ORTHOGONAL FAMILIES OF CURVES

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1 8 ORTHOGONAL CURVES Spring her winning the car is the probability of her initially choosing a door with a goat behind it, that is 66%! There is a very nice, complete discussion of this topic, and the controversy it engendered, under the heading Monty Hall Problem in Wikipedia. ORTHOGONAL FAMILIES OF CURVES JAMES FENNELL, UNIVERSITY COLLEGE CORK It is a well known fact from secondary school geometry that a straight line passing through the centre of a circle will intersect the circle at right angles. We say that the circle and the line are orthogonal. Orthogonal is a word derived from Greek (ortho meaning right and gonal angle) and is often used as a synonym for perpendicular. However it is more general: one wouldn t say the circle and the line are perpendicular as such. If we let C denote the set of all circles centred at the origin and L the set of all lines passing through the origin, then every line in L meets every circle in C at right angles (see figure 1 (a)). We call C and L families of curves and because every element of C is orthogonal to every element in L and vice-versa, we say that C and L are orthogonal families of curves. y y (a) The families C and L (b) The families E and L Figure 1: The first three families of curves

2 2012 ORTHOGONAL CURVES 9 These facts were known to Euclid in 300 BC when he compiled the Elements, though the terminology is modern. In Proposition 18 of Book 3 he proved what we would now call an eistence theorem: he showed that if we start with the family of circles C then it is possible to find a family of curves that is orthogonal to C; namely L. We may now ask if the family L unique. That is, is L the only family of curves that is orthogonal to C? Now, suppose while pondering this question you put down your copy of the Elements and go to make tea. When you return you find your dog sitting on Euclid s book, and upon shoving him off notice that he s squashed the whole thing so that the circles have become ellipses! But your not angry of course, because it immediately prompts another question: given the family of squashed circles E centred at the origin, is there a family of curves orthogonal to it? Clearly (figure 1(b)) lines centred at the origin are not orthogonal. So whereas for the circles we had a uniqueness question, here we have to start with an eistence question. It s worth remarking at this point that the ellipse problem probably wouldn t have occured to Euclid. Euclid studied a fied finite set of curves - straight lines, circles, conic sections, etc. - and the kind of general question we pose here - is there any kind of curve satisfying our conditions - wouldn t have fitted into this finite paradigm. 1. A solution To attempt to answer these questions we recall an elementary fact from analytic geometry (or coordinate geometry): if we have a line of slope m then any line perpendicular to it will have slope 1/m. Of course we re not dealing with two straight lines; we re dealing with more general curves as well. But this isn t an issue for we know how to get the slope of a general curve like a circle: we just differentiate it! We can describe the family of circles centred at the origin like so C = { 2 + y 2 = r2 r > 0} (1) We differentiate the epression 2 + y 2 = r 2 with respect to, being careful to differentiate y implicitly as we view it as a function of. So, d 2 + y 2 = d 2 + 2y d y = 0 d y = y r 2 (2)

3 10 ORTHOGONAL CURVES Spring We have thus determined an epression for the slope of the family C at any particular point (, y). But we have actually gotten a little more. Because of Picard s Uniqueness Theorem for ordinary differential equations, the differential equation we have arrived at has eactly one unique solution up to a constant. And, by the Fundamental Theorem of Calculus, we know that that one solution is given by the epression we started with, (1) integration being the inverse operation of differentiation. Now we wish to employ our fact about slopes. The family C has slope m = d y/ at every point (, y) and so any orthogonal family will have slope 1/m = 1/(d y/). So now to find any orthogonal family we must solve the epression dz = 1 m = 1 /z = z (3) where the y variable has been relabeled z to avoid confusion between the different families. Now we can solve this equation very easily as it s separable: dz z = log( z ) = log( ) + A = z = b for b any real number. This is precisely the equation describing the set of all straight lines passing through the origin! But not only have we recovered Euclid s fact that straight lines are orthogonal to circles, but we have also proved uniqueness. For the equation (3) is a differential equation just like (2) and so has the unique solution z = b. All parts of our derivation maintained uniqueness (that we can reverse the whole process confirms this) and so there is a one-to-one correspondence between the families C and L: straight lines passing through the origin are the only set of curves that are every perpendicular to the set of circles centered at the origin. With this technique we can tackle the ellipse problem too. There are an infinite number of families of ellipses centred at the origin, but we will select E = { 2 + 2y 2 = r 2 r > 0} This is the family of ellipses whose width is 2 time their height. We have, as

4 2012 ORTHOGONAL CURVES 11 before and so d 2 + 2y 2 = d 2 + 4y d y = d y = 2 y dz = 1 /2z = 2z dz z = 2 r 2 log( z ) = 2 log( ) + A = log( 2 ) + A = z = b 2 (4) This new family is illustrated in figure 2. y y = y 2 = 3 2 y = y = y 2 = 4 2 Figure 2: The family E, ellipses of the form 2 + 2y 2 = r 2, and its unique orthogonal family, curves of the form y = b 2. Now that we ve applied the technique twice we might search for general conditions under which a given family of curves F possesses a unique orthogonal

5 12 ORTHOGONAL CURVES Spring family. We assume the family is governed by an analytic epression of the form f (, y, c) = 0 where and y are co-ordinates and c a parameter that produces different curves when varied. For the family C the governing equation is 2 + y 2 c 2 = 0 However this curve is special in that when differentiate it the constant disappears. If instead we had the family y = e c after differentiation we would still have c in the epression. This needs to be removed by solving the equation for c and then subbing that into the derivative. So the full process is: first differentiate f implicitly to find d y/ as a function of, y and c. Then apply the inverse function theorem to ensure that we can solve f (, y, c) for c as a function of and y. We will then have a differential equation of the form d y = g(, y, c(, y)) giving the orthogonal equation dz = 1 g(, z, c(, z)) We then apply Picard s Uniqueness Theorem to ensure this has a unique solution which is precisely the orthogonal family of F. Theorem 1.1. Let F be an orthogonal family of curves governed by an equation of the form f (, y, c) = 0 in U, an open set in 3. Suppose that f / y eists and is never 0 so that, by the implicit function theorem, there eists a function g with d y = g(, y, c) and g never 0. Suppose also that f / c is never 0, so that we can write c = c(, y). Then F has a unique orthogonal family in U governed by the solution to the differential equation dz = 1 g(, z, c(, z))

6 2012 ORTHOGONAL CURVES Orthogonality in the comple plane Our analysis so far has been restricted to the real plane. However there is a strong argument that the best contet in which to eplore orthogonality is the comple plane. To see this, take any comple number z = a+ bi and plot it on an Argand diagram of the comple plane. Now multiply by i to get iz = b + ai and plot the resulting comple number on the same diagram. The result is shown in figure 2. We can see that geometrically, multiplying by i is the same as rotating the vector a + bi π/2 radians counter-clockwise to arrive at a perpendicular vector b + ai. So orthogonality is built-in to the comple numbers! Im bi a + bi b + ai ai b a Re Figure 3: The geometric effect of multiplying z = a + bi by i: rotation by π/2 radians. We wish to draw curves in the comple plane and see if we can find families of curves that are orthogonal. Let f (z) be a comple function. As we vary z the value of f (z) will change about the comple plane tracing out a curve. As z is comple there are an infinite number of ways of varying it - as opposed to a real number, where one can only increase it or decrease it in the one direction. We will restrict our attention to comple differentiable functions. Recall that f (z) = lim h 0 f (z + h) f (z) h When defining a real derivative h is a real number and so we can approach 0 from the negative side or the positive side. When defining a comple derivative h is a comple number and so can approach 0 from any direction in the comple plane. As a result, comple differentiable functions are a little rarer in a sense,

7 14 ORTHOGONAL CURVES Spring but this won t be an issue here. Now, for h small we can write f f (z + h) f (z) (z) h = f (z + h) f (z) + hf (z) This is also the first order Taylor epansion of f. As h is comple we can replace it by ih and so write f (z + ih) f (z) + ihf (z) We are interested in the local behaviour of f and in particular how f affects the natural orthogonality of comple numbers discussed above. We take any point z, a small comple vector h and the perpendicular vector ih. Then we consider f (z) and two, again perpendicular, vectors hf (z) and ihf (z), at f (z). The end-points of these vectors will be f (z) + hf (z) and f (z) + ihf (z). Given our approimations above, these new vectors roughly describe the result of applying f to the three points z, z + h and z + ih (see figure 4). Im Im z + ih f f (z) i f (z) f (z) + ihf (z) f (z + ih) z ih h z + h Re f (z) f (z) + hf (z) f (z + h) Re Figure 4: The geometric effect of applying a comple differentiable function: orthogonality preserved. As we take the limit h 0 these new vectors become tangents to the two curves generated as f travels in the perpendicular directions h and ih. Thus if we apply f to perpendicular lines we will get two curves that intersect at a right angle. Remember that we haven t said anything about f besides that it is differentiable, so this result applies to a large class of functions! In [1] Tristam Neehan produces a very nice application of this fact to prove two well known curves can form orthogonal families. We consider the comple cosine function cos(z) and in particular the identity cos( + i y) = cos() cosh(y) + i sin() sinh(y)

8 2012 ORTHOGONAL CURVES 15 where cosh and sinh are the hyperbolic trigonometric functions. We now form a grid in the comple plane formed of vertical lines and horizontal lines. Vertical lines occur when the real part of z is kept constant and horizontal lines when the imaginary part of z is kept constant. We take the latter case first. Holding Im(z) = b constant we have cos( + i b) = cos() cosh(b) + i sin() sinh(b) = Acos() + ib sin() for constants A and B. As f traverses a horizontal line it traces out the curve in the comple plane parameterized like so (Acos(t), B sin(t)) This is just the usual parameterization of an ellipse in the plane centred at the origin and with major ais on the horizontal ais. The distance from each foci to the origin is given by the usual formula A 2 B 2 = cosh 2 (b) sinh 2 (b) = 1 and so the foci for all of our ellipses are at ( 1, 0) and (+1, 0). The family of curves so generated (as b is varied) is then the family of all ellipses with centre (0, 0) and foci (±1, 0). Now, what if f traverses a vertical line? In this case Re(z) = = a is constant and we have cos(a + i y) = C cosh(y) + id sinh(y) When we write the curve traced by f in parametric form, as (C cosh(t), D sinh(t)), we that it is another conic section, the hyperbola! The distance from the origin to each of the foci is C 2 + D 2 = cos 2 (a) + sin 2 (a) = 1 and so the foci are again ( 1, 0) and (+1, 0). The family of curves generated by f traversing the vertical lines in the comple plane is then the family of hyperbolas with centre (0, 0) and foci at (±1, 0). But, of course, vertical lines and horizontal lines are perpendicular. We applied a comple differentiable function to each of these and got a family of hyperbolas and a family of ellipses with the same foci. By our reasoning above the orthogonality is preserved, and so we have the result that hyperbolas and ellipses with the same foci intersect at right angles. The two families are illustrated in figure 4. This method can be used to find a wealth of orthogonal families. Application to the comple eponential function yields another proof that circles and lines are orthogonal. Even applying it to f (z) = z 2 yields an interesting picture which is omitted intentionally!

9 16 ORTHOGONAL CURVES Spring y Figure 5: Ellipses and hyperbolas with the same foci. References [1] Tristam Needham. Visual Comple Analysis. Oford University Press, ISBN: [2] George F. Simmons. Differential Equations. Tata McGraw-Hill, ISBN:

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