Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.03 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. April 4 12.1, 12.2 Conic Sections p520: 1-4,7,8,(14),17-22,(26),31,41 p526: 1-4,(10),(14),17-19,31,38 Main Idea: The entire curve is determined by just three points. Key Words: Parabola, Hyperbola, Ellipse, Eccentricity. 2 2 2 2 x y x y y = 2 px ----- + ----- = 1 ------ - ------ = 1 2 2 2 2 a b a b Goal: Learn to relate the equations of the conics to their properties. ------------------------------------------------------------------ Previous assignment: March 28 11.1 Taylor Approximation p485:1,(3),(5),9-12,17,(18),20,27,28,(39) Page 485 Problem 3 Find the Maclaurin polynomial of order 4 for f(x) and use it to approximate f(0.12). f(x) = Sin[2x]. f(x) = (2x) - (2x)^3/3! f(0.12) = 0.237696 Sin[2*0.12] = 0.237703 -6 Error = -6.62643 10 -------------------------------------------------------- f[x_] := 2x -(2x)^3/3!; Print[f[0.12]]; a = Plot[f[x],{x,-Pi,Pi}]; b = Plot[Sin[2x],{x,-Pi,Pi},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P485 P3, red=Sin[2x],black=series"]; Display["3x.ps",c]; d = Plot[f[x]-Sin[2x],{x, -1/2,1/2},PlotRange->All]; e = Show[d,PlotLabel->"P485 P3, Series-Sin[2x]"]; Display["3y.ps",e]; ------------------------------------------------------------ Page 485 Problem 5 Find the Maclaurin polynomial of order 4 for f(x) and use it to approximate f(0.12). f(x) = ln(1+x) f(x) = x-x^2/2+x^3/3-x^4/4 f[0.12] = 0.113324 ln(1.12) = 0.113329 -6 error = -4.52531 10 ------------------------------------------ f[x_] = x-x^2/2+x^3/3-x^4/4; Log[1.12] f[x]-Log[1.12] a = Plot[f[x],{x,-Pi,Pi}]; b = Plot[Log[1+x],{x,-Pi,Pi},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P485 P3, red=Log[1+x],black=series"]; Display["5x.ps",c]; d = Plot[f[x]-Log[1+x],{x, -1/2,1/2},PlotRange->All]; e = Show[d,PlotLabel->"P485 P5, Series-Log[1+x]"]; Display["5y.ps",e]; ------------------------------------------------------------ Page 485 Problem 18 Find the Maclaurin polynomial of order n (n odd) for Sin[x]. Then use it with n=5 to approximate each of the following. (This example should convince you that the Maclaurin approximation can be exceedingly poor if x is far from zero.) Compare your answers with those given by your calculator. What conclusion do you draw? f[x] = x - x^3/3! + x^5/5! (a) Sin[0.1] = 0.0998334 (b) Sin[0.5] = 0.479427 (c) Sin[1.0] = 0.841667 (d) Sin[10.0] = 676.667 Here are the four errors. -11 -6 {1.98385 10 , 1.54473 10 , 0.000195682, 677.211} ----------------------------------------------- f[x_] := x - x^3/3! + x^5/5!; a = Plot[f[x],{x,-10,10}]; b = Plot[Sin[x],{x,-10,10},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P485 P18, red=Sin[x],black=series"]; Display["18x.ps",c]; --------------------------------------------------------- Page 485 Problem 39 Determine the order n of the Maclaurin polynomial for e^x that is required to approximate e to five decimal places, that is, so that |Rn[1]| < 0.000005. Going back to the definition of e^x, we have to look for the ln(x). The area under the curve y = 1/x is the ln(x). And e^x is the inverse function of the ln(x). | ` 1/2 | |`. | |__`_ 1/3 | |////|`. | |////|__`_ 1/4 | |////|////|`_._ | |////|////|///|` .... ----------------------+------------------------------------- | 1 2 3 4 5 6 | ln(4) >= 1/2 + 1/3 + 1/4 = 13/12 > 1 ln(4) 1 Therefore e > e so 4 > e. (n+1) 1 f (c) n+1 e 4 Error = ------------- x <= ----------- <= ------------ (n+1)! (n+1)! (n+1)! n = 1 2 3 4 5 6 7 8 9 n+1 = 2 3 4 5 6 7 8 9 10 (n+1)! = 2 6 24 120 720 5,040 40,320 362,880 3,628,800 | | 800,000 4 Need ------- < 0.000005 (n+1)! 4 Need ---------- < (n+1)! 0.000005 6 4*10 Need -------- = 800,000 < (n+1)! 5 So you have to use n = 9 f[x_] = Sum[x^n/n!,{n,0,9}]; 98641 f[1] =-------------- = 2.718281526 36288 e = 2.718281828 | five decimal places -7 Error = -3.028858528 10 -------------------------------------------------------------------------------------- New Material: Conics: Parabola, Hyperbola, Ellipse. Parabola First description. The trace when a cone is cut by a plane parallel to the edge of the cone. . ' '. \. ./ \ ' ' / \ / \ / /\ X / \ / \ / / / \ / / / * / / /* */ \ '/* */ .' / ''/''' / / \ / \/ Second description: Parallel light rays are focused at a point. * * *__________________/_________ * \ \ * \ * F * / */_______________________/__________ * \ * Third description: The locus of all points equidistant from a fixed point F (called the Focus) and a fixed line L (called the directrix). |___________P=(x,y) | c \ | \ | c\ | \ | F (p,0) | | | | |x=-p Equation: Let the directrix be x = -p. Let the focus be (p,0) The equation for the equality of the distances is x+p = Sqrt[ (x-p)^2 + y^2 ] x^2 + 2 p x + p^2 = x^2 -2 p x + p^2 + y^2 4 p x = y^2 ################################ # Equation of a parabola # # Focus (0,p) # # directrix x=-p # # # # 2 # # y = 4 p x # # # ################################ 2 Example 1 Page 518: Graph y = 12 x p = 3 * | * | | * | | * 6 |* | | | ---------------*-----3----F--------------- | | |* 6 | * | | * | | * | | * 2 Example 2 Page 519 Graph x = -y. Solution: p = 1/4 The curve opens downward | | | ----------------------*----------------------------- * | * * 1/4 * * | * *-1/2---F---1/2-* * | * | | | | Example 3 Page 519. Find the equation of the parabola with vertex at the origin and focus at (0,5). 2 p = 5 x = 20 y Example 4 Page 519. Find the equation of the parabola with vertex at the origin that goes through (-2,4) and opens left. | * | * | (-2,4)* | * | *| | ---------------------------------*------------------------ | *| * | * | * | * | | 2 y = -4 p x 16 = -4 p (-2) p = 2. 2 y = -8x solution. ============================================================== ELLIPSE: First description: The trace when a cone is cut by a plane that cuts completely across one nappe. . . ' ` \`. .'/ \ ' ' / \ / \ / \ / \ / X . / \ . . / \ . / . \ . / . \ . / . . \ . /' ` \ . . /` ` \ . . / ` `\ . . / ` ` . . ` ` . / . ` . .'\ . . ` ' . . . . . . . . Second description. Light rays emitting from one focus. pass through the other focus. _________________________________________________________ | | | | | . . . | | . ' ` o . | | . ' o o ` . | | . ' o o ` . | | . o o ` . | |. o o `. | |. F F' .| | . o o .| | o o . | | ` . o o . ' | | ` . o o . ' | | ` . o o . ' | | ` . o . ' | | ` ` ` ` | |_______________________________________________________| Third description: The locus of points the sum of the sum of the distances from two fixed points are constant. . . . P(x,y) . ' ` o . . ' o o ` . . ' o o ` . .' o o ` . . o o `. . F(-c,0) F'(c,0) . . . . . ` . . ' ` . . ' ` . . ' ` . . . . . ' |PF| + |PF'| = 2a Sqrt[( (x+c)^2 + y^2 ] + Sqrt[(x-c)^2 + y^2] = 2 a Sqrt[( (x+c)^2 + y^2 ] = 2 a - Sqrt[(x-c)^2 + y^2] x^2 + 2 c x + c^2 + y ^2 = 4 a^2 - 4 a Sqrt[(x-c)^2 + y^2) + x^2 -2 c x + c^2 + y^2 4 c x - 4 a^2 = -4 a Sqrt[(x-c)^2 + y^2 ] c x - a^2 = - a Sqrt[(x-c)^2 + y^2 ] c^2 x^2 - 2 a^2 c x + a^4 = a^2( x^2 - 2 c x + c^2 + y^2 ) a^4-a^2 c^2 = (a^2-c^2) x^2 + a^2 y^2 a^2(a^2-c^2) = (a^2-c^2) x^2 + a^2 y^2 x^2 y^2 1 = --------- + ------------ a^2 a^2-c^2 Letting b^2 = a^2 - c^2 we write the equation as ############################################################### # The equation of the ellipse with Foci (-c,0) (c,0) and # # semi major axis a and semi minor axis b. b^2 + c^2 = a^2 # # # # 2 2 # # x y # # 1 = --------- + ------------ # # 2 2 # # a b # # # ############################################################### 2 2 When x = c then a^2-b^2 y^2 1 = --------- + ------- a^2 b^2 So y^2 = b^4/a^2 or y = b^2/a Page 524 Example 1. Graph 2 2 x y --------- + --------- = 1 36 4 a = 6 b = 2 c = Sqrt[32] = 4 Sqrt[2] = 5.66 6 | 6 ____________________________________________________________ | o | o | | o | o | | o |2 | | | o | | 4/6| o | | o |4/6 | | | |o | | 6 | o| |-----|-----------------------+-----------------------|----| |o | 4 Sqrt[2] | 4 Sqrt[2] | o| | o |4/6 | 4/6| | | o | | | o | | o |2 o | | | | |__________________o__________|______________o_____________| 6 | 6 | ---------------------------------------------------------------- a = Plot[ Sqrt[ 4(1- x^2/36)],{x,-6,6}]; b = Plot[-Sqrt[ 4(1- x^2/36)],{x,-6,6}]; c = 4 Sqrt[2]; d = ListPlot[{{0,0},{0,2},{0,-2},{6,0},{-6,0},{c,0},{c,2/3},{c,-2/3},{-c,0},{-c,+2/3},{-c,-2/3}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; e = Show[a,b,d,PlotLabel->"x^2/36 + y^2/4 = 1",PlotRange->All,AspectRatio->Automatic]; Display["ex1.ps",e]; ==================================================================== HYPERBOLA First description: The trace when a cone is cut by a plane that cuts both nappes. \ _________________ / \ | | / \ | | / \ |* *|/ \| | | * */| |\ / | | \ * * / | | \ * / | | \ / | | \ / | | \ / | | \/ | | /\ | | / \ | | / \ | | / \ | | / * \ | | / * * \ | |/ \ | | * * \| /| | / |* * |\ / | | \ / | | \ / |_______________| \ Second description. The time delay from the two listening posts is a constant. \ / \ / \ /P(x,y) \ ./o . \ . /o . . \ /o . . \ /o . (-c,0) F X o F' (c,0) / \ o / \ o / \ o / \ o / \ / \ / \ | PF | - | PF' | = 2 a | Sqrt[(x+c)^2 + y^2] - Sqrt[(x-c)^2 + y^2] | = 2 a Expanding out gives x^2 y^2 ------- - ----------- = 1 a^2 b^2 ################################################################### # The equation of the hyperbola with foci (-c,0) and (c,0) with # # semi major axis a and semi minor axis b where c^2 + b^2 = a^2 # # # # x^2 y^2 # # ------- - ----------- = 1 # # a^2 b^2 # # # ################################################################### Page 525 Example 3 Graph x^2 y^2 -------- - ------ = 1 9 16 . | . . | . Draw the box. ________________________. | . | . | | . 4 | . | | . | . | ___|___3_______.____3______|___ | . | . | | . | . | | . 4 | . | |_______________________| . | . . | . The ellipse goes inside the box. The Hyperbola is tangent to the side of the box and has the extended diagonals as asymptotes. ---------------------------------------------------------------- a = 3; b = 4; c = 5; f = b^2/a; p = Plot[ Sqrt[16(x^2/9 - 1 )],{x,-8,-3}]; q = Plot[-Sqrt[16(x^2/9 - 1 )],{x,-8,-3}]; r = Plot[ Sqrt[16(x^2/9 - 1 )],{x, 3, 8}]; s = Plot[-Sqrt[16(x^2/9 - 1 )],{x, 3, 8}]; t = Plot[{-b/a x, b/a x},{x,-8,8},PlotStyle->{RGBColor[0,0,1]}]; u = Plot[{b,-b},{x,-3,3},PlotStyle->{RGBColor[0,1,0]}]; v = ParametricPlot[{-3,y},{y,-4,4},PlotStyle->{RGBColor[0,1,0]}]; w = ParametricPlot[{ 3,y},{y,-4,4},PlotStyle->{RGBColor[0,1,0]}]; d = ListPlot[{{a,0},{-a,0},{c,0},{-c,0},{c,f},{c,-f},{-c,f},{-c,-f}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; e = Show[p,q,r,s,t,u,v,w,d,PlotLabel->"x^2/9 - y^2/16 = 1",PlotRange->All,AspectRatio->Automatic]; Display["ex2.ps",e]; ==================================================================== Extra Credit Hand in home work assignment. The equation Sqrt[x] + Sqrt[y] = 1 is a parabola. Find the equation of the directrix and the focus.