next up previous
Next: Surfaces Up: Coordinate Geometry Previous: Conics


To emphasise the obvious, the fundamental difference between lines and curves is that the latter are ``curved''. How do we quantify this ``curvy-curvy''-ness or curvature? We shall give a definition and examine some properties.

Let C denote the curve given as the locus of points (x, y) in the coordinate plane which satisfy the equation f (x, y) = 0. (Here and elsewhere we will restrict ourselves to polynomial functions since these are the natural outcome of the operations so far defined geometrically). Let p = (x0, y0) be a point on the curve. For any line l (for example given by the equation a(x - x0) = b(y - y0)) that contains this point we can define the order of contact between l and C at p as follows. We substitute the parametric solution (x, y) = (bt + x0, at + y0) of the line in the equation of C to obtain F(t) = f (bt + x0, at + y0). The order of contact r(l, C, p) is then the ``largest'' power of t that divides F(t). If F(t) = 0 then the curve contains the line l and we define r(l, C, p) = $ \infty$ otherwise this order is bounded above by the degree of F which is a polynomial in t. Since t = 0 corresponds to the point p we always have r(l, C, p) $ \geq$ 1.

Definition 1   A line l with a higher order of contact with C at p than any other line is called an osculating line or tangent to C at p.

Exercise 2   Either r(l, C, p) $ \geq$ 2 for all l through p or there is a unique line l through p for which r(l, C, p) $ \geq$ 2. In the former case we say C is singular at p and in the latter case we say that C is non-singular or smooth at p.

picture(8273,5477)(886,-5425) (886,-5371)(0,0)[lb]Smooth curve and its tangent line (5281,-5341)(0,0)[lb]Singular curve and its tangent
Smooth and Singular Curves
In what follows we will concentrate our attention on non-singular curves. The singular case is very interesting but beyond the scope of our current discussion.

In order to motivate one definition of curvature, we first note that a circle is clearly curved. The curvature of a circle should be inversely proportional to the radius since the family of circles touching a line at a given point approach the line as the radius increases. Thus we could define

Definition 2   Given a point on a curve the osculating circle is the circle that has the highest order of contact with the curve at the given point. The curvature of the curve at the given point is the inverse of the radius of this circle.

Of course, we must first define the order of contact of a circle and a curve or more generally of two curves, then we must show that a unique circle as defined exists. Note also that once we use circles in a definition we can only make changes of coordinates with preserve circles and their radii; that is, rotations, reflections and translations. In other words, we must fix a notion of distance for curvature to make sense.

Exercise 3   Show that (u, v) $ \mapsto$ (x + y3, y) is a one-to-one and onto correspondence between the (u, v)-plane and the (x, y)-plane. What curves in the (u, v)-plane correspond to lines in the (x, y)-plane? Are these curved? Can we take these to be lines in a geometry?

picture(5626,4613)(548,-4846) (2431,-4846)(0,0)[lb]Curvilinear coordinates
To mimic the definition of the order r(l, C, p) of contact of a line and a curve in order to define r(C, D, p) we need to find a parametric solution of at least one of the curves. Such parametrisations are impossible in general. Instead, we have

Definition 3   We say that (x(t), y(t)) is a parametric solution of f (x, y) of order r if tr divides f (x(t), x(t)).

If C and D are curves defined by f (x, y) = 0 and g(x, y) = 0 and p = (x0, y0) is a point on both the curves then we say that C and D have order of contact (at least) r is there is a common parametric solution (x(t), y(t)) or order r for both curves so that (x(0), y(0)) = (x0, y0). We can define r(C, D, p) as the supremum of such r.

Exercise 4   Show that r(C, D, p) $ \geq$ 2 for smooth curves C and D if and only if the tangent lines at p to C and D respectively coincide.

In particular, it is always possible to find a circle D which has order of contact 2 at a given point p = (x0, y0) on C. In order to find a circle with order of contact 3 or more we need to find conditions so that the following equations have a common solution
t3 | (x0 + x1t + x2t2 - a)2 + (y0 + y1t + y2t2 - b)2 - r2  
t3 | f (x0 + x1t + x2t2, y0 + y1t + y2t2)  

We collect coefficients of 1, t and t2 to obtain the following system of equations
(x0 - a)2 + (y0 - b)2 - r2 = 0  
f (x0, y0) = 0  
(x0 - a)x1 + (y0 - a)y1 = 0  
fx(x0, y0)x1 + fy(x0, y0)y1 = 0  
(x0 - a)x2 + (y0 - a)y2 + x12 + y12 = 0  
fxx(x0, y0)x12 + 2fxy(x0, y0)x1y1 + fyy(x0, y0)y12 +      
fx(x0, y0)x2 + fy(x0, y0)y2 = 0  

Here we adopt the subscript notation gu to denote the (partial) derivative of a polynomial g in the variable u; we note that this derivative is defined formally (without the use of limits) and arises in the above equations due to the binomial expansion which is the Taylor expansion for polynomials.

Exercise 5   A solution for the above equations exists only if

r2 = $\displaystyle {\frac{(f_x^2 + f_y^2)^3}{(f_{xx} f_y^2 + 2 f_{xy} f_x f_y + f_{yy}
f_x^2)^2}}$$\displaystyle \mid_{(x_0,y_0)}^{}$

Moreover, if we can form square roots then a solution does exist in this case.

Thus we obtain the following formula for the curvature $ \kappa$ of a curve

$\displaystyle \kappa$ = $\displaystyle {\frac{(f_{xx} f_y^2 + 2 f_{xy} f_x f_y + f_{yy} f_x^2)}{(f_x^2+ f_y^2)^{3/2}}}$$\displaystyle \big\vert _{(x_0,y_0)}^{}$

Another way to approach curvature is by reversing Newton's law of motion. If a body experiences no acceleration then it must travel along a straight line at constant speed. Thus, we can define the curvature of a curve as (the magnitude of) the acceleration experienced by a body travelling along the curve at constant speed. This definition has the advantage of being applicable to space curves (x(t), y(t), z(t)) as well. We again have to surmount a difficulty that it may not be easy to find (x(t), y(t), z(t)) so that (xt(t), yt(t), zt(t)) is of length 1; certainly there will in general be no polynomial functions that will do the trick. However, we note that acceleration is just (xtt(0), ytt(0), ztt(0)) for such a curve so that it will be enough to find a parametric solution of order 3 which has constant speed.

We now apply this method to the curve defined by f (x, y) as above. The above considerations lead to the following pair of conditions

t3 | f (x0 + x1t + x2t2, y0 + y1t + y2t2)  
t2 | (x1 + 2x2t)2 + (y1 + 2y2t)2 - 1  

We then have to compute (x2, y2) or rather its magnitude. Now we collect coefficients of powers of t as before to obtain the following system of equations
f (x0, y0) = 0  
fx(x0, y0)x1 + fx(x0, y0)y1 = 0  
fxx(x0, y0)x12 + 2fxy(x0, y0)x1y1 + fyy(x0, y0)y12 +      
fx(x0, y0)x2 + fy(x0, y0)y2 = 0  
x12 + y12 = 1  
x1x2 + y1y2 = 0  

Exercise 6   Solve the above equations to obtain the magnitude $ \rho$ = $ \sqrt{x_2^2+y_2^2}$ of the acceleration as

$\displaystyle \rho$ = $\displaystyle {\frac{(f_{xx} f_y^2 + 2 f_{xy} f_x f_y + f_{yy} f_x^2)}{(f_x^2+ f_y^2)^{3/2}}}$$\displaystyle \big\vert _{(x_0,y_0)}^{}$

Thus we see that the notion of curvature can indeed be recovered from Newton's law as the acceleration of a particle moving along the curve at constant speed.

We now compute the curvature for any parametric curve (x(t), y(t), z(t) at the point p corresponding to t = t0. To compute curvature following the above definitions we need to re-parametrise the above curve by t = t0 + g(s) so that the parametric curve (x'(s), y'(s), z'(s)) = (x(t0 + g(s)), y(t0 + g(s)), z(t0 + g(s))) is traversed at constant speed (with respect to the parameter s). The acceleration (x'ss(0), y'ss(0), z'ss(0)) gives the curvature of the curve as its magnitude.

Exercise 7   Work with a solution g(s) = g1s + g2s2 of order 3 to obtain an expression for the curvature of this curve.

A useful notion is that of the directional derivative Dv(g)(p) of a function g at a point p = (x0, y0) along a direction v = (x1, y1). This is the derivative at t = 0 of the function G(t) = g(x0 + tx1, y0 + ty1); in other words

Dv(g)(p) = x1gx(x0, y0) + y1g(x0, y0) + z1g(x0, y0)

We consider the function which assigns to each point of a plane curve C its unit normal, that is

(x, y) $\displaystyle \mapsto$ n(x, y) = $\displaystyle {\frac{(f_x,f_y)}{(f_x^2+f_y^2)^{1/2}}}$

Exercise 8   Show that the directional derivative Dv(n)(x, y) of n(x, y) is a vector which is orthogonal to n(x, y).

In particular, we note that if v is a non-zero tangent vector to C at a point (x0, y0) on C then Dv(n)(x0, y0) = $ \alpha$ . v for some $ \alpha$.

Exercise 9   Show that $ \alpha$ is the same as the curvature of C upto sign.

An important aspect of the above calculations is that there is no ``intrinsic'' curvature to a curve--at least not in a local picture. Any curve can be parametrised in such a way that distance along the curve depends linearly on the parameter; exactly as for a line. The curvature occurs only in the manner in which the curve is embedded in the surrounding plane (or space). We shall see that the behaviour for surfaces is very different.

next up previous
Next: Surfaces Up: Coordinate Geometry Previous: Conics
Kapil H. Paranjape 2001-01-20