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 8 12.5, 12.6 Rotationn and Polar Coordinates p539: 1-4,(11) p544: 1,5,7,9,11-14,17-21,(30),33,(38) Main Idea: 10 miles that-a-way. Key Words: polar coordinates, limacon, cardioid, spiral Goal: learn rotation of axes and polar coordinates. ------------------------------------------------------------------ Previous assignment: Page 530 Problem 3 Find the equation of points P satisfying the difference of the distances of P from (+/- 7,0) is 12. o. .o o. .o o ._________________. o o| . . |o o . . o ----------------F--o--------0--------o--F-------------- (-7,0)o . . o(7,0) o| . . |o o |_________________| o o. .o o. .o c = 7 a = 6 b = Sqrt[49-36] = Sqrt[13] 2 2 x y ------- - ------ = 1 36 13 -------------------------------------------------------- Page 530 Problem 31 Listeners A(-8,0), B(8,0), and C(8,10) recorded the exact times at which they heard an explosion. If B and C heard the explosion at the same time and A heard it 12 seconds later, where was the explosion. Assume that distances are in kilometers and that sound travels 1/3 kilometer per second. Clearly, the point is on the perpendicular of the segment BC which means that the y coordinate of the location of the explosion has to be 5 kilometers | C (8,10) | | | . | . o | . o Sqrt[60] . o ____________________ | . | . | o | . | . | | . | . |o | . | . | -----A----------|---------+---------o----------B---------- (-8,0) (-2,0) . | . (2,0) (8,0) | . | . |o | . | . | | . | . | o |___________________| -Sqrt[60] . o . o . o . 30-18 = 12 seconds. c = 8 a = 2 b = Sqrt[64-4] = Sqrt[60] x^2 y^2 ---- - ----- = 1 4 60 when y = 5 x^2 25 ------ - ---- = 1 4 60 x^2 85 ----- = ---- 4 60 4*85 17 x^2 = ------ = ---- 60 3 x = Sqrt[17/3] = 2.38048 kilometers. y = 5 Answer (Sqrt[17/3], 5) Sqrt[ (Sqrt[17/3]+8)^2 + 25 ] = 11.5219 km or 34.5657 seconds Sqrt[ (Sqrt[17/3]-8)^2 + 25 ] = 7.5219 km or 22.5657 seconds ----------------------------------------------------------------------- Page 535 Problem 34 Find the focus and directrix of the parabola 2 x - 6 x + 4 y + 3 = 0 2 x - 6 x = -4 y - 3 2 x - 6 x + 9 = -4 y - 3 +9 2 (x-3) = -4(y-3/2) p = 1 The vertex is at (3,3/2) The focus is at (3,1/2) The directrix is at (3,5/2). --------------------------------------- p1 = Plot[ (x^2-6x+3)/(-4),{x,-2,8}]; p2 = ListPlot[{{3,1/2}, {5,1/2},{1,1/2}},PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; p3 = Plot[1/2,{x,1,5},PlotStyle->{RGBColor[1,0,0]}]; p4 = Plot[5/2,{x,-2,8},PlotStyle->{RGBColor[0,0,1]}]; p5 = Show[p1,p2,p3,p4,PlotLabel->"P535 P34",AspectRatio->Automatic,PlotRange->All]; Display["34.ps",p5]; ---------------------------------------- Page 535 Problem 36 Find the equation of the given conic. Hyperbola with center (2,-1), vertex at (4,-1) and focus at (5,-1) | | | | ---------------------------+-------------------------------- | (2,-1) (4,-1)F | | | | | | | | | C V (5,-1) | | a = 2 c = 3 b = Sqrt[9-4] = Sqrt[5] 2 2 (x-2) (y+1) ------ - -------- = 1 4 5 ----------------------------------------------------- Page 535 Problem 40 Ellipse with foci at (2,0) and (2,12) and a vertex at (2,14) | (2,14) V | | (2,12) F | | | | ----------------------------+------(2, 0)F-------------------- | | center at (2,6) c = 6 a = 8 b = Sqrt[64-36] = Sqrt[28] 2 2 (x-2) (y-6) -------- + ------- = 1 28 64 -------------------------------------------------------- Page 535 Problem 44 Hyperbola with foci (0,0) and (0,4) that passes through (12,9) Center at (0,2) c = 2 a^2 + b^2 = 4 2 2 x (y-2) - ----- + --------- = 1 2 2 b a - 144 49 = 1 ---- + ---- 2 2 b a 2 2 a + b = 4 Solve[{ -144/b^2 + 49/a^2 ==1, a^2 + b^2 ==4}] a = 1, b= Sqrt[3] 2 2 x (y-2) - ----- + ----- = 1 3 1 Check (12,9) -144/3 + 7^2 = -48+49 = 1. a^2 + b^2 = 4. ========================================================= New Material Rotation of Axes. x = u Cos[t] - v Sin[t] y = u Sin[t] + v Cos[t] Example 1. xy = 1. Rotate the axis by 45 degrees. 2 2 u /2 - v /2 = 1 -------------------------------------------------- The general quadratic 2 2 Ax + Bxy + Cy + Dx + Ey + F = 0 x = u Cos[t] - v Sin[t] y = u Sin[t] + v Cos[t] The new equation is 2 2 au + buv + cv + du + ev + f = 0 2 2 Where b = B(Cos[t] - Sin[t] ) - 2(A-C) Sin[t] Cos[t] To make b = 0 b = B Cos[2t] - (A-C) Sin[2t] If B Cos[2t] = (A-C) Sin[2t] then Ctn[2t] = (A-C)/B Example 2: What angle should be used to eliminate the xy-term in 2 2 4x + 2 Sqrt[3] xy + 2 y + 10 Sqrt[3] x + 10 y = 5 4-2 Ctn[2t] = ------------ = 1/Sqrt[3] 2t = 60 degrees so t = 30 degrees. 2 Sqrt[3] -------------------------------------------------------------------- 2 Show that B - 4 A C is invariant under rotation. This tells us what type of quadratic the curve is. ---------------------------------------------------------- Polar graphing. Graph theta = Pi/4 r = 3 r = theta r = Cos[t] r = Sin[t] r = 1 + Cos[t] r theta = 1