Question

A particle moves along the plane trajectory $y\left(x\right)$ with velocity $v$ whose modulus is constant. Find the curvature radius of the trajectory at the point $x=0$ if the trajectory has the form of an ellipse ${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2=1$; $a$ and $b$ are constants here.

• A. $R=\frac{a^2}{b}$.
• B. $R=\frac{a^3}{b^2}$.
• C. $R=\frac{a^3}{b}$.
• D. $R=\frac{2a^2}{b}$

Answer 1

Answer: (A)

$R=\frac{a^2}{b}$.

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2=1....\left(1\right)$
differentiating with respect to $x$.
$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{x{b}^2}{y{a}^2}......\left(2\right)$
Again differentiating with respect to $x$.
$\frac{d^{2}y}{d{x}^2}=-\frac{b^2}{a^2}\left[x\right(-\frac{1}{y^2})\frac{dy}{dx}+\frac{1}{y}]$
$\frac{d^{2}y}{d{x}^2}=-\frac{b^2}{a^2}[-\frac{x^2{b}^2}{y^3{a}^2}+\frac{1}{y}]$ using (2)
Radius of curvature , $R=\frac{{[1+(\frac{dy}{dx}{)}^2]}^{3/2}}{\frac{d^{2}y}{d{x}^2}}$
or, $R=\frac{{[1+(\frac{x^2{b}^4}{y^2{a}^4}{)}^2]}^{3/2}}{-\frac{b^2}{a^2}[-\frac{x^2{b}^2}{y^3{a}^2}+\frac{1}{y}]}$
At $x=0,R=-\frac{a^2}{b^2}y$
Using (1) for $x=0,y=b$ or $-b$
Thus , $R=\frac{a^2}{b}$