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A particle moves along the curve y=\ln x so that its abscissa is increasing at a rate of 2 units per second. At what rate is the particle moving away from the origin as it passes through the point (e,1)?

Solution

Let R be the distance between the particle and the origin, and t denote time in seconds. Using the distance formula, we can express R as a function of x:

\displaystyle R = \sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+\left(\ln x\right)^{2}}

We are to find \displaystyle \frac{dR}{dt} at (e,1) given that \displaystyle \frac{dx}{dt}=2. Differentiating the above equation implicitly with respect to t, we obtain

\displaystyle \frac{dR}{dt}=\frac{\dfrac{2\ln x}{x}\dfrac{dx}{dt}+2x\dfrac{dx}{dt}}{2\sqrt{x^{2}+\left(\ln x\right)^{2}}}.

Hence,

\displaystyle \left.\frac{dR}{dt}\right|_{(e,1)}=\frac{\dfrac{4}{e}+4e}{2\sqrt{1+e^{2}}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\left.\frac{dR}{dt}\right|_{(e,1)}}=\frac{2\left(\dfrac{e^{2}+1}{e}\right)}{\sqrt{1+e^{2}}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\left.\frac{dR}{dt}\right|_{(e,1)}}=\frac{2\sqrt{1+e^{2}}}{e}

This entry was posted on Sunday, March 23rd, 2008 at 3:19 pm and is filed under Calculus. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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