OPTCIRC BRIEF ABSTRACT Determine the largest circle which can be inscribed in the region bounded by y = cos(x), x = 0, y = 0. GENERAL INFORMATION FileName: OPTCIRC Full title: Determining the largest circle which we can inscribed in a geometric region. Developer: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM Determine the largest circle (its mathematical description and its area) which can be placed in the region R which is bound by an arc of the function y = cos(x) (from x = 0 to x = Pi/2), the line segment from (0, 0) to (Pi/2, 0), and the line segment (0, 0) to (0, 1). State your assumptions, define your terms, and show your analysis. KEYWORDS Derivatives, tangency, slope, two equations (nonlinear) in two unknowns, geometric optimization. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Derivatives, tangency, slope, two equations (nonlinear) in two unknowns, knowledge of geometric optimization. Time allotment - time management We have devoted 15 minutes in class to starting up this problem (and in some cases actually finishing it) with students writing up a full solution for homework. Further, from time to time we have given serious thought to putting this problem on an exam as an extended problem. Pictures help and we need to encourage them to sketch. Expectations Future payoffs Extensions Another possibility is to ask the students to determine the largest circle (its mathematical description and its area) which can be placed in the region R which is bound the arc of the function y = cos(x) (from x = a to x = Pi/2) , the line segment from (a, 0) to (Pi/2, 0), and the line segment (a, 0) to (a, cos(a)). Notice that if a < -.5 then the maximum circle does not touch the left hand boundary, but has radius .5 and center at (0, .5), touching the curve at the point (0, 1). But in all cases -.5 < a < Pi/2 the circle does touch the left hand boundary and has a radius less than .5 . *** A really challenging extension would be to add another dimension to this problem. If we revolved y = f(x) around the y-axis but kept the boundaries the same, then asked for the maximum volume of the sphere inside, would that be impossible to work out? NB: Rotating the region about the x-axis References and Sources POSSIBLE SOLUTION(S) We plot the region in which we intend to inscribe a circle. f[x_] = Cos[x]; q = Plot[f[x],{x,0, Pi/2},AspectRatio->Automatic] We examine the picture and see that such a circle would have radius r and would have center at (r, r). Basically the circle would sit in the origin's corner and as such be tangent to the x-axis and the y-axis. This means the circle would touch the y-axis at point (0, r) and the x-axis at point (0, r). Further, the circle and the function should be tangent at the point of contact. Thus if we find the upper branch of the circle and set the derivatives and the y values at the point of contact to be the same (yielding two equations in the two unkowns, r and xp - the x coordinate of the point of contact of the circle and the function) we should be able to determine the point of contact and the radius. Here we solve the implicit circle expression for the two branches and select our function, g(x), to be the upper branc for y. sol = Solve[(x-r)^2 + (y-r)^2 == r^2,y] 2 2 2 2 r - Sqrt[4 r - 4 (r - 2 r x + x )] {{y -> --------------------------------------}, 2 2 2 2 2 r + Sqrt[4 r - 4 (r - 2 r x + x )] {y -> --------------------------------------}} 2 We consider the upper section of the circle with center at (r, r). g[x_] = y/.sol[[2]] 2 2 2 2 r + Sqrt[4 r - 4 (r - 2 r x + x )] -------------------------------------- 2 At the point of contact the circle should be tangent to the curve y = f(x). We articulate that by setting the two derivatives (slopes) equal. eq1 = f'[x] == g'[x] -2 r + 2 x -Sin[x] == -(--------------------------------) 2 2 2 Sqrt[4 r - 4 (r - 2 r x + x )] At this same point of contact both the functional value of the circle and the function y = f(x) should be the same eq2 = f[x] == g[x] 2 2 2 2 r + Sqrt[4 r - 4 (r - 2 r x + x )] Cos[x] == -------------------------------------- 2 We determine the radius and the point of contact (xp, yp) between the circle and the curve y = f(x). solr = FindRoot[{eq1,eq2},{r,1},{x,1}] {r -> 0.428779, x -> 0.650993} radius = r/.solr[[1]] 0.428779 xp = x/.solr[[2]] 0.650993 yp=f[x]/.solr[[2]] 0.795483 We plot the circle, the function, and the point of contact to confirm our result. p = ParametricPlot[{radius + radius Cos[t], radius + radius Sin[t]},{t,0, 2 Pi}, AspectRatio->Automatic] s = Show[Graphics[{PointSize[.02],Point[{xp,yp}]}]] Show[p,q,s] Hence the maximum area is MaxArea = Pi radius^2//N 0.577585 ISSUES IN SOLUTION A plausible argument for why the circle of maximum area (in this case) would touch the left hand boundary and be tangent to the curve y = f(x) at the point of contact is to consider such maximum circles which touch the x-axis in the right hand end of the region for a given point of contact on the x-axis, say (a, 0). It is not hard to convince oneself that such a circle would be tangent to the curve y = f(x) at the point of contact with the curve. And as we move the point of contact, i.e. decrease a, along the x-axis the radius of the maximum circle grows. Once the circle, more specifically the left hand side of the circle, touches the y-axis we can no longer move the point of contact of the circle and the x-axis to the left and so we are at a maximum radius and hence a maximum area. Of course this is, in part, due to the "increasing" (as we move to smaller cx values along the x-axis) height of y = cos(x). At this point the distance from the center of the circle to the y-axis will be the radius of the circle and the vertical distance from the center to the point of contact on the x-axis will be the same radial distance.