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tangents with parametric curves.

2007-10-07 17:11:08 · 2 answers · asked by Robert M 2 in Science & Mathematics Mathematics

2 answers

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.

2007-10-07 22:20:10 · answer #1 · answered by Anonymous · 1 0

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)

2007-10-08 01:35:05 · answer #2 · answered by santmann2002 7 · 0 1

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