A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?
Let the woman be b feet from the pole, and let her shadow be a feet long.
Using similar triangles,
a/6 = (a+b)/16
so a = 3/5 b
da/dt = 3/5 db/dt = 3/5 * 4 = 12/5 ft/sec, a constant speed.
That seems odd. I expected her shadow's length to accelerate, but I don't see an error. These problems usually involve a steadily changing angle, and with the tangent function things speed up quickly. Not so here...
To find the rate at which the tip of the woman's shadow is moving, we can use similar triangles and the concept of related rates.
Let's denote:
- The height of the pole as H = 16 ft
- The height of the woman as h = 6 ft
- The distance between the woman and the base of the pole as x = 30 ft
Now, let's consider two similar triangles: the triangle formed by the pole, its shadow, and the ground, and the triangle formed by the woman, her shadow, and the ground.
The ratio of the height of the pole to the height of its shadow is constant and equal to the ratio of the height of the woman to the height of her shadow. So we have:
H / (H + s) = h / s
Where s represents the length of the woman's shadow.
Rearranging the equation, we get:
Hs = hs + sh
Hs - hs = sh
s(H - h) = sh
s = (sh) / (H - h)
Now, let's find the rate at which the tip of the woman's shadow is moving, which is represented by ds/dt. We are given that dx/dt (the rate at which the woman is moving away from the pole) is 4 ft/sec. We need to find ds/dt when x = 30 ft.
Differentiating the equation s = (sh) / (H - h) with respect to time (t), we get:
ds/dt = [(dsh)/(dt)(H - h)] - [s(dH/dt) / (H - h)]
Since dH/dt is 0 (the height of the pole is not changing), the equation simplifies to:
ds/dt = (dsh) / (dt)(H - h)
Now, we need to find dsh/dt. By taking the derivative of s = (sh) / (H - h) with respect to time (t), we get:
ds/dt = (h(ds/dt) + s(dh/dt)) / (H - h)
Rearranging the equation and substituting the known values, we get:
ds/dt = (h(4) + s(0)) / (16 - 6)
ds/dt = (4h) / 10
Now, substitute the value of h as 6 ft:
ds/dt = (4(6)) / 10
ds/dt = 24 / 10
ds/dt = 2.4 ft/sec
Therefore, the tip of the woman's shadow is moving at a rate of 2.4 ft/sec when she is 30 ft from the base of the pole.