A street light is mounted at the top of a 6 meter tall pole. A man 2m tall walks away from the top with a speed of 1.5 m/s along a straight path. How fast is the tip of his shadow moving when he is 10m from the pole?
The wording in this question is always tricky
Oobleck found the rate at which the shadow increasing,
the question was how fast is the tip of the shadow moving, that is,
d(x+y)/dt = .75 + 1.5 or 2.25 m/s
(just like if I am walking in the same direction as a train going at 80 km/h
and I am walking at 4 km/h, I am moving at 84 km/h)
dang! Way to watch, Reiny!
Although I'm sure that Erich caught that, too.
Well, let me shed some light on this! It's a shadowy situation indeed.
To find the speed at which the tip of his shadow is moving, we can use similar triangles. Let's call the length of the shadow x.
Now, if the man is 2m tall, and his shadow is x meters long, we can form a ratio: 2/x = 6/(x + 10). This is because the height of the man's shadow and the height of the pole form a similar triangle with the man and the pole.
Now, let's solve this equation for x, the length of the shadow. Cross-multiplying gives us 2(x + 10) = 6x, which simplifies to 2x + 20 = 6x. Rearranging, we get 4x = 20, so x = 5 meters.
Now that we know the length of the shadow is 5 meters, we can differentiate this equation with respect to time to find the rate of change. Differentiating 2/x = 6/(x + 10) with respect to time gives us -2(dx/dt)/x^2 = -6(dx/dt)/(x + 10)^2.
Simplifying, we get (dx/dt)/x^2 = 3(dx/dt)/(x + 10)^2. We can cancel out the dx/dt terms and set them equal to each other, yielding 1/x^2 = 3/(x + 10)^2.
Now, let's substitute x = 5 into this equation and solve for dx/dt, which is the rate at which the tip of the shadow is moving. We get 1/5^2 = 3/(5 + 10)^2.
Simplifying further, 1/25 = 3/225. Cross-multiplying gives us 225 = 75. Uh-oh, something seems to have gone awry here! My calculations are getting a bit shady.
Hmm, it appears there might have been an error in our calculations. I must apologize! As a clown, math is sometimes not my forte. It seems I've gotten caught up in the shadows and led us in the wrong direction. Let me rectify that by consulting my comedy book.
To find the speed at which the tip of the man's shadow is moving when he is 10m from the pole, we need to use similar triangles.
Let's define some key variables:
- Let h represent the height of the man (2m).
- Let x represent the distance between the man and the base of the pole.
- Let d represent the length of the shadow.
We can form a right triangle by connecting the top of the pole, the man's feet, and the tip of his shadow, with the pole being the hypotenuse. The similar triangle can be formed by connecting the top of the pole, the top of the man's head, and the tip of his shadow.
Based on the similar triangles, we can set up a proportion:
h / (d + x) = 6 / d
Cross multiplying this proportion, we get:
h * d = 6 * (d + x)
Simplifying:
2 * d = 6 * (d + x)
2d = 6d + 6x
4d = 6x
d = (3/2) * x
Now, we need to find the rate of change of the shadow length with respect to time. Let's differentiate both sides of the equation with respect to time:
d/dt (d) = d/dt (3/2 * x)
Now, let's substitute the given values into the equation:
d/dt (d) = d/dt (3/2 * 10)
d/dt (d) = d/dt (15)
Since the distance between the man and the pole is not changing, dx/dt = 0. Thus, we can ignore the term d/dt (x) in our equation.
d/dt (d) = 0
Therefore, the rate of change of the shadow's length is 0. This means that the tip of the shadow is not moving.
Let
x = distance of man from pole
s = length of shadow
Then
s/2 = (x+s)/6
3s = x+s
2s = x
2 ds/dt = dx/dt
ds/dt = 1.5/2 = 0.75 m/s
it does not matter how far away from the pole the man is.