The front wheel is free to turn but the back wheel is fixed in the rear part of the frame and always essentially points towards the front wheel
It can only be right to left in this case
Yes, this is what I was going for.
Spoiler!
The back wheel always points towards the front wheel which means if we draw a line tangent from the back wheel curve, it will intersect the front wheel curve. The endpoints of this line are the positions of the back and front wheel.
Since the distance between back and front wheel is constant, these tangent lines should all be the same length.
Tangents going left to right - lengths are all over the place
Spoiler!
Tangents going right to left - lengths are constant
Spoiler!
In the above images, I chose which curve was the back wheel curve by looking at the tangents. Switching curves, it's obvious that drawing tangents wouldn't always intersect the other curve.
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The back wheel always points towards the front wheel which means if we draw a line tangent from the back wheel curve, it will intersect the front wheel curve. The endpoints of this line are the positions of the back and front wheel.
Since the distance between back and front wheel is constant, these tangent lines should all be the same length.
Tangents going left to right - lengths are all over the place
Spoiler!
Tangents going right to left - lengths are constant
Spoiler!
In the above images, I chose which curve was the back wheel curve by looking at the tangents. Switching curves, it's obvious that drawing tangents wouldn't always intersect the other curve.
Neat, I knew the answer because I know what bike tracks should look like. I had no idea how to actually explain it.
I think you've come closest to really articulating why the bike has to be moving right-to-left. There is a particular concept/idea I'm going for here that could be applied to any such tracks to determine the bike's direction.
Agree that the graph cannot be a real life depiction. In either case (R to L) or (L to R) the trailing tire would never cross past the line of the steering tire, unless there was a counter steer to the other direction. Think of pulling a trailer, the trailer always goes to the inside of the curve and never outside.
I get the mathematical, but the graph is wrong.
In my earlier post I was wrong as I was assuming the turns on a straight line perpendicular to the x-axis and forgetting there is length to the bicycle.
^ for the tracks as drawn to make sense, there would have had to be a decent left turn prior to where the tracks start, then a reasonably hard left turn after they end
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New puzzle. This one a friend sent to me, and took me an hour this morning to figure out. This is solvable without pen/paper - just by looking at it and thinking for a bit.
Also, the dx being under the square root sign is a typo - it should be outside as with a normal integral.
New puzzle. This one a friend sent to me, and took me an hour this morning to figure out. This is solvable without pen/paper - just by looking at it and thinking for a bit.
Also, the dx being under the square root sign is a typo - it should be outside as with a normal integral.
Spoiler!
I'm just on mobile and don't have the pen and paper I'd need for it. Seems like integration by parts, us ILATE to decide that the square root is the "first" function.
The symmetry on the endpoints of the definite integral (and the fact that this is on a coffee shop wall) makes me think the answer is probably reducing to something obvious like 0 or pi.
I'm just on mobile and don't have the pen and paper I'd need for it. Seems like integration by parts, us ILATE to decide that the square root is the "first" function. Integrate that piece using the chain rule.
The symmetry on the endpoints of the definite integral (and the fact that this is on a coffee shop wall) makes me think the answer is probably reducing to something obvious like 0 or pi.
I hesitate to respond without giving hints, but I'll just reiterate that you don't need pen/paper to solve.
I hesitate to respond without giving hints, but I'll just reiterate that you don't need pen/paper to solve.
One comment I think is quite useful but isn't a solution.
Spoiler!
I think I can get part of the way there using the hint of no pen/paper as a hint, because there aren't many ways to solve an integral by inspection.
cos(x/2) is an even function. x^3 is an odd function. An even function multiplied by an odd function is odd, so cos(x/2)*x^3 is odd. Any definite integral of an odd function between a and -a is zero, which using integration by parts makes that whole term 0.
So if we distribute it that gets us down to the (1/2)*integral[sqrt(4-x^2)]
I don't think I can solve that piece - I suspect some sort of substitution is required.
One comment I think is quite useful but isn't a solution.
Spoiler!
I think I can get part of the way there using the hint of no pen/paper as a hint, because there aren't many ways to solve an integral by inspection.
cos(x/2) is an even function. x^3 is an odd function. An even function multiplied by an odd function is odd, so cos(x/2)*x^3 is odd. Any definite integral of an odd function between a and -a is zero, which using integration by parts makes that whole term 0.
So if we distribute it that gets us down to the (1/2)*integral[sqrt(4-x^2)]
I don't think I can solve that piece - I suspect some sort of substitution is required.
For a definite integral, don't you just sub in the upper and lower limits for X, and subtract the lower one from the upper? In this case, subbing in 2 for X and -2 for X both result in a zero for the square root term, so - is the answer just zero?
For a definite integral, don't you just sub in the upper and lower limits for X, and subtract the lower one from the upper? In this case, subbing in 2 for X and -2 for X both result in a zero for the square root term, so - is the answer just zero?
Spoiler!
No, you have to integrate that term before you sub in the 2 and -2.
You can't integrate it directly by any method I'm aware of. I think it probably requires a substitution for the inside of the square root. The fact that they're both squares makes me think it should be a trig substitution and then use the trig identities to make it something that can be integrated.
Another thought that is still not a solution but maybe helps someone?
Spoiler!
I mean, the remaining piece is even so you could re-write as
integral[sqrt(4-x^2)] (from 0 to 2)
That doesn't help me get any closer to an actual solution though, because I still don't know how to evaluate that integral.
I do still think trig substitutions could work, but I don't remember how to do that so I couldn't move forward without a reference which doesn't seem in the spirit of this.
full solution, I think, piggybacking on what bizaro86 has already explained about the first half of the function:
Spoiler!
A circle is x^2+y^2=r^2
y^2=r^2-x^2
y=sqrt(r^2-x^2)
y=sqrt(4-x^2) is a semi-circle of radius 2
integral of y=sqrt(4-x^2) from -2 to 2 is half of the area of a circle of radius 2 (because the bottom half is negative)
area of a circle is pi*r^2, in this case pi*4... half of that is pi*2
but the function we're integrating is y=1/2*sqrt(4-x^2), which is half as tall, so the area under the curve is also halved
so the answer to the full integral included the odd and even parts is pi
wifi password is 314 or something similar, though you still have to brute force it a bit because the cafe didn't say how many digits are required
which probably would've been in everyone's top few guesses anyways
psyang: What drives your interest in all this? Are you a math teacher professionally? I'm just curious.
We have a winner. Joint winner, I'd say, with Bizaro86 getting the first half, and SebC finishing it off. Bizaro86 did guess the right answer originally, but as always, you have to show your work
As for my interest? I just really enjoy math problems, and especially enjoy interesting problems with elegant solutions. I graduated with a math degree, though I work now as a software developer.
I loved high school and university math contests, and in my second or third year took a course called the 100 problems course. On the first day of class, we were given a sheet with 100 problems similar in nature to the ones I've posted in this thread. You tried to answer as many as you could by the end of the semester, and as long as you put in effort, you got an A, regardless of how many you solved (I think I ended up with solutions to about 60% of the problems). It was the first course where it clicked for me that university was a place to learn and explore interests, not just a place to try to get good grades and a degree.
The list of problems was lost during one of my many post university moves, something I still regret to this today. I guess this thread is my attempt to recreate the energy that was in those classes which consisted of about a dozen math geeks who had lots of fun discussing and solving problems together.
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Bizaro86 did guess the right answer originally, but as always, you have to show your work
I was just kidding about that - obviously me spitballing that the answer was probably something usable as a wifi password isn't a solution to a math problem. (Although during my academic career I received much higher grades than I deserved for content knowledge based on similar deductions while test taking).
That's a neat problem. I sort of of enjoy that it took two totally different approaches to complete the answer, but that it was doable completely without a pen and paper (as you noted).
Last edited by bizaro86; 10-03-2022 at 04:45 PM.
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Ha, I just took a look at the profs now. I graduated 30 years ago. There are only 4 profs I recognized from when I was an undergrad. But also two profs now who were undergrads with me.
I then tried to see if the 100 problems course is still offered, and I think it might be, though it is now called "Mathematical Discovery and Invention". I'll see if I can find the prof and get the problems list.
