posted by Sarah on .
One end of a long 26 foot ladder is on the floor and the other end rests on a verical wall the bottom end of the ladder is pulled away from the wall qt a rate of 4 feet per second. how fast is the angle formed by the ladder and the floor changing when the ladder is 10 feet away from the wall
"bepper" is not a subject I am familiar with.
Let x be the distance of the bottom from the wall.
x = (26 ft)*sin(theta)
dx/dt = 26 cos(theta)* d(theta)/dt = 4 ft/s
Solve for d(theta)/dt. It will be in radians per second.