A street light is at the top of a 16 ft pole. A 5 ft tall girl walks along a straight path away from the pole with a speed of 3 ft/sec.
At what rate is the tip of her shadow moving away from the light (ie. away from the top of the pole) when the girl is 36 ft away from the pole?
I know its last second but any help is appreciated
Make a sketch.
Let the length of the girl's shadow be x
let her distance from the lamp be y
so you have 2 similar triangles:
5/x = 16/(x+y)
5x + 5y = 16x
5y = 11x
5 dy/dx = 11 dx/dx
dy/dx = 11/5
so her shadow is increasing at 11/5 ft/sec or2.2 ft/sec
but it is moving at 11/5 + 3 or
26/5 ft/sec or 5.2 ft/s
To find the rate at which the tip of the girl's shadow is moving away from the light, we need to use similar triangles and related rates.
Let's denote the height of the girl as h and the distance between the girl and the pole as x. The shadow of the girl will then have a length of h + 16 ft (the height of the pole) and will be similar to the triangle formed by the girl, the pole, and the tip of the shadow.
Since the triangles are similar, we can set up a proportion:
(h + 16 ft) / h = (x + 16 ft) / x
Now, differentiate both sides of the equation with respect to time (t):
(d(h + 16) / dt) / dh = (d(x + 16) / dt) / dx
The derivative of the height of the girl h is given as -3 ft/sec (since she's walking away at a speed of 3 ft/sec).
So, substituting the given values and solving for (dx / dt), we have:
(-3) / h = (dx / dt) / 36 ft
We know that h = 5 ft and we need to find (dx / dt) when x = 36 ft. Substituting these values, we can solve for (dx / dt):
-3 / 5 = (dx / dt) / 36
Cross-multiplying:
-3 * 36 = (dx / dt) * 5
Simplifying:
-108 = (dx / dt) * 5
Now, solve for (dx / dt):
(dx / dt) = -108 / 5
Therefore, the tip of the girl's shadow is moving away from the light at a rate of -108/5 ft/sec when the girl is 36 ft away from the pole.
Note: The negative sign indicates that the shadow is moving away from the light in the opposite direction of the girl's movement.