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. How fast is the tip of her shadow moving when she is 50 ft from the base of the pole.
Let her distance from the pole be x ft
let the length of her shadow at that moment be y ft
by ratio:
17/(x+y) = 6/y
17y = 6x+6y
11y = 6x
11 dy/dt = 6 dx/dt
dy/dt = 6(6)/11 = 36/11 ft/s
The tip of her shadow is moving at
dy/dt + dx/dt
= 36/11 + 6
= 102/11 or appr 9.27 ft/s
notice that this is independent of where she is, that is the 50 ft did not even enter the picture.
To find the rate at which the tip of the shadow is moving, we can use similar triangles and the concept of related rates.
Let's consider two similar triangles formed by the woman, the pole, and the shadow:
Triangle 1: The woman, the pole, and the shadow.
Triangle 2: The woman's shadow, the pole, and the total shadow on the ground.
The height of the pole is given as 17 ft, and the height of the woman is given as 6 ft. Therefore, the height of the woman's shadow is given by the proportion:
(Height of woman's shadow) / (Height of pole) = (Height of woman) / (Height of woman's shadow + Height of pole)
Let's denote the height of the woman's shadow as "x." From the given information:
x / 17 = 6 / (x + 17)
To solve for x, we can cross-multiply and solve the resulting quadratic equation:
x(x + 17) = 6 * 17
x^2 + 17x = 102
x^2 + 17x - 102 = 0
Factoring the quadratic equation:
(x + 23)(x - 6) = 0
Since the height of the shadow cannot be negative, we have x = 6 ft.
Now, let's consider the rate at which the woman's shadow is changing as she walks away from the pole. We are given that the woman's walking rate is 6 ft/sec.
Let's denote the distance of the woman from the base of the pole as "d" and the length of the woman's shadow as "y." We want to find the rate at which y is changing with respect to time, or dy/dt.
From the similar triangles, we have:
(Length of woman's shadow) / (Distance of woman from pole) = (Height of woman's shadow) / (Height of woman)
Using this relationship and differentiating both sides of the equation with respect to time (t), we get:
(dy/dt) / d = (x') / 6
Simplifying, we have:
dy/dt = (x' * 6) / d
We are given that the woman walks away from the pole with a speed of 6 ft/sec, so x' (the rate at which the height of the woman's shadow changes) is also 6 ft/sec.
Substituting the given values into the equation, we have:
dy/dt = (6 * 6) / 50
Calculating, we get:
dy/dt = 36 / 50
Simplifying, dy/dt = 0.72 ft/sec
Therefore, the tip of the woman's shadow is moving at a rate of 0.72 ft/sec when she is 50 ft from the base of the pole.
To find the rate at which the tip of the shadow is moving, we can use related rates. Here's how we can approach this problem step-by-step:
Step 1: Identify the given information and what we need to find.
- The height of the pole is 17 ft.
- The height of the woman is 6 ft.
- The woman is walking away from the pole with a speed of 6 ft/sec.
- We need to find the rate at which the tip of her shadow is moving when she is 50 ft from the base of the pole.
Step 2: Set up a diagram and label the given quantities.
```
S --> X
| |
| |
--- |
pole |
--------------
height of pole = 17 ft
height of woman = 6 ft
```
Let's assume that the distance between the woman and the tip of her shadow is X, and the length of her shadow is S.
Step 3: Determine the relationship between the quantities and establish an equation.
From the diagram, we can see that similar triangles are formed between the pole, the woman, and her shadow. The proportions are:
X/S = (X+6)/(S+17)
Step 4: Differentiate the equation with respect to time (t).
Differentiating both sides of the equation implicitly with respect to time (t), we get:
d(X)/dt / S = (d(X)/dt + 6) / (S + 17)
Step 5: Substitute the given values and solve for d(X)/dt.
We are given that d(X)/dt is 6 ft/sec when X = 50 ft, so we substitute these values:
6 / S = (6 + 6) / (S + 17)
Step 6: Solve for S.
Now, we solve the above equation for S:
6(S + 17) = 12S
6S + 102 = 12S
102 = 6S
S = 17 ft
Step 7: Substitute the value of S back into the equation and solve for d(X)/dt.
Using the equation we established earlier:
6 / 17 = (6 + 6) / (17 + 17)
Multiplying both sides by 17, we get:
6 = 12 / (34)
6 = 0.3529 ft/sec
Therefore, the tip of her shadow is moving at a speed of approximately 0.3529 ft/sec when she is 50 ft from the base of the pole.