Calculus
posted by Lost student on .
a 16 foot ladder is leaning against a building. the bottom of the ladder os sliding along the pavement at 3 ft/s How fast is the top of the ladder moving down when the foot of the ladder is 2 ft from the wall

Let the distance of the bottom of the ladder from the wall be x, and let the distance of the top of the ladder from the floor be y.
x^2 + y^2 = 256
You want to know dy/dt when dx/dt = 3 and x = 2. At that time, y^2 = 252 and y = 15.87
2x *dx/dt + 2y*dy/dt = 0
dy/dt = (x/y)*dx/dt = 0.126 dx/dt
Use that equation to calculate dy/dt