Posted by Mike on Thursday, June 13, 2013 at 8:51pm.
A car leaves an intersection traveling west. Its position 5 sec later is 22 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 5 sec later is 27 ft from the intersection. If the speed of the cars at that instant of time is 15 ft/sec and 5 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.)
ft/sec

Calculus  Reiny, Thursday, June 13, 2013 at 11:48pm
let the distance of the westbound car from the intersection be x
let the distance of the northbound car from the intersection be y
and let the distance between them be d
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
d dd/dt = x dx/dt + y dy/dt
when x=22, y=27 , dx/dt = 15 ft/sec , dy/dt = 5 ft/sec
and d^2 = 22^2 + 27^2 = 1213
d = √1213
√1213 dd/dt = 22(15) + 27(5) = 465
dd/dt = 465/√1213 = appr 13.4 ft/s
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