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anonymous

  • one year ago

Last differentiation problem for the night (hopefully). I'm not sure if it's the way it's worded, but this problem is just downright confusing to me.

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  1. anonymous
    • one year ago
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    Refer to a 16-ft ladder sliding down a wall. The variable h is the height fo the ladder's top at time t, and x is the distance from the wall to the ladder's bottom. a) Assume that the bottom slides away from the wall at a rate of 3 ft/s. Find the velocity of the top of the ladder at t = 2 if the bottom is 5 ft from the wall at t = 0.

  2. anonymous
    • one year ago
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    |dw:1444353532318:dw|

  3. anonymous
    • one year ago
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    you are told \(x'=3\)

  4. anonymous
    • one year ago
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    when \(t=2\) how far is the ladder away form the wall?

  5. anonymous
    • one year ago
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    Is there a formula to use on how to calculate that answer?

  6. jim_thompson5910
    • one year ago
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    hints: ` the bottom slides away from the wall at a rate of 3 ft/s` `the bottom is 5 ft from the wall at t = 0`

  7. jim_thompson5910
    • one year ago
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    |dw:1444353987783:dw|

  8. jim_thompson5910
    • one year ago
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    |dw:1444354010477:dw|

  9. anonymous
    • one year ago
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    not really a formula...

  10. anonymous
    • one year ago
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    So x would equal 11 at t=2 and that means that h would be about 11.6.

  11. jim_thompson5910
    • one year ago
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    |dw:1444354236271:dw|

  12. anonymous
    • one year ago
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    put those in last use pythagoras to find the relation between \(x,h\) and \(16\)

  13. anonymous
    • one year ago
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    I ended up getting about 2.84

  14. jim_thompson5910
    • one year ago
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    I'm getting about the same \[\Large x^2 + h^2 = 16^2\] \[\Large x^2 + h^2 = 256\] \[\Large \frac{d}{dt}(x^2 + h^2) = \frac{d}{dt}(256)\] \[\Large 2x*\frac{dx}{dt}+ 2h*\frac{dh}{dt} = 0\] \[\Large 2*11*3+ 2*11.61895004*\frac{dh}{dt} = 0\] \[\Large 66+ 23.23790008\frac{dh}{dt} = 0\] \[\Large 23.23790008\frac{dh}{dt} = -66\] \[\Large \frac{dh}{dt} = \frac{-66}{23.23790008}\] \[\Large \frac{dh}{dt} = -2.840187787\]

  15. anonymous
    • one year ago
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    Oops! Forgot my negative sign. Thanks! I think it was just the way the problem was worded is what threw me off. Once you guys helped explain it, it actually made sense. Thank you! :D

  16. jim_thompson5910
    • one year ago
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    you're welcome

  17. anonymous
    • one year ago
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    This is the last part of this problem. Could you help me with it by chance? Show that the velocity dh/dt approaches infinity as the ladder slides down to the ground (assuming dx/dt is constant). This suggests that our mathematical description is unrealistic, at least for small values of h. What wound in fact, happen as the top of the ladder approaches the ground?

  18. anonymous
    • one year ago
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    @jim_thompson5910

  19. jim_thompson5910
    • one year ago
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    Let p = dx/dt q = dh/dt \[\Large x^2 + h^2 = 16^2\] \[\Large x^2 + h^2 = 256\] \[\Large \frac{d}{dt}(x^2 + h^2) = \frac{d}{dt}(256)\] \[\Large 2x*\frac{dx}{dt}+ 2h*\frac{dh}{dt} = 0\] \[\Large 2x*p+ 2h*q = 0\] \[\Large 2hq = -2xp\] \[\Large q = \frac{-2xp}{2h}\] \[\Large q = -\frac{xp}{h}\] \[\Large \frac{dh}{dt} = -\frac{xp}{h}\] We can see that as the ladder slides to the ground (h --> 0), the fraction dh/dt approaches -infinity. If you have dx/dt be some negative constant, then the fraction dh/dt approaches +infinity.

  20. jim_thompson5910
    • one year ago
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    Obviously in the real world, the speed would hit some limit just before the ladder hits the ground

  21. anonymous
    • one year ago
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    Thank you so much!

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