A street light is at the top of a 16.0 ft. tall pole. A man 5.9 ft tall walks away from the pole with a speed of 5.5 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 47 feet from the pole?
look at Steve's solution here
http://www.jiskha.com/display.cgi?id=1382123644
then change the numbers to fit your problem
To find the rate at which the tip of the man's shadow is moving, we need to use related rates. This involves finding an equation that relates the quantities involved and differentiating it with respect to time.
Let's define some variables:
y = the distance from the man to the pole (this is changing)
x = the length of the man's shadow (this is also changing)
We are given:
The height of the pole is 16 ft, so y = 16 ft.
The man's height is 5.9 ft.
The man's speed is 5.5 ft/s.
We need to find:
The rate at which the tip of the man's shadow is moving, which is dx/dt.
Now, let's consider similar triangles formed by the man, his shadow, and the pole. We have:
y / (x + y) = 5.9 / x
We can rearrange this equation to solve for x:
xy + y² = 5.9x
x = (y²) / (5.9 - y)
Next, we differentiate both sides of the equation with respect to time:
d(x)/dt = [ d(y²)/dt / (5.9 - y) ] - y² d(5.9 - y)/dt / (5.9 - y)²
Now, let's plug in the given values: y = 16 ft and (dy/dt) = -5.5 ft/s (negative because the man is moving away). We are looking for dx/dt when y = 47 ft.
d(x)/dt = [ d(16²)/ dt / (5.9 - 16) ] - 16² (-5.5) / (5.9 - 16)²
Simplifying and evaluating the expression:
d(x)/dt = [ 0 / (5.9 - 16) ] - 256 (-5.5) / (-10.1)²
d(x)/dt = 0 + 1184 / 102.01
d(x)/dt = 11.61 ft/s
Therefore, the tip of the man's shadow is moving at a rate of 11.61 ft/s when he is 47 ft from the pole.