Let x t = 2 2 cos t sin 2 t and y t = 2 2 sin t sin 2 t , t ∈ 0 , π 2 . Then 1 + d y d x 2 d 2 y dx 2 at t = π 4 is equal to

Step-by-step solution

Given x = 2 2 cos t sin 2 t Now differentiating w.r.t t both side we get, d x d t = 2 2 cos 3 t sin 2 t   . . . . . . 1 Also given y t = 2 2 sin t sin 2 t Again differentiating w.r.t t both side we get, d y d t = 2 2 sin 3 t sin 2 t   . . . . . . 2 Now dividing equation 2 from 1 we get, d y d x = tan 3 t Now finding the value of d y d x at t = π 4 we get, d y d x = - 1 Now finding d 2 y d x 2 we get, d 2 y d x 2 = 3 2 2 sec 2 3 t · sin 2 t cos 3 t Now finding value of d 2 y d x 2 at t = π 4