Question

find the radius of the curvature at


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Answer 1

Answer:

3a3a =6a hope u understand

Answer 2

The radius of curvature is defined as,

$\displaystyle\small\longrightarrow R=\dfrac{ds}{d\alpha}$

where,

• $\displaystyle\smallds=\sqrt{(dx)^2+(dy)^2}$

is an infinitesimal segment along the curve and,

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

is the angle made by the segment with the horizontal.

Here the equation of the curve is,

$\displaystyle\small\longrightarrow x^3+y^3=3axy$

Differentiating,

$\displaystyle\small\longrightarrow 3x^2dx+3y^2dy=3a\left(y\,dx+x\,dy\right)$

$\displaystyle\small\longrightarrow\dfrac{dy}{dx}=\dfrac{ay-x^2}{y^2-ax}\quad\quad\dots(1)$

At $\displaystyle\small(x,\ y)=\left(\dfrac{3a}{2},\ \dfrac{3a}{2}\right),$

$\displaystyle\small\longrightarrow\dfrac{dy}{dx}=-1\quad\quad\dots(2)$

Differentiating (1) wrt x,

$\displaystyle\small\text{ \longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{\left(a\,\dfrac{dy}{dx}-2x\right)\left(y^2-ax\right)-\left(ay-x^2\right)\left(2y\,\dfrac{dy}{dx}-a\right)}{\left(y^2-ax\right)^2} }$

At $\displaystyle\small(x,\ y)=\left(\dfrac{3a}{2},\ \dfrac{3a}{2}\right),$

$\displaystyle\small\longrightarrow\dfrac{d^2y}{dx^2}=-\dfrac{32}{3a}\quad\quad\dots(3)$

Then,

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

From (2),

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

$\displaystyle\small\longrightarrow ds=\sqrt{2}\ dx$

And,

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

Differentiating,

$\displaystyle\small\text{ \longrightarrow d\alpha=\dfrac{1}{1+\left(\dfrac{dy}{dx}\right)^2}\cdot\left(\dfrac{d^2y}{dx^2}\right)dx }$

From (2) and (3),

$\displaystyle\small\longrightarrow d\alpha=\dfrac{1}{1+\left(-1\right)^2}\cdot\left(-\dfrac{32}{3a}\right)dx$

$\displaystyle\small\longrightarrow d\alpha=-\dfrac{16}{3a}\,dx$

Then, radius of curvature,

$\displaystyle\small\longrightarrow R=\dfrac{ds}{d\alpha}$

$\displaystyle\small\text{ \longrightarrow R=\dfrac{\sqrt2\ dx}{\left(-\dfrac{16}{3a}\,dx\right)} }$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{R=-\dfrac{3a\sqrt2}{16}}} }$