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.

Course: Multivariable calculus>Unit 2

Lesson 6: Curvature

Curvature formula, part 4

After the last video made reference to an explicit curvature formula, here you can start to get an intuition for why that seemingly unrelated formula describes curvature. Created by Grant Sanderson.

Want to join the conversation?

• Isn't cross product only for 3D vectors?
• Yes, you're right, it's a 3D only club. Think of it this way; if you add the third component to the vectors (naturally being zeros) and then take there cross product, you'll see that the first two terms cancel out, and you're left with a z component that's identical to what Grant showed.

If it's confusing why the answer is now a vector in the third dimension, you can watch Sal' video on how to interpret this in terms of physics : https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/v/cross-product-and-torque
or if this doesn't make sense, then watch the video on cross products again and you'll see that by definition, the cross product produces a vector in 3d
• At Grant says the cross product is interpreted as the area of the parallelogram formed by the two vectors. I don't see how this corresponds to the area of the parallelogram without taking the magnitude of the cross product, because crossing the two vectors would result in another vector.
• You are right. The geometric meaning of the cross product is that the magnitude of it is equal to the area of a parallelogram formed by two vectors. I think Grant meant "the magnitude" of the cross product here.
• Why is curvature defined as the vector change between the unit vectors rather than the change in the angle between them? If it was the angle between them, then with a circle going 1/4 around the circumference would be a change of pi/2 over the arc length, (pi/2)*r, therefor satisfying our definition of curvature as 1/r.
• I think this is pretty confusing. The cross product is only defined for two vectors in three dimensions. When having this kind of operation extended to other dimensions, it should technically be called an exterior product. But more important: is the cross product of two vectors not supposed to be just another vector?
• Actually the cross product is defined for 2 dimensions as well. When taking the cross product of 2 2-dimensional vectors, you just take the discriminant of the vectors in matrix form, as shown below:
[a, b] x [c, d] = ad - bc

Also, yes, the cross product of two vectors is just another vector, but Grant takes the magnitude of the vector in the video, so it becomes a number.
• why the cross product of those two vectors represent the curvature of a short length of a curve
• Curvature is always positive, right? Shouldn't we take an absolute value of a numerator?