Chapter 5.6, Problem 28E

To determine

To find: the particle’s distance from the origin changing as it passes through the origin $\left(5,12\right)$ .

Answer to Problem 28E

The particle’s distance from the origin is $-5\text{ m}\text{./sec}\text{.}$

Explanation of Solution

Given information:

  $\frac{dx}{dt}=-1\text{ m}\text{./sec}\text{.}$

  $\frac{dy}{dt}=-5\text{ m}\text{./sec}\text{.}$

  $x=5$

  $y=12$

All variables are differentiable functions of t .

Calculation :

We have to calculate the particle’s distance from the origin changing as it passes through the origin $\left(5,12\right)$ .

Therefore,

To find the distance in the coordinate plane, we use Pythagoras theorem.

  $\text{D}=\sqrt{x^2+y^2}={\left(x^2+y^2\right)}^{\frac{1}{2}}$ .

Differentiate with respect to t using chain rule.

  $\begin{array}{l}\frac{d\text{D}}{dt}=\frac{1}{2}{\left[x^2+y^2\right]}^{\frac{1}{2}-1}\cdot \left(2x\frac{dx}{dt}+2y\frac{dy}{dt}\right)\\ \frac{d\text{D}}{dt}=\frac{1}{2}{\left[x^2+y^2\right]}^{-\frac{1}{2}}\cdot 2\left(x\frac{dx}{dt}+y\frac{dy}{dt}\right)\\ \frac{d\text{D}}{dt}=\frac{1}{\sqrt{x^2+y^2}}\left(x\frac{dx}{dt}+y\frac{dy}{dt}\right)\end{array}$

Putting the value of $x=5$ , $y=12$ , $\frac{dx}{dt}=-1$ and $\frac{dy}{dt}=-5$ in above equation,

  $\begin{array}{l}\frac{d\text{D}}{dt}=\frac{1}{\sqrt{{\left(5\right)}^2+{\left(12\right)}^2}}\left[\left(5\right)\left(-1\right)+\left(12\right)\left(-5\right)\right]\\ \frac{d\text{D}}{dt}=\frac{1}{\sqrt{25+144}}\left[\left(-5\right)+\left(-60\right)\right]\\ \frac{d\text{D}}{dt}=\frac{1}{\sqrt{169}}\left[-5-60\right]\\ \frac{d\text{D}}{dt}=\frac{1}{\sqrt{169}}\left[-65\right]\left[\text{Since }\sqrt{169}=13\right]\\ \frac{d\text{D}}{dt}=\frac{-65}{13}\\ \frac{d\text{D}}{dt}=-5\text{ m}\text{./sec}\text{.}\end{array}$

Hence, the particle’s distance from the origin is $-5\text{ m}\text{./sec}\text{.}$