Formulas you need to know for computing the curvature!
ฝัง
- เผยแพร่เมื่อ 30 ก.ย. 2024
- Here are all the formulas you should know for computing the curvature of a curve. These formulas will help you with your multi-variable calculus class. Some formulas might also help you with differential geometry.
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Me after watching his videos about curvature:
"Einstein, let me calculate this curvature of spacetime..."
This is like the 0th step of understanding GR😢
Fantastic. This really helps with different situations and saving time. Thanks.
(Hope youtube doesn't delete this)
Derivation of last formula:
Dependance tree: R -> s -> t
or x,y,z -> s -> t
R(s(t)) = xi+yj+zk = position vector
R' = dR/dt, x'= dx/dt, y'=...
R'' = dR'/dt = d^2R/dt^2
ds/dt = |R'| (magnitude of velocity, speed, like m/s)
dt/ds = 1/|R'|
d/dt(sqrt(x^2+y^2)) =
2xx'/2sqrt(x^2+y^2) + 2yy'/2sqrt(x^2+y^2)
= (xx'+yy')/sqrt(x^2+y^2)
...generalize!...
d/dt(|R'|) = •R'/|R'| (multivariable chain rule, chain rule, power rule)
T = dR/ds = dR/dt*dt/ds (chain rule)
T = R'/|R'|
lets consider K as a vector for now...
K = d/ds(T) = d/dt(R'/|R'|) * 1/|R'| (chain rule)
K = [R''/|R'| - R'/(|R'|^2) * d|R'|/dt] / |R'|
K = [R'' * |R'|^2 - R' * R''•R'] / [|R'|^4] (same denominator)
K = [R'' * R'•R' - R' * R''•R'] / [|R'|^4] (square of magnitude is the same as dot product with itself)
K = [R' x (R'' x R')] / [|R'|^4] (double cross product identity)
K = - R' x (R' x R'') / [|R'|^4] (swap cross product inputs for a (-) sign)
K = [|R'|^2] * R''_perp / [|R'|^4]
(After using the geometric interpretation of cross product, R''_perp is the part of R'' perpendicular to R')
As it turns out, | |R'| * R''_perp | = |R' x R''| !!! (In 2d and 3d)
So while everything is fine in 2d and 3d for
|R' x R''|, the cross product isnt well defined in other dimensions. In fact, even in 3d,
R' x R'' as a vector is facing the wrong way?? It's not even in the direction of curvature for a parametric 3d line, while R''_perp is.
K = R''_perp / [|R'|^2
R''_perp = R'' - R' * R''•R'/|R'|^2
R''_perp can be defined as R'' - parallel part of R'' (with dot product)
This gives the same formula as before, so nothing to simplify.
It's also possible to get the magnitude of K with the formula |K| = sqrt(K•K).
|K| = sqrt[|R''|^2 * |R'|^2 - (R''•R')^2]/|R'|^3 (n dimensions)
= |R' x R''| / |R'|^3 (in 2d and 3d)
I think it is now time to deal with the next concepts which are very very important to the study of motion along a curved path and how curvature comes into play. It's why Newton spent so much time and effort studying curvature in Fluents and Fluxions. I will use the 2-D case for simplicity and you can extend to a curved path in 3-D easily. These concepts are: (1) Radius of curvature, (2) Center of Curvature which lead directly to (3) the Circle of Curvature.
1) Take the point on the curve y = f(x) where x = a and calculate the curvature at that point. Use any method. We also know that the curvature of a circle is a constant. You can find that constant curvature by applying any of the methods to x^2+y^2 = R^2 to obtain : the "curvature of any circle" = 1/R where R is the radius of the circle ( = 1/R at every point on the circle)
2) Now, imagine a circle that has the SAME curvature as the curvature of the of the curve at x = a. MOVE that circle so that it is tangent to the curve at the point ( a, f(a) ) . There are 2 places where that circle might go. Concave up side and concave down side. Place the circle on the side of y = f(x) where the concavity circle generally matches that of the curve. We can now calculate the radius of that circle by setting 1/R = k(a) giving us the "radius of curvature" R = 1/k(a).
3a) Finding the "centre of curvature" requires a bit of work. Let the centre of our circle be C(h,k) and the point of contact be P(a,f(a)). Join CP which is the raduis, R. Next, sketch in the tangent to the curve at (a,f(a). We can find the slope of the tangent by finding f ' (a). Here's the key point - CP MUST be perpendicular to the tangent so we now know the slope of CP as the negative reciprocal of f '(a).
3b) We can now obtain the "equation of radius line CP". We know the slope of CP = -1/f '(a) and we know the point P(a,f(a)) on CP. Next, know that C(h,k) lies on CP. Substitute h and k into the radius equation. We now have a "linear equation" in h and k". The equation of the circle is : (x-h)^2 + (y=k)^2 = R^2. Substitute the known quantities x = a, y = f(a) and r = 1/k(a) to obtain a "quadratic equation in h and k".
4) Solve the linear/quadratic system to find h and k. You will get solutions. Pick C(h,k) on the side where the curve is concave toward C(h,k)
5) Now for the grand finale: You have all of these equations: Plot the curve, the tangent line, radius CP (the normal) and the circle on you favourite graphing software. It is a beautiful sight to behold.
6) I used f(x) = x^3 at the point (1,1) on the curve. You can check your answers as you go.
k(1) = 3/5*sqrt(5) ====> R = 3*sqrt(5)/3 =====> R^2 = 250/9
Equation of tangent: 3x - y = 2
Equation of radius CP (normal): x + 3y = 4
Quadratic: (1-h)^2 + (1-k)^2 = 250/9
Linear: h = 4 - 3k
(h,k) = ( -4 , 8/3) or ( 6 , -2/3) - we want ( -4 , 8/3 ) , concave up side of y = x^3 at (1,1)
Circle Of Curvature : (x+4)^2 + (y - 8/3)^2 = 250/9
Nice and thank you! Yes I do plan to cover that soon. I actually have planned out an example of y=x^2 at (1,1). I am reading/following the calculus book by Thomas.
@@bprpcalculusbasics Looking forward to it! I just made it up myself.
I took Multivariable & Vector Calculus this past year, and never had to calculate the curvature. Should I know how to do this heading into diff eq/future math courses?
It’s good to know, but doesn’t really come up in diff eqs or further math courses. You might see it applied in further physics or engineering classes though. I don’t know specifically though (I was a chemistry major, so I didn’t have to take courses in modern physics (general relativity) or aerospace engineering (avionics maybe? idk). Those are the only two direct applications I know of off the top of my head
If you’re a math major I’m sure it comes up again, but gets a more thorough treatment. Like studying Gaussian curvature in a 400-level differential geometry course
Good video 👍
Niceeeee
Does the 3D curve method also work for 4D?
I haven't seen the 3d yet but if it is related to matrices then yes
I think example 6 could be expanded to higher dimensions
It would be quite complicated, as we need to use tensors rather than cross products.
@@AlbertTheGamer-gk7snof you derive the formula with the cross product, it's actually composed of dot products, which exist in any dimention! I'll try post a derivation as a comment
A fun thing I came to know about curvature!
As we all know that the first derivative dy/dx can we written as tanθ. If we ever want to write the double derivative in terms of θ as as well we can do some basic calculus and get that the second order derivative d^2y/dx^2 = k sec^3 θ. Where k is the curvature kappa!