Parametric equation:

Question

help on finding 2nd derivative?

 

x=t+1, y=t^2+3t, t=-1

 

Answer 1

The parametric equation is x = t + 1, y = t² + 3t, where

x = t + 1

dx/dt = 1.

y = t² + 3t,

dy/dt = 2t + 3.

dy/dx = (dy/dt) / (dx/dt) = (2t + 3) / (1) = 2t + 3.

dy/dx = 2t + 3

dy/dx = 2(x - 1) + 3    (since x = t + 1)

dy/dx = 2x - 2 + 3

dy/dx = 2x + 1.

d²y/dx² = 2.

Therefore, the second derivative is two.