A street light is at the top of a 14 ft tall pole. A woman 6 ft tall 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 when she is 30 ft from the base of the pole?

Well, isn't this an enlightening question! Let's shed some light on the situation, shall we?

We have the height of the pole, which is 14 ft, and the height of the woman, which is 6 ft. As the woman walks away from the pole, we can imagine that she creates a shadow on the ground.

Now, when she is 30 ft from the base of the pole, we can create a right triangle with the pole, the woman, and her shadow. The length of the shadow is 30 ft, and the height of the woman is 6 ft.

To find out how fast the tip of her shadow is moving, we need to find the rate at which the length of the shadow is changing with respect to time. In other words, we need to find the derivative of the length of the shadow with respect to time.

Now, here comes the punchline! Since we're given that the woman is walking away with a speed of 7 ft/sec, we can differentiate the equation for the length of the shadow.

Using the Pythagorean theorem, we have:

(length of shadow)^2 = (distance from base of pole)^2 + (height of woman)^2

Taking the derivative of both sides with respect to time t, we get:

2(length of shadow)(rate of change of length of shadow) = 2(distance from base of pole)(rate of change of distance from base of pole)

Simplifying this, we have:

(length of shadow)(rate of change of length of shadow) = (distance from base of pole)(rate of change of distance from base of pole)

Now, we plug in the values we know:

30 ft (rate of change of length of shadow) = 30 ft (rate of change of distance from base of pole)

Canceling out the units, we find that the rate of change of the length of the shadow is equal to the rate of change of the distance from the base of the pole. And since the woman is walking away from the pole at a speed of 7 ft/sec, the rate of change of the length of the shadow is also 7 ft/sec!

So, to answer your question, when the woman is 30 ft from the base of the pole, the tip of her shadow is moving at a speed of 7 ft/sec. Shadows can really keep up with their shadowy tips, can't they?

To determine how fast the tip of the woman's shadow is moving, we can use similar triangles.

Let's denote:
- h₁ as the height of the woman
- h₂ as the height of the pole
- d₁ as the distance from the woman to the base of the pole
- d₂ as the distance from the tip of the shadow to the base of the pole

We are given that:
h₁ = 6 ft
h₂ = 14 ft
d₁ = 30 ft

To find the rate at which the tip of the shadow is moving (d₂/dt), we need to find a relationship between d₁ and d₂.

Using the similar triangles formed by the woman, her shadow, and the pole, we can write the following equation:

h₁ / d₁ = h₂ / d₂

Plugging in the given values, we have:

6 / 30 = 14 / d₂

We can solve for d₂ by cross-multiplying and simplifying:

6 * d₂ = 30 * 14
d₂ = (30 * 14) / 6
d₂ = 70 ft

Now we can differentiate both sides of the equation with respect to time t to find the rate of change of the shadow's distance with respect to time:

d/dt (6 / 30) = d/dt (14 / d₂)

To simplify, we can rewrite the equation as:

(1 / 5) * d₁' = (-14 / d₂²) * d₂'

Where d₁' represents the rate at which the woman is moving away from the pole (7 ft/sec) and d₂' represents the rate at which the tip of the shadow is moving (what we're trying to find).

Plugging in the known values, we have:

(1 / 5) * 7 = (-14 / 70²) * d₂'

Simplifying further, we can solve for d₂':

(1 / 5) * 7 = (-14 / 4900) * d₂'

(7 / 5) = (-14 / 4900) * d₂'

To isolate d₂', we can multiply both sides by (5 / 14) * 4900:

d₂' = (7 / 5) * (5 / 14) * 4900

Simplifying, we have:

d₂' = (1 / 2) * 4900
d₂' = 2450 ft/sec

Therefore, the tip of the woman's shadow is moving at a rate of 2450 ft/sec when she is 30 ft from the base of the pole.

To solve this problem, we need to use similar triangles and related rates.

Let's assume the length of the woman's shadow is x ft. Since the height of the pole is 14 ft, the length of the pole's shadow will be x+14 ft (including the length of the pole).

Now, let's set up a proportion using the similar triangles formed by the woman, her shadow, the pole, and its shadow:

(height of woman) / (length of woman's shadow) = (height of pole) / (length of pole's shadow)

Substituting the given values:

6 / x = 14 / (x + 14)

Now, we can cross-multiply and solve for x:

6(x + 14) = 14x

6x + 84 = 14x

84 = 8x

x = 10.5 ft

So, when the woman is 30 ft from the base of the pole, the length of her shadow is 10.5 ft.

Now, to find how fast the tip of her shadow is moving, we need to differentiate the equation we used to find x with respect to time:

6 / x = 14 / (x + 14)

Differentiating both sides of the equation:

(6 / x^2) * dx/dt = (14 / (x + 14)^2) * d(x + 14)/dt

Since dx/dt represents the woman's speed along the horizontal direction (which is given as 7 ft/sec), we substitute this value:

(6 / x^2) * 7 = (14 / (x + 14)^2) * d(x + 14)/dt

Now, we can solve for d(x + 14)/dt, which represents the rate at which the length of the pole's shadow is changing (i.e., how fast the tip of the shadow is moving).

(6 * 7 * x^2) = (14 * (x + 14)^2) * d(x + 14)/dt

Simplifying further:

42x^2 = 14(x + 14)^2 * d(x + 14)/dt

d(x + 14)/dt = (42x^2) / (14(x + 14)^2)

Substituting x = 10.5 ft (the length of the shadow) when the woman is 30 ft from the pole's base:

d(x + 14)/dt = (42 * (10.5)^2) / (14 * (10.5 + 14)^2)

Simplifying and calculating:

d(x + 14)/dt ≈ 1.092 ft/sec

Therefore, when the woman is 30 ft from the base of the pole, the tip of her shadow is moving at a speed of approximately 1.092 ft/sec.

Make your sketch, you will have a smaller right-angled triangle within a larger one.

They are similar.
At a time of t seconds, let the length of her shadow be x ft, let her distance from the lamppost be y ft
6/x = 14/(x+y)
14x = 6x + 6y
8x = 6y
4x = 3y
4 dx/dt = 3 dy/dt

given: dy/dt = 7 ft/s
find dx/dt

4dx/dt = 3(7)
dx/dt = 21/4 ft/s , notice that rate is constant, no matter where she is

So her shadow is lengthening at 5.25 ft/sec
but her shadow is moving at 5.25 + 7 or 12.25 ft/s

(think of a moving train at 60 km/h and you walking in the train in its direction
at 4 km/h, so you are moving at 64 km/h relative to the ground)