A 53 in ice hockey stick rests against a vertical wall. The bottom of the stick slides away from the wall. How fast does the distance, y change with respect to the angle, θ, which is made by the wall and the stick, when θ is equals 45^o?

Assumption:

• The angle made by the wall and the stick is θ.
• The slipped distance is y.

According to the diagram,

     sinθ = y/53

Thus

     y = 53 sinθ

Derivative y with respect to θ

     dy/dθ = d(53 sinθ)/dθ

              =53 cosθ

When θ = 45^o,

     dy/dθ = 53 cosθ = 37.5 in/rad