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

User Latheesan
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To determine how fast the tip of the man's shadow is moving when he is a certain distance away from the pole, we can use similar triangles and the concept of related rates.

Let's denote the distance of the man from the pole as x (in feet) and the length of his shadow as y (in feet). The height of the pole is given as 15 feet.

From the similar triangles formed by the man, his shadow, and the pole, we have the following relationship:

x/y = (man's height)/(pole's height)

x/y = m/15

Differentiating both sides of the equation with respect to time t, we get:

(d(x)/dt) / y = (d(m)/dt) / 15

We are given that (d(m)/dt) = 4 ft/s, as the man is walking away from the pole with a speed of 4 ft/s.

Now, let's find the value of y in terms of x using the Pythagorean theorem:

y^2 + x^2 = 15^2

y^2 = 15^2 - x^2

y = sqrt(225 - x^2)

Differentiating y with respect to time t, we get:

(d(y)/dt) = (d/dt)[sqrt(225 - x^2)]

To find (d(y)/dt) when the man is x feet from the pole, we substitute the given values into the equations above.

First, solve for y when x = 8 ft (the distance from the pole):

y = sqrt(225 - (8^2))

y = sqrt(225 - 64)

y = sqrt(161)

Now, substitute x = 8 ft and y = sqrt(161) into the equation:

(d(x)/dt) / sqrt(161) = 4 / 15

Solving for (d(x)/dt), we get:

(d(x)/dt) = (4 / 15) * sqrt(161)

Therefore, when the man is 8 ft from the pole, the tip of his shadow is moving at a rate of (4 / 15) * sqrt(161) ft/s.

User Charlotte Russell
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