A street light is at the top of a 15 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 50 ft from the base of the pole?
Did you make a sketch?
I called her distance from the pole x ft, and the length of her shadow y ft
So by similar triangles,
6/y = 15/(x+y)
15y = 6x + 6y
9y = 6x
3y = 2x
3dy/dt = 2dx/dt
we know that dx/dt = 5
3dy/dt = 10
dy/dt = 10/3 ft/sec
So her shadow is increasing at 10/3 ft/sec
BUT, her shadow is moving along with her speed of 5 ft/sec, so
the end of her shadow is moving at 5 + 10/3
or 25/3 ft/s
(that last part is the trick-part of this question)
15 ft 3 dy
15 ft
2 ps
2 ps
To find the rate at which the tip of the woman's shadow is moving, we can use similar triangles.
Let's denote:
- The height of the pole as h = 15 ft.
- The height of the woman as w = 6 ft.
- The distance between the woman and the base of the pole as x.
We can create a proportion with similar triangles:
(woman's height + length of her shadow) / length of pole = woman's height / length of her shadow
Substituting the known values:
(6 + x) / (h + x) = 6 / x
Next, we can cross-multiply to simplify the equation:
6x = 6(h + x)
Expanding the equation:
6x = 6h + 6x
Rearranging the equation:
6x - 6x = 6h
Simplifying:
0 = 6h
Since the height of the pole (h) is always constant, we can say that when x = 50 ft, the height of the shadow (6 + x) is also constant.
Therefore, the tip of the woman's shadow is not moving when she is 50 ft from the base of the pole.