A particle moves along the top of the parabola y2 = 2x from left to right at a constant speed of 5 units per second. Find the velocity of the particle as it moves through the point (2, 2).

A particle moves along the top of the parabola $y^2=2x$ from left to right at a constant speed of $5$ units per second.

We have to find the velocity of the particle as it moves through the point $\left(2, 2\right)$.

A particle moves along the top of the parabola $y^2=2x$ from left to right at a constant speed of $5$ units per second.

$y=\sqrt{2x}$ and $\frac{ds}{dt}=5$.

The speed is the magnitude of the velocity,

$\frac{ds}{dt}=\left|\frac{dr}{dt}\right|=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2} ....\left(1\right)$

Now,

$y=\sqrt{2x}$

Differentiate with respect to $t$,

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\sqrt{2x}\\ & =& \frac{1}{2\sqrt{2x}}\frac{d}{dt}\left(2x\right)\\ & =& \frac{1}{2\sqrt{2x}}\left(2\frac{dx}{dt}\right)\\ \frac{dy}{dt}& =& \frac{1}{\sqrt{2x}}\frac{dx}{dt}\end{array}$

Use $\frac{ds}{dt}=5$ and $\begin{array}{rcl}\frac{dy}{dt}& =& \frac{1}{\sqrt{2x}}\frac{dx}{dt}\end{array}$ in equation $\left(1\right)$,

$\begin{array}{rcl}5& =& \sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{1}{\sqrt{2x}}\frac{dx}{dt}\right)}^2}\\ 5& =& \sqrt{1+\frac{1}{2x}}\frac{dx}{dt}\\ 5& =& \sqrt{\frac{2x+1}{2x}}\frac{dx}{dt}\\ \frac{dx}{dt}& =& 5\sqrt{\frac{2x}{2x+1}}\end{array}$

The particle as it moves through the point $\left(2, 2\right)$,

$\begin{array}{rcl}\frac{dx}{dt}& =& 5\sqrt{\frac{2\left(2\right)}{2\left(2\right)+1}}\\ & =& 5\sqrt{\frac{4}{5}}\\ \frac{dx}{dt}& =& 2\sqrt{5}\end{array}$

Use the value of $\frac{dx}{dt}=2\sqrt{5}$ and $\left(x, y\right)=\left(2, 2\right)$ in $\begin{array}{rcl}\frac{dy}{dt}& =& \frac{1}{\sqrt{2x}}\frac{dx}{dt}\end{array}$,

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{1}{\sqrt{2\left(2\right)}}\left(2\sqrt{5}\right)\\ & =& \frac{1}{2}\left(2\sqrt{5}\right)\\ \frac{dy}{dt}& =& \sqrt{5}\end{array}$

The velocity of the particle through the point $\left(2, 2\right)$ is,

$$\begin{gathered} v=\frac{dx}{dt}i+\frac{dy}{dt}j \\ v=2\sqrt{5}i+\sqrt{5}j. \end{gathered}$$