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 west-bound car from the intersection be x

let the distance of the north-bound 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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