If a vector has constant length, then it's orthogonal to its derivative (proof & example, calc 3)

[@nekothecat]

I think the proof can be easier when you know the vector is on a sphere or circle in R2. You just need to proof r'(t) is tangent to r(t), which is powerful beginner equation for Multical

[@neilgerace355]

5:43 Not the Best Friend, but a good friend.

[@michaelmounts1269]

good post this morning!

[@kirtisoni5153]

Thankyou

[@hugodanis6144]

Thanks, and is the converse true?

[@sujitsivadanam]

I believe so. In fact, you can follow his steps backwards to prove the converse is true.

[@DeJay7]

I don't know why, this has never happened to me before I don't think, but the example given felt absolutely worthless. The example itself was completely fine, but we just proved that if a vector has constant magnitude then its derivative is orthogonal to it. Then we picked a vector with a constant magnitude. Then we calculated the dot product to verify that it is indeed 0, as expected, but it obviously would turn out to be 0, we JUST proved that it would be given the property that it has, and the proof had no mistakes or assumptions or anything, so why would you need to verify? Don't get my words twisted, this is THE FIRST TIME I think about a proof+example in this way, it has nothing to do with this specific one, it was great, but I just wanted to yap unnecessarily.

[@phiefer3]

This is literally true any time that an example is given after a proof. The purpose is not to verity the proof (that concept doesn't even make sense), the purpose is to demonstrate it.

[@DeJay7]

@@phiefer3 Yeah yeah exactly, the weird thing is that I noticed this just now for the first time, and I've seen A LOT of proofs, strange.

[@billycheung5114]

Wow! But why this important 🙏🏻🤔

[@alvargd6771]

sometimes you work in spheres or circles, and there all vectors and parametrisations of them have the same length