Math
posted by Anonymous .
A 15 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 3ft/sec.
Find the velocity of the top of the ladder at time t=3.

Let the foot of the ladder be x feet from the wall
let the height of the ladder be y feet
then x^2 + y^2 = 225
differentiate with respect to t, this is a "related rate problem"
2x dx/dt + 2y dy/dt = 0 (1)
when t=3 sec,
x = 9 feet (you knew the rate)
then 81+y^2=225
y=12 feet
sub x=9, y=12, dx/dt = 3 into (1) and solve for dy/dt
I got dy/dt = 9/4 ft/sec
(the negative value shows that y was decreasing at 9/4 ft/sec at that instant) 
a building is 3ft. from a 8ft. fence that surround propert a worker wants to wash a window in the building 13ft. from the ground. he will need what size ft. ladder