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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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.

Mathematics
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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.
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you are told \(x'=3\)

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when \(t=2\) how far is the ladder away form the wall?
Is there a formula to use on how to calculate that answer?
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`
|dw:1444353987783:dw|
|dw:1444354010477:dw|
not really a formula...
So x would equal 11 at t=2 and that means that h would be about 11.6.
|dw:1444354236271:dw|
put those in last use pythagoras to find the relation between \(x,h\) and \(16\)
I ended up getting about 2.84
I'm getting about the same \[\Large x^2 + h^2 = 16^2\] \[\Large x^2 + h^2 = 256\] \[\Large \frac{d}{dt}(x^2 + h^2) = \frac{d}{dt}(256)\] \[\Large 2x*\frac{dx}{dt}+ 2h*\frac{dh}{dt} = 0\] \[\Large 2*11*3+ 2*11.61895004*\frac{dh}{dt} = 0\] \[\Large 66+ 23.23790008\frac{dh}{dt} = 0\] \[\Large 23.23790008\frac{dh}{dt} = -66\] \[\Large \frac{dh}{dt} = \frac{-66}{23.23790008}\] \[\Large \frac{dh}{dt} = -2.840187787\]
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
you're welcome
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?
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.
Obviously in the real world, the speed would hit some limit just before the ladder hits the ground
Thank you so much!

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