anonymous
  • anonymous
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.
Calculus1
schrodinger
  • schrodinger
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zepdrix
  • zepdrix
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zepdrix
  • zepdrix
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? :)
anonymous
  • anonymous
Ya so x would then be 11 as well because 14^2-11^2=121 and the square root of that is 11

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zepdrix
  • zepdrix
k cool :D
anonymous
  • anonymous
wow lol my bad i mean it's the square root of 75 is x not 11 so like 8.66 is x :) miscalculated
zepdrix
  • zepdrix
Oh heh :3
anonymous
  • anonymous
Sorry about that :)
zepdrix
  • zepdrix
So we'll call it ... 5sqrt3 i guess :D in simplified form.
anonymous
  • anonymous
sounds good
zepdrix
  • zepdrix
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).
anonymous
  • anonymous
so the derivative of x^2+y^2=14^2 is -x/y?
zepdrix
  • zepdrix
\[\large 2x\frac{ dx }{ dt }+2y \frac{ dy }{ dt }=0\] 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.
anonymous
  • anonymous
yes yes I remember my teacher saying that!
anonymous
  • anonymous
so we plug in the dy/dt which is -4 and solve for dx/dt?
zepdrix
  • zepdrix
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 :)
zepdrix
  • zepdrix
from the earlier? what is wrong with me -_- ugh..
anonymous
  • anonymous
lol its fine :)
anonymous
  • anonymous
so it's 2(5sqrt3)(dx/dt)+2(11)(-4) is this right?
zepdrix
  • zepdrix
Hmm yes good good :) equals 0
anonymous
  • anonymous
yes of course lol :)
anonymous
  • anonymous
so my answer is 5.08?
anonymous
  • anonymous
yes it's right!!1 THANK YOU SOOOO MUCH!! :)
zepdrix
  • zepdrix
Yay team \c:/
anonymous
  • anonymous
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?
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
the gravel problem?
anonymous
  • anonymous
true true, and yes the gravel problem lol :)

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