Question

A particle moves in the $xy$ plane as $v=a\hat{i}+bx\hat{j}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. The particle starts from origin at $t=0$. Find the radius of curvature of the particle as a function of $x$.

• A. $\frac{a^2+b^{2}x}{ba}$
• B. $\frac{a}{b}{[1+{\left(\frac{bx}{a}\right)}^2]}^{3/2}$
• C. $\frac{a}{b}{[1+{\left(\frac{ax}{b}\right)}^2]}^{3/2}$
• D. None of these

Answer 1

The correct option is B $\frac{a}{b}{[1+{\left(\frac{bx}{a}\right)}^2]}^{3/2}$

We have, $v_x=a,$ $v_y=bx$
or, $\frac{dx}{dt}=a,$ $\frac{dy}{dt}=bx$
Thus, $x=\int adt=at$ (because $x=0$ at $t=0$)
and $y=\int bxdt=\int b\left(at\right)dt=ab\frac{t^2}{2}$ (because $y=0$ at $t=0$).

We can write this as $y=ab\frac{{\left(x/a\right)}^2}{2}=\frac{b{x}^2}{2a}$
Now, for a function $y=f\left(x\right)$, we have radius of curvature, $R=\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}{|\frac{d^{2}y}{d{x}^2}|}$
Now, $\frac{dy}{dx}=\frac{bx}{a}$ and $\frac{d^{2}y}{d{x}^2}=\frac{b}{a}$

Putting these values in formula for $R$, we get,
$R=\frac{a}{b}{[1+{\left(\frac{bx}{a}\right)}^2]}^{3/2}$