A street light is at the top of a 16ft tall pole. A woman 6ft tall walks away from the pole with a speed of 6ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30ft from the base of the pole?
If the woman is x feet from the pole, and her shadow is of length s, then
s/6 = (x+s)/16
s/6 = x/16 + s/16
5/48 s = x/16
s = 3/5 x
So, unlike the problem where some angle is steadily changing, here the shadow is always 3/5 as long as the distance from the pole.
Since dx/dt = 6, ds/dt = 18/5
No matter how far she is from the pole.
However, that is not the answer to the question. The tip of her shadow is moving 18/5 from the woman. Add to that her own walking speed, and we see that the tip of the shadow is moving at 18/5 + 6 = 48/5 ft/s
To find the rate at which the tip of the woman's shadow is moving, we can use similar triangles. Let's break down the problem and define some variables:
Let:
- h = height of the pole (16 ft)
- x = distance of the woman from the base of the pole (30 ft)
- y = length of the woman's shadow
- q = height of the woman (6 ft)
- p = rate at which the tip of the woman's shadow is moving (what we need to find)
- s = rate at which the woman is moving (6 ft/sec)
From similar triangles, we know that the ratio of the height of the pole to the woman's height is equal to the ratio of the length of the shadow to the distance from the base of the pole:
h / q = y / x
Plugging in the given values:
16 / 6 = y / 30
Simplifying the equation:
y = (16/6) * 30 = 80 ft
Now, to find the rate at which the tip of the woman's shadow is moving, we can differentiate both sides of the equation with respect to time:
d(y) / dt = d((16/6)*30) / dt
Since the height of the woman and the distance from the base of the pole are constant, we have:
d(y) / dt = d((16/6)*30) / dt = 0
Therefore, the rate at which the tip of the woman's shadow is moving is 0 ft/sec.