Posted by bryan on Sunday, February 3, 2013 at 1:29pm.
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 = (13-6√2)t^2
At t=15, x = 15√(13-6√2)
2x dx/dt = 2(13-6√2)t
2*15√(13-6√2) dx/dt = 30(13-6√2)
dx/dt = √(13-6√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 = 9-6√2+2+2 = 13-6√2
|dx|dt| = √(13-6√2)
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