Chapter 4.1, Problem 22E

To determine

To find: The rate at which the distance from the particle to the origin changing at this instant.

Answer to Problem 22E

The rate at which the distance from the particle to the origin changing at this instant is $3.02$ .

Explanation of Solution

Given information: The curve is $y=\sqrt{x}$ and the $x$ coordinate increases at $3\text{cm}/\text{s}$ .

Calculation:

Consider the Figure.

  

Consider the equation.

  $y=\sqrt{x}$

Differentiate above equation.

  $\begin{array}{c}\frac{dy}{dt}=\frac{1}{2\sqrt{x}}\frac{dx}{dt}\\ \frac{dy}{dt}=\frac{1}{2\sqrt{4}}\left(3\right)\\ =\frac{3}{4}\end{array}$

The distance of the particle from the original point is,

  $\begin{array}{c}OP=\sqrt{4^2+2^2}\\ =\sqrt{16+4}\\ =4.47\end{array}$

From the figure,

  $\begin{array}{l}O{P}^2=O{X}^2+P{X}^2\\ O{P}^2=x^2+y^2\end{array}$

Differentiate above equation.

  $\begin{array}{c}2OP\frac{dOP}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\\ OP\frac{dOP}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\\ \left(4.47\right)\frac{dOP}{dt}=\left(4\right)\left(3\right)+\left(2\right)\left(\frac{3}{4}\right)\\ \frac{dOP}{dt}=3.02\end{array}$