To find how fast the tip of her shadow is moving, we need to use similar triangles and related rates.
Let's denote the distance between the woman and the base of the pole as x (in feet). We also need to define y as the length of the shadow (also in feet).
First, let's set up similar triangles. We have two triangles: the triangle formed by the woman, her shadow, and the pole, and the larger triangle formed by the pole, the shadow, and the ground.
The ratio of the woman's height to her distance from the pole is constant and equal to the ratio of the length of her shadow to its distance from the pole:
6 ft / x = (15 ft + y) / y
By cross-multiplying and simplifying, we get:
6y = 15x + xy
Now, let's differentiate both sides of the equation with respect to time (t):
d/dt (6y) = d/dt (15x + xy)
6(dy/dt) = 15(dx/dt) + x(dy/dt) + y(dx/dt)
Since we want to find how fast the tip of the shadow (y) is moving, we can substitute dx/dt = -4 ft/sec (negative because the woman is walking away) and dy/dt = ? into the equation. We also know that when the woman is 30 ft from the base of the pole, x = 30 ft.
Plugging these values into the equation, we have:
6(dy/dt) = 15(-4) + 30(dy/dt)
Simplifying further:
6(dy/dt) = -60 + 30(dy/dt)
Combining like terms:
6(dy/dt) - 30(dy/dt) = -60
-24(dy/dt) = -60
Solving for dy/dt:
(dy/dt) = (-60) / (-24)
(dy/dt) = 2.5 ft/sec
Therefore, the tip of her shadow is moving at a rate of 2.5 ft/sec when she is 30 ft from the base of the pole.