A street light is at the top of a 20 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?
To find the rate at which the tip of the woman's shadow is moving, we can use similar triangles.
Let's call the distance between the woman and the base of the pole "x" and the length of her shadow "y." We are given that the woman is moving away from the pole at a speed of 6 ft/sec.
We can set up the following proportion:
(20 ft - 6 ft) / x = 6 ft / (x + 35 ft)
This is because the height of the pole minus the height of the woman is equal to the height of the shadow, and this height should be proportional to the distance between the woman and the base of the pole.
Now, we can solve this equation for x. Cross-multiplying gives us:
(20 ft - 6 ft) * (x + 35 ft) = 6 ft * x
(14 ft) * (x + 35 ft) = 6 ft * x
14x + 490 ft = 6x
8x = 490 ft
x = 61.25 ft
So, when the woman is 35 ft from the base of the pole, the distance between the woman and the base of the pole (x) is 61.25 ft.
To find how fast the tip of her shadow is moving, we can find the rate at which the shadow length (y) is changing with respect to time. We can differentiate the original equation with respect to time, t:
(20 ft - 6 ft) / x = 6 ft / (x + 35 ft)
Differentiating both sides:
(-8 ft / x^2) * dx/dt = (6 ft / (x + 35 ft))^2 * (d(x + 35 ft) / dt)
Now, we substitute the values we know:
(-8 ft / (61.25 ft)^2) * dx/dt = (6 ft / (61.25 ft + 35 ft))^2 * (d(61.25 ft + 35 ft) / dt)
Simplifying:
-8 ft / (61.25 ft)^2 * dx/dt = (6 ft / 96.25 ft)^2 * (d(96.25 ft) / dt)
-8 ft / (61.25 ft)^2 * dx/dt = (6 ft / 96.25 ft)^2 * 0 ft/sec (since the woman's speed is not given)
Now, we can solve for dx/dt (the rate at which the tip of the shadow is moving):
-8 ft / (61.25 ft)^2 * dx/dt = 0 ft/sec
dx/dt = 0 ft/sec
Therefore, when the woman is 35 ft from the base of the pole, the tip of her shadow is not moving.