Calculus
posted by bryan .
Two people start from rest at the same point. One walks east at 3 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changingg after 15 minutes?

one way, using the law of cosines. Draw a triangle. The angle between the two journeys is 45°. At time t, the distances traveled are 3t and 2t. So, the distance x is
x^2 = (3t)^2 + (2t)^2  2(3t)(2t)*1/√2
x^2 = 9t^2 + 4t^2  6√2 t^2
x^2 = (136√2)t^2
At t=15, x = 15√(136√2)
2x dx/dt = 2(136√2)t
2*15√(136√2) dx/dt = 30(136√2)
dx/dt = √(136√2)
using vectors,
x = (3t,0)  (√2t,√2t) = ((3√2)t,√2t)
dx/dt = (3√2,√2)
dx/dt^2 = (3√2)^2 + √2^2 = 96√2+2+2 = 136√2
dxdt = √(136√2)
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