Proving the curvature formula for a parametric plane curve

This Calculus 3 tutorial covers the derivation of the formula for finding the curvature of a plane curve. We start with a parametric equation and find the derivative, unit tangent vector and its derivative, and their magnitudes. The process is quite long but the result is very nice! I saw this result on Wikipedia here: en.wikipedia.org/wiki/Curvature
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#calculus #bprpcalculus #apcalculus #tutorial #math

[@bprpcalculusbasics]

Here's an example of finding the curvature of y=x^2 at (1, 1) kzbin.info/www/bejne/hIbKeJWwf8iBY9U

[@abacaabaca8131]

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.

[@tobybartels8426]

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).

[@algorithminc.8850]

I think these are some of my favorite topics in calculus ... Thanks. Cheers

[@jamescollier3]

That escalated quickly

[@sebastiangudino9377]

Can we calculate the curvature to see just how quickly did it escalate?

[@SterlingChaseStubblefield]

Might be a case where the “dot” notation for derivative is a little cleaner. Love your videos!

[@bprpcalculusbasics]

Thanks!

[@mugger8509]

Thank you so much! This is the only video I could find for this

[@nathanoher4865]

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…

[@ianfowler9340]

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!

[@urluberlu2757]

Thank you! 🤩👍

[@hassankhamis77]

I was searching for this ❤

[@Asiago9]

That's a nice new eraser

[@cdkw2]

That's a curved ball to my brain

[@ianfowler9340]

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)

[@bprpcalculusbasics]

Thank you very much for this wonderful method! And I just recorded a video today too! Thanks. kzbin.info/www/bejne/oYq9dIdmjbSMq7s

[@ianfowler9340]

@@bprpcalculusbasics You are very welcome. Thanks for the shout out. Very nice of you to do that in your new video! Cheers.

[@guilhermerocha2832]

Amazing. Now do the 3d cruve version hahaha

[@abacaabaca8131]

If its about geometry I am willing to listen.

[@fsisrael9224]

What field of math is that? Vector calculus?

[@Gremriel]

That's a lot of x's and y's.

[@bernarddoherty4014]

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,, 😂😂😂😂😩😩😩😳😳

[@cdkw8254]

I remember some guy from your Instagram reel saying that if this is math where are the numbers?

[@Alians0108]

What does numbers have to do with maths though?

[@windowsxpmemesandstufflol]

numbers are only for computation and calculation?