Question

a ladder 6 feet long leans against a vertical building. the bottom of the ladder slides away from the building horizontally at rate of 1/2 ft/sec. A) at what rate is the top of the ladder sliding down the wall when the bottom of the ladder is 3 feet from the wall? b)at what rate is the angle between the ladder and the ground changing when the bottom of the ladder is 3 feet from the wall?

Answer 1

since x^2+y^2 = 36
2x dx/dt + 2y dy/dt = 0
when x=3, y=√27, so we have

(3)(1/2) + √27 dy/dt = 0
so now you know how y is changing

Since
tanθ = y/x,
sec^2θ dθ/dt = 1/x dy/dt - y/x^2 dx/dt
Now just plug in your numbers again to find dθ/dt

Of course, you can also say

secθ = x/√27 so
secθ tanθ dθ/dt = 1/√27 dx/dt
but the same answer comes out.