Aladder 10 ft long rests against a vertical wall. if the bottom of the ladder slides away from the wall at a rate of 0.6 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (that is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

1) Set up an equation to model the situation.
    cos(Ф) = x/10

2) Take the derivative/gradient in terms of time (t) to find the related rates.
    d[cos(Ф)]/dt = d[x/10]/dt
    -sin(Ф)*dФ/dt = (1/10)*dx/dt
    dФ/dt = -(dx/dt)/(10*sin(Ф))

3) Substitute in the constants for the moment in time
    Given: dx/dt = 0.6; x=6 -> Ф = cos-1(6/10) = ~0.93rad
    Therefore: dФ/dt = -3/40 rad/sec

The slope of the perpendicular line  m =6

Step-by-step explanation:

Step(i):-

Given that the equation of the straight line

                                            2x + 12y =-192

                                        ⇒ 12 y = -192 -2x      

                                               

                                             

                                             y = m x +C

The slope of the given line

                                         

Step(ii):-

The slope of the perpendicular line

                                       

                                   =

The slope of the perpendicular line m =6