Asked by Marilyn
                A ladder 5m long is standing vertically, flat against a vertical wall, while its lower end in on the horizontal floor. The lower end moves horizontally away from the wall at a constant speed of 1m/s while the upper end stays in contact with the wall. Find the speed at which the upper end is moving down the wall 4 seconds after the lower end has left the wall.
            
            
        Answers
                    Answered by
            Steve
            
    If y is the height and x is the distance from the wall, then
x^2+y^2=25
x dx/dt + y dy/dt = 0
After 4 seconds, x=4 and y=3 so,
4 * 1 + 3 dy/dt = 0
dy/dt = -4/3
so the top is sliding down the wall at 4/3 m/s
    
x^2+y^2=25
x dx/dt + y dy/dt = 0
After 4 seconds, x=4 and y=3 so,
4 * 1 + 3 dy/dt = 0
dy/dt = -4/3
so the top is sliding down the wall at 4/3 m/s
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