# Thread: Second derivative of the equation of an ellipse

1. ## Second derivative of the equation of an ellipse

I've $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I'm asked to find $\frac{d^2y}{dx^2}$. I've implicitly differentiated it twice, but I can't get the required answer. What I did:

$y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{ dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2} +\frac{2y}{b^2}\right) = 0$

From this I got $\frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}$, whereas I was expected to get something quite different. So what has possibly gone wrong?

2. Originally Posted by Hardwork
I've $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I'm asked to find $\frac{d^2y}{dx^2}$. I've implicitly differentiated it twice, but I can't get the required answer. What I did:

$y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{ dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2} +\frac{2y}{b^2}\right) = 0$

From this I got $\frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}$, whereas I was expected to get something quite different. So what has possibly gone wrong?
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$

$b^2x^2 + a^2y^2 = a^2b^2$

$\displaystyle 2b^2 x + 2a^2 y \cdot \frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{x}{y}$

$\displaystyle \frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{y - x \cdot \frac{dy}{dx}}{y^2}$

$\displaystyle \frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{y + x \cdot \frac{b^2}{a^2} \cdot \frac{x}{y}}{y^2}$

multiply numerator and denominator by $\displaystyle \frac{y}{b^2}$ ...

$\displaystyle \frac{d^2y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{\frac{y^2}{b^2} + \frac{x^2}{a^2}}{\frac{y^3}{b^2}}$

$\displaystyle \frac{d^2y}{dx^2} = -\frac{b^4}{a^2y^3}$

3. I would say

$\displaystyle y = \sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}$

then differentiate explicitly twice.

4. Thanks guys. @ Pickslides, that's a wise suggestion, but it's from an implicit differentiation exercise.
Originally Posted by skeeter
$\displaystyle \frac{d^2y}{dx^2} = -\frac{b^4}{a^2y^3}$
That's indeed the expected answer, thanks. But I can't manage to find anything that's wrong with the one I got either.

5. Originally Posted by Hardwork
$y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{ dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2} +\frac{2y}{b^2}\right) = 0$

From this I got $\frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}$, whereas I was expected to get something quite different. So what has possibly gone wrong?
3rd step to 4th ...

$\frac{dy}{dx} \cdot \frac{dy}{dx} \ne \frac{d^2y}{dx^2}$

6. Originally Posted by skeeter
3rd step to 4th ...

$\frac{dy}{dx} \cdot \frac{dy}{dx} \ne \frac{d^2y}{dx^2}$
Oh, my, my! I can't thank you enough, Skeeter. That assumption was what frustrated me!

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