# Rate of Change of theta in ladder to wall.

1. Oct 13, 2008

### ryandamartini

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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 dont know what to do here.

2. Oct 13, 2008

### HallsofIvy

Staff Emeritus
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).

3. Jan 3, 2012

### Rayquesto

theta=tan(y/x) right
then what the formula to find the change in theta at any given point?