two ships leave port at nooon on perpendicular paths. one moves north at a constant 18 mph, and one moves east at 12 mph. how fast is the distance between them changing at 2:00 PM the same day?....
i got an answer completely different from my teacher, but sometimes he changes his questions without his answer sheets, soo...help! thanks
Related rates problem?
Let P be the starting point A and B the ships position at a given time PAB is a triangle rectangle so
(AP)^2 + (BP)^2 = (BA)^2
The distance is therefore
d = BA = sqrt((18*t)^2 +) 12*t)^2) = sqrt(324+144)*t
the speed at which distance changes is
d/t = sqrt(468)
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