# Evaluating $\frac{d^2y}{dx^2}$

I want to evaluate $\frac{d^2y}{dx^2}$ from the given data: $x=a\cos^3\theta$ and $y=b\sin^3\theta$.

I tried in this way- $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^6\theta+\sin^6\theta=1-3(\frac{xy}{ab})^\frac{2}{3}$. After that I seem to be lost and will appreciate some help in doing this efficiently.

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$$\large\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$ $$\large \frac{d^2y}{dx^2}=\frac{d\left(\frac{dy}{dx}\right)}{dx}=\frac{\frac{d\left(\frac{dy}{dx}\right)}{d\theta}}{\frac{dx}{d \theta}}$$

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$\displaystyle x=a\cos^3\theta \Rightarrow \frac {dx}{d\theta}=-3a\cos^2\theta\sin\theta$

$\displaystyle y=b\sin^3\theta \Rightarrow \frac {dy}{d\theta}=3b\sin^2\theta\cos\theta$

$\displaystyle \Rightarrow\frac {dy}{dx}=\frac{\frac {dy}{d\theta}}{\frac {dx}{d\theta}}=\frac{3b\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\frac b a \tan\theta$

$\displaystyle \Rightarrow \frac{d^2y}{dx^2}=-\frac b a \sec^2\theta=-\frac b a (\frac a x)^\frac 3 2=-\frac {\sqrt a b} {\sqrt{x^3}}$

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There are two errors here. First, $dx\over d\theta$ is missing a minus sign. A bigger error is in the last step, where the left side is a derivative with respect to $x$, but the right side is a derivative with respect to $\theta$. It's not true that ${d\over{dx}}\tan\theta = {\sec}^2\theta$. – Steve Kass Mar 3 '13 at 15:53