How to find the circle of curvature (osculating circle) of y=x^2 at (1,1)

We will write the equation for the circle of curvature (i.e. osculating circle) of y=x^2 at (1,1).
The curvature of y=x^2 at (1,1): • How to find the curvat...
6:24 How to solve that quadratic equation: • Solving a quadratic eq...
100 Series: • 100 series convergence...
Get the files to the review live stream: / 104031457
Get converge? t-shirt: a.co/d/03Lqw5s
-----------------------------
Support this channel and get my calculus notes on Patreon: 👉
/ blackpenredpen
Get the coolest math shirts from my Amazon store 👉 amzn.to/3qBeuw6
-----------------------------
#calculus #bprpcalculus #apcalculus #tutorial #math

[@bprpcalculusbasics]

6:24 How to solve that quadratic equation: kzbin.info/www/bejne/paiVlnZnprKdqZY

[@Gremriel]

I admire your skill in drawing circles.

[@Patrik6920]

kzbin.info/www/bejne/hYjPpaeQh9uEZrM

[@ianfowler9340]

Another cool observation. When you plot y = x^2 and the Circle of Curvature with graphing software - look at how well the circle approximates the curve in the neighbourhood of (1,1) Way, way better than the tangent does. Think of the implications for Euler's Method and calculating x_n+1.

[@ianfowler9340]

Bang on ! Well done. Often when we find the accealleration vector (I'll denote that upper case A) we break A down into 2 vertical components - one horizontal (A_x) and one vertical (A_y). In the physics of motion, we instead often break A up into 2 other vertical components - one in the direction of the tangent (A_t) and one perpendicular to the tangent (A_n) Remember that if a point mass is moving on a curved path it MUST be experiencing an net Force and hence a net accelleration. It's this net Force, and hence A_n, that actually makes the point move on a "curved" path. A_n is soley responsible for the the push that we feel when are turning. So P is the point of contact for (1) the curve and (2) the Circle of Curvature (and (3) the tangent). C is the Center of Curvature. Then the "vector PC is pointing in the direction of A_n". Imagine P as moving along the curved path with a changing position vector R(t), velocity Vector V(t) and accelleration Vector A(t). A_t accounts for the scalar change of speed along the curve while A_n accounts for the turning action and hence the centripetal force (accelaeration) actng at P. Newton knew this and it's why he spent so much time in Fluxions and Fluents on curvature. Knowing the center of curvature tells us how k is related to A_n. For example, if you are building a highway with a curve path and you know the speed of the cars moving along the highway. It is important to know the radius and center of curvature so as not to make magnitude of A_n too high - or the cars may flip or skid off the highway.

[@tomascosta9237]

That's very nice, but I feel like it's easier to find (h, k) using vectors, since you already know the norm (the radius) and a colinear vector ((2, -1), because the slope is -1/2).

[@jesusthroughmary]

It's a calculus course, not a linear algebra class

[@herbie_the_hillbillie_goat]

You THINK it's easier. You don't FEEL it.

[@joshuafenner8867]

​​@@herbie_the_hillbillie_goat to "feel like" something is a colloquialism that means to think a certain way

[@eduardofrancisquini249]

Thanks, You were the first one to explain clearly how to find the circumference center

[@Qermaq]

It's clear to me that the center must be on the perpendicular to the tangent. I am supposing that the radius is the largest possible radius that is centered on that line, intersects (1, 1) and does not overlap y = x^2 on either side of (1, 1) - do I have this correct?

[@josevidal354]

Some years ago, I found the formulas to find the equation of a circle tangent to any point in the curve y=f(x). Given (y-a)²+(x-b)²=r², and that we want the circle in the point (u,f(u)), then: r = (1+f'(u)²)^(3/2) / f''(u) a = (1+f'(u)²)/f''(u) + f(u) b = u - f'(u)(1+f'(u)²)/f''(u) Crazy thing is that I didnt know what Curvature is. I did the limit of h->0 for finding the circle in the points at x=u and x=u±h.

[@a.kofficial6140]

Bro you got my respect to find that years ago❤

[@tedhorton6315]

I don't think we can state "h has to be negative" without looking at a couple of other details, but we can say that "h has to be less than 1" based on the fact we know it lies to the left of the point (1,1). Then we know the negative answer is the one that makes sense, because the positive answer is greater than 1.

[@Patrik6920]

..it actually has to be on the left of x=0 in the coordinate sytem so, it has to be negative (x+h)² + (y+k)² = r² can be rewitten as 0= ( r)² - (x+h)² - (y+k)² -> 0 = r² + ( -(x) +(-h) ) + ( -(y) + (-k) )² [both x and y goes between 0 and ±1, x²+y²=1] 0 = x²+y² - 1 -> (x+a)² + (y+b)² = (c(1))² we usually write c(1) as r, in triangles we use a, b, c x has a max of 1 at cos(0π), and a min of -1 at cos(π).. y is zero at x max/min, y is max(1)/min(-1) when x=0 [ (h)² = r² + (±y-k)² ] think of it as polar coordinates (if it helps)

[@hallax]

If we don't take the absolute value when calculating the curvature then we would know in which direction (up or down) the circle curves and we wouldn't have to get (h, k) using the picture, but instead see which one of those centers gives the curvature needed (one would give + one -)

[@jamescollier3]

That escalated smoothly lol😅

[@lynnjones5587]

Why choice of 'a' instead of 'r' for radius?

[@justsaadunoyeah1234]

Bprp! Please see this comment! I have a question, what happens if we use transfinite numbers in normal everyday calculus? For example using omega, epsilon zero, zeta one, or eta zero, or maybe even SVO or LVO?

[@nizamkanchon3311]

I would be really grateful if u could provide me the pdf of aops introduction to algebra solutions manual

[@Ninja20704]

Is there a way to deduce which h value to take without the picture? Because we cannot always assume that the drawing/picture is accurate to scale, so we still need to justify it mathematically.

[@Momie_et_Masque]

I think that if you choose h=6 you will get the coordinates of the center of the circle that is symmetric to the one drawn relative to the tangent line, as they both have the same curvature and pass through (1,1) and are tangent to y=x². (Both circles have the same curvature because it's the absolute values of opposite values)

[@cyrusyeung8096]

I think you can take the second derivative, and see that y'' = 2 > 0. Since the curve is convex (concaves upward), (h, k) must lie above (1, 1). Therefore, we choose h = - 4 such that k = 7/2 > 1

[@z000ey]

@@Momie_et_Masque this. Both are correct, the other "h" is just for a circle symmetrical over the tangent, and thus you HAVE to look at the picture to choose the one required by the question. The scale or quality of the drawing/picture is unimportant, what matters is that h1 is to the left of the tangent, and the h2 is to the right, and in this case you choose the left one.

[@hassanhassane3663]

By computing distance center(h,k) to point(1,1) and must be equal to radius a

[@z000ey]

@@hassanhassane3663 you have a center of a circle that also has the same curvature and same tangent, but on the right side of that tangent, and the center is at (6, -3/2), thus the equation of that other circle is (x-6)^2+(y+3/2)^2=125/4

[@cdkw2]

Dude I respect you so much but you really saw that circle and was like that's ok. Fix the circle please 😭

[@silendil]

as I understand, all centers of curvature can be connected to continuous curve. Does it exist an equation of this curve? For example, for this y=x^2?

[@FedericoRulli]

Yes! It's the evolute of a parabola. The first example on the wikipedia page is exactly this one. en.m.wikipedia.org/wiki/Evolute

[@iAzazelHD]

And why is the 6 a solution? will it be the circle outside the parabola?

[@sawyerg8115]

it would probably be the x value of the center of the circle on the opposite side of the parabola yeah

[@jesusthroughmary]

Black pen red pen blue pen green pen yay