Question
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
Answers
drwls
"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.
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.