Math 545 Advanced Calculus Summer II, 2004 Irvin Roy Hentzel 432 Carver Hall Phone 515-294-8141 Fax 515-294-5454 hentzel@iastate.edu Test: Calculus Third Edition, Michael Spivak, Publish or Perish, Inc. 1436 West Gray, #377 Houston, Texas 77019 ISBN 0-914098-89-6 Editions 1, 2, and 3 are essentially the same Web Site: http://www.math.iastate.edu/hentzel/class.545 ============================================================== Assignment: Work ONE of the following (a) Page 89 problem 10 (b) Page 89 problem 8 (c) Page 88 problem 3 ============================================================= Appendix 3. Polar Coordinates Pages 84 - 87 ------------------------------------------------------------ Main idea: Key Words: (r,theta), limacon, cardioid, lemniscate, spiral of Archimedes, hyperbolic spiral, Goal: Learn to recognize r = Cos[theta] and understand how it is modified in the graphs of lots of other functions. ------------------------------------------------------------- Previous Assignment Work ONE of (a) Page 83 Problem 2 ***** * * * ______*_________ | * / | / |* / *| / ___|* / *| / /|\ |* / *| / | | * / o o | / | | * / o * o|/ | | * / o F' o/ | | / o****f /| 2a | | / f o/ | | | / f / | | | /f / | | | / f o/ | | | / f **** / | | |/o F o /* | | | * / * | | |o* o / * | \|/ |* o / *| ---|* / *| / |* / *| / | * / * | /_____|____*____/ * | **** Put two circles tangent to the cylinder and also tangent to the plane. Let the point of tangency with the plane be called F and F'. The distance from any point on the intersection of the plane and the cylinder to the two foci will equal the vertical distance between the equators of the two spheres. (b) Show that the intersection of a plane and a cone where the plane is parallel to one edge of the cone is a parabola. \ . / P /P' \ / / \ . / . / \ . F . / \ ./o. o / \ /o o / / o o/ Q' L-/-\-o---o/---- / \ / / \ / \/ Let F be the point of tangency of a sphere which is also tangent to the sides of the cone. Let L be the line of intersection of the horizontal plane containing the circle of tangency of the sphere and the sides of the cone. Then the distance of P to L is the same as the distance of P' to Q'where P' Q' is an element of the cone parallel to the plane. Again FP' = the distance alone the cone to the horizontal plane. (c) Show that the intersection of a plane and a cone where the plane intersects both nappes of the cone is a hyperbola. \ P | / \ x | / \ xx | / \ x | / \ o o x | / \ o x | / \ o x F| / \ |/ \o o/ \o o/| \ o ox / | \ / | \ / | \ / | \/ x | /\ | / \ | / \ | / \x | / o o \ | /o o\| / \ /o F|\ / | \ / o o | \ / o o | \ / | \ / | \ / | \ Again, place two spheres tangent to the sides of the cone and also tangent to the plane. The point of tangency with the plane are the foci. The differences of the distances PF and PF' is the slant length along the cone between the circles of tangency of the spheres and the cone. ============================================================ New Material Polar Coordinates. The Cartesian Coordinate system is not the only way to locate points on the earth's surface. The other common one is to give the distance and direction from a fixed location. "The storm is five miles south west of Ames." or "The moon is 200,000 miles from the Earth". or "See what is in the bottom 1/3 of the windshield. That is how far you can glide when the engine stops". Points are located as (r,theta) where r is the distance from the origin and theta is the angle. P | / | /__ | r/|`. | / . | / . |/theta . --------------------+------------------------------------- | | | | Given any r and any theta, the point P is uniquely determined. However, the same P can have lots of ways of being represented. For example, these all refer to the same point. (r,theta) (r, theta + 2 Pi) (r, theta + 4 Pi) (-r, theta + Pi) In general, when we are graphing, If a point is on the graph for one representation of (r,theta) then we say that the point is on the graph. It is possible that two graphs will appear to intersect, but if you try to find the (r,theta) where they intersect, there will not be any. The point has different coordinates on each graph. Never-the-less, we say the graphs intersect. As one of those inconsistencies that keep things from going smoothly. Polar coordinates seem most compatible with vectors because they specify a point by giving its "magnitude and direction". Yet one never uses vector addition (r1,theta1)+(r2,theta2) = (r1+r2,theta1+theta2) for anything. The correspondence with rectangular coordinates is r = Sqrt[x^2 + y^2] x = r Cos[theta] theta = ArcTan[y/x] y = r Sin[theta] But it is really rare that one wants to convert polar to rectangular. Ordinarily, if the problem has a lot of cylinders, or spheres, or circles, then one simply sets it up as a polar problem. It is unnecessarily complicated to go from one formulation to another. In fact, the usual process is to take the rectangular formulation, recreate the original problem, and then set the original problem into polar notation. Here are the basic equations. r = 3 Circle o theta = 30 Line through origin r = Cos[theta] Circle r = 1 + Cos[theta] Cardioid r = 2 + Cos[theta] Limacon 2 r = Cos[theta] r = Cos[2 theta] Four leaf rose r = Cos[3 theta] Three leaf rose r = theta Spiral of Archimedes r theta = 1 Hyperbolic spiral de r = ------------- Conic section 1+eCos[theta] ============================================= p[r_,t_] := {r Cos[t],r Sin[t]}; ans = ParametricPlot[ p[1+Cos[t],t],{t,0,2 Pi}]; bns = Show[ans,PlotLabel->"r = 1 + Cos[t]",AspectRatio->Automatic]; Display["p1.ps",bns]; ans = ParametricPlot[ p[2+Cos[t],t],{t,0,2 Pi}]; bns = Show[bns,PlotLabel->"r = 2 + Cos[t]",AspectRatio->Automatic]; Display["p2.ps",bns]; ans = ParametricPlot[ p[Sqrt[Cos[t]],t],{t,0,2 Pi}]; bns = ParametricPlot[ p[-Sqrt[Cos[t]],t],{t,0,2 Pi}]; cns = Show[ans,bns,PlotLabel->"r^2 = Cos[t]",AspectRatio->Automatic]; Display["p3.ps",cns]; ans = ParametricPlot[ p[Cos[2 t],t],{t,0,2 Pi}]; cns = Show[ans,PlotLabel->"r = Cos[2t]",AspectRatio->Automatic]; Display["p4.ps",cns]; ans = ParametricPlot[ p[Cos[3 t],t],{t,0,2 Pi}]; cns = Show[ans,PlotLabel->"r = Cos[3t]",AspectRatio->Automatic]; Display["p5.ps",cns]; ans = ParametricPlot[ p[t,t],{t,0,2 Pi}]; cns = Show[ans,PlotLabel->"r = theta]",AspectRatio->Automatic]; Display["p6.ps",cns]; ans = ParametricPlot[ p[1/t,t],{t,Pi/4,10 Pi}]; cns = Show[ans,PlotLabel->"r theta = 1]",AspectRatio->Automatic,PlotRange->All]; Display["p7.ps",cns]; d = 3; e = 1/2; ans = ParametricPlot[ p[d e/(1 + e Cos[t]),t],{t,0,2 Pi}]; cns = Show[ans,PlotLabel->"ellipse r = d e/(1+e Cos[t]) ]",AspectRatio->Automatic]; Display["p8.ps",cns]; d = 3; e = 3/2; ans = ParametricPlot[ p[d e/(1 + e Cos[t]),t],{t,0,2 Pi}]; cns = Show[ans,PlotLabel->"hyperbola r = d e/(1+e Cos[t]) ]",AspectRatio->Automatic]; Display["p9.ps",cns]; d = 3; e = 1; ans = ParametricPlot[ p[d e/(1 + e Cos[t]),t],{t,- 5 Pi/6 , 5 Pi/6 }]; cns = Show[ans,PlotLabel->"parabola r = d e/(1+e Cos[t]) ]",AspectRatio->Automatic]; Display["p10.ps",cns]; --------------------------------------------------- The key to the whole system is r = Cos[theta]. Understand why this is a circle. Then all the other graphs will follow. | o . o | o .'\ /. o | o r.' \/ `. o |o .' `.o o .t' `. --------------------------------o<----- 1 -----------o------ o o |o o | o o | o o o o This is a unit circle tangent to the y-axis. The angle inscribed in a semi-circle is a right angle. For points on this circle the right triangle has hypotenuse of length one and the adjacent side of theta is r. Therefore: r = Cos[t]. Now getting down the the fine points. As t goes from 0 to 2 Pi, one goes around the circle twice. At t=0, one is at the far right hand point. One travels along the top portion reaching the origin when t=Pi/2. After t=Pi/2, the cosine turns negative. the points which are generated for the second quadrant, will be plotted in the fourth quadrant. One reaches the far right hand side at when t = Pi. One reaches the origin for the second time when t = 3 Pi/2, and One finishes up back at the far right hand side with t = 2 Pi. If one thinks of starting and stopping at the origin, then one makes the first loop from -Pi/2 to Pi/2, and the second loop from Pi/2 to 3 Pi/2. --------------------------------------------------------------- Use the graph to r = Cos[t] to understand the other curves. Graph: r = 1 + Cos[t] We first sketch r = Cos[t] and then go around and add one to the radius all the way around. Try it. x x x x x | o o | o o x x | o o |o o x o o --------------------------------x---------k----------o------ x x o o |o o x | o o | o o x x o o x x x x This is a Cardioid. Notice that the curve just touches the origin on the back side. That is because 1+Cos[t] can be zero only for one value, namely when t = 3 Pi/2 and the Cos[t] is -1. --------------------------------------------------------------------- If we used another value other than 1, the graph r = a + Cos[t] is called a limacon. If a > 1, the graph will never pass through the origin. The graph will look somewhat like a squashed circle. ------------------------------------------------------------------ When 0 < a < 1, the graph will have an inner loop. ---------------------------------------------------------------- 2 Graph: r = Cos[t] Modify r = Cos[t] by taking the square root of r. ------------------------------------------------------------------- Graph: r = Cos[2 t] Identify the beginning and end of each loop. The loops begin and end when r = 0. This happens when 2t = Pi/2 and 2t = 3 Pi/2 and 2t = 5Pi/2 and 2t = 7 Pi/2. Thus the curve r = Cos[2t] reaches the origin at Pi/4, 3Pi/4, 5Pi/4, 7Pi/4. 9 Pi/4 There is a loop between each of these angles. Four loops in all. 3Pi/4 \ ... / Pi/4 \ . | . / \. | ./ . . \.|./ . . . .\|/. . -------------------.-------+--------.------------------ . ./|\. . . . /.|.\ .. /. | .\ / . | . \ / ... \ 5 Pi/4 7 Pi/4 This is a four leaf rose. ------------------------------------------------------------- Where will the leaves take place for r = Cos[3 t]? Draw these angles and make the graph. | | | 2Pi/3 | \ | / Pi/3 .''. | / 5 Pi/6. . \ .| / . Pi/6 ' . .\.| / .. ' ' .\|/..'' ' ---------------------------+---------)----------------- . './|\ ''... ' .' .' /.| \ ' . . ' . / .| \ ' 11 Pi/6 7Pi/6 . . | \ /' | \ 4 Pi/3 | 5 Pi/3 | | | | ------------------------------------------------------------ The two spirals are r = theta and r theta = 1. Graph them.