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How curvy is a curve? Intro to Curvature & Circles of Curvature | Multi-variable Calculus

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Dr. Trefor Bazett

Dr. Trefor Bazett

4 жыл бұрын

How curvy is a curve? In this video we define and come up with a formula for curvature and see how this relates to unit tangent and unit normal vectors. Finally we describe the circle of curvature. While this is a multivariable calculus topic in reference to vector valued functions, the question of curvature of a curve can be asked for any planar curve from single variable calculus as well!
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Пікірлер: 115
@whoatemywendys
@whoatemywendys 3 жыл бұрын
I like how this guy’s passionate about what he talks about
@BoZhaoengineering
@BoZhaoengineering 4 жыл бұрын
Cool!. The animation is great. This is how to describe how curvy is a curvature mathematically.
@loloferr7074
@loloferr7074 3 жыл бұрын
seriously you are a lifesaver!
@erikchen8897
@erikchen8897 2 жыл бұрын
these videos are great. still 3 years after they were made there helping student like me understand this crucial content.
@panchui4
@panchui4 Жыл бұрын
I really love your shirts! Thank you very clear, glad I found this channel!
@DRMath
@DRMath 4 жыл бұрын
Well explained and very impressive demo of the topic. Thank you sharing the knowledge!! Peace ✌️!!
@allison5169
@allison5169 2 жыл бұрын
Thanks for these videos! They are so amazing! I know so much now! And hahahahah my friend regular wants me to tell you that he loves math!!!
@sdsa007
@sdsa007 2 жыл бұрын
great stuff! i finally understand curvature in relation to radius and T.. the circle of curvature is really good visual aid.
@TrampTuts
@TrampTuts 4 жыл бұрын
Another stellar video!
@HosRo4161
@HosRo4161 4 жыл бұрын
Excellent! Thank you!
@Festus2022
@Festus2022 4 жыл бұрын
Well explained. Great video! Thx.
@vaibhavdlxit1050
@vaibhavdlxit1050 3 жыл бұрын
This is only way one needs to teach the topic, u really dumbified it for me! Big thanks!
@DrTrefor
@DrTrefor 3 жыл бұрын
Glad it was helpful!
@balacharya1395
@balacharya1395 2 жыл бұрын
your explanation is very effective, thank u so much
@michealjameshermano6003
@michealjameshermano6003 3 жыл бұрын
Your videos are amazing, it is my first time to subscribe to an educational channel where I only just watch 2 of your videos. Because I am a type of learner that learns formulas by knowing how it is derived.
@DrTrefor
@DrTrefor 3 жыл бұрын
Thank you so much
@fugoogle_was_already_taken
@fugoogle_was_already_taken 3 жыл бұрын
Dude, you rock! Love your videos
@chinmaypradhan7680
@chinmaypradhan7680 3 жыл бұрын
Love your content sir.
@askmathsboard
@askmathsboard 2 жыл бұрын
Your explanation is super Sir... Thank you for sharing your knowledge...
@ianfowler9340
@ianfowler9340 3 жыл бұрын
Another cool thing. I had always thought of curvature as d(theta)/ds where theta is the angle between the tangent and the + x-axis - kind of a turning rate of the tangent in radians per unit of arc length. Then when I saw your definition I realized that they are exactly the same thing! That was a good moment for me.
@DrTrefor
@DrTrefor 3 жыл бұрын
Nice, that’s a great observation:)
@shim-kun2538
@shim-kun2538 Жыл бұрын
Thanks for making these videos. They've been helping me a lot! I also really appreciate the graphs since they make it easier to visualize what's going on. What software do you use to make those? It would come in handy in my assignments
@hungtuan9347
@hungtuan9347 4 жыл бұрын
Thanks for the great explanation and animation. How about the singed curvature, do you have any explanation about this? I am looking for a relation between the curvature defined in your presentation and that with signed curvature. Thank you.
@LarryChen-bh8uo
@LarryChen-bh8uo 26 күн бұрын
LOVED THIS VIDEO
@lailazannat5989
@lailazannat5989 2 жыл бұрын
Amazing Explanation.
@filomenagoncalves2768
@filomenagoncalves2768 3 жыл бұрын
I think you explain very well. I am looking for material on composite functions, higher derivatives. Do you have a video on in?
@saikumar-xh9vc
@saikumar-xh9vc 4 жыл бұрын
Thank you so much from India
@continnum_radhe-radhe
@continnum_radhe-radhe 2 жыл бұрын
Thank you sir 🔥🔥🔥
@mohanprasad6373
@mohanprasad6373 3 жыл бұрын
Well explore
@joaomattos9271
@joaomattos9271 Жыл бұрын
Excellent!!!!
@alial3802
@alial3802 4 жыл бұрын
Great job.
@nestorwalters8079
@nestorwalters8079 3 жыл бұрын
Super useful graphics. Thank you!
@DrTrefor
@DrTrefor 3 жыл бұрын
Glad it was helpful!
@talhamahamud3176
@talhamahamud3176 Жыл бұрын
you are best !!!
@alperensonmez6875
@alperensonmez6875 3 жыл бұрын
Thank you Dr. Trefor Bazett
@erockbrox8484
@erockbrox8484 2 жыл бұрын
Thanks so much, I didn't know you could use chain rule to convert the dT/ds into the much more friendly form of (1/v)(dT/dt).
@aalaaula
@aalaaula 2 жыл бұрын
Why is the velocity defined on arc length S instead of t itself? Is it because we are trying to align it with a circle? so that we traverse same distance per unit?
@nahiananis5995
@nahiananis5995 Жыл бұрын
from which book I can practice all the concepts you taught in the playlist?
@ukbasak
@ukbasak 4 жыл бұрын
I see that the curve is crossing itself in your animation. How would you define curvature at the intersection points?
@yasirbilal3971
@yasirbilal3971 4 жыл бұрын
Good explained
@surtvalheim
@surtvalheim 3 жыл бұрын
The normal is the second derivative of the curve ?
@MrMarkgyuro
@MrMarkgyuro 4 жыл бұрын
where did the principle unit normal vector formula come from? :O
@rf-iu5jj
@rf-iu5jj 3 жыл бұрын
Starting with T(t) * T(t) = 1, where * is the dot product, since both are unit vectors. Then getting the derivative of both sides: ( T(t)^2 )' = ( 1 )' 2T(t) * T'(t) = 0 In other words T'(t) is perpendicular to 2T(t). Then T'(t)/|T'(t)| must be the unit vector perpendicular to T(t) Not the best derivation, but it's one possible way.
@cory99998
@cory99998 2 жыл бұрын
Ty!
@Lionheart683
@Lionheart683 4 жыл бұрын
Nice video.
@akashsunil7464
@akashsunil7464 Жыл бұрын
Could you show the proof of normal vector its kinda hard to develop the intuition for it like unlike curvature cuz overtime I think I might forget about the normal vector
@muhammadabuzarjanjoa4174
@muhammadabuzarjanjoa4174 3 жыл бұрын
Just love it. 👍
@DrTrefor
@DrTrefor 3 жыл бұрын
Thank you! Cheers!
@ganli1034
@ganli1034 3 жыл бұрын
Great job!
@DrTrefor
@DrTrefor 3 жыл бұрын
Thanks!
@theleviathan3902
@theleviathan3902 Жыл бұрын
Why do we evaluate in terms of the unit tangent instead of the velocity vector at the beginning? Like I get why we evaluate in terms of arclength since different parameriterizations have different speeds but why measure in the unit tangent?
@mnada72
@mnada72 3 жыл бұрын
Thank you for the explanation. Although this is a hard , I had to repeat it several times to internalize the concept. Yet I am missing where we proved the concept of velocity at 2:04
@aryanthacker
@aryanthacker 2 жыл бұрын
The change in arc length(ds) is the scalar distance travelled by the curve, so ds/dt=|v| because ds/dt is scalar distance travelled/time, if I've understood correctly.
@ranam
@ranam 4 жыл бұрын
Can we create a differential equation out of a curvature that is can we equate a equation equal to curvature weather it's 2d or 3d and if we solve it for arbitrary curvature is it possible to get any interesting curve without any variational method and so pure in nature could say the abstraction of the curve and if so how to form it I hope you understand I have been searching for this question for a long time
@ranam
@ranam 4 жыл бұрын
@@DrTrefor yes sir thank you but I have been using your explanation throughout years unknowing that there is a software called xplore math 3d where we just give curvature and it plots but it does not seem to work for arbitrary curvature because I think do I did not follow the rules that you have said before I think or guess so now it's night in India but IAM going to give it a try and the curvature plotter in this software also can plot 3d surfaces ok let me summarize so 1.) Line is made up of bunch of points and to draw a line we use points as generators in its location and to draw a circle we use radius by using compass by the idea you just gave above can I make any curve or surface just knowing tangent and normal vectors only if it happens so what all I should know going to higher dimensions and shortest distance between any two points in a plane 2.) Curve when you zoom it still has a nature of straight line and I just conclude for a joke that it's made up of infinitesimal lines 3.) I can not conclude with above points to a new one just puzzled Now if we no both tangent curvature vectors and normal curvature vectors we can make a curve in 2d with ode but what about 3d can I generalise them in the same manner then I would be needing a extra curvature vectors I think so and just waiting for your answers sir your information is highly valuble please check this software link below sir I know some dont really recommend or believe in softwares but I have no choice rather than this to explain my point of view on this topic which has deep roots in optics and quantum mechanics and also A.I 3d-xplormath.org/ And sir please make a video video physics modeling based on causality defining homogeneous or non-homogeneous equations which could model experimental results and without any linear logic in physics or other tools could make assumptions to create a theory or formulas based on it these are.my ideas and I even have the idea of parameterizing fractals and use scwartz cristofels conformal mapping for some advenced parametric oscillation theory and an analytical symbolic method to compute the equation or any shape you imagine or see and taking to the higher levels of octanion number system for surface modeling which also has deep roots in.many sciences
@abacaabaca8131
@abacaabaca8131 3 ай бұрын
@3:24 May I ask a question? Why is that the velocity vector need to be calculated from the transformation of a point in a circular motion ? Because the thing is, the way we draw a circle is by transforming it by multiplying with rotation matrix. Is not velocity vector is something that we can decide how much. For example, in the context of programming, I can move or draw the circle slowly with respect to the time given by the operating system to my app to draw. Meaning, for every frames I got, I can multiply the velocity vector with the angle together with time given by os to operate.
@jitendrasingh-nd7kq
@jitendrasingh-nd7kq 4 жыл бұрын
Sir, I have read that magnitude of derivative of r(s) is 1 but why???and what is mean of smooth curve?
@HT-rq5pi
@HT-rq5pi 4 жыл бұрын
Suppose we choose "s" as the parameter for the curve. Then if we change the parameter by ds, then we have traversed a length of "ds" along the durve. Now, we can also label this change length as |dr|, the magnitude of the change in position vector. Thus |dr| = ds, so |dr/ds| = 1, meaning that the magnitude of the vector dr/ds is 1. The main takeaway is that "s" is a parameter which represents length traversed along the curve. Everything else is a consequence of this.
@fardin5059
@fardin5059 3 жыл бұрын
You are doing this well. maybe another Walter Luwin you are. thank you.
@ianfowler9340
@ianfowler9340 3 жыл бұрын
I love your videos TB! A question. Why not define curvature as a vector? Since dT/dt is perp. to v then dT/dt will give a direction vector which will point to the centre of the circle of curvature. Knowing that r = 1/k and having a vector pointing from the point of contact to the center, we could easily find the centre of curvature. Just a thought.
@DrTrefor
@DrTrefor 3 жыл бұрын
Nothing wrong with that, but I think we commonly care what the curvature is (example designing roads that can’t have too high a curvature at a given speed limit) without caring precisely how that curve is oriented, ie that direction part isn’t useful. So nice to have a scalar quantity separately
@erockbrox8484
@erockbrox8484 2 жыл бұрын
What about the curvature for a sphere? Is that the same formula of that of a circle? Also what about curvature in higher dimensions?
@duckymomo7935
@duckymomo7935 4 жыл бұрын
are we doing differential geometry now?
@sarithamithun3640
@sarithamithun3640 2 жыл бұрын
Nice demo.does it have long curve
@arghyasantra4300
@arghyasantra4300 4 ай бұрын
I learned the term 'signed curvature ' in the book by A.Pressley. Is the concept of curvature and signed curvature same???
@learnmath4268
@learnmath4268 3 жыл бұрын
Sir make videos on differential geometry
@jamesking2439
@jamesking2439 2 жыл бұрын
I think the derivative of T is normal to the curve because T has constant length, meaning the tangential component of its derivative is zero.
@user-ww1rr2up2i
@user-ww1rr2up2i 5 ай бұрын
I'm assuming curvature can be negative. Am I wrong? Also, if curvature is negative, what does this say about my curve at the point?
@thecarlostheory
@thecarlostheory 2 жыл бұрын
how can we prove the formula for the kurvature kappa?
@muhammadabuzarjanjoa4174
@muhammadabuzarjanjoa4174 3 жыл бұрын
Very informative video. Can you make a video on "oscillating sphere"?
@armaniqbal4712
@armaniqbal4712 11 ай бұрын
Yes sir plz .....your animated videos makes things understandable..
@pacchutubu
@pacchutubu 3 жыл бұрын
This is really good. For first time I came across, where someone say that derivative of unit tangent w.r.t 's' is indeed normal to tangent, needs proof. Everywhere else they take it for granted, like it is obvious, and give and example for circle. Can you please point to some source which gives this proof? You mention this @5:50 Thanks
@pacchutubu
@pacchutubu 3 жыл бұрын
I think, I got it. Unit vectors always move in circular way. So its derivative is normal.
@vaibhavdlxit1050
@vaibhavdlxit1050 3 жыл бұрын
Just take a vector product of B and dT/dS. Hence show that B and dT/dS are perpendicular to each other. Hence you will have B and T and N -all of them to be mutually orthogonal! Thus dT/dS becomes Principal normal vector!
@omermuharremyagcoglu6398
@omermuharremyagcoglu6398 3 жыл бұрын
Sir, What is the curve on your screen..I wonder this.
@atarsingh8214
@atarsingh8214 3 жыл бұрын
Sir can you derive the derivation for unit normal vector and explain that why unit normal vector is derivative of tangent vector please🙏
@papkenhartunian186
@papkenhartunian186 5 ай бұрын
Is a line a circle with a radius of infinity?
@user-xl9zc5fg2b
@user-xl9zc5fg2b Жыл бұрын
Can a curve in the plane transition from having curvature of 0 ( ie a straight line) to some non-zero curvature in a continuous manner? If we consider the circle of curvature for the straight line, it has a radius of infinite length. Whereas any section of the curve with non-zero curvature has a circle of curvature with a radius of finite length. How does one go from the infinite radius to finite radius without a jump in value?
@luminousvalentine8011
@luminousvalentine8011 4 ай бұрын
I think your curve would be a piece wise function, in that case
@emonsinghrana8363
@emonsinghrana8363 3 жыл бұрын
r(t)... Whether that t is angle or time!??? Sin (t) for angle and ds/dt=v for time?? I am confused
@okoyoso
@okoyoso 11 ай бұрын
Is there any name for a surface where you glue all of the circles of curvature together? For the trefoil knot curve, it would look like three hourglasses glued together.
@arockdurai5110
@arockdurai5110 4 жыл бұрын
Could you please do a video on Geodesic?
@nesslange1833
@nesslange1833 3 жыл бұрын
I wonder: what are the units of all this stuff? Concerning the curvature: is it s/m...? Apparently it has to be 1/m so it matches the radius, but on the other hand 1/v results to s/m...? Also I had have been more happy to see a connection between dT/dt = acceleration and unit normal vector. After all such a great video tho!!!
@abhayranjith
@abhayranjith 3 жыл бұрын
You got your answer for the units in your own comment I see 😂
@alexmaciver-redwood3081
@alexmaciver-redwood3081 2 жыл бұрын
why is ds/dt the magnitude of velocity an not the velocity. Why does it lose its ability to have a direction when differentiated?
@IqbalKhan-uo8kg
@IqbalKhan-uo8kg Жыл бұрын
Sir,plz give any example of osculating geometry like osculating plane,or circle in daily life . Example of in human body like how how tumor growth show geometry.
@ianfowler9340
@ianfowler9340 3 жыл бұрын
For a surface, S, F(x,y,z) = 0 in R^3, is there such a thing as the "sphere" of curvature which would be a sphere that is tangent to the surface on the convcave (or maybe convex) side of the surface and therefore there would exist a centre of curvature and the curvature of the surface at the point (x,y,z) on the surface. And if we we use such a thing, how do we define it? Maybe it's easier to express the surface in vector form: R(s,t) = ? Is the curvature of a sphere still 1/R ?
@DrTrefor
@DrTrefor 3 жыл бұрын
Unfortunately it’s a little more complex for surfaces because the curvature can be different in different directions so aligning to a sphere not appropriate
@Han-ve8uh
@Han-ve8uh 3 жыл бұрын
​@@DrTrefor Aren't curvature different in different direction for 2d too? Like it always gets more, or less curvy as we go along, rather than stay the same, so even.a circle never perfectly aligns.
@ianfowler9340
@ianfowler9340 Жыл бұрын
@@Han-ve8uh There is a unique circle for any point on a curve in 2-space called the "circle of curvature", it's center called the "center of curvature" and its radius called the "radius of curvature". So given a curve y = f(x) and a point P(a,b) on that curve, calculate the curvature, k, at that point. The curvature of a circle is constant (= 1/R) so a circle with radius = 1/k will have the same curvature as the curve at P(a,b). Now imagine moving this circle so that it is now tangent to the curve at P(a,b). There are 2 possible circles - each opposite the other at P(a,b). Choose the circle for which the curve is concave toward the center and ignore the other. We now have a unique "circle of curvature" with a unique "center of curvature" that has the same curvature as the curve at P(a,b) and is tangent to the curve at P(a,b). So why is this important? As you zoom in further and further on the point on contact , the circle of curvature becomes a better and better approximation of the curve itself - better than the tangent. Next, how do you find the center of this circle (m,n)? (1) We can find the equation of the radius at P(a,b) by using the fact that the slope of this radius is perpendicular to the slope of the tangent (which is easy to find at P(a,b)) and using P(a,b) as a point on the radius. Note that (m,n) lies on the radius. We also have the equation of the circle which is (x-m)^2 + (y-n)^2 = (1/k)^2. (2) Solve the linear/quadratic system to find m and n. Out of the 2 solutions pick the one that the curve is concave toward. Try y = x^3 at P(1,1). You should get r = 5*sqrt(10)/3 and (m,n) = (-4,8/3) -- discard (6,-2/3) -- and (x+4)^2 + (y-8/3)^2 = 250/9 as the circle of curvature. Newton had this bang on in Fluents and Fluxions.
@maedehsanei7368
@maedehsanei7368 Жыл бұрын
rate of the change of tangent vector =curveture
@nafeesalam7929
@nafeesalam7929 3 жыл бұрын
Sir i want to know , how curvature is large when circle is small, please explain me
@DrTrefor
@DrTrefor 3 жыл бұрын
Imagine driving down a very tight curve on a road vs a very big curve. WHich feels more g-forces? The smaller it is, the more it is curving per unit of arclength.
@tomatrix7525
@tomatrix7525 3 жыл бұрын
Also, look at 4:34 . See how a is the radius. If you increase a, given 1/a is instantaneous curvature here, then the value gets smaller, so a larger R gives smaller curvature, the opposite also applies.
@huevosxklfados1098
@huevosxklfados1098 Жыл бұрын
wht does ds means, what is ds??
@kepler22abc
@kepler22abc Жыл бұрын
I think the proof of this will be: since the unit tangent vector is constant in length, then Dt/ds should be normal to T
@kadiryaman7815
@kadiryaman7815 4 жыл бұрын
you can upload videos in order.
@toomanycharacter
@toomanycharacter Жыл бұрын
5:49 Well, I have a pretty easy visual explanation as to why the derivative of a unit vector is going to be perpendicular to the vector. First, define vector T as a function of t, so really T = T(t) Take an increasingly small dt, and consider what happens to T(t + dt) - T(t) T(t + dt) and T(t) are both unit vectors. Define a triangle ABC, where AB = T(t), AC = T(t + dt), BC = T(t + dt) - T(t). We'd like to find angle ABC. AB = AC = |T| = 1, so angle ABC = angle ACB = 90 - (angle BAC)/2. Notice that as dt becomes smaller and smaller, angle BAC must become smaller and smaller as well, since vectors T(t) and T(t + dt) must become closer and closer to each other (and eventually become same vector). So as dt becomes infinitesimally small, angle BAC becomes infinitesimally small, so angle ABC becomes 90. So BC = T(t + dt) - T(t) is perpendicular to AB = T(t). Therefore dT/dt = (T(t + dt) - T(t))/dt must also be perpendicular to T(t), since scalar multiplication of a vector does not change the fact that it is perpendicular to another vector.
@user-kc2nu6ue9f
@user-kc2nu6ue9f 4 ай бұрын
Oh, so the curvature of curve having infinite r is 0 means it is a line or plane on what we see either 2d or 3d means the Euclid geometry is valid in that condition where r is infinite
@ishraqhossain1235
@ishraqhossain1235 4 жыл бұрын
can someone please add the english subtitle
@henrymiller5709
@henrymiller5709 3 жыл бұрын
from the derivation, v= ds/dt, but in the first example, you use v=dr/dt to calculate v, why???
@DrTrefor
@DrTrefor 3 жыл бұрын
Careful, |v|=ds/dt, hence he difference. I normally define v=dr/dt and then derive that |v|=ds/dt from the definition of arclength.
@henrymiller5709
@henrymiller5709 3 жыл бұрын
@@DrTrefor thx, I got it. Since the arc length s(t) is the integral of ||dr/dt||, the differentiation of s(t), i.e. ds/dt is just the differentiation of the integration of ||dr/dt||, which is just ||dr/dt||, so we can say ds/dt=||dr/dt||, i.e. ||v||. Am I right?
@vaibhavdlxit1050
@vaibhavdlxit1050 3 жыл бұрын
@@henrymiller5709 Yeah right! What you have just showed is the fact: Magnitude of Velocity is speed!!
@bingusiswatching6335
@bingusiswatching6335 2 жыл бұрын
Why can't you take the average of the second derivative?
@culturecreating
@culturecreating 2 ай бұрын
@5:53 how to prove that its indeed a normal? anyone with the short answer.
@culturecreating
@culturecreating 2 ай бұрын
there is already few comments on this, thanks
@rumansaha6910
@rumansaha6910 3 жыл бұрын
Sound is not clear
@bowenchamp7162
@bowenchamp7162 3 жыл бұрын
Now I can tell my gym bros how curvy their dumptrucks are.
@user-vo6oq1bv8x
@user-vo6oq1bv8x Жыл бұрын
Only one question: if we have parallel displacement AROUND THE (INFINITESIMAL) LOOP - TRIANGLE, - YES, ONLY TRIANGLE, OUR TANGENT VECTOR DO NOT CHANGE! SO , GENERALLY, THE DEFINITION OF CURVATURE OF LINE IS NOT THERE.
@fibroidss1194
@fibroidss1194 4 күн бұрын
I actually didn’t understand why you wrote dT/ds as dT/dt*dt/ds actually s is a function of t since it’s a parameter defined by an integral function t is NOT a function of s so I don’t get how you would just introduce the variable t and say that it’s by the chain rule since the chain rule applies if the variable that you are integrating with respect to is actually a function of another variable
@MrArtod
@MrArtod 3 жыл бұрын
I don't like that you show formulas without providing where from they derive. When watching your videos, on each new statement I have to go to google and look for an explanation :(
@DrTrefor
@DrTrefor 3 жыл бұрын
It is a balance. Some of my videos are more theory, some emphasize the geometry and intuition, some purely calculation based. When I teach a full semester long course, it is a balance to introduce all of these into math and I choose what to emphasize in part based on the pedagogical value for each topic...i.e. which topics are best to show derivations for, which get cumbersome and unhelpful?
@mindloop4070
@mindloop4070 3 жыл бұрын
the equation is wrong ...u must mention omega t infront of cos and sin as we have to assume that angular velocity here is 1 rad per sec then only this equation fits correctly..otherwise magnitude of velocity is a times omega if omega not equal to 1
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