Rate of Change of theta in ladder to wall.

ryandamartini
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Homework Statement


A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically.


Homework Equations



a^2+b^2=c^2 ?
Theta(dot) is change of time respect to theta.

The Attempt at a Solution



In class he went over finding the rate of theta, getting the equation Theta(dot)=X/(10cos(45) I tried plugging in 6 for X, I just really don't know what to do here.
 
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I would suggest that instead of using a2+ b2= c2, which does not involve an angle, you use a= bcos(\theta) where a is the "hypotenuse" (the length of the ladder) and a is the "near side" (the distance from the foot of the ladder to the wall).
 
theta=tan(y/x) right
then what the formula to find the change in theta at any given point?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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