The speed of the car is 5 m/sec.The height of the pole and the car is 4 and 1.5 meters respectively. How fast does the the angle θ change 1 second after the car starts to move.

Let the angle made by the pole, the light and the car be θ. Let the distance from the pole to the car be x. Let the length of the shadow be y.

ΔABC is similar to ΔADE, so

     1.5/4 = y/(x + y)

     y = 0.6x

     x + y = 1.6x

In trigonometry, the relationship between x + y and θ is:

     tanθ = (x + y)/4

     = 1.6x/4

     = 0.4x

Differentiate this relationship with respect to time:

     sec²θ(dθ/dt) = 0.4dx/dt

     dθ/dt = 0.4(dx/dt)/sec²θ

Notice that the rate of change of the angle, dθ/dt is a function of both the car speed, dx/dt and the current angle. In this case dx/dt = 5 is given, but sec²θ must be determined.

1 second after the car starts,

     x = (1)(5) = 5

     x + y = 1.6(5) = 8

     sec²θ(AD/4)²

              = (<(x + y)² +4²)/16

              = 5

Substituting sec²θ = 5 into θ^'_t = 0.4x^'_t/sec²θ gives


     dθ/dt =(0.4)5/5 = 0.4 radians/sec