A street light is at the top of a 27 ft pole. A 6 ft tall girl walks along a straight path away from the pole with a speed of 8 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 23 ft away from the pole?
how fast is her shadow lengthening?
To solve this problem, we can use similar triangles and differentiate with respect to time.
Let's denote:
- h as the height of the girl's shadow
- x as the distance between the girl and the base of the pole
- y as the distance between the tip of the girl's shadow and the top of the pole
We know that the height of the pole is 27 ft and the height of the girl is 6 ft. Therefore, the height of the girl's shadow can be calculated as follows:
h = (27 - 6)x/27
To find the rate at which the tip of her shadow is moving away from the light, we need to differentiate the equation with respect to time (t), assuming x is also a function of t.
dh/dt = [(27 - 6)(dx/dt) - (27 - 6)x(d(27 - 6)/dt)]/27
Now, let's calculate dx/dt. We are given that the girl's speed is 8 ft/sec, and since x represents her distance from the base of the pole, dx/dt = 8 ft/sec.
Now, let's calculate d(27 - 6)/dt:
d(27 - 6)/dt = d(21)/dt = 0
Therefore, the equation can be simplified to:
dh/dt = [(27 - 6)(8) - (27 - 6)x(0)]/27
Now, let's calculate y when the girl is 23 ft away from the pole:
x = 23 ft
Now, substituting the values into the equation:
dh/dt = [(27 - 6)(8) - (27 - 6)(23)(0)]/27
Simplifying further:
dh/dt = [(21)(8)]/27
dh/dt = 168/27
So, the rate at which the tip of the girl's shadow is moving away from the light when the girl is 23 ft away from the pole is approximately 6.22 ft/sec.
To find how fast her shadow is lengthening, we need to find dy/dt, which is the rate at which y is changing with time.
Since y is the distance between the tip of the girl's shadow and the top of the pole, the shadow's length is given by y + h.
Therefore, dy/dt = dh/dt + dx/dt.
Substituting the values we already know:
dy/dt = 168/27 + 8
dy/dt = 6.22 + 8
So, the shadow is lengthening at a rate of approximately 14.22 ft/sec.
To find the rate at which the tip of the girl's shadow is moving away from the light, we can use similar triangles. Let's denote the length of the girl's shadow as "x" and the distance of the girl from the light pole as "y".
We are given the following information:
Height of the light pole (h) = 27 ft
Height of the girl (g) = 6 ft
Rate of the girl's movement (dy/dt) = 8 ft/sec
Distance of the girl from the pole (y) = 23 ft
Using similar triangles, we can set up the following proportion:
(g + x) / x = h / y
Now, let's differentiate both sides of the equation with respect to time (t) to find the rate at which x is changing with respect to y (dx/dy):
(d/dt)(g + x) / x = (d/dt)(h / y)
Now, let's substitute the given values into the equation:
(dy/dt)(x)^-1 = (dh/dt)(y)^-1
We can rearrange the equation to solve for dx/dy:
(dy/dt) / x = -(dh/dt) / y
Now, substitute the given values:
8 / x = 0 / 23
Since the rate of change of the height of the pole (dh/dt) is 0 (it remains constant), the rate at which the tip of the girl's shadow is moving away from the light is also 0.
Therefore, the tip of the girl's shadow is not moving away from the light when the girl is 23 ft away from the pole. The shadow length remains constant.
If her shadow has length s, then when she is x feet from the pole,
s/6 = (s+x)/27
s = 2/7 x
so,
ds/dt = 2/7 dx/dt = 2/7 * 8 = 16/7 ft/s
regardless of how far she is from the pole.
The tip of the shadow is moving at dx/dt + ds/dt = 8 + 16/7 ft/s