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shann803

  • 2 years ago

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.

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  1. zepdrix
    • 2 years ago
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    |dw:1352140346872:dw|

  2. zepdrix
    • 2 years ago
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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? :)

  3. shann803
    • 2 years ago
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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

  4. zepdrix
    • 2 years ago
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    k cool :D

  5. shann803
    • 2 years ago
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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

  6. zepdrix
    • 2 years ago
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    Oh heh :3

  7. shann803
    • 2 years ago
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    Sorry about that :)

  8. zepdrix
    • 2 years ago
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    So we'll call it ... 5sqrt3 i guess :D in simplified form.

  9. shann803
    • 2 years ago
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    sounds good

  10. zepdrix
    • 2 years ago
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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).

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

  12. zepdrix
    • 2 years ago
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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.

  13. shann803
    • 2 years ago
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    yes yes I remember my teacher saying that!

  14. shann803
    • 2 years ago
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    so we plug in the dy/dt which is -4 and solve for dx/dt?

  15. zepdrix
    • 2 years ago
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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 :)

  16. zepdrix
    • 2 years ago
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    from the earlier? what is wrong with me -_- ugh..

  17. shann803
    • 2 years ago
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    lol its fine :)

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

  19. zepdrix
    • 2 years ago
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    Hmm yes good good :) equals 0

  20. shann803
    • 2 years ago
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    yes of course lol :)

  21. shann803
    • 2 years ago
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    so my answer is 5.08?

  22. shann803
    • 2 years ago
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    yes it's right!!1 THANK YOU SOOOO MUCH!! :)

  23. zepdrix
    • 2 years ago
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    Yay team \c:/

  24. shann803
    • 2 years ago
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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?

  25. zepdrix
    • 2 years ago
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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.

  26. zepdrix
    • 2 years ago
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    the gravel problem?

  27. shann803
    • 2 years ago
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    true true, and yes the gravel problem lol :)

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