1. The problem statement, all variables and given/known data PQ represents a 40 ft ladder with end P against a wall and end Q on level ground. R represents the angle where the wall meets the ground. If the ladder is slipping down the wall, what is the distance RQ at the instant when Q is moving along the ground 3/4 as fast as P is moving down the wall? 2. Relevant equations a^2 + b^2 = c^2 3. The attempt at a solution PR^2 + RQ^2 = PQ^2 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt) = 2(40)0 = 0 x(dx/dt) + y(dy/dt) = 0 What do I do now?!