Does degree of an equation represent degree of curvature of the curve given by the equation?

blackhash wrote:Does degree of an equation represent degree of curvature of the curve given by the equation?

I never mix spinach and mustard.

The degree of an equation is just the highest power that appears on one of its variables. The degree of an equation in x and y can tell you something about the shape of the graph. So degree 1 equations graph as straight lines, and such equations are called "linear", while degree 2 equations graph as parabolas.

VazScep wrote:The degree of an equation is just the highest power that appears on one of its variables. The degree of an equation in x and y can tell you something about the shape of the graph. So degree 1 equations graph as straight lines, and such equations are called "linear", while degree 2 equations graph as parabolas.

That is the point. From straight line to curves(parabola,ellipse,hyperbola etc). From curves to...........?

blackhash wrote:

VazScep wrote:The degree of an equation is just the highest power that appears on one of its variables. The degree of an equation in x and y can tell you something about the shape of the graph. So degree 1 equations graph as straight lines, and such equations are called "linear", while degree 2 equations graph as parabolas.

That is the point. From straight line to curves(parabola,ellipse,hyperbola etc). From curves to...........?

The parabola, ellipse and hyperbola are the three kind of curves you get by taking a cross-section of an infinite cone. But, in general, you're looking at all sorts of curve, subject to the constraint that if you draw a straight line across, it'll only cut the curve a finite number of times, corresponding to the degree.

It is the number of solutions of the equation if set to zero, but in the complex plane generally. To go to zero at some number of points does imply curvature dependence to some degree. I hope I'm not bollixing that up, it's highschool maths.

See also, the fundamental theorem of algebra:

crank wrote:It is the number of solutions of the equation if set to zero, but in the complex plane generally. To go to zero at some number of points does imply curvature dependence to some degree. I hope I'm not bollixing that up, it's highschool maths.

You'd have to count "repeated roots" with that definition. x^2 = 0 has only one solution, but degree 2.

The curvature of a function isn't usually constant - the principal exception being the curve x2+y2=r2, which is the equation of a circle. Consequently, a polynomial may have a large curvature for some values of the independent variable, but a small curvature for other values of the independent variable.

The curvature of a function f(x) is the curvature of the largest circle that is tangent to the function curve at a given point, and is inversely proportional to the radius of that circle. If the circle in question has radius r, then the curvature k is 1/r.

An explicit expression for k is as follows:

k = (x'y'' - y'x'')/([x']2+[y']2)½

where x and y are functions of another parameter (say, t) such that they generate the curve f(x), and the primes indicate differentiation by the parameter. So, for the parabola y2 = 4ax, a suitable parameterisation is:

x = t2, y = 2at

Differentiating these gives:

dx/dt = 2t, d2x/dt2 = 2
dy/dt = 2a, d2y/dt2 = 0

Therefore, for the parabola,

k = [(2t)(0) - (2a)(2t)]/[4t2+4a2]½

= -4at/2(a2+t2)½

= -2at/(a2+t2)½

At t=0 (which corresponds to x=0), the curvature k is zero, which means that at x=0, the parabola has the same curvature as a straight line (but only at that point). At t=1, which corresponds to x=1, the curvature is -2a/(a2+1)½. If a=1, then the curvature is -√2.

The curvature is a signed entity, because the sign tells you which direction a unit tangent vector rotates as it moves along the curve in the direction of increasing values of t. If the tangent rotates in the anticlockwise direction, k>0 (i.e., is positive), whilst if the tangent rotates in the clockwise direction, k<0 (i.e., is negative).

I leave it as an exercise for the readers of this post, to derive the parametric expression above for k from the expression in terms of arc length.

ERRATUM

The power of the term in the denominator in the above expression for k should be 3/2, not 1/2. Adjust the results accordingly.

I was thinking of derivations wrt an independent variable. We get the slope for an equation with degree 2. For the same equation second derivative gives us rate of change of slope(I hope I am right here). It is a constant.
So by induction I assumed for equations with degree more than 2 for independent variables we might be having successive derivations till we get a constant.
So I reached the conclusion that the degree of independent variable tells us about degree of curvature.

VazScep wrote:

crank wrote:It is the number of solutions of the equation if set to zero, but in the complex plane generally. To go to zero at some number of points does imply curvature dependence to some degree. I hope I'm not bollixing that up, it's highschool maths.

You'd have to count "repeated roots" with that definition. x^2 = 0 has only one solution, but degree 2.

What about 0 + 0i and 0 - 0i? That's not entirely facetious, but the details of such things, while learned, are mostly forgotten or at best faded into a hazy and confused mess. I'm about to watch that video, it should disperse some of the haze.

blackhash wrote:I was thinking of derivations wrt an independent variable. We get the slope for an equation with degree 2. For the same equation second derivative gives us rate of change of slope(I hope I am right here). It is a constant.
So by induction I assumed for equations with degree more than 2 for independent variables we might be having successive derivations till we get a constant.
So I reached the conclusion that the degree of independent variable tells us about degree of curvature.

No it doesn't. Because as I stated earlier, for any general function, the curvature at a given point on the curve depends upon the value of the curvature expression I supplied at that point, which in the case of a polynomial, need not have any connection to the degree of that polynomial. For example, the curvature of y=x2-5x+2 at x=0 is 0.015. Meanwhile, the curvature of y=x3-x2+3x+6 at x=0 is -0.063. But the curvature of y=3x7-5x3+10 at x=0 is zero. On the other hand, the curvature of y=x4+100x3-20x2+400x is nonzero - it's -6.2×10-7.

Now if you can find a simple relationship between those curvature values, and the degrees of those polynomials, be my guest.

Incidentally, the curvature for a second degree polynomial isn't a constant. Here's the full set of values for the curvature k, that I tabulated in Excel for x=0 to x=10, for the polynomial y=x2-5x+2:

x = 0, k = 0.015
x = 1, k = 0.063
x = 2, k = 0.707
x = 3, k = 0.707
x = 4, k = 0.063
x = 5, k = 0.015
x = 6, k = 0.0056
x = 7, k = 0.0029
x = 8, k = 0.0014
x = 9, k = 0.00090
x = 10, k = 0.00058

Of course, the whole question becomes null and void the moment one starts applying the curvature formula to functions that aren't finite polynomials, such as y=ex or y=sin(x), but that's an entire new ball game.