A point moves along the arc of parabola \(y=2 x^2\). Its abscissa increases uniformly at the rate of 2 units \(/ \mathrm{sec}\). At the instant, the point is passing through \((1,2)\), its distance from origin is increasing at the rate of

Given, \(\frac{\mathrm{d} x}{\mathrm{dt}}=2\) units/sec
Given equation of parabola is \(y=2 x^2\)
Differentiating w.r.to \(t\), we get
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{dt}}=4 x \times \frac{\mathrm{d} x}{\mathrm{dt}} \\
& \frac{\mathrm{~d} y}{\mathrm{dt}}=8 x
\end{aligned}\)
\(\ldots\) (i) \(\left[\frac{\mathrm{d} x}{\mathrm{dt}}=2\right]\)
\(\therefore \quad\) The distance of point from origin is given by \(\sqrt{x^2+y^2}\)
\(\therefore \quad\) The rate of increasing distance of point from origin
\(\begin{aligned}
& =\frac{\mathrm{d}}{\mathrm{dt}}\left(\sqrt{x^2+y^2}\right) \\
& =\frac{1}{2 \sqrt{x^2+y^2}} \cdot \frac{\mathrm{~d}}{\mathrm{dt}}\left(x^2+y^2\right) \\
& =\frac{1}{2 \sqrt{x^2+y^2}} \cdot\left[2 x \times \frac{\mathrm{d} x}{\mathrm{dt}}+2 y \times \frac{\mathrm{d} y}{\mathrm{dt}}\right] \\
& =\frac{1}{2 \sqrt{x^2+y^2}} \cdot[4 x+2 y \times 8 x] ...[From(i)\\
& =\frac{(2 x+8 x y)}{\sqrt{x^2+y^2}}
\end{aligned}\)
\(\therefore \quad\) Since point is passing through \((1,2)\)
\(\therefore \quad\) Rate of increasing distance of point from origin
\(\begin{aligned}
& =\frac{2(1)+8(1)(2)}{\sqrt{1^2+2^2}} \\
& =\frac{18}{\sqrt{5}} \text { units } / \mathrm{sec}
\end{aligned}\)