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eva12

A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 7 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 10 ft/min, at what rate will the boat be approaching the dock when 100 ft of rope is out?

  • one year ago
  • one year ago

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  1. baldymcgee6
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    |dw:1350689236706:dw|

    • one year ago
  2. eva12
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    how do you get the answer or solve it?

    • one year ago
  3. baldymcgee6
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    You must find a way to relate rate of the pully rope to the rate of the boat using the right angle triangle

    • one year ago
  4. eva12
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    i do not understand this problem or how to approach it to solve it

    • one year ago
  5. eva12
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    can someone help me

    • one year ago
  6. baldymcgee6
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    |dw:1350690450622:dw|

    • one year ago
  7. baldymcgee6
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    you can express the triangle as 7^2 + x^2 =r^2

    • one year ago
  8. eva12
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    then what will x be and would r will stand for rate rigth

    • one year ago
  9. baldymcgee6
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    \[\frac{ dr }{ dt } = -10ft/\min\]\[\frac{ dx }{ dt }|_{r=100}\]

    • one year ago
  10. baldymcgee6
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    I'm just calling r the length of ROPE

    • one year ago
  11. eva12
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    so the equation 7^2+x^2=r^2 will be dr/dt=2x+2r

    • one year ago
  12. baldymcgee6
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    when we differentiate 7^2 + x^2 =r^2, we will get: 49 + 2x* dx/dt = 2r* dr/dt

    • one year ago
  13. eva12
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    then i pug in 100 for r and 10 for x in the equation

    • one year ago
  14. baldymcgee6
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    no, dr/dt = -10 ft/min.. we don't know what x is

    • one year ago
  15. eva12
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    49 +2x=2(100)= will = x is 75.5

    • one year ago
  16. baldymcgee6
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    no, we don't solve for x as x is always changing. that is essentially what the question is asking.

    • one year ago
  17. eva12
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    so then how do i approach then

    • one year ago
  18. baldymcgee6
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    okay... so here's what we do, we have to equations that will help us solve this problem, namely: 7^2 + x^2 =r^2 49 + 2x* dx/dt = 2r* dr/dt are you with me so far?

    • one year ago
  19. eva12
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    do i solve for 1 equation then the 2 one

    • one year ago
  20. baldymcgee6
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    First lets deal with this equation: 49 + 2x* dx/dt = 2r* dr/dt what are we looking for, well the question asks, "at what rate will the boat be approaching the dock?" The rate at which the boat is approaching the dock is represented by dx/dt (the rate of change of x), so we need to re arrange this equation for dx/dt

    • one year ago
  21. eva12
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    dx/dt=2r*dr/dt-49/2x

    • one year ago
  22. baldymcgee6
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    good.. what do we know, we know what dr/dt is, and we know what that r=100 at the point of interest. but we don't know what x is at this given instance, so what we need to do is express x in terms of r in the other other equation: 7^2 + x^2 =r^2 x = sqrt(r^2-49) now substitute that into our change of rates equation, and we know that r =100, and dr/dt= -10: dx/dt = (2(100)*(-10)-49)/2(sqrt(100^2-49) = your answer

    • one year ago
  23. eva12
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    -102198.6893

    • one year ago
  24. baldymcgee6
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    no, it shouldn't be that big

    • one year ago
  25. baldymcgee6
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    you must have typed it in wrong in your calculator

    • one year ago
  26. eva12
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    -10.27

    • one year ago
  27. eva12
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    thanks

    • one year ago
  28. baldymcgee6
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    you're very welcome

    • one year ago
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