A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)

Answer 1

we have
x^2 + y^2 = 100
when x=8, y=6
Now, consider
tan θ = y/x

sec^2 θ dθ/dt = (x dy/dt - y dx/dt)/x^2
Now just plug in your numbers and solve for dθ/dt