Second derivative of the equation for an elipse.

[agent_509]

Homework Statement

Determine the second derivative of y with respect to x when 2x²+3y²=0

Homework Equations

possible answers include:
2/(3y²)
-2/(9y³)
2/(3y³)
-2/(3y²)
-2/(3y³)

The Attempt at a Solution

I took the first derivative with respect to x implicitly and came up with
-2x/3y

I then took the second derivative, but here's where I get stuck.
I wind up with:
(-2y+2x(y'))/(3y^2)

The only possible answers have only y in them, while I see no way to get rid of x in my solution. I've done this several times and come up with the same answer each time, what am I doing wrong?

 

[vela]

Use the equation for the ellipse to eliminate x.

 

[SammyS]

Homework Statement

Determine the second derivative of y with respect to x when 2x²+3y²=0

Homework Equations

possible answers include:
2/(3y²)
-2/(9y³)
2/(3y³)
-2/(3y²)
-2/(3y³)

The Attempt at a Solution

I took the first derivative with respect to x implicitly and came up with
-2x/3y

I then took the second derivative, but here's where I get stuck.
I wind up with:
(-2y+2x(y'))/(3y^2)

The only possible answers have only y in them, while I see no way to get rid of x in my solution. I've done this several times and come up with the same answer each time, what am I doing wrong?

Did you plug your result for y' into you result for y": y" = (-2y+2x(y'))/(3y²)

 

[agent_509]

Yes I did,but you'll notice that still doesn't get rid of x.

Vela, what do you mean by that?

 

[vela]

Oh, sorry, it's not an equation of an ellipse. You can use the original equation $2x^2+3y^2=0$ to solve for $x$ in terms of $y$.

 

[SammyS]

What did you get when you plugged your result for y' into your result for y": y" = (-2y+2x(y'))/(3y²) ?