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if $$ \begin{cases} x= 2 \cos \theta - \cos 2\theta \\ y = 2 \sin \theta - \sin 2\theta \end{cases} $$ find $d^2y/dx^2$ at $\theta = \frac{\pi}{2}$

I have solved $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ and from this also solved $$ \frac{dy}{dx}= \frac{\cos {\theta}-\cos{ 2\theta}}{-\sin {\theta} +\sin{2\theta}}$$

after this I am confused.

Arctic Char
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diwakar yadav
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    https://math.stackexchange.com/a/4021938/688539 – Clemens Bartholdy Aug 17 '21 at 07:20
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    Since you have found the first derivative, you can find the second derivative by differentiating the first derivative. You will need to use the quotient and the chain rules. You can then substitute π/2 in the result. – MathGeek Aug 17 '21 at 16:41

1 Answers1

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To solve the problem there a couple of ways that you solve the problem:

\begin{align} \require{cancel} x(\theta)&=2\cos(\theta)-\cos(2\theta)\\ y(\theta)&=2\sin(\theta)-\sin(2\theta)\\ \frac{dx}{d\theta}&=-2\sin(\theta)+2\sin(2\theta)\\ \frac{dy}{d\theta}&=2\cos(\theta)-2\cos(2\theta)\\ \frac{dy}{dx}&=\frac{dy}{\cancel{d\theta}}\times\frac{\cancel{d\theta}}{dx}=\frac{2\cos(\theta)-2\cos(2\theta)}{-2\sin(\theta)+2\sin(2\theta)} \end{align}

Resulting in the following: \begin{align} \frac{d^2y}{dx^2}&=\frac{d}{d\theta}\frac{dy}{dx}\times\frac{d\theta}{dx} \\ \frac{d^2y}{dx^2}&=\frac{((-2\sin(\theta)+2\sin(2\theta))(-2\sin(\theta)+4\sin(2\theta))-((2\cos(\theta)-2\cos(2\theta))(-2\cos(\theta)+4\cos(2\theta)))}{(-2\sin(\theta)+2\sin(2\theta))^3} \end{align}

$$\frac{d^2y}{dx^2}\Bigg\vert_{\theta=\frac{\theta}2}=-\frac{3}2$$

Jose M Serra
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    Your chain rule is incorrect. See https://math.stackexchange.com/questions/49734/taking-the-second-derivative-of-a-parametric-curve?noredirect=1&lq=1 for the correct expression. – Gonçalo Nov 29 '24 at 03:15
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    @Gonçalo I adjusted the post with the necessary updates. Is it in the right method now with the proper chain rule because I copied the post format. – Jose M Serra Nov 29 '24 at 16:21