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Question
I have this old, but very difficult problem I just could never figure out the answer to. Can you help with this Derivative-related problem?

A railroad track and a road cross at a right angle. An observer stands on the road 90 meters south of the crossing and watches an eastbound train traveling at 60 meters per second. At how many meters per second is the train moving away from the observer 2 seconds after it passes through the intersection?

Answer
Initially, let the observer be at the point O = (xO,yO) = (0,-90), and the train at the intersection at point T = (xT,yT) = (0,0). These points define a vector R pointing from the observer to the train. The goal is to calculate how the magnitude of R is changing at time t = 2 seconds.

R = sqrt( x^2 + y^2 ).

From the geometry of the problem, we know that the derivative of the x component of R is dx/dt = 60 m/s and dy/dt = 0. The total derivative of R is

dR/dt = (dR/dx)・(dx/dt) + (dR/dy)・(dy/dt)

which becomes

dR/dt = (1/2)(x^2 + y^2)^(-1/2)・2x・(dx/dt).

We also need the value of x after 2 seconds (y doesn't change); this is (60m/s)・(2 sec) = 120m

Now its just a matter of plugging in the numerical values. I get

dR/dt = 48 m/s.

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randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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