Let x = at^2 - 1 , where a > 0 and y = t^3 + 1 . If at t = 1 , d^2y dx^2 = 3 16 , then the value of a is...

Step-by-step solution

Given x = at^2 - 1 and y = t^3 + 1 Differentiating with respect to t , we get dx dt = 2at and dy dt = 3t^2 dy dx = dy dt dx dt = 3t^2 2at = 3t 2a Differentiating again with respect to x , we get d^2y dx^2 = d dt ( 3t 2a ) dt dx d^2y dx^2 = 3 2a 1 2at = 3 4a^2t At t = 1 , d^2y dx^2 = 3 4a^2 Given that at t = 1 , d^2y dx^2 = 3 16 3 4a^2 = 3 16 4a^2 = 16 a^2 = 4 Since a > 0 , we get a = 2 Answer: 2