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anonymous

  • one year ago

I need help again! A 15-ft ladder is leaning against one wall of an alley 9 ft wide. the lade slips, its top sliding down the wall, its foot sliding across the alley and striking the opposite wall at a speed of 4ft/sec. how fast is the top of the ladder falling at that instance? calculus makes me want to cry!

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  1. Astrophysics
    • one year ago
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    |dw:1438840509295:dw|

  2. Astrophysics
    • one year ago
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    Do you agree with the diagram

  3. anonymous
    • one year ago
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    i think so... i suck at calculus.

  4. Astrophysics
    • one year ago
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    |dw:1438841158267:dw| as it's sliding so we use the Pythagorean theorem \[x^2+y^2=r^2\]

  5. anonymous
    • one year ago
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    i got that part

  6. Astrophysics
    • one year ago
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    \[x^2+y^2=15^2 \implies x^2+y^2=225\] now differentiate \[2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt } = 0\] Finish it off

  7. Astrophysics
    • one year ago
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    `how fast is the top of the ladder falling at that instance` this tells you what you're solving for, should be clear

  8. Astrophysics
    • one year ago
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    |dw:1438841472668:dw| just making sure you know it's a right triangle :P

  9. anonymous
    • one year ago
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    i just confused myself even further. I just can't figure this out.

  10. Astrophysics
    • one year ago
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    Ok, so given the equation what are we solving for?

  11. anonymous
    • one year ago
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    dt ??

  12. Astrophysics
    • one year ago
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    You are solving for dy/dt as we're looking for how fast the top of the ladder is falling

  13. Astrophysics
    • one year ago
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    So solve for dy/dt first, as you would for x, don't plug anything just isolate dy/dt.

  14. anonymous
    • one year ago
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    \[\frac{ dy }{dt }=\frac{ -x \frac{ dx }{ dt } }{ y }\] ??

  15. Astrophysics
    • one year ago
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    Yeah :) good \[\frac{ dy }{ dt }= -\frac{ x }{ y }\frac{ dx }{ dt }\]

  16. Astrophysics
    • one year ago
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    So we know x = 9, and r = 15, what is y?

  17. anonymous
    • one year ago
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    12

  18. Astrophysics
    • one year ago
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    Good

  19. Astrophysics
    • one year ago
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    x = 9, y = 12, r = 15, and we know dx/dt = 4 ft/s \[\frac{ dy }{ dt } = - \frac{ 9 }{ 12 }(4) ft/s\]

  20. anonymous
    • one year ago
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    so 3ft/s ?

  21. Astrophysics
    • one year ago
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    - 3ft/s the negative just means the distance from the top of the ladder is decreasing

  22. anonymous
    • one year ago
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    oh okay i get it now!

  23. anonymous
    • one year ago
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    thank you so much

  24. Astrophysics
    • one year ago
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    No problem

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