Related Rate: Ladder Problem

Question

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 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.)

Answer 1

The ladder, wall, and ground form a triangle. The length of the ladder (constant 10 ft) forms the hypotenuse of the triangle.

 

When the ladder slides down the wall, the bottom of the ladder moves away from the wall, in this case at a given constant 0.6 ft/s.

 

The cosine of the angle between the ladder and the wall is equal to the adjacent length (x for length between ladder and wall at any point in time) divided by the hypotenuse (10 ft, for the length of ladder).

 

cos(θ) = x / 10

 

       /|

      / |

10 /  |

    /   | y

   /    |

  /θ__|

     x

 

 

(dθ/dt) is the rate of change of the angle between with respect to time

(dx/dt) is the rate of change of the distance of the bottom of the ladder to the wall

 

Take the derivative of both sides of cos(θ) = x / 10 with respect to time:

 

cos(θ) = x / 10

-sin(θ) (dθ/dt) = (1/10)(dx/dt)    (Derivative)

           (dθ/dt) = (0.1) (0.6 ft/s) / (-sin(θ))

 

Now the foot of the ladder is sliding away from the wall, and when it is 6 ft from the wall:

x = 6ft,

y = sqrt(10^2 - 6^2) = 8ft

 

sin(θ) = 8/10

 

Plug your variables into the equation below and solve:

 

(dθ/dt) = (0.1) (0.6) / (-sin(θ))

           = (0.1) (0.6) / (-8/10)

           = -0.075 rad/s