A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 minutes after the woman starts walking

Answer:Both are moving apart with the rate of 8.99 feet per sec.Step-by-step explanation:From the figure attached,Man is walking north with the speed = 4 ft per second[tex]\frac{dx}{dt}=4[/tex] feet per sec.Woman starts walking due south with the speed = 5ft per second[tex]\frac{dy}{dt}=5[/tex] ft per sec.We have to find the rate of change in distance z.From the right angle triangle given in the figure,[tex]z^{2}=(x+y)^{2}+(500)^{2}[/tex]We take the derivative of the given equation with respect to t,[tex]2z.\frac{dz}{dt}=2(x+y)(\frac{dx}{dt}+\frac{dy}{dt})+0[/tex] -----(1)Since distance = speed × timeDistance covered by woman in 15 minutes or 900 seconds = 5(900) = 450 fty = 4500 ftAs the man has taken 5 minutes more, so distance covered by man in 20 minutes or 1200 sec = 4×1200 = 4800 ftx = 4800 ftSince, z² = (500)² + (x + y)²z² = (500)² + (4500 + 4800)²z² = 250000 + 86490000z = √86740000z = 9313.43 ftNow we plug in the values in the formula (1)2(9313.43)[tex]\frac{dz}{dt}[/tex] = 2(4800 + 4500)(4 + 5)18626.86[tex]\frac{dz}{dt}[/tex] = 18(9300)[tex]\frac{dz}{dt}=\frac{167400}{18626.86}[/tex][tex]\frac{dz}{dt}=8.99[/tex] feet per sec.Therefore, both the persons are moving apart by 8.99 feet per sec.