Equations for the Paths of Four Lady Bugs
Date: 12/02/98 at 15:06:54 From: Corey Morcombe Subject: Calculus and Polar Equation Four ladybugs are placed at the four corners of a square with sides of length S. The bugs crawl counterclockwise at the same speed and each bug crawls directly toward the next bug at all times. They approach the center of the square along spiral paths. Find the polar equation of a bug's path assuming that the pole is at the center of the square. Also find the distance traveled by the bug by the time it meets the others in the center. I have no idea even where to get started on this problem, and could use some help. Thank you.
Date: 12/02/98 at 17:17:08 From: Doctor Anthony Subject: Re: Calculus and Polar Equation I shall be using the 'pedal' equation of the curve to answer this. Pedal Equation of a Curve -------------------------- Let O be the usual pole of co-ordinates with OX the initial line, and p the perpendicular ON to the tangent at P (r, theta) to the curve. The relation between p and r only is known as the 'pedal equation' of the given curve. If phi is the angle that the tangent at P makes with OP, then (if you draw a diagram) it is easy to see that p = r.sin(phi) Also by consideration of elementary triangle formed by neighbouring points on the curve, (r, theta) and (r+dr, theta+d(theta)) we get sin(phi) = r.d(theta)/ds cos(phi) = dr/ds ds^2 = (r.d(theta))^2 + (dr)^2 (Pythagoas) [ds/d(theta)]^2 = r^2 + [dr/d(theta)]^2 ds/d(theta) = sqrt[r^2 + (dr/d(theta))^2] Now turning to the question of the ladybugs, we take the origin at the centre of the square and OX the initial line from O to one of the corners. By symmetry it is easy to see that as the bugs spiral in they continue to form a square (decreasing in size) and the direction of the path keeps a constant angle of 45 degrees with the radius vector r joining O to one of the bugs. This means that the angle phi (as described above) is a constant 45 degrees. So p = r.sin(45) = r/sqrt(2) We have also r.d(theta)/ds = sin(45) and dr/ds = -cos(45) From the second of these ds = -dr/cos(45) = -dr.sqrt(2) From r.d(theta)/ds = sin(45) r.d(theta)/[-dr.sqrt(2)] = sin(45) -r. d(theta)/dr = sqrt(2).sin(45) -r. d(theta)/dr = 1 and separating the variables; -d(theta) = dr/r Integrating -theta = ln(r) + constant and then, a = 0 when r = S.cos(45) = S/sqrt(2) so 0 = ln[S/sqrt(2)] + constant constant = -ln[S/sqrt(2)] The polar equation is -theta = ln[r/(S/sqrt(2))] -theta = ln[r.sqrt(2)/S] which can be written: r.sqrt(2)/S = e^(-theta) r = S.e^(-theta)/sqrt(2) and this is the required polar equation of the curve. As theta -> infinity, r -> 0 To find the distance travelled by a bug we can use dr/ds = -cos(45) So ds = -dr/cos(45) ds = -sqrt(2).dr INT[ds] = -sqrt(2).INT(S/sqrt(2) to 0)[dr] s = -sqrt(2)[0 - S/sqrt(2)] s = S So each bug walks a distance equal to the side S of the square. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.