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shann803 Group Title

A 14 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 4ft/s, how fast will the foot be moving away from the wall when the top is 11 feet from the ground.

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    |dw:1352140346872:dw|

    • one year ago
  2. zepdrix Group Title
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    So our first step is to find x. The length of that side. I think we'll end up needing that :o Remember your Pythagorean Theorem for finding that side? :)

    • one year ago
  3. shann803 Group Title
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    Ya so x would then be 11 as well because 14^2-11^2=121 and the square root of that is 11

    • one year ago
  4. zepdrix Group Title
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    k cool :D

    • one year ago
  5. shann803 Group Title
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    wow lol my bad i mean it's the square root of 75 is x not 11 so like 8.66 is x :) miscalculated

    • one year ago
  6. zepdrix Group Title
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    Oh heh :3

    • one year ago
  7. shann803 Group Title
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    Sorry about that :)

    • one year ago
  8. zepdrix Group Title
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    So we'll call it ... 5sqrt3 i guess :D in simplified form.

    • one year ago
  9. shann803 Group Title
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    sounds good

    • one year ago
  10. zepdrix Group Title
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    Hmm I'm trying to remember how to do this type of problem XD Lol. I think we can do it like this. \[\large x^2 + y^2 = 14^2\] Taking the derivative from this point. (With respect to t, time).

    • one year ago
  11. shann803 Group Title
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    so the derivative of x^2+y^2=14^2 is -x/y?

    • one year ago
  12. zepdrix Group Title
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    \[\large 2x\frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] Since our derivative is with respect to t, Every time we differentiate a variable that is NOT t, we have to apply the chain rule multiplying by the d/dt term. For example when we differentiate x^2, we get 2x, but a dx/dt will pop out also.

    • one year ago
  13. shann803 Group Title
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    yes yes I remember my teacher saying that!

    • one year ago
  14. shann803 Group Title
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    so we plug in the dy/dt which is -4 and solve for dx/dt?

    • one year ago
  15. zepdrix Group Title
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    Heh :3 Ok cool. So now if we look at our problem. We now have 4 variables! Eeek! But we already KNOW 3 of them from the earlier! Yes plug in dy/dt, and also plug in y and x that we set up earlier :)

    • one year ago
  16. zepdrix Group Title
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    from the earlier? what is wrong with me -_- ugh..

    • one year ago
  17. shann803 Group Title
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    lol its fine :)

    • one year ago
  18. shann803 Group Title
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    so it's 2(5sqrt3)(dx/dt)+2(11)(-4) is this right?

    • one year ago
  19. zepdrix Group Title
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    Hmm yes good good :) equals 0

    • one year ago
  20. shann803 Group Title
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    yes of course lol :)

    • one year ago
  21. shann803 Group Title
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    so my answer is 5.08?

    • one year ago
  22. shann803 Group Title
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    yes it's right!!1 THANK YOU SOOOO MUCH!! :)

    • one year ago
  23. zepdrix Group Title
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    Yay team \c:/

    • one year ago
  24. shann803 Group Title
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    I have another question going it's a problem solving like this but in the mathematics section do you think you can help me with that one too please?

    • one year ago
  25. zepdrix Group Title
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    When you do these problems, one way to check your work is this... You should get an answer that MAKES SENSE. If the ladder is sliding down the wall in the y direction at 4 ft/s Then our slide along the x direction should be a similar number.

    • one year ago
  26. zepdrix Group Title
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    the gravel problem?

    • one year ago
  27. shann803 Group Title
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    true true, and yes the gravel problem lol :)

    • one year ago
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