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### Course: Multivariable calculus>Unit 2

Lesson 6: Curvature

# Curvature of a cycloid

An example of computing curvature with the explicit formula. Created by Grant Sanderson.

## Want to join the conversation?

• The curvature formula reminds me of the quotient rule for derivatives. Is that coincidence or is there something to it?
• I think it could just be a coincidence, if you use algebra and calculus to solve for dT/ds (if ds = the vector [x(t), y(t)] it doesn't come out to equal the quotient rule (which has a denominator of g(x)^2). However, I do agree that the similarities are eerily close so it couldn't hurt to ask someone wiser than I.
• How does the formula look if you also have a z variable?
• This question has been bugging me for some time as well, and I can't seem to find it anywhere, so I tried to derive it myself.

I took the cross product of s' and s'' (where s was a three-dimensional vector instead of a two-dimensional one) and divided it by the magnitude of the vector cubed (just like in one of the earlier videos).

The result was this: (y'z'' - z'y'' - x'z'' + z'x'' + x'y'' - y'x'')/((x')^2+(y')^2+(z')^2)^3/2

This is probably one of the most hideous equations I've ever seen, and I'm likely to have made a mistake in the cross product or the derivation in general. It could be tested by inputting the helix from the two previous videos, but I don't trust myself to do that either.

• If it's a cycloid, then didn't the circle have to b a constant size?
(1 vote)
• The circle that is related to the curvature changes depending on where you are on the curve.
The circle that generated the cycloid (not discussed in this video) would be a constant size
• Reminds me of your video on the route that'd give the shortest time featuring Newton, Bernoulli, etc.
• Now that I have finished watching the videos on curvature, I need to ask something! In one of the formulas, curvature was linked to cross-product. However, I find it a little strange to talk about the cross-product of two two-dimensional vectors! I understand the reasoning and the explanations, but how can the cross-product be defined for vectors in R^2?
• I had the same question too I think these are still in a 3 dimensional space but their z components are supposed to be zero, I mean the way of calculating the cross product between 2 3d vectors was to imagine them in a single plane and find a vector that was perpendicular to that plane and in here its like that too, just the plane that we choose is parallel to the xy plane
• In the final result, k = ((1-cos(t)) cos(t) - sin(t)^2)/((1-cos(t))^2 + sin(t)^2)^(3/2)
where ((1-cos(t)) cos(t) - sin(t)^2 = cos(t) - cos(t)^2 - sin(t)^2 = cos(t) - 1 is non-positive, and the denominator is non-negative.
So, k is non-positive. Is this possible?
• I think he needs absolute values on the numerator of this formula. I checked his cross product formula and that should also be magnitude of the cross product