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 40 ft from the base of the pole?
L = pole to shadow tip
x = pole to walker
so (L-x) = walker to shadow tip
L/16 = (L-x)/6 (similar triangles)
6 L = 16 L - 16 x
16 x = 10 L
L = 1.6 x
dL/dt = 1.6 dx/dt
To find the rate at which the tip of the woman's shadow is moving, we can use the concept of similar triangles. Let's consider two similar triangles: one formed by the woman, her shadow, and the direction of the light, and the other formed by the pole, the woman's height, and the length of her shadow.
We know that the woman's height is 6 ft and the pole's height is 16 ft. Now, let x represent the length of the woman's shadow.
Since the triangles are similar, we can set up a proportion:
(woman's height) / (length of her shadow) = (pole's height) / (total length of the shadow)
Using the given values, we have:
6 / x = 16 / (x + 40)
Now, we can cross-multiply and solve for x:
6(x + 40) = 16x
6x + 240 = 16x
10x = 240
x = 24 ft
Now that we know the length of the woman's shadow is 24 ft, we can find how fast the tip of her shadow is moving when she is 40 ft from the base of the pole.
Let's differentiate the proportion we found earlier with respect to time:
6 dx/dt = 16 (dx/dt)
We can simplify this equation:
6 dx/dt = 16 dx/dt
Now, let's plug in the values we know:
6 dx/dt = 16 * 4 (since the woman's speed is given as 4 ft/sec)
Simplifying further:
6 dx/dt = 64
Finally, we can solve for dx/dt (the rate at which the tip of the shadow is moving):
dx/dt = 64 / 6
dx/dt = 10.67 ft/sec (approximately)
Therefore, the tip of the woman's shadow is moving at a rate of approximately 10.67 ft/sec.