Both Curvature Formulas Derivation :: Vector Calculus
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- Опубліковано 19 вер 2024
- In this video we derive both curvature formulas from the basic definition of what curvature is.
Curvature is the rate of change of the unit tangent vector with respect to arclength.
The first curvature formulas derivation starts with that definition.
The second curvature formulas derivation is much less intuitive, but it is a great exercise in keeping track of useful things we know or can show.
I hope this video helps clear up from where both curvature formulas arise.
Thanks for watching!
-dr. Dub
This is the clearest derivation of the second formula Ive found on the internet. The random cross product coming from nowhere made me totally unable to understand the formula.
Thanks! I'm happy to have done this one! It's a classic example of "Where does that come from??!?!?!" Ha! =P
Thanks. You're better than my university prof at explaining
I’m happy this has helped!
i thought the random cross product has to come with the binormal vector - now it makes so much sense
Really helped a lot! Thanks for the clear explanation !
I’m glad this helped! You’re welcome!
Nice. I was trying to get this but could only get the curvature vector but by a much bigger path. This techinque of starting at the cross product helps a lot. Ty
Glad I could help!
Jonathon i really enjoyed this video great derivation and filled in some details
Thanks for your support! Glad this helped!
Perfect.
I'm glad this helped!
this was amazing thank you!
Why the curvature is determine by respect to the arc length ? Why not direcly in the function!? This confuse me all the time
I wouldn't exactly say curvature is defined by arclength. It just so happens that |r'(t)| appears in both. =)
Curvature by definition is basically the rate of change of the unit tangent vector.
Thanks, Jonathan.
You’re welcome!
thank you ，this problem confusing me for a long time
I'm so happy this helped! Thanks for watching!
Sweet vid
Thanks Leo!
Thanks for this bro
Always happy to help! Thanks for watching!
thank you!!
Glad to help!
thanku so much
You’re welcome!
I don't know if this comment will reach out to you but I have a question. The chain rule in the first proof states that there should be two functions T = f(t) and t = g(s) that are differentiable. Well I want to know if the chain rule still applicable given that we don't know about t = g(s) yet
I see. This question is a little more on the side of functions. s(t) is and increasing function meaning it’s one to one. Therefore it’s invertible. ie. we know t(s) exists as a function.
@@JonathanWaltersDrDub Thanks for your prompt reply! Another question from me if you don't mind. Is every vector-valued function one to one? Or are there some certain criteria for this?
Good question. The arc length function s(t) is one to one. Not the vector function itself.
Basically, if you know a value of s, then you know exactly where you are on the curve.
This is really helpful and clear, but I'm still confused about ds/dt in the second proof. It seems like it's treated like a variable when you find the derivative of [ r' = T(ds/dt) ] but a scalar when you find the cross product of r' and r''. I'm self-taught to the degree I'm taught at all and I know there are mysteries I'd understand if only I was a better mathematician, so I'm probably making a mistake with this simpler route but I don't know what it is: Keep |r'| instead of replacing it with ds/dt to emphasize it is always a scalar. Then taking the derivative of both sides of r' = T|r'| means r'' = T'|r'|. [I mean, T(d^2s/dt^2) = T(ds/dt)|r'| = 0, right?] Then r' x r'' = |r'|^2 [ T x T'] immediately, and etc.
r''(t) = d/dt (T|r'(t)|) which would require a product rule since T and |r'(t)| are both functions of t. So yes ds/dt is itself a scalar function of t still.
@@JonathanWaltersDrDub First part face-palm convincing; second sentence I'll understand with a little more work. Thank you very much; you've been really helpful.
Let me know if you want to discuss further.
If the unit tangent vector is perpendicular to its derivative, can it be said for a non-unit tangent vector? Also what if the tangent vector is a vector with variable length?
Yes, the magnitude of the tangent vector may be different but it’s direction is still the same.
The length of the tangent vector r’(t) is fixed for any value of the parameter t.
But the length of r’(t) does change with t.
That’s why we define the unit tangent vector T(t) to always have into length (note T is a function of t but always has unit length).
@@JonathanWaltersDrDub That explanation was so simple! I was too focused on the variation of r'(t) wrt to t, I didn't consider it was instantaneous.
Thanks for the reply. It was really helpful!
You’re welcome! I’m glad it’s clear!
Why did we take the magnitude of the cross product of T and T’ when the formula for cross products is a cross b=|a||b|sinx. Why didn’t we use this formula and instead took the magnitude of both the sides
That formula is |a x b| = |a| |b| sin (t) . So we needed to take the magnitude first to apply that formula.
4:21 Shouldn't *r' = T' |r'|?*
Isnt there a modulus around ds/dt?
Good question! Note that ds/dt = |r'(t)| already (see discussion at 0:44). Thus, ds/dt is already a scalar quantity. =)
@@JonathanWaltersDrDub thanks for reply! Great content!
@@quantised1703 No Problem! Thanks for watching and for your feedback and questions!
Thank you very much for this synthetic, complex «not that much really» yet simple and very objective video... MUCH better than 3Blue1Brown for Khan Academy «which uses a series of cumbersome analogies or explanations». I will use this and some of the other authors mentioned as blueprints for videos of calculus. The only problem is that you were a bit fast, maybe there is a limit to video time, as I once thought there was.
I would just use coupled with that one graphic pointing out what are the tangents «or derivatives» to curves r(t) and what are tangents of tangents of vector of curves «the second derivative of the curve with respect to an infinitesimal arc length of the curve», id est, the normal vector of that curve; this last of which the magnitude of that proportion is the Curvature K of the curve.
I prefer to approach these problems as a mathematical intuition of Physics or Natural Philosophy. The Unit Tangent Vector is the Velocity/Speed and the Curvature of the Curve is the Magnitude of the Acceleration or the instantaneous rate of change of the velocity of that curve with respect to an infinitesimal arc length of the curve.
Thank you very much again for this!
Thank you for your feedback!
My goal has always been to be clear and concise in my explanations!