Find ${d}^{2}y/d{x}^{2}$ for the curve given by $x=2\mathrm{cos}\theta \text{}and\text{}y=\mathrm{sin}\theta $

Nelson Jennings
2022-07-28
Answered

Find ${d}^{2}y/d{x}^{2}$ for the curve given by $x=2\mathrm{cos}\theta \text{}and\text{}y=\mathrm{sin}\theta $

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Hassan Watkins

Answered 2022-07-29
Author has **18** answers

$\mathrm{cos}(\theta )=\frac{x}{2}$

$\mathrm{sin}(\theta )=y$

${\mathrm{cos}}^{2}(\theta )+{\mathrm{sin}}^{2}(\theta )=1$

$\Rightarrow (\frac{x}{2}{)}^{2}+{y}^{2}=1$

$\therefore {x}^{2}+4{y}^{2}=4$ [equation of curve]

Differentiating both sides with respect to x,

$2x+4\times (2y\frac{dy}{dx})=0$

$\therefore \frac{dy}{dx}=-\frac{x}{4y}$

$\therefore \frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}(-\frac{x}{4y})$

$\Rightarrow \frac{{d}^{2}y}{d{x}^{2}}=-\frac{y\times 1-x\times \frac{dy}{dx}}{4{y}^{2}}$

Now, put value of dy/dx.

$\mathrm{sin}(\theta )=y$

${\mathrm{cos}}^{2}(\theta )+{\mathrm{sin}}^{2}(\theta )=1$

$\Rightarrow (\frac{x}{2}{)}^{2}+{y}^{2}=1$

$\therefore {x}^{2}+4{y}^{2}=4$ [equation of curve]

Differentiating both sides with respect to x,

$2x+4\times (2y\frac{dy}{dx})=0$

$\therefore \frac{dy}{dx}=-\frac{x}{4y}$

$\therefore \frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}(-\frac{x}{4y})$

$\Rightarrow \frac{{d}^{2}y}{d{x}^{2}}=-\frac{y\times 1-x\times \frac{dy}{dx}}{4{y}^{2}}$

Now, put value of dy/dx.

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