I had to back to my book for that one, quote:

"In general, you can can show that if $P_n$ is the nth degree Taylor polynomial of f at c, then the derivatives of $P_n$ at c agree with the derivatives of f at c up to and including those of order n."

Which is what you mean, I believe.

I agree then that the osculating circle would share the same curvature as the second degree Taylor polynomial at that point.

Reference for later: en.wikipedia.org/wiki/Osculating_circle

"Without loss of generality, we can apply an affine transformation (translation and rotation of the plane) to move the point on the curve to the origin so that the tangent vector there is on the positive x-axis; this greatly simplifies the upcoming algebra."

Seems reasonable.

Reference: en.wikipedia.org/?title=Affine_transformation

So just move the graph of the curve so that the point we're interested in is at the origin.

mmhmm

"The tangent line, i.e. the x-axis itself, interpreted as the graph of a (constant) function, already shares the same first derivative as the curve at the origin."

Or move it such at the x-axis becomes the tangent at the origin.

"We need to find a circular arc (we only consider the arc of a whole circle so that technically we are talking about a function, at least near the origin) that shares the same second derivative as the curve."

We need a function that approximates the curve of an osculating circle local to that point.

"Without loss of generality, suppose the curve, interpreted locally as a graph of y=f(x) near (0,0), has a positive second derivative f′′(0)>0. Now place a circle on the plane of radius R tangent to the x-axis at the origin whose center is above the x-axis. The center must be at (0,R). "

So far so good, looks like that picture I posted:

mathwiki.ucdavis.edu/@api/deki/files/702/…

Oh cool, how did you do that?

I could not get the normal picture insertion tags to work (above)

deleted the stuff after .gif in your url

> We need a function that approximates the curve of an osculating circle local to that point.

Oh okay

not exactly sure what you're saying there

A function that would mimic the curve of the circle near P.

we have two things: the curve itself, and a circular arc, both situated nicely at the origin. from there, we're going to describe the circular arc by a function, and then conclude things from there using math

Right.

its graph wouldn't just mimic or approximate the circle, it would be the circle, or at least that arc of it

Right, we wouldn't need the entire circle, just the curve nearest P. We don't necessarily need a function that would be a circle.

Just the curve nearby P

no circle can be the graph of a function, because circles fail the vertical line test

but yeah

Right

You might use the function for a circle as convenience, I just meant wasn't absolutely necessary. ;)

I just meant it wasn't*

You might actually use $P_2$ like we saw above, that would be a good fit.

no, you cannot use the function for a circle, because there is none

that's what my parenthetical comment is noting

Oh okay, so

"we only consider the arc of a whole circle so that technically we are talking about a function, at least near the origin"

yes

only talking about the lower half or so of the circle, and that does pass the vertical line test

Would be a half circle (or some piece) that we could get with a function that would intersect P and be a good local representation of the curve.

Got it. Good. =D

now, that lower half or so of the circle is the graph of $y=R-\sqrt{R^2-x^2}$

Ahhh okay, I was just about to figure out how you derived that. That makes more sense.

if the circle were centered at the origin, it would be defined by $x^2+y^2=R^2$, yielding $y=\sqrt{R^2-x^2}$ for the upper half and $y=-\sqrt{R^2-x^2}$ for the lower half. since our circle touches the origin on its circumference, its center is $R$ units up, and so the lower half's graph is $y=R-\sqrt{R^2-x^2}$

Right, I got that. I just needed context of how you had arrived at that equation.

With the hint of "it's half of a circle" it was immediately apparent where you got it. ;)

"Compute $g′′(0)=\frac{1}{R}$."

doh

Wish we had a preview on that, so I can fix the latex. ><

you can edit messages within a 2 min time window

hover over your message so the "down arrow" on the left appears, and click it for a drop-down menu

among the options will be "edit"

There we go.

That one was pretty simple

Reference:

wolframalpha.com/input/?i=d^2%2Fdx^2+R+-+sqrt%28R^2+-+x^2%29+if+x%3D0

"In order for f and g to have the same second derivative at 0, we must have $f′′(0)=\frac{1}{R}$."

Makes sense. They should have the same second derivative at that point, so it must be ^.

mmhmm

This notation I'm not familiar with (writing)

"Lo and behold... Parameterize $r(t)=(t,f(t))$"

So $r(t)$ is our parameterization equation, fine.

$f(t)$ seems to be the function it's using the parameterize

So $t$ would be the independent variable? In that notation?

yes

a particle at time t is located at (t,f(t))

according to this parametrization

(which is not unit-speed)

Oh okay, just using a "point notation" so to speak. Okay.

vector function, yes

that's how you write down parametrizations

r(t)=(x(t),y(t)), generally

My book is divergent in quite a few ways, I'm guessing. ><

aight, I'm going on a bike ride, you can post more questions here for later

of course.

I wasn't expecting real time help.

This has been really helpful, thank you. =D

"so that $\|\frac{dr}{dt}\| = \sqrt{1 + f'(t)^2}$"

That one is just the magnitude formula:

khanacademy.org/math/precalculus/vectors-precalc/…

So $ \frac{d}{dt}(t) = 1 $ (first term)

And $\frac{d}{dt} (f(t)) = f'(t)$

(second term)

And so the magnitude of any vector of the derivative of $r(t)$ would be as above.

"A tangent vector is given by $(1,f′(t))$"

So any tangent vector should have that form, since this is essentially $\frac{dr}{dt}$

no

every tangent vector is parallel to that one

(which is why we normalize to get the unit tangent vector)

My book skips over introducing unit tangent vectors in any rational way. I need a bit more explanation. I'm not sure what you mean. =/

Wikipedia isn't much help (since they skip over the properties of $r'(t)$ and go straight to the unit tangent vector. en.wikipedia.org/wiki/Tangent_vector

I'm guessing because the unit tangent vector is much more desirable than plain $r'(t)$.

if v is a tangent vector, then any scalar multiple of v is also a tangent vector

Okay, I agree

(Don't let me keep you from your ride, I was just posting for later consideration)

so if you multiply by the reciprocal of v's modulus, you get a new vector with unit size

k

So doing $\frac{v}{\| v \|}$ gets you a unit vector of v

Following so far.

(back later, lunch)

Oh and another reference from Paul of Internet fame:

tutorial.math.lamar.edu/Classes/CalcIII/…

(About this topic)