Answer:
0.17 °/s
Step-by-step explanation:
Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.
We differentiate the above expression with respect to time to have
dsinθ/dt = d(D/L)/dt
cosθdθ/dt = (1/L)dD/dt
dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.
We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft
We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]
dθ/dt = (1/Lcosθ)dD/dt
dθ/dt = (1/L√[1 - (D/L)²])dD/dt
dθ/dt = (1/√[L² - D²])dD/dt
Substituting the values of the variables, we have
dθ/dt = (1/√[20² - 16²]) 2 ft/s
dθ/dt = (1/√[400 - 256]) 2 ft/s
dθ/dt = (1/√144) 2 ft/s
dθ/dt = (1/12) 2 ft/s
dθ/dt = 1/6 °/s
dθ/dt = 0.17 °/s
