Unformatted text preview: the remaining one. As in the case when there are just two variables, take the derivative of both sides of the equation relating all of the variables, and then substitute all of the known values and solve for the unknown rate. EXAMPLE 6.18 A road running north to south crosses a road going east to west at the point P . Car A is driving north along the ﬁrst road, and car B is driving east along the second road. At a particular time car A is 10 kilometers to the north of P and traveling at 80 km/hr, while car B is 15 kilometers to the east of P and traveling at 100 km/hr. How fast is the distance between the two cars changing? Let a ( t ) be the distance of car A north of P at time t , and b ( t ) the distance of car B east of P at time t , and let c ( t ) the distance from car A to car B at time t . By the Pythagorean Theorem, c ( t ) 2 = a ( t ) 2 + b ( t ) 2 . Taking derivatives we get 2 c ( t ) c ± ( t ) = 2 a ( t ) a ± ( t )+2 b ( t ) b ± ( t ),...
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 Spring '07
 JonathanRogawski
 Math, Calculus, Derivative, Pythagorean Theorem, 1 sec, 15 kilometers, 4 rad, 10 kilometers, 43 deg

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