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Nicholas · 2008-04-24 00:27:55

I am posing this as a question. I have never seen it addressed anywhere else. If the circle is a curve how do we calculate its derivative?

Mitch Raemsch; falling light changes colour

Nicholas · 2008-04-24 00:30:47

I just realized a circle is an unchanging curve. But it can have a different radius. I don't think a derivative applies.

Thanks anyway.

Mitch Raemsch

M@Man · 2008-04-24 05:17:25

I don't know what you mean by "an unchanging curve," but the derivative of any smooth curve (or trajectory) is the tangent vector to the curve. Therefore, the derivative of the curve that is a circle is a vector (strictly speaking, a vector field) that is tangent to the circle at every point.

The fact that you can make circles of different radii has nothing to do with the ability to differentiate the curve. Circles with different radii are different circles.

Nicholas · 2008-04-25 02:12:53

There is no tangent to a circle. Calculus needs to touch two points to derive the derivative.

Chris · 2008-04-25 16:23:43

There is no tangent to a circle.

There is no single tangent to a circle. That is why Matt said "strictly speaking, a vector field". The derivative of the circle is a vector field, meaning many, many tangent vectors around the curve.

Chris · 2008-04-25 16:25:15

how do we calculate its derivative?

BTW, the derivative with respect to what?

Nicholas · 2008-04-25 23:24:37

In order to find a slope you need two tangents points.

Mitch Raemsch

Chris · 2008-04-25 23:51:08

Nicholas wrote:
In order to find a slope you need two tangents points.

No. You do not. You are fundamentally misunderstanding what a tangent line is.

http://en.wikipedia.org/wiki/Tangent

http://www.mathopenref.com/tangent.html

Nicholas · 2008-04-26 23:46:55

It takes two points to derive a slope.

Mitch Raemsch

Chris · 2008-04-28 15:06:21

You can keep saying something over and over, but that doesn't make it true.

Nicholas · 2008-04-29 04:25:22

There is no slope to a zero dimensional point. Not even if its a nonzero infinitely small point.

Mitch Raemsch

M@Man · 2008-04-29 05:15:23

Chris wrote:
how do we calculate its derivative?
BTW, the derivative with respect to what?

To answer this question, I should be a bit more precise. A circle is merely a set of points. Since these points are ordered in no particular fashion, there is nothing to take a derivative with respect to. A curve, on the other hand, is a mapping from the set of real numbers to a particular trajectory throughout space. Thus the curve is a function of a real number, called the curve parameter. Since the curve itself is a function, it can be differentiated. For the case of vector-valued functions (i.e., curves in more than 1 dimension), the derivative of the curve is itself a vector-valued curve, and is tangent to the original curve.

In our case, one parameterization of the unit circle is

$\vec{r}(\theta)=\hat{x} cos(\theta) + \hat{y} sin(\theta)$

where $\theta$ is any real number. This parameterization of the curve runs counterclockwise, as does its tangent vector:

$\vec{T}(\theta)=\hat{x}(-sin(\theta)) + \hat{y} cos(\theta)$

An alternate parameterization of the same trajectory is a different curve, with a different derivative. If we instead parameterized the unit circle by $\theta --> -\phi$ , the curve function would be

$\vec{r}(\phi)=\hat{x} cos(\phi) - \hat{y} sin(\phi)$

with derivative

$\vec{T}(\phi)=\hat{x}(-sin(\phi)) - \hat{y} cos(\theta)$

so the tangent field now runs clockwise. We can also modify the length of the tangent vector by rescaling the curve parameter: if we change $\theta --> 2\lambda$ , then we double the length of the (counterclockwise) tangent vector:

$\vec{T}(\theta)=\hat{x}(-2 sin(\theta)) + \hat{y} 2 cos(\theta)$

A smooth curve is a curve whose function is analytic (i.e., all derivatives exist at every value of the curve parameter), and a smooth curve can always be differentiated (any number of times), but it's also worth saying that even a "smooth-looking" curve can fail to have its derivatives exist. This can happen if an ordinary-looking curve is parameterized in an awkward way. For example, we could also cover all the points on the circle with the parameterization

$\vec{r}(t)=\hat{x} cos(x D(x)) + \hat{y} \sin(x D(x))$

where $D(x)$ is the Dirichlet condition,

[tex] D(x) = 1 if x is rational
D(x) = 0 if x is irrational [/tex]

but, since $D(x)$ is nowhere differentiable, the tangent is not defined anywhere either.

So, precisely speaking, the tangent vector to the circle depends entirely on its parameterization.

Chris · 2008-04-29 17:12:51

Nicholas wrote:
There is no slope to a zero dimensional point. Not even if its a nonzero infinitely small point.

You did not ask about a "zero dimensional point". You asked about a circle.

Nicholas · 2008-04-30 03:32:46

Pick one point on the circle. It is zero dimensional and therefor no slope can be defined.

M@Man · 2008-04-30 14:53:04

The entire formalism of differential calculus is aimed at treating the instantaneous rate of change of functions. The crucial point is that this is accomplished by considering the behavior of points within a small neighborhood of your desired point. Within this neighborhood, you can approximate the slope of the tangent line by a secant line (which intersects the curve in at least two places) passing through the point you want and a second point somewhere in the neighborhood. If the function is well-behaved in this neighborhood, then the slope of the secant line converges to a well-defined value in the limit as you bring the second point close to the first. If this limit exists, it is the slope of the tangent line at that point.

The concept of the derivative is a bit awkward at first (what does it mean to have an instantaneous rate of change?), but it's really not that strange a concept when you get used to it.

For our example in question:

Take a circle of any radius $R$ . Let's describe the curve using the polar angle $\theta$ so that

$x = R cos \theta$
$y = R sin \theta$

This is a parameterization of the circle into two ordinary functions

$x(\theta)$ and $y(\theta)$ .

We can calculate the slope of the tangent to this parametric curve using ordinary differential calculus, without appealing to vector derivatives, using

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{R cos \theta}{- R sin \theta } = - cot \theta$

This is the slope of the tangent at angle $\theta$ on the circle, which agrees with the slope of the tangent vector I calculated in an earlier post.

Clearly the derivative of the curve exists, and its tangent line has a well-defined value. For instance, at $\theta = 45^{o}$ , the slope of the tangent line is $-1$ , so that the tangent line is parallel to $y = -x$ .

The derivative can be calculated in this point because the curve parameter $\theta$ orders the points on the circle so that there is a neighborhood of well-behaved points near $(R, \theta)$ for any value of $\theta$ . It is this neighborhood of points from which the derivative is calculated.

None of this is any stranger than is normally covered in a first course on differential calculus.

What does the 0-dimensional nature of a point have to do with this discussion? The relevant feature is that there is a neighborhood of points - infinitely many points, to be exact. You can prove that there are as many points in a finite arc (or any finitely-long smooth curve) as there are on the entire real line. Therefore the dimension (strictly speaking, the cardinality of the set of points) of the segment of the circle is actually 1 (like the real line) and not 0 (like a point in isolation). The behavior of functions in small, but 1-dimensional neighborhoods is the entire basis of differential calculus.

Chris · 2008-04-30 15:32:19

What does the 0-dimensional nature of a point have to do with this discussion?

Matt, you will soon learn that our friend Mitch/Nicholas (who claims to be a Nobel laureate), typically hasn't the slightest clue what he is talking about.

However, from time to time he does force enlightening discussions such as this one.

Nicholas · 2008-05-01 22:29:45

You can't calculate a tangent without using the calculus. You must pick two points on a changing curve to estimate the tangent line.

Mitch Raemsch

M@Man · 2008-05-02 07:21:59

Nicholas,

I would agree with both of your statements:

1.) You cannot calculate a tangent line without using calculus
2.) You must pick two points on the curve to estimate the tangent line

- I would add that you also need to be able to take the limit in which the second point approaches the first, which means you actually need infinitely many points in the neighborhood of where you're calculating the tangent line.

So yes, both of these statements are true, and we use both of these facts to actually calculate the tangent line to the circle, which does exist.

Chris,

Yes, I've mostly been pressing on in the hopes that a sufficient amount of factual discussion would amount to actual progress, willing audience or otherwise.

Nicholas · 2008-05-02 08:26:52

What I am saying is that a tangent line is really the shortest cord. The Infinitely small nonzero cord is required to be a precise tangent line.

Mitch Raemsch

M@Man · 2008-05-02 19:50:53

Nicholas,

I think both of your statements here are true as well:

1.) The secant line on a circle is also a chord of the circle (or an extension of that line segment beyond the radius of the circle)

2.) The slope of the chord, in the limit where the chord length goes to zero, is the slope of the tangent line for a circle.

Can you rephrase your earlier question about the existence of a derivative on the circle in this language? Or are you convinced that the derivative does exist, and that you can calculate it by taking the limit in which the chord length goes to zero?

Nicholas · 2008-05-03 00:13:46

I recoginse that the curvature of a circle is unchanging. Calculus calculates changing curves.

Mitch Raemsch

M@Man · 2008-05-03 18:40:55

Nicholas wrote:
I recoginse that the curvature of a circle is unchanging. Calculus calculates changing curves.

Mitch Raemsch

Ah, you're saying that the curvature of the circle is unchanging. This is a very different statement from your original claim, that the circle is an unchanging curve.

The following is adapted from my old calculus book:

Stewart, James. Multivariable Calculus; Concepts and Contexts, 2nd Ed. Brooks/Cole, Pacific Grove, CA, 2001, pp. 719-720.

The curvature $\kappa$ of at a point on a curve is defined to be

$\kappa = \| \frac{d \hat{T}}{ds} \|$

where $\hat{T}$ is the unit tangent vector at the point and $ds$ is the differential arc length of the curve at that point. Since the unit tangent vector is already a derivative of the curve, we can write the curvature explicitly in terms of a parameterization of the curve as

$\kappa (t) = \frac{ \| \hat{T}^{\prime} (t) \| }{ \| \vec{r}^{\prime} (t) \|}$

The important conclusion from this is that the curvature of a curve comes from the second derivative of the curve at that point. This means that, if the derivative does not exist, then neither does the curvature. So if the circle has any curvature at all, then its tangent line must exist!

The next example in my calculus text is:

Example 3 Show that the curvature of a circle of radius $a$ is $1/a$ .

Solution

We can take the circle to have the origin as its center, with the particular parameterization

$\vec{r}(t) = a cos t \hat{x} + a sin t \hat{y}$ .

Therefore the (unnormalized) tangent vector is

$\vec{r}^{\prime}(t) = -a sin t \hat{x} + a cos t \hat {y}$

and its magnitude is

$\| \vec{r}^{\prime}(t) \| = a$

Then the unit tangent vector $\hat{T}(t)$ is

$\hat{T}(t)=\frac{\vec{r}^{\prime}(t)}{\| \vec{r}^{\prime}(t) \|}= -sin t \hat{x} + cos t \hat{y}$

Then the derivative of the unit tangent (i.e., the normalized second derivative of the curve) is

$\hat{T}^{\prime}(t)= -cos t \hat{x} - sin t \hat{y}$

and its magnitude is

$\| \hat{T}^{\prime}(t) \| = 1$

So the curvature of the circle is therefore

$\kappa = \frac{\| \hat{T}^{\prime} (t) \|}{\| \vec{r}^{\prime} (t) \|} = \frac{1}{a}$

Thus we have proved that the curvature of a circle is indeed constant (as you claimed), but the important conclusion is that this result, rather than implying that the derivative of the curve does not exist, actually requires the existence of the derivative in order to be true! Therefore, if you claim that the circle has constant curvature, you must also concede that the tangent line to the circle exists.