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anonymous

  • 5 years ago

Related Rates Problemin Calculus: 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

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  1. anonymous
    • 5 years ago
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    finish asking this question.

  2. anonymous
    • 5 years ago
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    oops . how fast is the distance between the ships changing at 4:00pm?

  3. anonymous
    • 5 years ago
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    ok, so you know the distance formula & how to find derivatives right?

  4. anonymous
    • 5 years ago
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    yes

  5. anonymous
    • 5 years ago
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    ok, in 4 hourse, ship b will have traveled 160 miles due north, and ship a will have traveled 80 miles due south, there are still 180 miles between them east to west, right?

  6. anonymous
    • 5 years ago
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    oh yeah okayy

  7. anonymous
    • 5 years ago
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    z^2=180^2 + (x+y)^2..z=diagonal distance between ship a & ship b, x = ship a's distance south & y = shipb's distance north...

  8. anonymous
    • 5 years ago
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    implicit diff. w/respect to t...you should knon the distances for x, y, z after 4 hrs, dx/dt = 20 kmh, dy/dt = 40 kmh, so just sub & solve.

  9. anonymous
    • 5 years ago
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    wait where do i use the implicit differentiation? on the equation: 180^2 + (x + y)^2? because i'm still kind of confused

  10. anonymous
    • 5 years ago
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    the whole thing...so you get... 2z dz/dt = 2(x+y) (dx/dt + dy/dt) *chain rule & 180 is constant so its der. is 0. then just sub & solve.

  11. anonymous
    • 5 years ago
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    okay i think i might get it now

  12. anonymous
    • 5 years ago
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    from z^2 = 180^2 + (x + Y)^2 (this was from pythag. theorem)

  13. anonymous
    • 5 years ago
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    kay thanks for the help

  14. anonymous
    • 5 years ago
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    just another question. is x = 20 and y = 40 and z = 180?

  15. anonymous
    • 5 years ago
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    oh wait nvm thats wrong i think

  16. anonymous
    • 5 years ago
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    sorry nat, but no x, will be 80 (the distance that ship a has sailed south after 4 hrs), y will be 160 (the distance that ship b has sailed north in 4 hrs), and z will be 300 (z^2 = 180^2 +(x+y)^2)...those are your distances. 20 and 40 are your rates (km/h), so dx/dt=20 and dy/dt = 40.

  17. anonymous
    • 5 years ago
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    you begin with z^2 = 180^2 + (x+y)^2 (pythag. theorem*distance) and RELATE rates to it using implicit diff., helps to draw it sometimes as well.

  18. anonymous
    • 5 years ago
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    ohhh okay that makes more sense

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