Question

A man starts walking north at 5 ft/s from a point P. Five minutes later a woman starts walking south at 6 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.)

Answer 1

Check the picture below.

the man is going North, and he has a rate of 5 f/s, she's going south, 500 miles from P, the green distance, and she's going 6 f/s.

by the time she started walking, he already had been walking for 5 minutes, or 300 seconds, since he's going 5 f/s, then he had already covered 1500 feet by then.

after that, they both walked for 15 minutes more or 900 seconds, and for her that means 5400 feet, whilst for him going slower means 4500 feet, as you see in the  picture.

since he's going North and she's going South, dy/dt is the combined rates, as you see there.

now, let's use the pythagorean theorem, bearing in mind that the green 500 feet are constant all the while, that matters because the derivative of a constant is 0.

$\bf r^2=x^2+y^2\stackrel{chain~rule}{2r\cfrac{dr}{dt}}=0+\stackrel{chain~rule}{2y\cfrac{dy}{dt}}\implies \cfrac{dr}{dt}=\cfrac{2y}{2r}\cdot \cfrac{dy}{dt} \\\\\\ \cfrac{dr}{dt}=\cfrac{y}{r}\cdot \cfrac{dy}{dt}$

so, what is "r" value 15 minutes later anyway?

$\bf r=\sqrt{(1500+4500+5400)^2+500^2}\implies r=\sqrt{11400^2+500^2} \\\\\\ r=\sqrt{130210000}\implies r=100\sqrt{13021}$

then we just plug those folks in the derivative,

$\bf \cfrac{dr}{dt}=\cfrac{y}{r}\cdot \cfrac{dy}{dt}\qquad \begin{cases} r=100\sqrt{13021}\\ \frac{dy}{dt}=11\\ y=11400 \end{cases}\implies \cfrac{dr}{dt}=\cfrac{11400}{100\sqrt{13021}}\cdot 11 \\\\\\ \cfrac{dr}{dt}=\cfrac{114\cdot 11}{\sqrt{13021}}\implies \cfrac{dr}{dt}=\cfrac{1254}{\sqrt{13021}}\implies \cfrac{dr}{dt}\approx 10.99~\frac{f}{s}$