A cleaning man is cleaning Koerner library with long cleaning equipment. The cleaner puts the equipment on the ground and the equipment 5 meters long leans against the library. If the bottom of the equipment slides away from the library horizontally at a rate of 0.5 m/s, how fast is the ladder sliding down the house when the top of the ladder is 3 meters from the ground?

What we know in this question is y=3, z=5, dx/dt=0.5 and the equation we have to use is       (x^2)+(y^2)=z^2

if we differentiate the equation, 2x(dx/dt)+2y(dy/dt)=2z(dz/dt) and we plug the numbers we already know.

2x(0.5)+2(3)(dy/dt)=2(5)(0), dz/dt equals 0 because it does not change.

To find x, we use (x^2)+(y^2)=z^2 and plug y=3, and z=5

Then x^2=25-9=16. Therefore x=4 and we plug 4 to 2x(0.5)+2(3)(dy/dt)=2(5)(0).

2(4)(0.5)+2(3)(dy/dt)=2(5)(0)

4+(6)(dy/dt)=0

(6)(dy/dt)=-4

dy/dt=-2/3