A street light is at the top of a 15 foot tall pole. A 6 foot tall woman walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving (away from the woman) when she is 50 feet from the base of the pole?
The tip of the shadow is moving at ____ ft/sec.
To determine how fast the tip of the woman's shadow is moving, we'll need to use similar triangles and the concept of related rates.
Let's assume that the woman's height is h, the height of the pole is p, the distance between the woman and the base of the pole is x, and the length of the shadow is s.
We can set up a proportion between the similar triangles formed by the woman, the pole, and its shadow:
h / s = (h + p) / x
Taking the derivative of both sides with respect to time:
(dh/dt) / s = [(dh/dt) + (dp/dt)] / x
We're given that the woman's height (h) is 6 ft and is not changing. The height of the pole (p) is 15 ft and also not changing. The distance from the woman to the base of the pole (x) is changing, and we need to find the rate at which the tip of the shadow (s) is moving.
Let's plug in the given values and solve for the unknown rate, which is (ds/dt):
6 / s = (6 + 15) / x
6 / s = 21 / x
Cross-multiply:
6x = 21s
Now, let's differentiate both sides with respect to time (t):
6(dx/dt) = 21(ds/dt)
Now we have an equation relating dx/dt (the rate at which the woman is moving away from the pole) and ds/dt (the rate at which the tip of her shadow is moving, which is what we want to determine).
We're given that dx/dt = 7 ft/sec, and we're asked to find ds/dt when x = 50 ft. We can plug in these values:
6(7) = 21(ds/dt)
42 = 21(ds/dt)
ds/dt = 42 / 21
ds/dt = 2 ft/sec
Therefore, the tip of the woman's shadow is moving away from her at a rate of 2 ft/sec when she is 50 feet from the base of the pole.
To solve this problem, we can use similar triangles. Let's denote the distance from the base of the pole to the woman as x, with x increasing as the woman walks away from the pole.
The height of the woman's shadow can be denoted as h, and the height of the pole can be denoted as H.
We are given:
- The height of the pole H is 15 feet.
- The height of the woman h is 6 feet.
- The rate at which the woman is moving away from the pole, dx/dt, is 7 ft/sec.
We need to find the rate at which the tip of the shadow, y, is moving away from the woman, dy/dt when x = 50 ft.
Now, let's set up the similar triangles:
First Triangle:
Side 1: woman's height h
Side 2: height of the pole H
Side 3: distance from the woman to the pole x
Second Triangle:
Side 1: height of the woman's shadow h
Side 2: height of the pole's shadow H + y (where y is the length of the shadow cast by the top of the pole)
Side 3: distance from the woman to the tip of the shadow x + y
Since the triangles are similar, we can set up the ratio:
h / H = (h + y) / (H + y)
Cross-multiplying, we get:
h(H + y) = H(h + y)
Expanding, we have:
hH + hy = hH + Hy
Simplifying, we obtain:
hy - Hy = hH - hH
This reduces to:
hy - Hy = 0
Now, let's differentiate both sides of the equation with respect to time, t:
d(yh)/dt - d(Hy)/dt = 0
Using the product rule and chain rule, we have:
y * dh/dt + h * dy/dt - H * dy/dt = 0
Since we know dh/dt (the rate at which the woman's height changes) is 0 (the woman's height remains constant), we can remove the y * dh/dt term:
h * dy/dt - H * dy/dt = 0
Factoring out dy/dt, we get:
( h - H ) * dy/dt = 0
Since dy/dt cannot be zero (it represents the rate at which the tip of the shadow is moving), we have:
h - H = 0
Substituting the given values:
6 - 15 = -9
Therefore, when the woman is 50 feet from the base of the pole, the tip of her shadow is not moving away from her. Thus, the tip of the shadow is moving at 0 ft/sec.
draw a diagram. Let
z = distance from pole
x = length of shadow
Then using similar triangles,
x/6 = (z+x)/15
15x = 6z+6x
z = 3/2 x
so, dz/dt = 3/2 dx/dt
since dz/dt = 7, dx/dt = 14/3
so the tip of the shadow is moving at 7 + 14/3 = 35/3 ft/s
the distance from the pole does not matter for the speed.