calculus
One bicycle is east of an intersection , and it is travelling towards the intersection of 9 miles per hour. at the same time a second bicycle is south of the intersection , and it is travelling away from the intersection at the rate of 10 miles per hour. Is the distance between the bicycles increasing or decreasing when the first is 4 miles east and the second is 4 miles south of the intersection? at what rate?

let u be the distance of the first bike
let v be the distance of the second bike
u^2+v^2=x^2
given du/dt=9
and dv/dt=10
2x dx/dt=2u du/dt + 2v dv/dt
dx/dt= (u du/dt +v dv/dt)/x
now solve for dx/dt when u=4,v=4, and du/dt and dv/dt given as above.
if it is increasing, dx/dt will be +posted by bobpursley

e = distance east
s = distance south
h = hypotenuse, what we want
at start
e = 4
s = 4
de/dt = 9
ds/dt = +10
h = 4 sqrt 2
h^2 = e^2 + s^2
2 h dh/dt = 2 e de/dt + 2 s ds/dt
4 sqrt 2 (dh/dt) = 4 (9) + 4 (+10)
sqrt 2 (dh/dt) = +1
dh/dt = 1/sqrt 2 = +.707 miles/hour
obviously 10 is bigger than 9 :)
posted by Anonymous
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