# A man 5.00 ft tall approaches a street light 17.0ft above the ground at the rate of 4.00 ft/s. How fast is the end of the man's shadow moving when he is 9.0 ft from the base of the light? The end of the man's shadow is moving at a rate of_ft/s

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

Let's denote the height of the man as h and the height of the streetlight as H. Also, let's denote the distance between the man and the base of the light as x.

From the given information, we have:
h = 5.00 ft
H = 17.0 ft
dx/dt = 4.00 ft/s
We need to find dH/dt when x = 9.0 ft.

Using similar triangles, we can set up the proportion: h / x = H / (x + D), where D represents the length of the shadow.

Rearranging the equation, we can solve for D: D = (H * x) / h

Now, we can differentiate both sides of the equation with respect to time t:

dD/dt = [(H * dx) / h]

Now, plug in the known values:

dD/dt = [(17.0 ft * 4.00 ft/s) / 5.00 ft]

Simplifying:

dD/dt = 13.6 ft/s

Therefore, the end of the man's shadow is moving at a rate of 13.6 ft/s when he is 9.0 ft from the base of the light.