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# Curves

Now that we have a basic understanding of differentiability of functions we can begin the study of plane curves which can be defined parametrically or as the locus of vanishing of a function of two variables. At the very least we need the function to be (piecewise) differentiable with good'' first order properties. We shall see later that for any interesting (and æsthetic!) study we will need second derivatives as well.

A plane curve is defined (locally) as the locus of points where a good'' function f of two variables vanishes. In particular, let p = (a, b) be a point where f vanishes, then we assume that f (x, y) = f1, 0(x - a) + f0, 1(y - b) + o(x) is continuously differentiable at this point. The curve is said to be singular at p if both the above coefficients are 0; otherwise we call the curve non-singular or smooth. For a smooth curve through p the line f1, 0(x - a) + f0, 1(y - b) = 0 is called the tangent line. It is the best'' linear approximation to the curve in an obvious way:

Exercise 57   Consider the natural parametrisation of the line (x - a) + (y - b) = 0. The restriction of the function f to this line can then be thought of as a function of one variable. Show that this function vanishes to order 2 if and only if the line is the one above or the curve f = 0 is singular at the point p = (a, b).

In what follows we restrict our attention to smooth curves. Singular curves are very interesting and are studied extensively in algebraic geometry.

A different way of representing curves is to think of a curve as a moving point''. A curve can be given in parametric form by writing a pair of functions (x(t), y(t)) so that as t varies we will trace out a curve. As before we will insist on the two functions being continuously differentiable. We say that our curve is non-singular at time'' t = t0 we need at least of the pair (x'(t0), y'(t0)) to be non-zero; otherwise we call the curve singular.

Exercise 58   Consider the function of t given by substituting the above pair of functions in the linear form l (x, y) = x + y + . Show that this function vanishes to order 2 at t = t0 if and only if the curve is singular or the form l (x, y) is the tangent form y'(t0)(x - x(t0)) - x'(t0)(y - y(t0)).

We need to have some way of going from the parametric form of a curve to the equation and vice verse. For the first problem, let us assume (without loss of generality) that x'(t0) 0. Then, by the inverse function theorem, we have g(x) so that g(x(t)) = t, so that we can re-parametrise the curve to get (x, y(g(x)). The curve is the (locally) given by the equation y - h(x) = 0 where h(x) = y(g(x).

To go from the equation to the parametric form we need to show that lines parallel to a line which is not tangent to the curve will meet the curve in exactly one point near the given point. This done through the implicit function theorem. (Note to author: Exercises to be added here).

One of the advantanges of working with orders'' of vanishing is that make these theorems explicit'' if we only need our equations to be satisfied upto terms of some order. For example, we say that (x(t), y(t)) is a parametrisation at t = t0 upto order r of the curve f (x(t), y(t)) = o((t - t0)r). Similarly, two curves, f and g are said to osculate upto order r if f - g = o(xr). In particular, any curve osculates upto order 2 with a conic; thus it is possible to write a parametrisation upto order two quite explicitly.

Finally, there is one distinguished parametrisatisation. Let (x(t), y(t)) be a curve. Thinking of this as a moving point we have not only a tangent line but a tangent vector (called the velocity vector) (x'(t), y'(t)). It is thus natural to define the speed of the curve as length of this vector. We can ask for a constant speed (or more accurately constant energy) parametrisation. In other words, can we find t = u(s) so that

(x'(t)2 + y'(t)2) | t = u(s)u'(s)2 = constant

Exercise 59   Use the inverse function theorem to show that the function s(t) = has an inverse. Show that this inverse function satisfies the above equation.

We will see that such a parametrisation called parametrisation by arc-length plays an important role in geometry. Meanwhile,

Exercise 60   Show that to obtain such a parametrisation for the circle, we need to solve the equation

u'(t)2 = 1

(Hint: Use the following parametrisation.)

(x(t), y(t)) = ,

We will study the solution of this and related equations in the next section.

Next: Elementary functions Up: Pre-requisites Previous: Integration
Kapil H. Paranjape 2001-01-20