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October 20, 2014

October 20, 2014

Posted by **kchik** on Thursday, November 22, 2012 at 12:36am.

a) 2 hours after it leaves the point P

b) 7 hours after it leaves the point P

- calculus -
**drwls**, Thursday, November 22, 2012 at 8:31amThe distance from point P after t hours is

d = sqrt[30^2 + (10 t)^2]

= sqrt(900 + 100 t^2)

dd/dt = [ (1/2)/( sqrt(900 + 100 t^2)]*200 t

Plug in the t value and compute the corresponding rate of change of d

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