A street light is at the top of a 17 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 40 ft from the base of the pole?
Draw a right-angled triangle ABC, B the right angle, and the vertical AB = 17 feet
Somewhere between B and C draw a vertical DE = 6 feet, the height of the woman
let BD = x and DC = y, the length of her shadow
given: dx/dt = 6 ft/s
find: dy/dt
(actually d(x+y)/dt, I'll come back to that later)
by similar triangles 6/y = 17/(x+y)
.
.
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11y = 6x
then 11dy/dt = 6dx/dt
dy/dt = 6(6)/11 = 36/11 ft/s
so no matter where she is, the length of her shadow is increasing at 36/11 ft/s
but the woman herself is moving at 6 ft/sec
so her shadow is moving at 6+36/11 or
102/11 ft/s
To determine how fast the tip of the woman's shadow is moving, we can use similar triangles. Let's break down the problem and establish some variables.
Let:
- The height of the pole be represented by "h" = 17 ft.
- The distance between the woman and the pole be represented by "x" = 40 ft.
- The height of the woman be represented by "y" = 6 ft.
We can set up a proportion between the height of the pole, the height of the woman, and the height of their respective shadows:
h + y x
------- = ---
h s
Where "s" represents the height of the shadow.
By cross-multiplying, we can rewrite the equation:
(h + y)s = hx
Now, we'll differentiate both sides of the equation with respect to time (t) and apply the Chain Rule:
(d/dt)[(h + y)s] = (d/dt)[hx]
(h + y) * ds/dt = h * dx/dt
We are given that dx/dt = 6 ft/sec (since the woman is moving away at a speed of 6 ft/sec). Substituting all the known values into the equation:
(17 + 6) * ds/dt = 17 * 6
23 * ds/dt = 102
Finally, we can solve for ds/dt, the rate at which the tip of the shadow is moving:
ds/dt = 102 / 23 ≈ 4.43 ft/sec
Therefore, the tip of the woman's shadow is moving at a rate of approximately 4.43 ft/sec when she is 40 ft from the base of the pole.