Determine the concavity of the curve with parametric equations x = t2, y = t3 - 3t at t = 2. Justify your answer.

Question

Determine the concavity of the curve with parametric equations x = t², y = t³ - 3t at t = 2. Justify your answer.

Answer 1

call x'=dx/dt, call y'=dy/dt, sure dy/dx=y'/x'

why, because ;;;want dy/dx =dy/dt*dt/dx same as dy/dx=[dy/dt]/[dx/dt]

but for d²y/dx² the derivative is [d(y'/x')/dt]*dt/dx=[{(x'*y"-y'*x"]/(x')²]/dx/dt = [x'y''-y'x'']/(x')³where x"=dx'/dt and y"=dy'/dt (note the denominator is x' cubed not sqaured)

my calculation is d²y/dx²=[(6t²+6)/4]*(1/(2t)) =[6t²+6]/8t at t=2, it is positive so curve concave up (the first derivative is increasing)

bold underlime is correction