Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 22

At what rate is the distance from he particle to the origin changing at this moment?

Given:

$\qquad $ equation of the curve $y = \sqrt{x}$

$\qquad $ Point $(4,2)$

$\qquad \displaystyle \frac{dx}{dt} = 3 cm/s $

Required: Rate of change of distance of the particle to the origin

Solution:

Using the distance formula from point $(x, y)$ on the curve to the origin will result to..

$\displaystyle d = \sqrt{(y - 0)^2 + (x - 0)^2} \qquad$ Equation 1

Taking the derivative of the equation with respect to time we get


$
\begin{equation}
\begin{aligned}

\frac{dd}{dt} =& \frac{\displaystyle \cancel{2}y \frac{dy}{dt} + \cancel{2} x \frac{dx}{dt}}{\cancel{2} \sqrt{y^2 + x^2}} \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


To get $\displaystyle \frac{dy}{dt}$, we take the derivative of the given curve with respect to time.


$
\begin{equation}
\begin{aligned}

y =& \sqrt{x}
\\
\\
\frac{dy}{dt} =& \frac{\displaystyle \frac{dx}{dt}}{2 \sqrt{x}} = \frac{3}{2 \sqrt{4}} = \frac{3}{4} cm/s

\end{aligned}
\end{equation}
$


Now, to get the required answer, we use equation 2 and substituting all the informations that we obtain


$
\begin{equation}
\begin{aligned}

& \frac{ddt}{dt} = \frac{\displaystyle 2 \left( \frac{3}{4} \right) + 4 (3)}{\sqrt{2^2 + 4^2}}
\\
\\
& \boxed{\displaystyle \frac{dd}{dt} = 3.02 cm/s}
\end{aligned}
\end{equation}
$


To get the unknown, we derive equation 1 with respect to time


$
\begin{equation}
\begin{aligned}

z^2 =& x^2 + y^2
\\
\\
\cancel{2} z \frac{dz}{dt} =& \cancel{2}x \frac{dx}{dt} + \cancel{2} y \frac{dy}{dt}
\\
\\
\frac{dz}{dt} =& \frac{\displaystyle x \frac{dx}{dt} + y \frac{dy}{dt} }{z}
\\
\\
\frac{dz}{dt} =& \frac{50(25) + 120 (60)}{130}
\\
\\
&\boxed{\displaystyle \frac{dz}{dt} = 65 mi/h}

\end{aligned}
\end{equation}
$