a woman 6ft tall walks away from a light 10ft above the ground. if her shadow lengthens at the rate of 2ft/sec. how fast is she walking?

Well, well, well, it seems we have a tall lady taking a stroll! Let's shine some light on this question, shall we?

We know her height is 6ft, and the light is 10ft above the ground. So, when she stands beside the light, her total height and the length of her shadow would be 16ft (6ft + 10ft).

Now, to find out how fast she's walking, we need to relate the rate of shadow lengthening to her walking speed. The shadow lengthens by 2ft/sec, so every second, the total height of the woman and her shadow increases by 2ft.

Since the length of her shadow is increasing at the same rate she's walking, we can say that her walking speed is also 2ft/sec.

So, drumroll please! The woman is walking at a speed of 2 feet per second. Keep striding, tall lady!

To find how fast the woman is walking, we need to use similar triangles. Let's define some variables:

Let x be the distance between the woman and the light.
Let y be the length of the woman's shadow.

From the given information, we know that:

The height of the woman, h, is 6ft.
The height of the light above the ground, h', is 10ft.
The rate at which the shadow is lengthening, dy/dt, is 2ft/sec.

Using similar triangles, we can set up the following equation:

y/x = h'/h

Substituting the known values, we have:

y/x = 10/6

Simplifying, we find:

y = (10/6)x

Now, let's differentiate both sides of the equation with respect to time (t):

(dy/dt) = (10/6)(dx/dt)

We know that (dy/dt) is equal to 2ft/sec, so we can substitute that in:

2 = (10/6)(dx/dt)

Simplifying, we find:

(dx/dt) = (2 * 6)/10
= 12/10
= 1.2 ft/sec

Therefore, the woman is walking at a speed of 1.2 ft/sec.

To solve this problem, we can use similar triangles and the concept of proportions. Let's break it down step by step:

1. First, let's draw a diagram to visualize the situation. We have a woman who is 6 feet tall walking away from a light source that is 10 feet above the ground. As she moves, her shadow lengthens.

(Light Source)
-------------
| |
---------------- <- Ground
| Woman |
| |
----------------

2. Since the sun's rays are parallel, we have similar triangles formed by the woman, her shadow, and the ground. Let's denote the height of the woman as h_w, the length of her shadow as s, and the distance she walks as x. The height of the light source is h_l.

L____________
| |
| |
| _______ |
|| |
|| s |
||_________|
H x

3. By using similar triangles, we can establish the following proportion:

(h_w + h_l) / s = h_w / x

The numerator (h_w + h_l) represents the total height of the woman and the light source, while the denominator represents the distance the woman walks. Since the shadow lengthens at a rate of 2 ft/s, the derivative of s with respect to time (ds/dt) is equal to 2 ft/s.

4. Now, we can differentiate the proportion with respect to time (t):

d/dt [(h_w + h_l) / s] = d/dt (h_w / x)

5. We can simplify the left side of this equation by applying the quotient rule:

[s * (0 - (h_w + h_l)) / s^2] * ds/dt = (-h_w / x^2) * dx/dt

Simplifying further, we get:

- (h_w + h_l) * ds/dt / s^2 = -h_w * dx/dt / x^2

6. Since ds/dt = 2 ft/s, we can substitute it into our equation:

- (h_w + h_l) * 2 / s^2 = -h_w * dx/dt / x^2

7. Rearranging the equation to solve for dx/dt (the rate at which the woman is walking):

dx/dt = (s^2 * h_w * 2) / (x^2 * (h_w + h_l))

8. Now we can substitute the given values into the equation. The woman's height (h_w) is 6 feet, the light source's height (h_l) is 10 feet, and ds/dt is 2 ft/s:

dx/dt = (s^2 * 6 * 2) / (x^2 * (6 + 10))

9. We also know that s = x, because the triangles are similar. Therefore, we can simplify the equation further:

dx/dt = (x^2 * 6 * 2) / (x^2 * (6 + 10))

10. Canceling out the x^2 terms, we get:

dx/dt = 12 / 16

Simplifying, we have:

dx/dt = 3 / 4 ft/s

Therefore, the woman is walking at a rate of 0.75 ft/s.

did you make a diagram?

let her distance from the lightpole be x ft
let the length of her shadow be y ft
by similar triangles ...
6/y = 10/(x+y)
6x + 6y = 10y
6x = 4y
6 dx/dt = 4 dy/dt

we are given: dx/dt + dy/dt = 2 ft/sec
dx/dt = 2 - dy/dt
so
6(2 - dy/dt) = 4 dy/dt
12 - 6dy/dt = 4dy/dt
12 = 10dy/dt
dy/dt = 12/10 = 1.2 ft/sec