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## Multivariable calculus

### Course: Multivariable calculus>Unit 2

Lesson 6: Curvature

# Curvature formula, part 3

Here, this concludes the explanation for how curvature is the derivative of a unit tangent vector with respect to length. Created by Grant Sanderson.

## Want to join the conversation?

• Why is 1/r also equaled to the derivative of the unit tangent vector with respect to arc length? It seems like this was just stated in the last video without any explanation. It makes sense to me that they would be related, but what proves that they are equal? • It is precisely that which this series of videos is trying to answer.
If you're wondering why the math aligns so nicely, that's because all circles have the same curvature, so it naturally becomes a constant!

If you're having problems with the explanation, a recap might help:
He started with the parameterization of a circle, then to find how it changes while incrementing 't' he differentiated it. He continued by normalising it (since cos(t) and cos(10t) would give you the same curvature even if you're going faster around). Then he differentiated again, because the question is not how it's changing at certain points on the circle, but /how/ it's changing from point to point.
Finally, we want our function to represent the change with respect to distance travelled (ie 's') and not increments of t. We divide dT/dt by ds/dt to cancel the dt and get dT/ds. Find the magnitude (see video for why) and voila! For any circle, this will always turn out to be one over the radius (1/R).
• At it's mentioned that there is an article that explains the steps that isn't shown in the video. Where might one find that article? • At , how is the derivative of the function with respect to the parameter equal to the derivative of the arc length with respect to the parameter? • Where's the article that shows the derivation of the equation? • The curvature formula can also be derived through the centripetal acceleration equation: v^2/r = a. a is the component of acceleration towards the center of the circle. V= <x'(t), y'(t)>.

The vector perpendicular to V points towards the center of the circle, so taking the dot product of V' (<x''(t),y''(t)>) and <-y'(t), x'(t)>, then dividing by the length of the perpendicular vector (sqrt(y'(t)^2 + x'(t)^2)) gives you the projection of V' onto that perpendicular vector, the acceleration towards the center. a = (x'(t)y''(t) - y'(t)x''(t)) / sqrt(y'(t)^2 + x'(t)^2)

We can plug ||V||, sqrt(y'(t)^2 + x'(t)^2), as v into the centripetal equation formula. k=1/r = a/v^2

k = (x'(t)y''(t) - y'(t)x''(t)) / (sqrt(y'(t)^2 + x'(t)^2) * sqrt(y'(t)^2 + x'(t)^2)^2)

k = (x'(t)y''(t) - y'(t)x''(t)) / (y'(t)^2 + x'(t)^2)^3/2 • So is ds arc length or is it differential of the function? Also what is the connection between defining curvature as 1/r and as dT/ds? Can they be derived from each other? • The differential of the function and the arc length are ds. The differential is a small change in s. This, geometrically, would look like vector subtraction and would result in a line connecting the tips of the two vectors. This line would be ds. When you think about things microscopically, a large arc length can be approximated by a large change in s and an infinitely small change in arc length is equivalent to ds.
• why ds/dt equal to R? • I can't understand in this Video what for we are computing second deriative for T. • The first derivative regards to the tangent of the curve, and this is not what we are looking for; we are looking for the change of the direction. In other words: the change of the slope (first derivative). So, derivating the tangent (first derivative), you get an idea about the change in the direction of the slope. Try to see the second derivative like a tangent of the tangents, this is what we are looking for. Sorry about my english.  