The pythagoras triple connection comes out by viewing the problem in 3D. If you take the surface of a cylinder of radius a and cut it from the side with a plane with gradient c/a the intersection is an ellipse. Algebraically the cylinder is {(x,y,z): x^2 + y^2 = a^2} and the plane is {(x,y,z): x/a = z/c}. Rolling the cylinder "on the ground" and taking a trace of a point on the circumference of the ellipse with "the ground" you get a sine curve. There's a few animations around on the internet with it. It's also clear that the ellipse we have constructed will make a full rotation going the length of the sine curve. With the above construction the ellipse has width 2a, and with a bit of pythagoras the height of the ellipse is sqrt( (2a)^2 + (2c)^2) = 2b.

@drpeyam

2 жыл бұрын

Super interesting, thank you!!

@sukhps

2 жыл бұрын

I did not understand but anyway i will approve👍

@eliyasne9695

2 жыл бұрын

This is a brilliant construction!

@ahsan4306

2 жыл бұрын

Can u send the link to animation video

@rede1195

Жыл бұрын

That's pretty cool man

What i found amazing was that you could compute b without needing to evaluate the actual integral. It's like messing around inside (4-4)/(6-6) and getting something useful, seeing as elliptic integrals are famously unsolvable in elementary terms

Brillant, cause you hadn't solved the unsolveables elliptics integrals , but just compared them

At the end of the video, you showed that b^2 = a^2 + c^2 for the general sinewave y = c*sin(x/a). Earlier, you found b for the particular case of y = 4sin(x/3), where c = 4 and a = 3. These are 2 values in the Pythagorean triplet (a,b,c) = (3,b,4), so the 3rd value (b = 5) of the triplet is determined.

What a fun problem!

@drpeyam

2 жыл бұрын

Awwwww thanks for being a member 😁

thanks, DR> PEYAM

The distance between the two peaks is 2πa. Now, suppose that 2b > 2πa, say b = 4a. This makes the ellipse getting stuck when rolling down the hill. Does the formula still hold in that case?

@drpeyam

2 жыл бұрын

I don’t see why not, at least mathematically

@anmoldeepsingh9281

2 жыл бұрын

yeah rolling is a physical phenomenon, we are just mapping the perimeter of the ellipse between the peaks

impressive gong ^v^

how would we really check if the ellipse rolls perfectly without clipping through the curve while rolling? Like at any instance the ellipse is locally rolling on the curve, but any other part of the ellipse might intersect the curve. Checking for solutions of points of intersection seems tedious.

@eliyasne9695

2 жыл бұрын

That the ellipse is rolling means it is of the same length as the curve regardless of whether or not there's clipping going on.

@anmoldeepsingh9281

2 жыл бұрын

@@eliyasne9695 Nah i understand that,.I look at it as a perimeter mapping of the ellipse to the curve, I just wanted to ask how we would check if it clips or not

will the distance covered by a revolution not be proportional to how far the centre has travelled? I remember some SAT problem got this wrong, but i could be misremembering stuff.

It's confusing to refer to x and y for the ellipse and for the curve with the same notation, because they are different curves, and you're only equating the arc length.

@alexdemoura9972

2 жыл бұрын

It is not a problem "only equating the arc length". The ellipse and the sine curve are mathematical concepts only, they are NOT physical/real solid objects. In real world probably the ellipse would be "wedged" between the two hills of sine curve before reach the bottom of the valley. As mathematical objects they have no such commitment to behave as real solids, they have only one mandatory rule to abide to: keep both objects touching at one common tangential point movable along the curve (and around the ellipse) all the time during the ellipse revolution(s), and that is all... no matter what about the rest of the both geometric bodies... touch, overlap, cross, or whatever.

@drpeyam

2 жыл бұрын

The ellipse is on the curve though

Nice 👍

please do this for any y(x) and ellipse with a and b

@drpeyam

2 жыл бұрын

Well just watch the end of the video

@andrewkarsten5268

2 жыл бұрын

If you consider a generic function instead of a sine curve, you may run into problems since solving it won’t be as straightforward, if solvable at all in explicit terms. A main issue is that the ellipse has no circumference formula for a general ellipse, in this problem we were able to solve by comparing the integrals to each other, but if you consider a function like y=x, then the answer effectively comes down to finding the formula for the circumference of an ellipse in generic terms, which is currently unknown.

Love your videos. Go bears!

@drpeyam

2 жыл бұрын

Go bears 😁

Easy. First, we'll just start with the formula for the perimeter of an ellipse. 🙂 ☹️ Oh no...

This is calculus

🙃

🙄🙄🙄

What makes me a little uncomfortable is knowing that Ramanujan may have wasted a part of his precious and short time to make the best possible formula to approximate the perimeter of the ellipse, since until today there is no exact formula to calculate such perimeter... if he and all Mathematicians knew that it would be enough to parameterize both axes as: x = a cos(t) y = b sin(t) And just apply an Integral from 0 to 2π over the square root of sum of squares of x and y... Ironies aside, if it's so easy why isn't there an official formula for the perimeter of the ellipse?

@MushookieMan

2 жыл бұрын

It can't be written as a finite algebraic expression, without defining new functions like the elliptic integral

@_Blazing_Inferno_

2 жыл бұрын

They both produce an integral of the same form, which was the problem in the first place. If there were a second equation that the parameterization produced without an integral, then it might be more helpful.

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