A man 5.50 ft tall approaches a street light 18.0 ft above the ground at a rate of 4.0 ft/s. at what rate is the end of the man's shadow moving when he is 9.0 ft from the base of the light?
wrong answer
To find the rate at which the end of the man's shadow is moving, we can use similar triangles. Let's denote the length of the man's shadow as x ft.
From the given information, we can form the following proportion:
(man's height + length of the shadow) / (height of the street light) = length of the shadow / distance from the street light
Using the given values, we have:
(5.50 ft + x) / 18.0 ft = x / 9.0 ft
Cross-multiplying, we get:
9.0 ft * (5.50 ft + x) = 18.0 ft * x
Expanding this equation, we have:
49.5 ft + 9.0 ft * x = 18.0 ft * x
Rearranging the terms, we get:
9.0 ft * x - 18.0 ft * x = -49.5 ft
Simplifying further:
-9.0 ft * x = -49.5 ft
Dividing both sides by -9.0 ft:
x = 5.5 ft
Now, we can find the rate at which the end of the man's shadow is moving by differentiating the equation with respect to time. Let's denote the rate at which the shadow is moving as y ft/s.
Differentiating the equation (9.0 ft * (5.50 ft + x) = 18.0 ft * x), we get:
9.0 ft * dy/dt = 18.0 ft * dx/dt
Since dx/dt is given as 4.0 ft/s, we have:
9.0 ft * dy/dt = 18.0 ft * 4.0 ft/s
Simplifying, we get:
9.0 ft * dy/dt = 72.0 ft²/s
Dividing both sides by 9.0 ft, we find:
dy/dt = (72.0 ft²/s) / 9.0 ft
Simplifying further:
dy/dt = 8.0 ft/s
Therefore, the end of the man's shadow is moving at a rate of 8.0 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 related rates.
Let's define:
x = distance from the man to the base of the light (in feet)
y = length of the man's shadow (in feet)
We can set up a proportion using the similar triangles formed by the man, his shadow, and the streetlight:
y / (18 + y) = 5.5 / x
Now, we need to find the rate at which y is changing with respect to x. To do this, we can take the derivative of both sides of the equation with respect to time t (since the distances are changing over time):
(d/dt) [y / (18 + y)] = (d/dt) [5.5 / x]
Let's differentiate each term:
(d/dt) [y / (18 + y)] = (d/dt) (5.5 / x)
[(d/dt) y / (18 + y)] - [y (d/dt) (18 + y) / (18 + y)^2] = [-(5.5/x^2) (dx/dt)]
Now, let's plug in the given values:
The man's height, which we'll call h, is 5.5 ft.
The height of the light, which we'll call L, is 18.0 ft.
The rate at which the man is approaching the light, which we'll call dx/dt, is 4.0 ft/s.
So, x = 9.0 ft (distance from the man to the base of the light)
And dx/dt = 4.0 ft/s (rate at which the man is approaching the light)
Substituting these values into the equation, we get:
[(dy/dt) / (18 + y)] - [y (dy/dt) / (18 + y)^2] = [-(5.5/9^2) (4.0)]
Now, we need to solve for (dy/dt), which represents the rate at which the end of the man's shadow is moving when he is 9.0 ft from the base of the light.
Make a sketch.
Let the distance between the man and the post be x ft
let the length of his shadow be y ft
by similar triangles:
5.5/y = 18/(x+y)
18y = 5.5x + 5.5y
12.5y = 5.5x
12.5dy/dt = 5.5dx/dt
given: dx/dt = 4 ft/s
then dy/dt = 5.5/12.5 or .44 ft/s
So the shadow is lengthening at .44 ft/s
and the shadow is moving at 4+.44 or 4.44 ft/sec
(The end of the shadow is both lengthening and being moved along by the movement of the man.
Similar to somebody walking in a moving train.
Suppose a train is going at 60 mph, but somebody is walking inside the train at 4 mph in the same direction as the train. To an observer at a crossing, the person would be moving at 64 mph )