A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 7 ft/s along a straight path. How fast is the tip of his shadow moving when he is 45 ft from the pole?
ft/s
To find how fast the tip of the man's shadow is moving when he is 45 ft from the pole, we can use similar triangles.
Let's consider the following diagram:
O (Street Light)
|\
| \
| \
| \
h| \ x
| \
| \
M|-------Q (Man)
| p|
| |\
| | \
| | \
S|------|---\ T (Shadow)
| |
| |\
| | \
| | \
W|------|---\ Y (Ground)
In this diagram, O represents the street light at the top of the pole, Q represents the position of the man, and T represents the tip of the shadow on the ground.
We are given that the height of the pole is 15 ft, the height of the man is 6 ft, and the man is walking away from the pole at a speed of 7 ft/s.
Let's define the following variables:
- h: height of the pole = 15 ft
- x: distance of the man from the pole = 45 ft
- p: distance from the pole to the tip of the man's shadow = ?
- s: distance from the man to the tip of his shadow = ?
- w: width of the man's shadow = ?
- y: distance from the tip of the shadow to the pole = ?
We can set up a proportion using similar triangles:
h / p = (h + s) / w
Substituting the given values:
15 / p = (15 + s) / w
We need to find the rate of change of s with respect to time (ds/dt) when x = 45 ft. To do this, we can differentiate both sides of the equation with respect to time:
d(15) / dt / p = d(15 + s) / dt / w
Since the height of the pole (15 ft) and the width of the man's shadow (w) are constant, their derivatives with respect to time are zero:
0 = ds / dt / w
Simplifying the equation, we get:
ds / dt = 0
Therefore, the tip of the man's shadow is not moving.