If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Curvature formula, part 2

A continuation in explaining how curvature is computed, with the formula for a circle as a guiding example. Created by Grant Sanderson.

Want to join the conversation?

  • blobby green style avatar for user Nazim Nazie
    Why do we need unit tangent ?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Caroline
      I think it's for the same reason that we take the derivative with respect to arc length (ds) instead of time (dt) in the definition of curvature from part 1. When defining curvature, we only care about how the direction of the tangent vectors change, not their size.
      Why don't we care about size? Because the size of the tangent vector at a point depends on the parameterization of the curve, but the direction of the tangent vector does not.

      One way to think about this is to imagine a parameterization as a way of drawing the curve (on a sheet of paper). Imagine yourself taking a pencil and tracing out the curve-- for example, a circle. The parameterization of the curve is a function which tells you where your pencil is at the time t, and the derivative (or tangent vector) of the parameterization at time t tells you how fast you were drawing the curve at time t. When you are drawing the curve, you can choose to draw it faster or slower. For example, drawing the circle in two seconds is a different parameterization from taking your time, and drawing it in twenty seconds. The speed at which you draw corresponds to the size of the tangent vector--so you can see that the tangent vector's size changes a lot depending on how you draw. But no matter how fast or slow you draw the curve, the direction of the tangent vector stays the same for all points on the curve*-- it has to be tangent to the curve!

      When we study curvature, we only care about the curve, and not the way that the curve is parameterized. So by normalizing the tangent vectors to have unit length, we're basically 'throwing out' the size information of the tangent vectors. This gives us exactly what we want -- a formula for curvature that does not depend on how we 'draw' the curve!

      *I lied a little here. If the curve starts at a and ends at b, and you draw starting from b instead of a, then the tangent vector will be backwards from if you started at a and ended at b. If we force all parameterizations to start at a and end at b, though, this stops being a problem. (This has to do with something we call the orientation of a curve.)
      (27 votes)
  • aqualine ultimate style avatar for user Sq20021210
    Is K the Second derivative of S vector
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user kluntaf
    Is a vector's derivative always orthogonal to the vector? Grant seems to imply this in the video.

    It turns out the derivative of a vector is orthogonal to the vector iff the vector has constant magnitude (a circle in R2, a sphere in R3...)
    (3 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Caroline
      The answer is: "not exactly!"

      Actually, it might be more helpful to move away from thinking about 'the vector which draws the curve' to instead thinking about the curve itself, and the parameterization. The reason for this is because the statement you made depends on where on the plane we draw the curve. For example, if I move the circle along the x axis, the vectors which 'draw out' the circle aren't orthogonal to the derivatives anymore. (Another reason is that the derivative with respect to a vector does not make a whole lot of sense. Actually, in the videos we are taking the derivative with respect to either t, which parameterizes our curve, or else the arc length of the curve).

      It is true that the derivative of a curve's parameterization is always tangent to the curve, though. There's also a neat fact which is related to the observation you made:

      let z = f(x,y) = x^2 + y^2. This is a function that 'generates circles' in the sense that the set of solutions (x,y) to f(x,y) = r^2 is precisely the equation of the circle of radius r: r^2 = x^2 + y^2. We call these the level sets of the function f(x,y).
      Then the gradient of f at each (x,y) is always orthogonal to the circle level set at that point.

      It turns out that any 'generator function' always has gradient vectors which are orthogonal to the tangent vectors of its level sets (this becomes useful when we are studying lagrange multipliers, another topic in multivariate calculus).
      (3 votes)
  • leaf green style avatar for user Antea S
    To me this whole concept of curvature sounds like second derivative/'acceleration'. Is there relationship between the two?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user declanhoban
    At , did Grant mean to put an arrow over the S in the denominator?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Ojus Singhal
    How is arc length (small 's') same as capital 'S', the curve?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leafers sapling style avatar for user Mohammed Ghaïth
    I think that, prior to computing the magnitude of s'(t), it should be indicated that R (implicitly thought of as the radius of the parametrised circle) is a positive number! Otherwise, when taking the square root, we should have the absolute value of R, |R|, rather than R.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • female robot grace style avatar for user Rushali
    Is R a scaler or a vector? Is it even relevant to ask what R is?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Mez Cooper
    k is basically how the tangent changes.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Tarang Janawalkar
    Is there a difference between the vector arrow above the “s” Grant wrote and the proper vector arrow? (He drew it with only one half of the head)
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] In the last video I started to talk about the formula for curvature. Just to remind everyone of where we are you imagine that you have some kind of curve in let's say two dimensional space just for the sake of being simple. Let's say this curve is parameterized by a function S of T. So every number T corresponds to some point on the curve. For the curvature you start thinking about unit tangent vectors. At every given point what does the unit tangent vector look like? The curvature itself which is denoted by this sort of Greek letter Kappa is gonna be the rate of change of those unit vectors kind of how quick they're turning in direction not with respect to the parameter t but with respect to arc length d s. What I mean by arc length here is just a tiny step you can think the size of a tiny step along the curve would be d s. You're wondering, as you take a tiny step like that does the unit tangent vector turn a a lot or does it turn a little bit? The little schematic that I said you might have in mind is just a completely separate space where for each one of these unit tangent vectors you go ahead and put them in that space saying okay so this one would look something like this this one is pointed down and to the right so it would look something like this. This one is pointed very much down. You're wondering basically as you take tiny little steps of size d s what is this change to the unit tangent vector and that change is gonna be some kind of vector. Because the curvature's really just a value a number that we want all we care about is the size of that vector. The size of the change to the tangent vector as you take a tiny step in d s. Now this is pretty abstract right? I've got these two completely separate things that are not the original functions that you have to think about. You have to think about this unit tangent vector function and then you also have to think about this notion of arc length. The reason, by the way, that I'm using an s here as well as here for the parameterized of the curve is because they're actually quite related. I'll get to that a little bit below. To make it clear what this means I'm gonna go ahead and go through an example here where let's say our parameterized with respect to t is a cosine sine pair. So we've got cosine of t as the x component and then sine of t as the y component sine of t. Just to make it so that it's not completely boring let's multiply both of these components by constant r. What this means, you might recognize this, cosine sine pair what this means is that in the x y plane you're actually drawing a circle with radius r. This would be some kind of circle with the radius r. While I go through this example I also want to make a note of what things would look like a little bit more abstractly. If we just had s of t equals not specific functions that I laid down but just any general function for the x component and for the y component. The reason I want to do this is because the concrete version is gonna be helpful and simple and something we can deal with but almost so simple as to not be indicative of just how complicated the normal circumstances but the more general circumstances so complicated I think it will actually confuse things a little bit too much. It'll be good to kind of go through both of them in parallel. The first step is to figure out what is this unit tangent vector? What is that function that at every given point gives you a unit tangent vector to the curve? The first thing for that is to realize that we already have a notion of what should give the tangent vector the derivative of our vector valued function as a function of t the direction in which in points is in the tangent direction. If I go over here and if I compute this derivative and I say s prime of t which involves just taking the derivative of both components so the derivative of cosine is negative sine of t multiplied by r and the derivative of sine is cosine of t multiplied by r. More abstractly, this is just gonna be anytime you have two different component functions you just take the derivative of each one. Hopefully you've seen this, if not maybe take a look of videos on taking the derivative of a position vector valued function. This we can interpret as that tangent vector but it might not be a unit vector right? We want a unit tangent vector and this only promises us the direction. What we do to normalize it and get a unit tangent vector function which I'll call capital t of lowercase t that's kind of confusing right? Capital t is for tangent vector lowercase t is the parameter. I'll try to keep that straight. It's sort of standard notation but there is the potential to confuse with this. What that's gonna be is your vector value derivative but normalized. So we have to divide by whatever it's magnitude is as a function of t. In this case, in a specific example that magnitude, if we take the magnitude of negative sine of t r multiply by r and then cosine of t multiplied by r so we're taking the magnitude of this whole vector what we get make myself even more room here is the square root of sine squared negative sine squared is just gonna be sine squared so sine squared of t multiplied by r squared and then over here cosine square times r squared cosine squared of t times r squared we can bring that r squared outside of the radical to sort of factor it out turning it into an r and on the inside we have sine squared plus cosine squared I'm being too lazy to write down the t's right now because no matter what the t is that whole value just equals one. This entire thing is just gonna equal r. What that means is that our unit tangent vector up here is gonna be the original function but divided by r. It happens to be a constant usually it's not but it happens to be a constant in this case. What that looks like given that our original function is negative sine of t times r and cosine of t times r we're dividing out by an r the ultimate function that we get is just negative sine of t and then cosine of t. For fear of running a little bit long I think I'll call it an end to this video and continue on with the same argument in the next video.