# A particle moves along the curve y = \ln(\frac{x}{2}) , where x > 0. As the particle passes...

## Question:

A particle moves along the curve {eq}y = \ln(\frac{x}{2}) {/eq}, where x > 0. As the particle passes through the point (2e,1), its x-coordinate increases at a rate of {eq}\sqrt {4e^2 +1} {/eq} cm/s. How fast is the distance of the particle from the origin changing at this instant?

## Derivative:

We will find the rate of change of distance by firs finding the distance between the two points that is the origin and another point say x,y and then differentiate.

## Answer and Explanation:

To find the distance we will find the distance between the point (0,0) and (x,y):

{eq}D=\sqrt{x^{2}+y^{2}} {/eq}

Now let us find the derivative:

{eq}\frac{\mathrm{d} D}{\mathrm{d} t}=\frac{1}{2\sqrt{x^{2}+y^{2}}}\left ( 2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right ) {/eq}

Now let us find the rate of change of y:

{eq}y=\ln (\frac{x}{2})\\ \frac{\mathrm{d} y}{\mathrm{d} t}=\frac{2}{x}\frac{\mathrm{d} x}{\mathrm{d} t}\\ =\frac{\sqrt{4e^{2}+1}}{e} {/eq}

Now we will write the rate of change of distance:

{eq}\frac{\mathrm{d} D}{\mathrm{d} t}=\frac{1}{\sqrt{4e^{2}+1}}\left ( 2e\sqrt{4e^{2}+1}+\frac{\sqrt{4e^{2}+1}}{e} \right )\\ =\frac{2e^{2}+1}{e} {/eq} 