A street light is at the top of a 15.000 ft. tall pole. A man 6.300 ft tall walks away from the pole with a speed of 6.000 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 45.000 feet from the pole?
using similar triangles, we see that if the man is x from the pole, and his shadow is s long,
(x+s)/15 = s/6.3
6.3x + 6.3s = 15s
6.3x = 8.7s
So, if x is increasing and s is increasing, then the tip of the shadow is moving at speed dx/dt + ds/dt.
Or, dx/dt (1 + 6.3/8.7)
Note that it does not matter how far from the pole the man is.
To find how fast the tip of the man's shadow is moving, we can use similar triangles and the concept of rates.
Let's assume that the length of the man's shadow at any given position is x ft. Since the street light is on top of a 15.000 ft pole, the total length from the top of the pole to the tip of the shadow is 15.000 + x.
We are given that the man's height is 6.300 ft, and he is walking at a speed of 6.000 feet/sec. This means that dx/dt (the rate of change of the length of the shadow with respect to time) is 6.000 ft/sec.
Using similar triangles, we can set up the following proportion:
(height of man) / (length of man's shadow) = (total height) / (total length of shadow)
6.300 / x = 15.000 / (15.000 + x)
To get rid of the fractions, we can cross-multiply:
6.300 * (15.000 + x) = 15.000 * x
Now, let's simplify this equation:
94.500 + 6.300x = 15.000x
Rearranging the terms:
8.700x = 94.500
Divide both sides by 8.700:
x = 10.862 ft
Now, we can differentiate the equation with respect to time:
d(6.300) / dt = d(15.000)/dt - d(x)/dt
0 = 0 - dx/dt
Therefore, the rate at which the length of the shadow is changing is 0 ft/sec. This is because the length of the shadow remains constant when the man is at a distance of 45.000 ft from the pole.
In conclusion, when the man is 45.000 ft away from the pole, the tip of his shadow is not moving, and its speed is 0 ft/sec.