Curvature formula, part 3

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

[@jasonli1060]

Many people are asking why he uses the position function as opposed to arc length in the calculation. This is because he actually is using arc length though it may not look like it. ds/dt is the rate of change of arc length. Using the formula for arc length, integral of speed, we can see that see that it's derivative is equal to speed or, in other words, the magnitude of the rate of change of the position function, |ds/dt|. This means it doesn't matter which you use, position or arc length. Put another way, the magnitude of rate of change of position is the same as the rate of change of arc length. They will lead to the same result.

[@AndreKoop-nu6vk]

I know you posted this a while ago, but that was super helpful, thanks!

[@atriagotler]

Grant is a gift.

[@isaackoerner1599]

Great explanation, but where's the article? A link would be nice.

[@qinlun6859]

Did you find that article？

[@That_One_Guy...]

Use Khan Academy app/website if you want to access the articles

[@qinlun6859]

@@That_One_Guy... Wow, thank you

[@jd-gw4gr]

excellent but what software is khan academy using to illustrate and annotation? i'm not having success on zoom using my pad for student's notes.

[@Reallycoolguy1369]

So I went ahead and derived the curvature formula (for this case of a 2 dimensional vector value function) and got what Grant gave us. x'(t)y"(t)-y'(t)x"(t)...etc... but after going through that process, I don't see why the equation couldn't be an equivalent equation except negative. For example, y'(t)x"(t)-x'(t)y"(t)...etc... seems just as valid to me. Is curvature as described here an absolute value? I had read somewhere that you could use curvature with its sign for convenience in engineering applications, but this derivation seems to "destroy" information about the sign.

[@raviyadav2552]

u are awesome sir

[@maunil108]

So nice,intuitive channel . But why people don't like this video? So strange.....

[@ManpreetKaur-ry3vz]

Is there any special reason to take the function as s(t) ?

[@IanCarl37]

No, it can be any letter

[@khalilmohammed2297]

where is the article

[@allywordee]

u used chain rule right @2:14?

[@YuvrajSingh-ci8nb]

Yaa

[@LaureanoLuna]

The equation at 2:35 is never explained. Furthermore, s with vector hat is not the length of the curve (as he says) but the position vector, which need not be the same in any obvious way.

[@stephenbeck7222]

He explained where it came from in the earlier parts.

[@adarshdomala6895]

But an infinitesimal arc length at a point is the magnitude of infinitesimal change in position vector at that point. Think about it

[@LaureanoLuna]

@@adarshdomala6895 I know what you mean but that seems to me lack of rigor: it is by no means clear what an infinitesimal arc length is.

[@lordspongebobofhousesquare1616]

@@LaureanoLuna this is a vector calculus playlist, not an analysis playlist. Don't expect rigour here

[@cauchyschwarz3295]

@@lordspongebobofhousesquare1616 why not? Nothing about loosely inserting and cancelling dts is helping real understanding.
That is nothing but an arithmetic device to use for people who don't actually understand what it is they are doing. And these can be found in large quantities in school books.
This lack of rigorour and proper explanation is not typical to 3b1b. In his own Analysis series he does it a lot better.

[@nileshpandey8032]

Why can't we simply say that the curvature will be the inverse of the magnitude of the derivative of the parametric surface vector with respect to t? I think this is what we were getting at the end. |Unit Vector|/|Vector|=1/|Vector|

[@juliuscampecino3579]

I second. I thought about this as well.

[@brokenwings6925]

This happens only in special case of circle where |dT/dt| =1 where T is unit tangent vector. But in general case |dT/dt| need not be equal to 1 where T is unit tangent vector.

[@SaadKhan-pn9rt]

why was the derivative of the function with respect to t treated the same as the change in arc length with respect to t?? like the change in arc length certainly isnt the same as the change in the value of the function with a small change in the parameter....atleast not always

[@adarshdomala6895]

Change in arc Len is not change in function but infinitesimally small change in vector valued func is a small arc Len at the specific point

[@effy1219]

what's the connection of T(t) and ||dT / ds|| i understand both, why you switch from explaining T(t) to ||dT / ds|| ??????????????????????????????

[@shahidakhanom4208]

Are u Sal?

[@priyankkharat7407]

No he's Grant from another awesome Math channel : 3Blue1Brown.

[@kennethcaron572]

What he is trying to say is don,t look its to hard? EXAMPLE standing at the beach water level .HOW can I see 18 meters around the curve in the road just 9 miles out ! thats 9 miles squared=81 X 8 inchs =648 inchs =54ft or 18 m

[@lj7817]

why would you multiply it by 8 inches? you're awful at math

[@cauchyschwarz3295]

This was unusually unhelpful. dT/dS is the derivative of a vector valued function with respect to another vector valued function. Nothing about the key equality you used is obvious. And cancelling out dt's is nothing but an algebraic rule for people who don't actually know what they are doing. Maybe good for engineers in practice but not helpful at all when explaining the concept.

[@benjaminojeda8094]

don't get so complicated, just make the limit and think of the dt as (t+h)-t

[@dr.muhammadrabiulalam7616]

I love this discussion because it is a sign for every man(slave) who turns to Allah in repentance. Dr Muhammad Rabiul Alam

[@Mariowatchesyt45]

How?

[@Anandhu-X]

What

[@bmbiz]

lolz I ain't turning to your imaginary sky friend for anything, "doc".