# 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.

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}