Question

Find radius of curvature of​

Answer 1

Answer:

The radius of curvature of a curve at a point M(x,y) is called the inverse of the curvature K of the curve at this point: R=1K. Hence for plane curves given by the explicit equation y=f(x), the radius of curvature at a point M(x,y) is given by the following expression: R=[1+(y′(x))2]32|y′′(x)|.

Step-by-step explanation:

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Answer 2

Given,

$\small\longrightarrow y=a\sec^3\theta$

Taking reciprocal,

$\small\longrightarrow \dfrac{1}{y}=\dfrac{1}{a}\cos^3\theta\quad\dots(1)$

But given,

$\small\longrightarrow x=a\cos^3\theta$

$\small\longrightarrow\dfrac{x}{a}=\cos^3\theta$

Then (1) becomes,

$\small\longrightarrow \dfrac{1}{y}=\dfrac{1}{a}\cdot\dfrac{x}{a}$

$\small\longrightarrow y=\dfrac{a^2}{x}$

$\small\longrightarrow dy=-\dfrac{a^2}{x^2}\,dx$

Now,

$\small\longrightarrow ds=\sqrt{(dx)^2+(dy)^2}$

$\small\text{ \longrightarrow ds=\sqrt{(dx)^2+\left(-\dfrac{a^2}{x^2}\,dx\right)^2} }$

$\small\text{ \longrightarrow ds=\sqrt{(dx)^2+\dfrac{a^4}{x^4}\,(dx)^2} }$

$\small\text{ \longrightarrow ds=dx\sqrt{1+\dfrac{a^4}{x^4}} }$

$\small\text{ \longrightarrow ds=\dfrac{\sqrt{x^4+a^4}}{x^2}\,dx }$

and,

$\small\longrightarrow\alpha=\tan^{-1}\left(\dfrac{dy}{dx}\right)$

$\small\longrightarrow\alpha=\tan^{-1}\left(-\dfrac{a^2}{x^2}\right)$

$\small\text{ \longrightarrow d\alpha=\dfrac{\dfrac{2a^2}{x^3}}{1+\left(-\dfrac{a^2}{x^2}\right)^2}\,dx }$

$\small\longrightarrow d\alpha=\dfrac{2a^2x}{x^4+a^4}\,dx$

Then radius of curvature is given by,

$\small\longrightarrow\rho=\dfrac{ds}{d\alpha}$

$\small\text{ \longrightarrow\rho=\dfrac{\left(\dfrac{\sqrt{x^4+a^4}}{x^2}\,dx\right)}{\left(\dfrac{2a^2x}{x^4+a^4}\,dx\right)} }$

$\small\text{ \longrightarrow\underline{\underline{\rho=\dfrac{(x^4+a^4)^{\frac{3}{2}}}{2a^2x^3}}} }$