If x=a [ t+ 1 2 ( ^2 t 2 ) ] and y=a t then d y d x =

Step-by-step solution

Given x = a [ t + 1 2 ( ^2 t 2 ) ] = a [ t + ( t 2 ) ] . Differentiating x with respect to t : dx dt = a [ - t + 1 (t/2) ^2(t/2) 1 2 ] Since 1 (t/2) ^2(t/2) 1 2 = (t/2) (t/2) 1 ^2(t/2) 1 2 = 1 2 (t/2) (t/2) = 1 t . Therefore, dx dt = a [ - t + 1 t ] = a [ 1 - ^2 t t ] = a ^2 t t . Given y = a t , differentiating with respect to t : dy dt = a t . Using the chain rule, dy dx = dy/dt dx/dt = a t a ^2 t / t = t t ^2 t = t t = t . Answer: t