Find y'' for: 4x^2 + 3y^2 = 4?

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Find y'' for: 4x^2 + 3y^2 = 4?

2006-11-01 17:25:44 · 4 answers · asked by soccerman2431 1 in Science & Mathematics ➔ Mathematics

4 answers

4x^2 + 3y^2 = 4
8x x' + 6y y' = 0
8 x '' + 6 y'' = 0
6y'' = -8x ''
y '' = -4x / 3

2006-11-01 17:29:41 · answer #1 · answered by Crellos 2 · 0⤊ 0⤋

Y=0
and X=1

2006-11-02 01:46:59 · answer #2 · answered by Anurag Bhatia 4 · 0⤊ 0⤋

4x^2 + 3y^2 = 4
8x + 6y y' = 0
8 + 6 y' + 6y y'' = 0
8 + 6(8x/6y) + 6y y'' = 0
y'' = -8/(6y) - 48x/(36y^2)
= -4/(3y) - 4/(3y^2)
and substitute for y and y^2 to get the long expression.

2006-11-02 03:02:47 · answer #3 · answered by back2nature 4 · 0⤊ 0⤋

3y^2=4-x^2
y=[(4-x^2)/3]^.5
y' = .5*[(4-x^2)/3]^(-.5)*(-2x)/3
y'' = -.25[(-x^2)/3]^(-1.5)*(-2x/3)^2 + .5*[(4-x^2)/3]^(-.5)*(-2/3)

2006-11-02 01:53:58 · answer #4 · answered by Frog Head Hunter 2 · 0⤊ 0⤋