If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Multivariable calculus

### Course: Multivariable calculus>Unit 2

Lesson 6: Curvature

# Curvature 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.

## Want to join the conversation?

• at grant wrote ||dT/ds||=||dT/dt||/||ds/ds|| instead he should worte ||dT/ds||=||dT/dt||/||ds/dt||
• At , 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"?
• At , 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'|| thusdT/dt = d(s' / ||s'||) / dt    = (s'' . ||s'||  - s' . ||s''||) / ||s'||²``

which has little to do with `s'(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
• 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 at , 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'|.
(1 vote)
• At , the "another formula for curvature" is incorrect? Curvature kappa (Greek k) is a magnitude/scalar, but the formula at shows 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 at is 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
• 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 at to be correct.
• Why do we scale down the derivative vector with ||S'|| and not ||S"|| at ?
• 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 at , 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'|.
• At , 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?
• 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
• At , Grant meant to right ds/dt, right? Not ds/ds?