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gcuwagirl

Here is the related rate problem: At noon Ship A is 150 km west of Ship B. Ship A is sailing east at 35km/h and Ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 p.m.? I can't figure the diagram to use to help with the equation. I think it is 10^2 + 100^2 = z^2 where dx/dt is 35 and dy/dt is 25. Is this correct? I was thinking I should land up with a negative answer as the distance is being reduced at this time.

  • one year ago
  • one year ago

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  1. sumanth4phy
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    you are on right track and calculate dz/dt....but it would be definitely positive The distance was initially decreasing and then at a point reached it's minimum seperation and then started to increase. You cannot c all this if u think mathematically. But physics explains all this by just a imagining the figure:)

    • one year ago
  2. amistre64
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    At (12) Ship A is 150 km west of Ship B. A <------- 150 km ---------- B Ship A is sailing east at 35km/h A <------- 150 km ---------- B --> 35km/h Ship B is sailing north at 25 km/h. ^ | 25 km/h A <------- 150 km ---------- B --> 35km/h How fast is the distance between the ships changing at 4:00 p.m.? a formula defining placement will give us a formula to derive to find speed. Looks to be the pythag thrm for a right triangle a^2 + b^2 = d^2 , implicit derivative gives 2a a' + 2b b' = 2d d' ; dividing off the common 2 and dividing by d solves for d' d' = (a a' + b b')/d

    • one year ago
  3. amistre64
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    a = 150 - 4(35) b = 4(25) d = sqrt(a^2+b^2) a' is the speed of a b' is the speed of b everything is known to find d'. the speed at which the ships are moving with resoect to each other

    • one year ago
  4. gcuwagirl
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    Thanks for your affirmations. That is very helpful

    • one year ago
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