Aman starts walking north at 2 ft/s from a point p. five minutes later a woman starts walking south at 3 ft/s from a point 500 ft due east of p. at what rate are the people moving apart 15 min after the woman starts walking? (round your answer to two decimal places.)

4.98 ft/s Let's determine the distance between the man and the woman for the moment that she's been walking 15 minutes. For this you can create a right triangle where one leg is 500 ft long (the east west difference between their locations) and the other leg is (distance man walked for 20 minutes + distance woman walked for 15 minutes). So Distance man walked = 20 min * 60 s/min * 2 ft/s = 2400 ft. Distance woman walked = 15 min * 60 s/min * 3 ft/s = 2700 ft. So the north south different in the man and woman's location is 2400+2700 = 5100 ft and will be increasing by 5 ft/sec. Creating a function of time (in seconds) for the distance the two people are apart is f(t) = sqrt(500^2 + (5100 + 5t)^2) where t = number of seconds from the 15 minutes the woman has been walking. For rate of change, you want the first derivative of the function. So let's calculate it. f(t) = sqrt(500^2 + (5100 + 5t)^2) f(t) = sqrt((5100 + 5t)^2 + 250000) f'(t) = d/dt[ sqrt((5100 + 5t)^2 + 250000) ] f'(t) = 0.5((5t + 5100)^2 + 250000)^(-0.5) * d/dt[ (5t + 5100)^2 + 250000 ] f'(t) = d/dt[ (5t + 5100)^2 ] / (2 * sqrt((5t + 5100)^2 + 250000)) f'(t) = 2(5t + 5100) * d/dt[ 5x + 5100 ]/(2 * sqrt((5t + 5100)^2 + 250000)) f'(t) = 5(5t + 5100/sqrt((5t + 5100)^2 + 250000) f'(t) = (25t + 25500)/sqrt((5t + 5100)^2 + 250000) Now calculate f'(t) for t = 0. So f'(t) = (25t + 25500)/sqrt((5t + 5100)^2 + 250000) f'(0) = (25*0 + 25500)/sqrt((5*0 + 5100)^2 + 250000) f'(0) = 25500/sqrt((5100)^2 + 250000) f'(0) = 25500/sqrt(26010000 + 250000) f'(0) = 25500/sqrt(26260000) f'(0) = 25500/5124.45119 f'(0) = 4.976142626 ft/sec So the man and woman are moving away from each other at the rate of 4.98 ft/s.