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Curvature of a cycloid

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

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  • winston baby style avatar for user Andrew
    The curvature formula reminds me of the quotient rule for derivatives. Is that coincidence or is there something to it?
    (20 votes)
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    • leaf green style avatar for user Joe Kern
      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.
      (14 votes)
  • leaf green style avatar for user Isaac Welch
    How does the formula look if you also have a z variable?
    (8 votes)
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    • blobby green style avatar for user dbhattaram
      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.

      Hope this was helpful.
      (6 votes)
  • leafers ultimate style avatar for user prasadshivshankar5
    If it's a cycloid, then didn't the circle have to b a constant size?
    (1 vote)
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  • leafers sapling style avatar for user Mohammed Ghaïth
    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?
    (2 votes)
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    • blobby green style avatar for user pb
      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
      (5 votes)
  • aqualine tree style avatar for user White
    Reminds me of your video on the route that'd give the shortest time featuring Newton, Bernoulli, etc.
    (4 votes)
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  • piceratops tree style avatar for user ewdriver
    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?
    (3 votes)
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  • male robot hal style avatar for user Adam Thai
    I think K must be positive because by k = 1/R.
    In addition, the final result can be furthermore derived into 1/(2*square root of (2 - 2*cos(x))
    (2 votes)
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  • blobby green style avatar for user M
    is there an intuitive explanation as to why the formula isn't symmetric with respect to x and y?
    (2 votes)
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  • starky sapling style avatar for user thegreatmagemerlin
    When you graph the curvature, it is seem to be negative. Is there a reason for this? Also when I did it the long method without the shortcut, I get a different answer that when graphed is shown to be positive.
    (1 vote)
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  • blobby green style avatar for user june.wang05
    Could you do a series for C2/G2 C3/G3 curvature continuity?
    (1 vote)
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Video transcript

- [Voiceover] So let's do another curvature example. This time, I'll just take a two-dimensional curve, so it'll have two different components, x of t and y of t and the specific components here will be t minus the sine of t, t minus sine of t, and then one minus cosine of t. One minus cosine of t. This is actually the curve, if you watched the very first video that I did about curvature introducing it, this is that curve. This is the curve that I said, "Imagine that it's a road and you're driving "along it, and if your steering wheel gets stuck, "you're thinking of the circle that you trace out "as a result and it various different points "you're gonna be turning it various different amounts," so the circle that your car ends up tracing out would be of varying sizes. So if the curvature's high, if you're steering a lot, radius of curvature is low and things like that. So here, let's actually compute it. And in the last example I walked through thinking in terms of the derivative of the unit-tangent vector with respect to arc length but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double-prime, minus y prime times x double-prime, and then all of that divided by x prime squared plus y prime squared. And then I'm writing x prime and y prime and such and all of these you should think of as taking in the variable t. I'm must being a little too lazy to write it. And you take that to the three-halves power. So, this was the formula and I'm not a huge fan of memorizing formulas and then hoping to apply them later. I really do think the one thing you should take away from curvature is the idea that it's the derivative of the unit-tangent vector with respect to arc length and if you need to, you can just look up a formula like this but it's worth pointing out that it makes some things easier to compute because finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the results here. So, first thing to do is just find x prime, y double-prime, y prime and x double prime. So let's go ahead and write those out. So the first derivative of x of t if we go up here that's t minus sine of t. So its derivative is one minus cosine of t. And the derivative of the y component of one minus cosine t, y prime of t, is gonna be, derivative of cosine is negative sine so negative derivative of that is sine, and that one goes to a constant, and then when we take the second derivatives of those guys, so maybe change the color for the second derivative here, x double prime of the, so now we're taking the derivative of this, which actually we just did because by coincidence the first derivative of x is also the y component so that also equals sine of t. And then y double prime is just the derivative of sine here so that's just gonna be cosine, cosine of t. So now, when we just plug those four values in for kappa, for our curvature, what we get is x prime was one minus cosine of t, multiplied by y double prime is cosine of t. Cosine of t. We subtract off from that y prime, which is sine of t, sine of t, multiplied by x double prime. So x double prime is also sine of t, so I could just say sine of t squared, and the whole thing is divided by x prime squared so x prime was one minus cosine of t, minus cosine of t, squared, plus y prime squared so y prime was just sine, so that's just gonna be sine squared of t. And that whole thing to the power three halves. And that's your answer, right? You apply the formula, you get the answer. So for example, when I was drawing this curve and kind of telling the computer to draw out the appropriate circle, I didn't go through the entire "find the unit tangent vector, "differentiate it with respect with arc length" process, even though that's decently easy to do in the case of things like circles or helixes but instead I just went to that formula. I looked it up because I had forgotten and I found the radius of curvature that way.