MHB Why does the Durango curve problem puzzle mathematicians?

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The Durango curve problem intrigues mathematicians due to its relationship with shadow projections and ellipses. Participants discuss the mathematical modeling of the shadow cast by a circular barrier, suggesting it may relate to the properties of ellipses, particularly under varying sun angles. The discussion includes the derivation of equations for the shadow's length and its geometric implications, with a focus on how the curve behaves under different conditions. Observations from real-life experiences with shadows further enrich the conversation, though some participants express uncertainty about the effects of different shapes on shadow formation. Overall, the problem highlights the complex interplay between geometry and real-world phenomena.
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Could be mistaken, but I believe the shadow curve is part of the ellipse ...

$\dfrac{(x-h)^2}{(r+s)^2} + \dfrac{(y-k)^2}{r^2} = 1$

... where $s$ represents the shadow length at the start of the curve.

https://www.physicsforums.com/attachments/6635
 
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I believe so too, but how do we prove it? It must have something to do with the projection of the circular barrier onto the freeway. A special case would be when the sun is directly overhead. Then it would be the quadrant of a circle (special case of an ellipse).

Other thoughts are:
1. It is a smooth curve
2. It's slope is zero at the south end and infinite on the west end.
Are there any other curves that meet those parameters?

Rick
 
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Let's use a simple model where the sun rises due east and sets due west. Let $\theta$ be the angle of the sun above the horizon, and only consider $$0\le\theta\le\frac{\pi}{2}$$ so that the shadow is on the outside of the described curved barrier. We then find the length of the shadow of an object of height $h$ to be:

$$s=h\cot(\theta)$$

Now suppose $\beta$ is the measure of the angle we have traveled through on the curve, where $\beta=0$ corresponds to first entering the curve and $\beta=\dfrac{\pi}{2}$ corresponds to first exiting the curve. Then the length $\ell$ of the shadow cast by the barrier as measured along the radius of curvature would be:

$$\ell=s\cos(\beta)=h\cot(\theta)\cos(\beta)$$
 
So, what's the equation of the ellipse and where are the foci? The barrier is 3 feet high.
 
rick123 said:
So, what's the equation of the ellipse and where are the foci? The barrier is 3 feet high.

We would have an ellipse with horizontal semi-major axis $r+s$ and vertical semi-minor axis $r$, exactly as described by skeeter.

If we orient our coordinate axes such that the center of the circle on which the curved road lies, then we would have:

$$\frac{x^2}{(r+s)^2}+\frac{y^2}{r^2}=1$$

And the foci would be located at:

$$\left(\pm\sqrt{(r+s)^2-r^2},0\right)=\left(\pm\sqrt{s(2r+s)},0\right)$$
 
Wouldn't it be a circle with the same radius as the turn? (Wondering)

With the sun due east and 45 degrees high, every point on the turn casts a shadow of 3 feet to the west.
This corresponds to a translation.
If the sun is at any other altitude, then that only affects the magnitude of the translation.
And if the sun is at a different azimuth, then that only affects the angle of the translation.
In all cases the curve of the shadow throughout the turn is a quarter circle with radius $r$.
 
I like Serena said:
Wouldn't it be a circle with the same radius as the turn? (Wondering)

With the sun due east and 45 degrees high, every point on the turn casts a shadow of 3 feet to the west.
This corresponds to a translation.
If the sun is at any other altitude, then that only affects the magnitude of the translation.
And if the sun is at a different azimuth, then that only affects the angle of the translation.
In all cases the curve of the shadow throughout the turn is a quarter circle with radius $r$.

Good question ... makes the sketch in the problem statement seem deceiving.

I tried to get a shadow picture of a cylindrical lid illuminated by my skylight, unfortunately it's overcast & raining today. I want to repeat this observation outside on a sunny day using my bride's circular cake pan.
 
Sun came out and took this pic on the back porch ... note the interior and exterior shadows form a circle. Interesting.
 
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Interesting! I wonder if the pan being flared out (instead of being cylindrical) effects the shadow? I've driven around the Durango curve many times, but my observations were always pretty casual.

Rick
 
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rick123 said:
Interesting! I wonder if the pan being flared out (instead of being cylindrical) effects the shadow? I've driven around the Durango curve many times, but my observations were always pretty casual.

Rick

I wouldn't think so ... a vertical obstruction and a slanted one with equal vertical heights should cast the same shadow.
 
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