Calculus
posted by Lereesa .
A 5m long ladder rests on a vertical wall. if the bottom of the ladder slides away from the wall at 1m/s, how fast is the angle between the lader and the ground changing when the bottom of the ladder is 3m from the wall?

Let X be the distance of the bottom from the wall. Let A be the angle
X = 5 cos A
dX/dt = (dX/dA)*(dA/dt)
= 5 sin A* dA/dt
dA/dt = (1 m/s)(1/5)(1/sinA)
In this case, sin A = 4/5 when X =3 m.
dA/dt will be in radians per second. 
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