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.

- anonymous

- schrodinger

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- zepdrix

|dw:1352140346872:dw|

- 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

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

k cool :D

- 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

Oh heh :3

- anonymous

Sorry about that :)

- zepdrix

So we'll call it ... 5sqrt3 i guess :D in simplified form.

- anonymous

sounds good

- 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

so the derivative of x^2+y^2=14^2 is -x/y?

- 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

yes yes I remember my teacher saying that!

- anonymous

so we plug in the dy/dt which is -4 and solve for dx/dt?

- 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

from the earlier? what is wrong with me -_- ugh..

- anonymous

lol its fine :)

- anonymous

so it's 2(5sqrt3)(dx/dt)+2(11)(-4) is this right?

- zepdrix

Hmm yes good good :) equals 0

- anonymous

yes of course lol :)

- anonymous

so my answer is 5.08?

- anonymous

yes it's right!!1 THANK YOU SOOOO MUCH!! :)

- zepdrix

Yay team \c:/

- 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

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

the gravel problem?

- anonymous

true true, and yes the gravel problem lol :)

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