A man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the ground. at what rate is the tip of his shadow moving? at what rate is his shadow lengthening?

General
1

a man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the ground. at what rate is the tip of his shadow moving? at what rate is his shadow lengthening?

2

What is the rate at which the tip of the man’s shadow is moving and the rate at which his shadow is lengthening?

Answer:
To find out the rates at which the tip of the man’s shadow is moving and his shadow is lengthening, we can use related rates and similar triangles.

Let:

  • Height of the man (h): 1.6 m
  • Rate at which the man is walking (dx/dt): 0.3 m/sec
  • Height of the street light (a): 4 m

The speed at which the tip of the man’s shadow is moving (ds/dt) can be calculated using the formula:

ds/dt = [(Height of the man)/(Distance from man to light)] * dx/dt

  1. Calculating the distance from the man to the light:
    This forms a right triangle where the height of the man forms one side, the shadow forms the other side, and the line from the light to the tip of the shadow is the hypotenuse. Using similar triangles, the distance from the man to the light can be found:

h / x = a / (x + d)
1.6 / x = 4 / (x + d)
1.6(x + d) = 4x
1.6x + 1.6d = 4x
1.6d = 2.4x
d = 1.5x

  1. Calculating the speed at which the tip of the shadow is moving:
    ds/dt = [(1.6)/(1.5x)] * 0.3
    ds/dt = 0.32 m/s

Therefore, the tip of the man’s shadow is moving at a rate of 0.32 m/s.

  1. Calculating the rate at which the shadow is lengthening:
    The rate at which the shadow is lengthening (dy/dt) can be calculated using the formula:

dy/dt = (h/x) * dx/dt
dy/dt = (1.6/x) * 0.3
dy/dt = 0.48 m/s

Hence, the shadow is lengthening at a rate of 0.48 m/s.