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eva12 Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. baldymcgee6 Group Title
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    |dw:1350689236706:dw|

    • 2 years ago
  2. eva12 Group Title
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    how do you get the answer or solve it?

    • 2 years ago
  3. baldymcgee6 Group Title
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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

    • 2 years ago
  4. eva12 Group Title
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    i do not understand this problem or how to approach it to solve it

    • 2 years ago
  5. eva12 Group Title
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    can someone help me

    • 2 years ago
  6. baldymcgee6 Group Title
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    |dw:1350690450622:dw|

    • 2 years ago
  7. baldymcgee6 Group Title
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    you can express the triangle as 7^2 + x^2 =r^2

    • 2 years ago
  8. eva12 Group Title
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    then what will x be and would r will stand for rate rigth

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

    • 2 years ago
  10. baldymcgee6 Group Title
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    I'm just calling r the length of ROPE

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

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

    • 2 years ago
  13. eva12 Group Title
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    then i pug in 100 for r and 10 for x in the equation

    • 2 years ago
  14. baldymcgee6 Group Title
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    no, dr/dt = -10 ft/min.. we don't know what x is

    • 2 years ago
  15. eva12 Group Title
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    49 +2x=2(100)= will = x is 75.5

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

    • 2 years ago
  17. eva12 Group Title
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    so then how do i approach then

    • 2 years ago
  18. baldymcgee6 Group Title
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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?

    • 2 years ago
  19. eva12 Group Title
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    do i solve for 1 equation then the 2 one

    • 2 years ago
  20. baldymcgee6 Group Title
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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

    • 2 years ago
  21. eva12 Group Title
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    dx/dt=2r*dr/dt-49/2x

    • 2 years ago
  22. baldymcgee6 Group Title
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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

    • 2 years ago
  23. eva12 Group Title
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    -102198.6893

    • 2 years ago
  24. baldymcgee6 Group Title
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    no, it shouldn't be that big

    • 2 years ago
  25. baldymcgee6 Group Title
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    you must have typed it in wrong in your calculator

    • 2 years ago
  26. eva12 Group Title
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    -10.27

    • 2 years ago
  27. eva12 Group Title
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    thanks

    • 2 years ago
  28. baldymcgee6 Group Title
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    you're very welcome

    • 2 years ago
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