A point moves along the arc of parabola $y=2 x^2$. Its abscissa increases uniformly at the rate of 2 units…
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
$\frac{36}{\sqrt{5}}$ units/sec.
$\frac{18}{\sqrt{5}}$ units $/ \mathrm{sec}$.
$\frac{36}{5}$ units/sec.
$\frac{18}{5}$ units / sec.
Solution
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
$\begin{aligned}
& =\frac{2(1)+8(1)(2)}{\sqrt{1^2+2^2}} \\
& =\frac{18}{\sqrt{5}} \text { units } / \mathrm{sec}
\end{aligned}$