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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 $ \alpha$(x - a) + $ \beta$(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) = $ \alpha$x + $ \beta$y + $ \gamma$. 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) $ \neq$ 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) = $ \sqrt{x'(t)^2+y'(t)^2}$ 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$\displaystyle {\frac{1}{(1+t^2)^2}}$ = 1

(Hint: Use the following parametrisation.)

(x(t), y(t)) = $\displaystyle \left(\vphantom{ \frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} }\right.$$\displaystyle {\frac{2t}{1+t^2}}$,$\displaystyle {\frac{1-t^2}{1+t^2}}$$\displaystyle \left.\vphantom{ \frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2} }\right)$

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

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