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jarp0120

  • one year ago

a ladder 6 m long leans against a vertical wall. The lower end of the ladder is moved away from he wall at the rate of 2 m/min. Find the rate of change of the area formed by the wall, the floor and the ladder when the lower end is 4 m from the wall. please help

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  1. Jarp0120
    • one year ago
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    |dw:1443859889821:dw|

  2. Jhannybean
    • one year ago
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    It doesnt seem right for some reason

  3. IrishBoy123
    • one year ago
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    can you think of an equation that connects l and b in your drawing?

  4. Jarp0120
    • one year ago
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    |dw:1443860176242:dw|

  5. Jarp0120
    • one year ago
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    A=1/2 bh

  6. Jhannybean
    • one year ago
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    Using the pythagorean theorem, we can find the rate of change for everything moving

  7. dan815
    • one year ago
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    oh boy a ladder problem :)

  8. Jarp0120
    • one year ago
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    what is the area if the base?

  9. Jarp0120
    • one year ago
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    is 4

  10. Jhannybean
    • one year ago
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    You don't necessarily need to know that. You're simply looking for \(\dfrac{dl}{dt}\) when \(x=4\)

  11. Jarp0120
    • one year ago
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    how?

  12. IrishBoy123
    • one year ago
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    Pythagoreas

  13. Jhannybean
    • one year ago
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    The area is constantly changing w.r.t time as the ladder moves down the wall

  14. dan815
    • one year ago
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    This question you are doing right now, will make or break your mathematical career

  15. Jhannybean
    • one year ago
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    Dan... -_-

  16. Jarp0120
    • one year ago
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    there's no change in y right? the ladder?

  17. Astrophysics
    • one year ago
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    |dw:1443860459057:dw| \[\frac{ dx }{ dt } = 2\] I thought this

  18. Jarp0120
    • one year ago
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    so i'll just substitute dy/dt and dx/dt?

  19. Jarp0120
    • one year ago
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    to find dl/dt?

  20. dan815
    • one year ago
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    |dw:1443860606641:dw|

  21. Jhannybean
    • one year ago
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    Yeah you're right, it `moved away from the wall` at that rate, whoops.

  22. Astrophysics
    • one year ago
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    :)

  23. Jarp0120
    • one year ago
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    y=2 sqrt of 5?

  24. dan815
    • one year ago
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    yes that is sqrt(2)

  25. dan815
    • one year ago
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    sqrt(20)*

  26. dan815
    • one year ago
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    but solve for dy/dt

  27. Astrophysics
    • one year ago
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    yes dy/dt is what you need

  28. Jhannybean
    • one year ago
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    I think i read the question wrong, fml.

  29. dan815
    • one year ago
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    |dw:1443861020387:dw|

  30. Jhannybean
    • one year ago
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    Oh okay, as the ladder moves down the wall, the ladder increases the base of the triangle but shortens the height of the triangle, therefore you're looking for the rate of change of the height

  31. Astrophysics
    • one year ago
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    \[x^2+y^2=36\] \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] \[\frac{ dy }{ dt }=-\frac{ x }{ y }\frac{ dx }{ dt }\]

  32. Astrophysics
    • one year ago
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    So x here is the distance from the bottom of the ladder to the wall and y is the distance of the top of the ladder to the ground, as the image shows. The second part of the equation we simply use chain rule, \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] and we just solve for dy/dt then

  33. Jarp0120
    • one year ago
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    wait is this right? dh/dt =-2sqrt5 / 5 that means the Area now is 6/sqrt5?

  34. Astrophysics
    • one year ago
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    The rate should be negative because the ladder is decreasing

  35. Astrophysics
    • one year ago
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    it just defines the direction

  36. Jarp0120
    • one year ago
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    or dy/dt= -2/sqrt5

  37. Astrophysics
    • one year ago
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    Well what did Pythagorean theorem give you? x=4, r=6 \[x^2+y^2=r^2 \implies y=\sqrt{36-16} = \sqrt{20}\]

  38. Astrophysics
    • one year ago
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    And we know \[\frac{ dy }{ dt }=-\frac{ x }{ y }\frac{ dx }{ dt }\] just plugging in the values at this point

  39. Astrophysics
    • one year ago
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    Yes dy/dt = - 2/sqrt(5)

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