A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 7 ft/s along a straight path. How fast is the tip of his shadow moving when he is 45 ft from the pole?

Question

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 7 ft/s along a straight path. How fast is the tip of his shadow moving when he is 45 ft from the pole?

Answer 1

Thank you very much!

By the way, 14/3+7 equals 35/3, not 35/7.

Answer 2

Well, let me shed some light on this question for you! The key here is to use similar triangles. The man's height and the length of his shadow create two similar triangles.

We have a 6-ft tall man, and his shadow is x ft long. Since the street light is at the top of a 15-ft tall pole, we can use the ratio of corresponding sides to set up the following proportion:

(6 ft)/(15 ft) = x/45 ft

Cross-multiplying and solving for x, we get:

x = (6 ft)(45 ft) / 15 ft
x = 18 ft

Now that we have the length of the shadow, we can differentiate both sides of the equation with respect to time to find its rate of change:

d(x)/dt = (d/dt) [18 ft]

Since the man is walking away at a constant speed of 7 ft/s, the rate of change of his shadow's length is also 7 ft/s.

So, the tip of his shadow is moving at a speed of 7 ft/s when he is 45 ft from the pole. Shadow racing, anyone?

Answer 3

To find the rate at which the tip of the man's shadow is moving, we need to use similar triangles.

Let's denote the distance of the man from the street light as x ft and the length of his shadow as y ft.

We have two similar triangles: one formed by the man, his shadow, and the distance he is from the pole, and the other formed by the street light, the top of the pole, and the length of the pole.

The ratio of the lengths of the corresponding sides of similar triangles is the same. So, we can write:

(man's height + length of his shadow) / distance to the man = height of the pole / height of the pole's shadow

Plugging in the given values, we get:

(6 + y) / x = 15 / y

Now, we can solve this equation to find y in terms of x:

(6 + y) * y = 15 * x

Simplifying, we have:

y^2 + 6y - 15x = 0

To find the length of the shadow, we use the quadratic formula:

y = (-6 ± √(6^2 - 4 * 1 * (-15x))) / (2 * 1)

y = (-6 ± √(36 + 60x)) / 2

Since we are interested in the rate at which the tip of the shadow is moving, we differentiate this equation with respect to time t:

dy/dt = (1/2) * [(± d/dt √(36 + 60x)) / √(36 + 60x)]

Now, we need to find dx/dt, which is the rate at which the man is moving away from the pole. dx/dt is given as 7 ft/s.

Substituting the given values into the derivative equation:

dy/dt = (1/2) * [(± d/dt √(36 + 60x)) / √(36 + 60x)] * dx/dt

dy/dt = (1/2) * [(± (30 / √(36 + 60x))) * 7]

Simplifying further:

dy/dt = ± (105 / √(36 + 60x)) ft/s

Now we can find the rate of the tip of the shadow when the man is 45 ft from the pole (x = 45):

dy/dt = ± (105 / √(36 + 60 * 45))

Calculating this expression will give us the desired answer.

Answer 4

make a sketch

label the man's distance from the post as x
and the length of his shadow y
by similar triangles and ratios
6/y = 15/(x+y)
15y = 6x + 6y
9y = 6x
3y = 2x
3dy/dt = 2dx/dt
dy/dt = (2/3)(7) = 14/3 ft/s
so the length of his shadow is lengthening at 14/3 ft/s
but it is moving at 14/3 + 7 or 35/7 ft/sec