# A man 6.5 ft tall approaches a street light 17.0ft above the ground at the rate of 8.00 ft/s. How fast is the end of the man's shadow moving when he is 5.0 ft from the base of the light? The end of the man's shadow is moving at a rate of_ft/s

## Draw a diagram. If the man is x feet from the pole, and his shadow has length s, then using similar triangles, we see that

s/6.5 = (x+s)/17

2s/13 = (x+s)/17

34s = 13x+13s

21s = 13x

s = 13/21 x

So,

ds/dt = 13/21 dx/dt

So, it does not matter how far away he is, his shadow's length is changing 13/21 as fast as his distance from the pole.

## To find the rate at which the end of the man's shadow is moving, we can use similar triangles and related rates.

Let's denote the rate at which the end of the man's shadow is moving as dy/dt. We want to find this rate when the man is 5.0 ft from the base of the light.

First, let's set up a proportion based on the similar triangles formed by the man, the street light, and the shadow:

(6.5 ft + y) / y = (17.0 ft + dy) / x

Where:

- (6.5 ft + y) is the height of the man and the length of the shadow (since they are proportional)

- (17.0 ft + dy) is the height of the street light and the length of the shadow (since they are proportional)

- x is the distance from the man to the base of the street light

The next step is to differentiate both sides of the equation with respect to time t:

d/dt [(6.5 ft + y) / y] = d/dt [(17.0 ft + dy) / x]

To solve for dy/dt, we need to rearrange the equation and isolate dy/dt:

(6.5 / y) * dy/dt = (17.0 / x) * dx/dt

Now, let's substitute the given values:

(6.5 / y) * dy/dt = (17.0 / 5.0) * 8.00 ft/s

Simplifying the equation further:

(6.5 / y) * dy/dt = 27.2 ft/s

Finally, we need to find the value of y when the man is 5.0 ft from the base of the light. Using the Pythagorean theorem:

y^2 + x^2 = (6.5 ft)^2

y^2 + (5.0 ft)^2 = (6.5 ft)^2

y^2 = (6.5 ft)^2 - (5.0 ft)^2

y^2 = 19.25 ft^2 - 25.00 ft^2

y^2 = -5.75 ft^2 (we discard the negative value since y represents a positive length)

Taking the square root of both sides:

y ≈ 2.39 ft

Substituting y back into the equation:

(6.5 / 2.39 ft) * dy/dt = 27.2 ft/s

Solving for dy/dt:

dy/dt ≈ (27.2 ft/s) * (2.39 ft / 6.5 ft)

dy/dt ≈ 9.97 ft/s

Therefore, the end of the man's shadow is moving at a rate of approximately 9.97 ft/s.