Differentiate Eq. (ii) w.r.t. t , we get aligned & d y d x =a t & d y d x = d y / d t d x / d t = a t a(1+ t) = t (1+ t) & Now, d^2 y d x^2 = d d x ( d y d x )= d d t ( d y d x ) d t d x & = d d t ( t 1+ t ) 1 a(1+ t) & = (1+ t) t- t(- t) (1+ t)^2 1 a(1+ t) & d^2 y d x^2 = ( t+1) a(1+ t)^3 = 1 a(1+ t)^2 & At t=2 / 3 & d^2 y d x^2 = 1 a ( 1 2 )^2 = 4 a & aligned