find dy/dx and d^2y/dx^2 for x= 3t^2 -9t +2, y= t^2 +3t.?

Question

tangents with parametric curves.

Answer 1

dy/dt = 2t + 3
dx/dt = 6t - 9.

dy/dx = (dy/dt ) / ( dx/dt )
= (2t + 3) / (6t - 9)
= (2t + 3) / 3(2t - 3).

Putting h = dy/dx:
d^2y/dx^2 = dh/dx
= (dh/dt) / (dx/dt)
= [(d/dt)(dy/dx)] / (dx/dt)
= [3(2t - 3)2 - (2t + 3)3(2)] / [ 9(2t - 3)^2 * (6t - 9) ]
= (12t - 18 -12t - 18) / 27(2t - 3)^2 (2t - 3)
= - 36 / 27(2t - 3)^2 (2t - 3)
= - 4 / 3(2t - 3)^3.

Answer 2

x=3t^2-9t
3y=3t^3+9t
3y-x=18t so t = (3y-x)/18 and x= (3y-x)^2/108 -(3y-x)/2
implicit derivate
1= (3y-x)/54 * (3y´-1) -1/2(3y´-1) so1=( 3y´-1) *[(3y-x )/54-1/2]
54=(3y´-1)(3y-x-27)
y´=1/3 * (3y-x+27)/(3y-x-27) (a)
Now derivate again and wherever y´appears put in the value given by (a)