Hello Mr, do you have a video about Newton-Raphson method which is one of the ways to get the root of a complicated equation. The output does not have to be a number, but it can also be an expression. Because the other day I talk to AI how do I convert a single parameter function into a function that relies on two dependent variables. This is because I have a Bezier equation which is dependent on parameter t which can also be expressed as f(x,y). But to do this it says I need to use Newton-Raphson method which is also an iterative method and is one of the numerical method to find the root of an equation. I want to do this, because then I want to apply Green's theorem to find the area of a closed curvature. From there maybe I can find the volume by applying triple integral with parameter z. From volume, I can find mass, if I know the density. From, mass I can find force. From force, I can find Power and Work done and also momentum. If I know all that I can create a football game. Woo wee. Haha lol.
Here's a fun problem I cooked up. Find the differential equation that must be satisfied to find the point on y = f(x) where the curvature is a maximum. It involves only y ' , y ' ' and y ' ' '. Next, use this DE to find the exact x-value where the curve f(x) = x^3 - 3x reaches maximum curvature. Very cool answer!
Same! I always wondered but I feel like I have so many questions that have definitely been answered somewhere… but nowhere that I can access! Until now…
Here is a way of developing this formula from scratch - 2D. No vector functions needed. Only basic calculus. 1) Sketch a general curve in the first Quadrant and label a point P(x,y) on that curve. 2) Sketch in a tangent line at P and sketch a horizontal line from P to the right. (// x-axis) 3) You now have defined an angle A between the tangent and the horizontal. 4) The Curvature at P can be defined the following way: Remember, we are trying to obtain a measure of how "bent" the curve is at P independent of it's orientation. As you move along the curve you want to measure how fast A changes per unit of ARC LENGTH. So k = dA/ds. You can use absolute value if you insist on a positive curvature. But negative curvature is O.K. - just indicates concave up or down. 5) Tools: k = dA/ds ds^2 = dx^2 + dy^2 ====> ds/dx = sqrt[ y ' )^2 + 1] y ' = tan(A) =====> sec^2(A) = ( y ' )^2 + 1 dA/dx = [dA/ds)] [ds/dx] y ' = tan(A) y ' ' = sec^2(A) * dA/dx y ' ' = [ (y ' )^2 + 1 ] * [ (dA/ds) * (ds/dx) ] Now solve for dA/ds using ds/dx = sqrt[ (y ' )^2 + 1] and you will get: k = [y '' ] / [ (y ' )^2 + 1] ^ (3/2)] Parametize: x. = dx/dt and y. = dy/dt giving: y ' = y. / x. (Newton's x_dot and y_dot) d( y ' )/dt = [ y..*x. - x..*y.] / [ (x.)^2 AND d(y ' )/dt = [d(y' )dx]*[dx/dt] = (y '')( *x.) Equating, substituting, re-arranging and a little work gives your formula. k = [x.y.. - y.x..] / [x.^2 + y.^2]^(3/2)
Holy cow! After “eating’ all of that ‘Math meal’ my mind feels like my belly would feel after eating 2 large pizza’s’ a bucket of KFC chicken and a liter of coke! I don’t have to ‘eat’ anymore math for 2 days,, 😂😂😂😂😩😩😩😳😳
Here's an example of finding the curvature of y=x^2 at (1, 1)
th-cam.com/video/NUgGbzJcN-s/w-d-xo.html
Hello Mr, do you have a video about Newton-Raphson method which is one of the ways to get the root of a complicated equation.
The output does not have to be a number, but it can also be an expression.
Because the other day I talk to AI how do I convert a single parameter function into a function that relies on two dependent variables.
This is because I have a Bezier equation which is dependent on parameter t which can also be expressed as f(x,y).
But to do this it says I need to use Newton-Raphson method which is also an iterative method and is one of the numerical method to find the root of an equation.
I want to do this, because then I want to apply Green's theorem to find the area of a closed curvature. From there maybe I can find the volume by applying triple integral with parameter z.
From volume, I can find mass, if I know the density. From, mass I can find force. From force, I can find Power and Work done and also momentum.
If I know all that I can create a football game. Woo wee. Haha lol.
If you leave out the absolute value, the sign tells you which way it's curving, to the right or to the left (as the parameter increases).
I think these are some of my favorite topics in calculus ... Thanks. Cheers
Here's a fun problem I cooked up.
Find the differential equation that must be satisfied to find the point on y = f(x) where the curvature is a maximum.
It involves only y ' , y ' ' and y ' ' '. Next, use this DE to find the exact x-value where the curve f(x) = x^3 - 3x reaches maximum curvature. Very cool answer!
Thank you so much! This is the only video I could find for this
Same! I always wondered but I feel like I have so many questions that have definitely been answered somewhere… but nowhere that I can access! Until now…
I was searching for this ❤
Here is a way of developing this formula from scratch - 2D. No vector functions needed. Only basic calculus.
1) Sketch a general curve in the first Quadrant and label a point P(x,y) on that curve.
2) Sketch in a tangent line at P and sketch a horizontal line from P to the right. (// x-axis)
3) You now have defined an angle A between the tangent and the horizontal.
4) The Curvature at P can be defined the following way: Remember, we are trying to obtain a measure of how "bent" the curve is at P independent of it's orientation.
As you move along the curve you want to measure how fast A changes per unit of ARC LENGTH. So k = dA/ds. You can use absolute value if you insist on a positive curvature. But negative curvature is O.K. - just indicates concave up or down.
5) Tools: k = dA/ds
ds^2 = dx^2 + dy^2 ====> ds/dx = sqrt[ y ' )^2 + 1]
y ' = tan(A) =====> sec^2(A) = ( y ' )^2 + 1
dA/dx = [dA/ds)] [ds/dx]
y ' = tan(A)
y ' ' = sec^2(A) * dA/dx
y ' ' = [ (y ' )^2 + 1 ] * [ (dA/ds) * (ds/dx) ]
Now solve for dA/ds using ds/dx = sqrt[ (y ' )^2 + 1] and you will get:
k = [y '' ] / [ (y ' )^2 + 1] ^ (3/2)]
Parametize: x. = dx/dt and y. = dy/dt giving: y ' = y. / x. (Newton's x_dot and y_dot)
d( y ' )/dt = [ y..*x. - x..*y.] / [ (x.)^2 AND
d(y ' )/dt = [d(y' )dx]*[dx/dt] = (y '')( *x.)
Equating, substituting, re-arranging and a little work gives your formula.
k = [x.y.. - y.x..] / [x.^2 + y.^2]^(3/2)
Thank you very much for this wonderful method! And I just recorded a video today too! Thanks.
th-cam.com/video/kYZCT0XOYuY/w-d-xo.html
@@bprpcalculusbasics You are very welcome. Thanks for the shout out. Very nice of you to do that in your new video! Cheers.
That escalated quickly
Can we calculate the curvature to see just how quickly did it escalate?
Thank you! 🤩👍
Might be a case where the “dot” notation for derivative is a little cleaner. Love your videos!
Thanks!
That's a nice new eraser
That's a curved ball to my brain
What field of math is that? Vector calculus?
Amazing. Now do the 3d cruve version hahaha
If its about geometry I am willing to listen.
That's a lot of x's and y's.
Holy cow! After “eating’ all of that ‘Math meal’ my mind feels like my belly would feel after eating 2 large pizza’s’ a bucket of KFC chicken and a liter of coke! I don’t have to ‘eat’ anymore math for 2 days,, 😂😂😂😂😩😩😩😳😳
I remember some guy from your Instagram reel saying that if this is math where are the numbers?
What does numbers have to do with maths though?
numbers are only for computation and calculation?