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### Course: Multivariable calculus>Unit 2

Lesson 6: Curvature

# 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?

• Why do we need unit tangent ?
• 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.)
• Is K the Second derivative of S vector
• No. Second derivative is the rate of change of curve with respect to t, while curvature is with respect to arc length.
• 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...)
• 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).
• To me this whole concept of curvature sounds like second derivative/'acceleration'. Is there relationship between the two?
• I think they mean the same thing: they are measures of the rate of how much a function is changing.
(1 vote)
• At , did Grant mean to put an arrow over the S in the denominator?
(1 vote)
• How is arc length (small 's') same as capital 'S', the curve?
(1 vote)
• 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)
• Is R a scaler or a vector? Is it even relevant to ask what R is?
(1 vote)
• k is basically how the tangent changes.
(1 vote)
• 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)
• No, both are different ways of representing the same thing, a vector.
(1 vote)