How do I solve this? (question too long)

Question

The top of a 15-foot-long ladder rests against a vertical wall with the bottom of the ladder on level ground, as shown above (it's just a ladder leaning against a wall). The ladder is sliding down the wall at a constant rate of 2 feet per second. At what rate, in radians per second, is the acute angle between the bottom of the ladder and the ground changing at the instant the bottom of the ladder is 9 feet from the base of the wall? (The last part confused me...)

Answer 1

I make a drawing with the ladder in first quadrant, forming a right triangle with the axes.'

x²  + y² = 15²  and when x=9, y = 12.

 

acute angle θ is formed by side x and hypotenuse 15.

you WANT dθ/dt at the instant when x = 9, at which time y = 12.

 

sin θ = y/15 at any time during the sliding.

cos θ (dθ/dt) = (1/15)(dy/dt) by differentiation of the above.

dy/dt = -2 (given rate of slippage)

(dθ/dt) = (1/15)(-2) / cos θ.  Now all you need to do is to substitute cos θ at this instant.  Got it?