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Course: Multivariable calculus > Unit 2
Lesson 6: CurvatureCurvature formula, part 5
Here, we finish the intuition for how the curvature relates to the cross product between the first two derivatives of a parametric function. Created by Grant Sanderson.
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at5:31grant wrote ||dT/ds||=||dT/dt||/||ds/ds|| instead he should worte ||dT/ds||=||dT/dt||/||ds/dt||(87 votes)
At5:13, isn't there a mistake, when the narrator writes "d s (arrow)/d s" in the denominator, when it should be "d s (arrow)/ dt"?(29 votes)
At4:45, why is it that:s'(t) / ||s'(t)|| x s''(t) / ||s'(t)|| = dT / dt
?
I don't get where this equality derives from.
If I start back from the definition of T, I get (I'll omit the(t)'s for simplicity):T = s' / ||s'|| thus
dT/dt = d(s' / ||s'||) / dt
= (s'' . ||s'|| - s' . ||s''||) / ||s'||²
which has little to do withs'(t) / ||s'(t)|| x s''(t) / ||s'(t)||
For one thing, where has the term||s''||gone? How about the sum in the numerator?
Thanks(23 votes)
we first defined s(t) = xi + yj right? From that T turns out to be {1/(sqrt(x'(t)^2+y'(t)^2)}*(x'(t)i+y'(t)j) from which you can confirm s'(t) / ||s'(t)|| x s''(t) / ||s'(t)|| = dT / dt.
If you see the videos again it implicitly shows.(2 votes)
- At6:09, the "another formula for curvature" is incorrect? Curvature kappa (Greek k) is a magnitude/scalar, but the formula at6:09shows a cross-product divided by a magnitude, which is a scaled vector. Throughout this video and the last video, the narrator repeatedly says "cross-product", but the cross-product is a vector quantity, not a pure magnitude.
The narrator attempts to match the numerator of the curvature with the appearance of the expression of the cross-product, but the narrator does not clarify that it is NOT actually a cross-product, which is a vector. The numerator of the curvature is NOT a vector.
Although the cross-product and the "two-dimensional" curl relate how perpendicular two entities are, they are still vectors. Admittedly, this video and the last are informal intuition-type explanations, but that formula for curvature at6:09is incorrect since it is a vector, while curvature is a magnitude/scalar.
By the way, the numerator of the curvature formula would be more accurately described as the Wronskian of the components x'(t), y'(t) of the vector s'(t), which measures linear independence of x', y'. Perhaps discussing cross-product (with curl in the background) is more intuitive of curvature and what the narrator is attempting to intuitively explain, compared to using the Wronskian and the linear independence of the components of the vector s'(t)
See: https://en.wikipedia.org/wiki/Wronskian(16 votes)
When I first saw this and previous videos, I was very confused about a cross product of two-dimensional vectors because as far as I know the cross product is defined only in R^3. But after some reflection I came up with an interesting way of thinking about this. What if we imagine that our position vector S(t) has 3 components, with z-component always equal to zero? Seems completely legitimate to do this. Then the cross product of S'(t) and S''(t) will be equal to another 3 dimensional vector [ 0, 0 , x'(t)y''(t) - y'(t)x''(t) ] which is perpendicular to xy plane and has a length that equal to the value of his 3rd component. This is exactly numerator in our formula and this length shows how perpendicular the 1st and 2nd-derivatives vectors are.
So I agree with @hemantjoshi that we should take the length of the whole cross-product to make the formula at6:09to be correct.(6 votes)
Why do we scale down the derivative vector with ||S'|| and not ||S"|| at3:15?(12 votes)
I was wondering the same thing and its rare that Grant's vids leave such a big hole in terms of intuition. From what I can understand about the graph he is talking about at1:10, if the r' vectors are of unequal length, then the second derivative would have to take into account the different magnitudes and directions of r'. But as Grant said, we don't want to account for different magnitudes of r'. So, if the pink vectors drawn just represented r'/|r'|, then the blue second derivative vectors would purely measure change in direction, which is what we want. So we are really differentiating r'/|r'| to get second derivative r''/|r'|. (Now a possible mathematical flaw in this intuition is that when you do differentiate r'/|r'|, it might not be r''/|r'| )
Another thing I was wondering was why didn't we just use the normalized version of r"? If you consider 2 upside down parabolas, 1 being very broad (lets say y=-x^2 ) and other being very narrow (y=-100x^2). Both graphs curve in the same direction and if you consider the uniform circular motion parallels, the acceleration vector or the second derivative would point in the same direction for both graphs. However, the narrow one has a relatively sharper curve and hence greater second derivative magnitude. Since its second derivative is larger, then its curvature must be greater as well which makes sense. However, if we had normalized or made second derivative a unit vector, then this fact might have been overlooked.
Now I must say you should not take these explanations at face value because I literally came up with this stuff and I am just a regular KA kid going through the playlist ,having no experience with multi var. calculus. But I hope this provides some sense as to why Grant happened to divide the s'' vector by |s'|.(5 votes)
At5:13, Grant meant to right ds/dt, right? Not ds/ds?(7 votes)
At1:39, why do we need a measure of how perpendicular the first and second derivative are to each other? If we are measuring how much the first derivative is turning, shouldn't the largest value be when the second and first derivative (put head-to-tail as in the video) make a very acute angle, and the smallest when they make a very obtuse angle? With the perpendicular case just being the middle way, indicating "moderate" rotation?(3 votes)- No, if you break the second derivative vector down into components that are parallel and perpendicular to the first derivative, the component that is parallel corresponds to a growing or shrinking of the first derivative and only the component perpendicular to the first derivative will correspond to turning. The second derivative has to be scaled according to the length of the first derivative though.
Hope this helps(8 votes)
This is a pretty nice interpretation, but I think there's a bit of missing explanation regarding the original dT/ds thing itself, and why the whole thing is related to circles being drawn at that point -- you can justify this from pointing out that dT/dt represents the change in the direction of velocity/movement over time. If the thing were moving in uniform circular motion, that would just be the rate of change of velocity over time, divided by the magnitude of the velocity vector (the speed), i.e. dT/dt = a/|v|. ds/dt is just the speed, so dT/ds = (dT/dt)/(ds/dt) = a/|v|^2, and since we know that a = v^2/R, this is just 1/R.(5 votes)- That was what I was building in the back of my mind while watching this playlist! It is actually kinda fun if you're already comfortable with the idea of circular motion.
Actually, it doesn't even have to be uniform, the tangential acceleration would correspond here, to increasing the speed!(2 votes)
How can you compute curvature (or can you) for a normal function ie. y=x^2? Does the function have to be parametrized? Does it have to be a vector?(2 votes)
Im not quite sure I understand why when normalising, we don't normalise by the norme of ||S''|| ? Why is it we just need to normalise by ||S'|| ?(2 votes)
5:31Video transcript
- [Voiceover] So, let's
sum up where we are so far. We're looking at this formula and trying to understand why
it corresponds to curvature. Why it tells you how much
a curve actually curves. And the first thing we did is
we noticed that this numerator corresponds to a certain cross-product. The cross-product between
the first derivative and the second derivative of the function parameterizing the curve. And the way we started to understand that is we say, well, the function
parameterizing the curve s of t produces vectors
whose tips trace out that curve itself. And now if you think about
how one tip moves to the next the direction that it needs to go for that tip to move to the next one, that's what the first
derivative tells you. And when you treat that
in an infinitesimal way this is why you get tangent
vectors along the curve and there's an entire
series of videos on that for the derivative of a
position vector valued function, and they explain why the first derivative of a parametric function, gives you tangent vectors to that curve, but if we draw all of
those tangent vectors just in their own space, their own little s prime of t space, you get all of these vectors, and we're rooting them at the same point to be able to relate them more easily. The way that you move from
the tip of one of those to the next one, is given
by the second derivative, it kind of plays the same
roll for the first derivative as the first derivative plays for the original parametric function. And specifically if
you have a circumstance where the tangent
vectors are just turning, the only thing they're doing
is purely turning around, which is when your curve
is actually curving, this corresponds to a case when the second derivative function
is pretty much perpendicular as vector. The vector it produces is perpendicular to the first derivative vector. So, this is loosely why the cross-product is kind of a good measurement of curving cause it tells you how
perpendicular these guys are. But, there is a bit of a catch. The original formula for curvature, the whole reason we're doing
it with respect to arc length and not with respect to the parameter t, is that curvature
doesn't really care about how you parameterize the function. If you imagine zipping
along it really quickly, so your first derivative
vectors are all super long, it shouldn't matter, compared to crawling
along it like a turtle. The curvature should
just always be the same. But, this is a problem
if you think back to the cross-product that
we're now looking at, where you're taking the
cross-product between the first derivative and
the second derivative, because if you were
traveling along this curve twice as quickly, what that would mean is your first derivative vector, so I'll kind of draw it again over here, would be twice as long, to indicate that your going twice as fast. And similarly your
second derivative vector, to kind of keep up with
that changing rate, would also be twice as long. And as a result, the
parallelogram that they trace out, I should actually, kinda
going off screen here. The parallelogram that they trace out would be actually four times as big. Right? Because both of the vectors get scaled up. So, the way that we
really want to be thinking about this is not the tangent
vector due to the derivative, but normalizing this. And this should kind of make sense, because we are thinking in
terms of unit tangent vectors for the curvature as a whole. So, if you imagine instead
kind of cutting of the vector to make sure that it's got a unit length, a length of one. And what that means is, you're taking the derivative vector and dividing it by it's own magnitude, by the magnitude of
that derivative vector, and then similarly we'll want
to scale everything else down, so you're taking this
and kinda scaling it down by what you need. The resulting parallelogram they trace out is a more pure measurement of
how perpendicular they are, without caring how long they are. And with this guy would
be, by the way then, this is the second derivative vector not normalized with respect to itself, but we're still dividing, you know the thing we're diving by as we scale everything down, is still just the size of
that first derivative vector. So, this cross-product, if
we take the cross-product between s prime
normalized, s double prime, oh no, no, no, sorry, s prime itself, and I should be saying
vectors for all of these, these are all vectors. If we take the cross-product between that and s double prime, scaled down by that same
value that's still s prime, so it's not normalized, this
is just scaled down by s prime. This here is a more pure
measurement of how perpendicular the second derivative vector is to first. And the reason that we
don't really care about the second derivative being normalized is, if it was the case, that you know, the second derivative
was really, really strong and wasn't necessarily a unit vector, that's fine that's just telling us that the tangent vector turns much more quickly and the curvature should be higher. And in fact, it turns out
that this whole expression is the derivative of the
unit tangent vector t, that unit tangent vector
that I've talked about a lot, with respect to the parameter t. So, whatever the parameter
of your original function is. And now, if you think back to the, I'm not sure if it was the last video or the one before that, but I talked about how, when you take, when we're looking for this derivative of the tangent vector, with
respect to arc length, the way that you compute this is to first take its derivative
with respect to the parameter, which is something we can actually do, cause everything is expressed
in terms of that parameter, and then dividing it by the, basically, the change in arc length with respect to that parameter, which is the size of that
first derivative function. So, if this whole thing is the derivative of the tangent vector with respect to t, what that means is when we take this, and we divide that whole thing by, the derivative, by s prime, that should give us curvature, and in fact that's just worth
writing in its own here. That's curvature. Curvature is equal to, and what I'm going to do
is, I'm going to take, since we three different times, we see the magnitude of s prime here, magnitude of s prime here, magnitude of s prime here. Since we see that three times, I'm going to go over here, and I'm going to put
that on the denominator but cubing it, so s prime, that derivative vector, the magnitude of that
derivative vector cubed. And then on the top,
we still have s prime, that vector s prime cross-product
with s double prime, that vector. And this, I mean, you could
think of this as yet another formula for curvature, I think I've given you
like four at this point. Or, you could think of it as
just kind of the same thing, and if we look back up
at our original one, that I was trying to justify. This is just the spelled
out version of it, because what is this
bottom component here? If we take x prime squared
plus y prime squared, and if we think of the
square root of that, so kinda taking it to the one half power, that would be the magnitude
of the derivative, and I kind of showed that in
some of the previous videos. And what we're doing is we're cubing that, so that whole formula
that's the very explicit, you know, in terms of x and y's, what's going on is really
just expressing this idea, you take the cross-product
between the first derivative and the second derivative, and then because you're normalizing it, you know normalizing with
respect to the first derivative, you want to scale down
the second derivative by that same amount, just so the parallelogram we're thinking, kinda shrinks and everything
stays in proportion. And then once again we're
dividing about that s prime, basically because
curvature is supposed to be with respect to s, and
not with respect to t. So, that's a way of, kinda
getting a correction factor for how wrong you're going
to be if you just think in terms of the parameter
t instead of steps in terms of the arc length. So, hopefully this makes
the original formula a little bit less random, seeming, you know, in this two dimensional case. And it also gives another
strong conceptual tool for understanding yet another
way that you can think about how much a curve itself actually curves. And I think I've probably
done enough videos here going through all the different formulas, for what curvatures should be. And in the next one or
two I'll go through some specific examples, just
to see what it looks like to compute that.