Mth 211/213 Clculus -V Kenneth Mssey Crson-Newmn University September 17, 2018 C-N Mth 211 - Mssey, 1 / 1 Directions You re t the origin nd giving directions to the point (4, 3). 1. n Mnhttn: go est 4 blocks, then north 3 blocks. 2. Supermn: ngle nd distnce C-N Mth 211 - Mssey, 2 / 1 Converting between (x, y) C nd (θ, r) P. r 2 = x 2 + y 2 x = r cos θ y = r sin θ tn θ = y x C-N Mth 211 - Mssey, 3 / 1
Alises Consider the point ( π 6, 8) P 1. Crtesin coordintes (unique) 2. List some lises in polr coordintes. 3. The set of ll lises is ( π 6 + kπ, ( 1)k 8 ) P, k Z C-N Mth 211 - Mssey, 4 / 1 Conversion When converting to polr, mke sure the qudrnts mtch. 1. ( π 4, 6) P 2. (2, 1) C 3. ( 1, 1) C 4. ( π 2, 1) P 5. ( 5π 6, 3) P 6. ( 3, 1) C C-N Mth 211 - Mssey, 5 / 1 Polr Grphs The grph of n eqution/inequlity is the set of points tht stisfy it. (Geogebr or Desmos) 1. r = 2 2. θ = 2 3. r < 3 4. 1 r 2 π 5. 6 θ π 3 6. r = θ 7. r = 1 θ, θ [1, 6π] 8. r = 2 sin(θ) 9. r = sin(2θ) 10. r = 1 sin θ+2 cos θ (convert to Crt.) 11. r 2 sin θ cos θ = 1 (convert to Crt.) 12. (x 2) 2 + y 2 = 4 (convert to polr) C-N Mth 211 - Mssey, 6 / 1
Nme tht Shpe Substitute nd complete the squre to identify the grph of r = cos θ + b sin θ. r r 2. = (x /r) + b(y/r) = x + by C-N Mth 211 - Mssey, 7 / 1 Tngents By the chin rule, dy dy dx = dθ. dx dθ Given point (x 0, y 0 ), nd slope m = dy dx, θ the tngent line is y = y 0 + m(x x 0 ) horizontl tngent = dy dθ = 0 (my need WA to solve) verticl tngent = dx dθ = 0 C-N Mth 211 - Mssey, 8 / 1 Crdioid The grph of r = 1 sin(θ) is [crdioid]. 1. Plot points nd sketch. 2. Corresponding prmetric equtions: x = (1 sin(t)) cos(t) y = (1 sin(t)) sin(t) 3. Find the tngent line when θ = π 4. 4. Find ll verticl or horizontl tngents. C-N Mth 211 - Mssey, 9 / 1
Lineriztion Consider polr function r(θ). Suppose r(1) = 8 nd dr θ=1 = 3. 1. Find dx 2. Find eqn tngent line where θ = 1. 3. Estimte the point (x (1.05), y(1.05)). r dr dθ θ dθ, dy dθ dθ, nd dy dx. C-N Mth 211 - Mssey, 10 / 1 Prbol y = x 2 for x [0, 2] just sttic pth x indep., y dep. x = t, y = t 2 for t [0, 2] dynmic motion long pth t indep., both x nd y dep. mke it go bckwrds, nd tke 5 seconds (2.4t, (2.4t) 2 ) mke it strt t (0, 0), nd oscillte bck-nd-forth forever C-N Mth 211 - Mssey, 11 / 1 Crdiod r = 1 cos(θ) for θ [0, 2π] just sttic pth nchored to origin x = (1 sin(t)) cos(t) y = (1 sin(t)) sin(t) dynmic motion, esier to modify move cusp to (3, 4), nd t [0, 1] no longer function r(θ) (RFOT) x (t) = (1 sin(2πt)) cos(2πt) + 3 y(t) = (1 sin(2πt)) cos(2πt) + 4 C-N Mth 211 - Mssey, 12 / 1
t D is the independent prmeter in the domin D (think time or thet) x (t ) gives dependent x s function of t y(t ) gives dependent y s function of t s coordintes (x (t ), y(t )) x (t ) or vector y(t ) generlizes y(x ) nd r (θ), nd to higher dimensionl spce C-N Mth 211 - Mssey, 13 / 1 Unit Circle Describe the unit circle 1. implicitly x 2 + y 2 = 1 2. two functions y = ± 1 x 2 3. polr r = 1 4. prmetriclly [x = cos(t ); y = sin(t )] independent vrible t [0, 2π] initil/terminl point (0, 1) counter-clockwise ngulr velocity ω = 1 rdin/sec. How would you chnge rdius? center? C-N Mth 211 - Mssey, 14 / 1 Vritions 1. Experiment with softwre to control rdius center ω period 2π, frequency 2π, speed ωr ω strting position clockwise / counter-clockwise 2. ellipse 3. spirl x h 2 C-N Mth 211 - Mssey, 15 / 1 + y k b 2 =1
Prmetric Grphs Any function y = f (x ) cn be prmeterized by x = t, y = f (t). f x = cos(t), y = sin 2 (t), write y s function of x nd stte the domin. GeoGebr: x = sin(2t), y = cos(t) for 1± t [0, 2π]. y = ± 1 x 2 2 Mke tble of points to sketch by hnd. e.g. (cos(3t), sin(t)) C-N Mth 211 - Mssey, 16 / 1 Chnging Forms prmetric x = t + 1, y = t eliminte prmeter solve for t, then substitute function y = x 2 1 implicit x 2 y 2 = 1 prmetric x = t 2 + 1, y = t or x = t, y = t 2 1 or x = sec(t), y = tn(t) C-N Mth 211 - Mssey, 17 / 1 Exmples Describe the pth s function y = f (x ), nd stte the domin. [ ] [ ] t t 2 1. 3. for t 2 t t [ ] [ ] t + 1 cos(t 2. t 3 4. 2 ) sin(t 2 ) C-N Mth 211 - Mssey, 18 / 1
Polr to Prmetric Exmple: r = θ, θ [0, ) Often r is written s function of θ. Corresponding prmetric equtions: x (θ) = r(θ) cos θ y(θ) = r(θ) sin θ Sketch by plotting points. symmetric bout the x -xis if r(θ) is even symmetric bout the y-xis if r(θ) is odd C-N Mth 211 - Mssey, 19 / 1 Lines 1. Prmeterize y = 3 + 2(x 1) x = t, y = 3 + 2(t 1) x = t + 1, y = 3 + 2t x = t 2 + 1, y = 3 + 2t 2 x = cos(t) + 1, y = 3 + 2 cos(t) if y = t, wht is x? 2. The line segment (x 0, y 0 ) (x 1, y 1 ) cn be prmeterized for t [0, 1] by x (t) = x 0 + (x 1 x 0 )t y(t) = y 0 + (y 1 y 0 )t C-N Mth 211 - Mssey, 20 / 1 Cycloid [Pth] of point on wheel with rdius r feet, rolling horizontlly t h feet per second. center (ht, r) frequency h 2πr 2πr ; period h sec. ngulr velocity ω = h r rdins per sec. ngle θ = ωt = ht/r point reltive to center ( r sin(θ), r cos(θ)) prmetric equtions x (t) = ht r sin(ht/r) y(t) = r r cos(ht/r) C-N Mth 211 - Mssey, 21 / 1
Prmetric Clculus llustrte with x (t) = cos(t), y(t) = sin(2t). nottion ẋ = dx chin rule dy d 2 y dx 2 = d dx ( dy dx = dy dx dx ) = d ( dy dx ) dx, so dy dx = dy dx re A = x 1 x 0 ydx = t 1 t 0 y dx C-N Mth 211 - Mssey, 22 / 1 Length of Curves Segment: distnce over time is Tking limits, define speed s(t) = (dx rc length L = Specil cses: y(x ), L = r(θ), L = b b C-N Mth 211 - Mssey, 23 / 1 t1 ) ( ) 2 2 + dy t 0 s(t) 1 + (dy/dx )2 dx r 2 + (dr/dθ) 2 dθ ( x )2 + ( y) 2 t Are Enclosed by Curves Tke cre for ± re, intersections, overlp. b For simple closed curve: A = ydx 1 Polr with boundry rys: A = b 2 r 2 dθ (integrte circulr sectors with re θ 2 r 2 ) e.g. between x -xis nd r = θ. Show tht re works out to π3 6 both wys. C-N Mth 211 - Mssey, 24 / 1
Exmples 1. perimeter of region bdb y = 1 x 2, y = 0 2. perimeter of r = 5 sin(3θ) 3. x = t 2, y = t 3 4t, t [0, 2] Use the speed (squred) function to find fstest, slowest. Find pth length nd re between curve nd x -xis. 4. show tht circumference of unit circle is 2π 5. re of ellipse is πb 6. [x = e t cos(t), y = e t sin(t)], t [0, T ] find speed, rc length s functions of T 7. cycloid rc length is 8r, enclosing re 3πr 2 (indep. of h) C-N Mth 211 - Mssey, 25 / 1 Fishing t [0, 10] x = 4t 2 40t + 100 y = t 3 15t 2 + 59t 1. tble/plot points t = 0, 1, 2, 5, 8, 9, 10 2. find dx, dy, dy dx, nd d 2 y dx 2 3. tngent line nd concvity where t = 2 4. where is tngent either horiz or verticl? 5. where does tngent hve slope 1? 6. enclosed re 7. speed when t = 2 8. totl rc length C-N Mth 211 - Mssey, 26 / 1 Polr Are/Length C-N Mth 211 - Mssey, 27 / 1
Length Theorem 1 Arc length of the curve r = f (θ), θ b b ( ) 2 dr L = r 2 + dθ dθ llustrte with picture, or derive vi prmetric eqns. Verify re nd circumference of circle. Perimeter of crdioid r = 1 sin θ. C-N Mth 211 - Mssey, 28 / 1 Are Theorem 2 The re bounded by r = f (θ), θ =, θ = b b 1 A = 2 r 2 dθ llustrte with picture. Find re of crdioid r = 1 sin θ. C-N Mth 211 - Mssey, 29 / 1 Crft Your Own 1. Design prmetric equtions to produce this horseshoe pth for t [0, 10]. (sketch x nd y s functions of t) 2. Let R be region below curve nd bove x -xis. Find the perimeter nd re of R. (note +/ re just works!) 3. Tn. line thru (0, 12) with negtive slope. C-N Mth 211 - Mssey, 30 / 1
For Everything Else Find the re of intersection of the two circles: r = 1 nd r = 2 cos θ. ( ) π π 2 1 2 (2 cos θ)2 dθ 2 6 + [WA] π 3 π 3 1 π π 2 (2 cos θ)2 1 2 (1)2 dθ [WA] 3 Wht bout the re of the circles union? C-N Mth 211 - Mssey, 31 / 1