• anonymous
At noon, ship A is 180 km west of ship B. Ship A is sailing south at 20 km/h and ship B is sailing north at 40 km/h. How fast is the distance between the ships changing at 4:00 PM?
  • Stacey Warren - Expert
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  • chestercat
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  • anonymous
You'll want to sketch this one... unfortunately, I can't sketch the triangles on here. Draw the ships 180 km away from each other with Ship A sailing south and Ship B sailing north. Now, draw Ship A's path as if it was starting at point B. Draw the 180 km segment connecting the two hypothetical Ship A paths. A right triangle is formed. A function for the distance between the two ships in the y-direction is y = 40t + 20t = 60t. Use the pythagorean theorem: c^2 = (60t)^2 + 180^2 = 3600t^2+32400 c = sqrt(3600t^2+32400)=sqrt(3600(t^2+9))=60sqrt(t^2+9) Take the derivative of c with respect to t to find: \[dc/dt=60*1/[2\sqrt{t^2+9}]*2t=60t/\sqrt{t^2+9}\] Then, put t = 4 into the derivative: \[dc/dt(4)=60(4)/\sqrt{4^2+9}=240/5=48\] So, at t = 4 hours or 4:00 p.m., the distance between the ships is changing at 48 km/h.

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