# 6.1: Polárne súradnice - matematika

Učebné ciele

• Rozlišujte medzi pravouhlým súradnicovým systémom a polárnym súradnicovým systémom a chápte ich.
• Vyneste body s polárnymi súradnicami na polárnu rovinu.

Viac ako (12 ) kilometrov od prístavu plachetnica narazí na nepriaznivé počasie a je odfúknutá z kurzu vetrom uzlom (16 ) - uzol (pozri obrázok ( PageIndex {1} )). Ako môže námorník naznačiť svoju polohu pobrežnej stráži? V tejto časti budeme skúmať metódu reprezentácie polohy, ktorá sa líši od štandardnej súradnicovej mriežky.

## Vynesenie bodov pomocou polárnych súradníc

Keď uvažujeme o vykreslení bodov v rovine, zvyčajne to myslíme obdĺžnikové súradnice ((x, y) ) v Karteziánska súradnicová rovina. Existujú však aj iné spôsoby zápisu súradnicového páru a iných typov sieťových systémov. V tejto časti si predstavíme polárne súradnice, čo sú body označené ((r, theta) ) a vynesené do polárnej mriežky. Polárna mriežka je reprezentovaná ako séria sústredných kružníc vyžarujúcich z pólu alebo počiatku súradnicovej roviny.

Polárna mriežka je zmenšená ako jednotkový kruh s kladným (x )-os sa teraz pozerá ako na polárnu os a počiatok ako na pól. Prvá súradnica (r ) je polomer alebo dĺžka smerovaného úsečky od pólu. Uhol ( theta ) meraný v radiánoch označuje smer (r ). Pohybujeme sa proti smeru hodinových ručičiek od polárnej osi o uhol ( theta ) a zmeriame smerovaný úsečku s dĺžkou (r ) v smere ( theta ). Aj keď najskôr zmeriame ( theta ) a potom (r ), polárny bod sa napíše ako prvá pomocou súradnice (r ). Napríklad na vykreslenie bodu ( left (2, dfrac { pi} {4} right) ) by sme presunuli ( dfrac { pi} {4} ) jednotky proti smeru hodinových ručičiek. a potom dĺžka (2 ) od pólu. Tento bod je zakreslený do mriežky na obrázku ( PageIndex {2} ).

Príklad ( PageIndex {1} ): Vynesenie bodu na polárnej mriežke

Vyneste bod ( doľava (3, dfrac { pi} {2} doprava) ) na polárnu mriežku.

Riešenie

Uhol ( dfrac { pi} {2} ) sa zistí zametaním proti smeru hodinových ručičiek (90 ° ) od polárnej osi. Bod sa nachádza v dĺžke (3 ) jednotiek od pólu v smere ( dfrac { pi} {2} ), ako je to znázornené na obrázku ( PageIndex {3} ).

Cvičenie ( PageIndex {1} )

Vyneste bod ( doľava (2, dfrac { pi} {3} doprava) ) do polárnej mriežky.

Odpoveď

Príklad ( PageIndex {2} ): Vynesenie bodu v polárnom súradnicovom systéme s negatívnou zložkou

Vyneste bod ( doľava (−2, dfrac { pi} {6} doprava) ) na polárnu mriežku.

Riešenie

Vieme, že ( dfrac { pi} {6} ) sa nachádza v prvom kvadrante. Avšak (r = −2 ). K vykresleniu bodu s negatívom (r ) môžeme pristupovať dvoma spôsobmi:

1. Vynesieme bod ( doľava (2, dfrac { pi} {6} doprava) ) posunutím ( dfrac { pi} {6} ) proti smeru hodinových ručičiek a predĺžením smerovaného úsečky (2 ) jednotiek do prvého kvadrantu. Potom vysledujte nasmerovaný úsečku späť cez pól a pokračujte (2 ) jednotkami do tretieho kvadrantu;
2. Posuňte ( dfrac { pi} {6} ) v smere proti smeru hodinových ručičiek a nakreslite smerovaný úsečku z pólových (2 ) jednotiek v zápornom smere do tretieho kvadrantu.

Viď obrázok ( PageIndex {5a} ). Porovnajte to s grafom polárnej súradnice ( left (2, dfrac { pi} {6} right) ) zobrazenom na obrázku ( PageIndex {5b} ).

Cvičenie ( PageIndex {2} )

Nakreslite body ( doľava (3, - dfrac { pi} {6} doprava) ) a ( doľava (2, dfrac {9 pi} {4} doprava) ) na rovnaká polárna mriežka.

Odpoveď

Keď dostali sadu polárne súradnice, možno ich budeme musieť previesť na obdĺžnikové súradnice. Za týmto účelom si môžeme pripomenúť vzťahy, ktoré existujú medzi premennými (x ), (y ), (r ) a ( theta ).

( cos theta = dfrac {x} {r} rightarrow x = r cos theta )

( sin theta = dfrac {y} {r} rightarrow y = r sin theta )

Vypustenie kolmice z bodu v rovine na X-os tvorí pravý trojuholník, ako je to znázornené na obrázku ( PageIndex {7} ). Ľahký spôsob, ako si spomenúť na vyššie uvedené rovnice, je myslieť na ( cos theta ) ako na susednú stranu cez preponu a ( sin theta ) ako na opačnú stranu nad preponou.

PREVÁDZKA Z POLÁRNYCH KOORDINÁTOV DO OBDĺžnikových súradníc

Ak chcete previesť polárne súradnice ((r, theta) ) na obdĺžnikové súradnice ((x, y) ), nechajte

[ cos theta = dfrac {x} {r} rightarrow x = r cos theta ]

[ sin theta = dfrac {y} {r} rightarrow y = r sin theta ]

1. Vzhľadom na polárnu súradnicu ((r, theta) ) napíšeme (x = r cos theta ) a (y = r sin theta ).
2. Vyhodnoťte ( cos theta ) a ( sin theta ).
3. Vynásobte ( cos theta ) znakom (r ), aby ste našli (x )-súradnica obdĺžnikového tvaru.
4. Vynásobte ( sin theta ) znakom (r ), aby ste našli (y )-súradnica obdĺžnikového tvaru.

Polárne súradnice ( doľava (3, dfrac { pi} {2} doprava) ) napíšte ako obdĺžnikové súradnice.

Riešenie

Použite rovnocenné vzťahy.

[ begin {align *} x & = r cos theta x & = 3 cos dfrac { pi} {2} & = 0 y & = r sin theta y & = 3 sin dfrac { pi} {2} & = 3 end {zarovnať *} ]

Obdĺžnikové súradnice sú ((0,3) ). Viď obrázok ( PageIndex {8} ).

Riešenie

Viď obrázok ( PageIndex {9} ). Zapisovanie polárnych súradníc ako obdĺžnikových, máme

[ begin {align *} x & = r cos theta x & = -2 cos (0) & = -2 y & = r sin theta y & = -2 sin ( 0) & = 0 end {zarovnať *} ]

Obdĺžnikové súradnice sú tiež ((- 2,0) ).

Cvičenie ( PageIndex {3} )

Napíšte polárne súradnice ( doľava (-1, dfrac {2 pi} {3} doprava) ) ako obdĺžnikové súradnice.

Odpoveď

((x, y) = left ( dfrac {1} {2}, - dfrac { sqrt {3}} {2} right) )

Na prevod obdĺžnikových súradníc na polárne súradnice použijeme ďalšie dva známe vzťahy. Pri tomto prepočte si však musíme uvedomiť, že sada obdĺžnikových súradníc prinesie viac ako jeden polárny bod.

Prepočet z obdĺžnikových súradníc na polárne súradnice si vyžaduje použitie jedného alebo viacerých vzťahov znázornených na obrázku ( PageIndex {10} ).

( cos theta = dfrac {x} {r} ) alebo (x = r cos theta )

( sin theta = dfrac {y} {r} ) alebo (y = r sin theta )

(r ^ 2 = x ^ 2 + y ^ 2 )

( tan theta = dfrac {y} {x} )

Riešenie

Vidíme, že pôvodný bod ((3,3) ) je v prvom kvadrante. Ak chcete nájsť ( theta ), použite vzorec ( tan theta = dfrac {y} {x} ). Toto dáva

[ begin {align *} tan theta & = dfrac {3} {3} tan theta & = 1 { tan} ^ {- 1} (1) & = dfrac { pi } {4} end {align *} ]

Aby sme našli (r ), dosadíme hodnoty za (x ) a (y ) do vzorca (r = sqrt {x ^ 2 + y ^ 2} ). Vieme, že (r ) musí byť kladné, pretože ( dfrac { pi} {4} ) je v prvom kvadrante. Teda

[ begin {align *} r & = sqrt {3 ^ 2 + 3 ^ 2} r & = sqrt {9 + 9} r & = sqrt {18} & = 3 sqrt {2 } end {zarovnať *} ]

Takže (r = 3 sqrt {2} ) a ( theta = dfrac { pi} {4} ), čo nám dáva polárny bod ((3 sqrt {2}, dfrac { pi} {4}) ). Viď obrázok ( PageIndex {11} ).

Analýza

Existujú aj ďalšie sady polárnych súradníc, ktoré budú rovnaké ako naše prvé riešenie. Napríklad body ( ľavý (−3 sqrt {2}, dfrac {5 pi} {4} pravý) ) a ( ľavý (3 sqrt {2}, - dfrac { 7 pi} {4} right) ) sa bude zhodovať s pôvodným riešením ( left (3 sqrt {2}, dfrac { pi} {4} right) ). Bod ( doľava (−3 sqrt {2}, dfrac {5 pi} {4} doprava) ) označuje posun ďalej proti smeru hodinových ručičiek o ( pi ), ktoré je priamo oproti ( dfrac { pi} {4} ). Polomer je vyjadrený ako (- 3 sqrt {2} ). Uhol ( dfrac {5 pi} {4} ) sa však nachádza v treťom kvadrante a keďže (r ) je záporné, predĺžime nasmerovaný úsečku v opačnom smere, do prvého kvadrantu. . Toto je rovnaký bod ako ( left (3 sqrt {2}, dfrac { pi} {4} right) ). Bod ( doľava (3 sqrt {2}, - dfrac {7 pi} {4} doprava) ) je o krok ďalej v smere hodinových ručičiek o (- dfrac {7 pi} {4} ), z ( dfrac { pi} {4} ). Polomer (3 sqrt {2} ) je rovnaký.

## Extra prax

1. Zostrojte bod s polárnymi súradnicami ( doľava (3, frac { pi} {6} doprava) ).
2. Vyneste bod s polárnymi súradnicami ( doľava (5, - frac {2 pi} {3} doprava) )
3. Preveďte ( doľava (6, - frac {3 pi} {4} doprava) ) na obdĺžnikové súradnice.
4. Preveďte ( doľava (-2, frac {3 pi} {2} doprava) ) na obdĺžnikové súradnice.
5. Prevod (7, -2) na polárne súradnice.
6. Prevod (-9, -4) na polárne súradnice.

## Kľúčové rovnice

 Prepočítavacie vzorce ( cos theta = dfrac {x} {r} rightarrow x = r cos theta ) ( sin theta = dfrac {y} {r} rightarrow y = r sin theta ) (r ^ 2 = x ^ 2 + y ^ 2 ) ( tan theta = dfrac {y} {x} )

## Kľúčové koncepty

• Polárna mriežka je reprezentovaná ako séria sústredných kruhov vyžarujúcich z pólu alebo počiatku.
• Ak chcete vykresliť bod v tvare ((r, theta) ), ( theta> 0 ), posuňte sa od polárnej osi proti smeru hodinových ručičiek o uhol ( theta ) a potom predĺžte nasmerovaný úsečka od pólu dĺžky (r ) v smere ( theta ). Ak je ( theta ) záporné, posuňte sa v smere hodinových ručičiek a predĺžte nasmerovaný úsečkový segment o dĺžku (r ) v smere ( theta ). Viď príklad ( PageIndex {1} ).
• Ak je záporné (r ), roztiahnite nasmerovaný úsečku v opačnom smere ( theta ). Viď príklad ( PageIndex {2} ).
• Na prevod z polárnych súradníc na obdĺžnikové súradnice použite vzorce (x = r cos theta ) a (y = r sin theta ). Viď Príklady ( PageIndex {3} ) a Príklad ( PageIndex {4} ).
• Ak chcete previesť z pravouhlých súradníc na polárne súradnice, použite jeden alebo viac vzorcov: ( cos theta = dfrac {x} {r} ), ( sin theta = dfrac {y} {r} ), ( tan theta = dfrac {y} {x} ) a (r = sqrt {x ^ 2 + y ^ 2} ). Viď príklad ( PageIndex {5} ).

## 6.1: Polárne súradnice - matematika

Súradnicové systémy sú nástroje, ktoré umožňujú využívať algebraické metódy na pochopenie geometrie. Kým obdĺžnikový (tiež nazývaný Karteziánsky) súradnice, ktoré sme používali, sú najbežnejšie, niektoré problémy sa dajú ľahšie analyzovať v alternatívnych súradnicových systémoch.

Súradnicový systém je schéma, ktorá nám umožňuje identifikovať ľubovoľný bod v rovine alebo v trojrozmernom priestore pomocou množiny čísel. V obdĺžnikových súradniciach sa tieto čísla zhruba interpretujú ako dĺžky strán obdĺžnika. V polárne súradnice bod v rovine je identifikovaný dvojicou čísel $(r, theta)$. Číslo $theta$ meria uhol medzi kladnou osou $x$ a lúčom, ktorý prechádza bodom, ako je znázornené na obrázku 12.1.1, číslo $r$ meria vzdialenosť od začiatku k bodu. Obrázok 12.1.1 zobrazuje bod s obdĺžnikovými súradnicami $ds (1, sqrt3)$ a polárnymi súradnicami $(2, pi / 3)$, 2 jednotky od počiatku a $pi / 3$ radiány od kladného čísla $x$ - os.

Rovnako ako popisujeme krivky v rovine pomocou rovníc zahŕňajúcich $x$ a $y$, môžeme tiež opísať krivky pomocou rovníc zahŕňajúcich $r$ a $theta$. Najbežnejšie sú rovnice v tvare $r = f ( theta)$.

Príklad 12.1.1 Vytvorte graf krivky zadanej pomocou $r = 2$. Všetky body s $r = 2$ sú vo vzdialenosti 2 od počiatku, takže $r = 2$ popisuje kruh s polomerom 2 so stredom v počiatku.

Príklad 12.1.2 Vytvorte graf krivky zadanej pomocou $r = 1 + cos theta$. Najprv zvážime $y = 1 + cos x$, ako na obrázku 12.1.2. Keď $theta$ prechádza hodnotami v $[0,2 pi]$, hodnota $r$ sleduje hodnotu $y$ a vytvára „kardioidný“ tvar z obrázka 12.1.2. Napríklad keď $theta = pi / 2$, $r = 1 + cos ( pi / 2) = 1$, tak nakreslíme bod vo vzdialenosti 1 od počiatku pozdĺž kladnej osi $y$, čo je v uhle $pi / 2$ od kladnej osi $x$. Keď $theta = 7 pi / 4$, $ds r = 1 + cos (7 pi / 4) = 1 + sqrt2 / 2 približne 1,71$ a zodpovedajúci bod sa objaví vo štvrtom kvadrante. To ilustruje jednu z potenciálnych výhod použitia polárnych súradníc: rovnica pre túto krivku v obdĺžnikových súradniciach by bola dosť komplikovaná.

Každý bod v rovine je spojený s presne jedným párom čísel v obdĺžnikovom súradnicovom systéme. Každý bod je spojený s nekonečným počtom párov v polárnych súradniciach. V kardioidnom príklade sme uvažovali iba s rozsahom le theta le2 pi $a už tu bol duplikát:$ (2,0) $a$ (2,2 pi) $sú rovnaký bod. Každá hodnota$ theta $mimo interval$ [0,2 pi) $skutočne duplikuje bod na krivke$ r = 1 + cos theta $, keď le theta Obrázok 12.1.3. Bod$ (- 2, pi / 4) = (2,5 pi / 4) = (2, -3 pi / 4) $v polárnych súradniciach. Vzťah medzi obdĺžnikovými a polárnymi súradnicami je celkom ľahko pochopiteľný. Bod s polárnymi súradnicami$ (r, theta) $má obdĺžnikové súradnice$ x = r cos theta $a$ y = r sin theta $, čo vyplýva bezprostredne z definície sínusových a kosínusových funkcií. Na príklade obrázku 12.1.3 má zobrazený bod obdĺžnikové súradnice$ ds x = (- 2) cos ( pi / 4) = - sqrt2 približne 1,4142 $a$ ds y = (- 2) sin ( pi / 4) = - sqrt2 $. Takto je veľmi ľahké prevádzať rovnice z obdĺžnikových na polárne súradnice. Príklad 12.1.3 Nájdite rovnicu priamky$ y = 3x + 2 $v polárnych súradniciach. Iba nahradíme:$ r sin theta = 3r cos theta + 2 $, alebo$ ds r = <2 over sin theta-3 cos theta> $. Príklad 12.1.4 Nájdite rovnicu kružnice$ ds (x-1/2) ^ 2 + y ^ 2 = 1/4 $v polárnych súradniciach. Opäť nahradenie:$ ds (r cos theta-1/2) ^ 2 + r ^ 2 sin ^ 2 theta = 1/4 $. Trocha algebry to zmení na$ r = cos (t) $. Mali by ste sa pokúsiť vykresliť niekoľko hodnôt$ (r, theta) $, aby ste sa presvedčili, že to má zmysel. Príklad 12.1.5 Vytvorte graf polárnej rovnice$ r = theta $. Tu sa vzdialenosť od počiatku presne zhoduje s uhlom, takže trochu premýšľania objasní, že keď$ theta ge0 $dostaneme špirálu Archimedes na obrázku 12.1.4. Keď$ theta Obrázok 12.1.4. Špirála Archimeda a celý graf $r = theta$.

Prevod polárnych rovníc na obdĺžnikové môže byť o niečo zložitejší a priame vytváranie grafov polárnych rovníc tiež nie je vždy ľahké.

Príklad 12.1.6 Graf $r = 2 sin theta$. Pretože sínus je periodický, vieme, že dostaneme celú krivku pre hodnoty $theta$ v $[0,2 pi)$. Pretože $theta$ beží od 0 do $pi / 2$, $r$ sa zvyšuje z 0 na 2. Potom, ako $theta$ pokračuje v $pi$, $r$ opäť klesá na 0. Keď $theta$ beží od $pi$ do $2 pi$, $r$ je záporné a nie je ťažké vidieť, že prvá časť krivky je jednoducho opäť vysledovateľná, takže v skutočnosti dostaneme celú krivku pre hodnoty $theta$ v $[0, pi)$. Krivka teda vyzerá ako na obrázku 12.1.5. Teraz to naznačuje, že krivka môže byť kruh, a ak je, musel by to byť kruh $ds x ^ 2 + (y-1) ^ 2 = 1$. Po vykonaní tohto odhadu ho môžeme ľahko skontrolovať. Najskôr nahradíme $x$ a $y$, aby sme dostali $ds (r cos theta) ^ 2 + (r sin theta-1) ^ 2 = 1$ rozširovanie a zjednodušenie to skutočne premení na $r = 2 sin theta$.

## Technická matematika, šieste vydanie, Paul A. Calter, Michael A. Calter Ph.D.

Získajte Technická matematika, šieste vydanie teraz s online učením O’Reilly.

Členovia O’Reilly zažijú živé online školenie, plus knihy, videá a digitálny obsah od viac ako 200 vydavateľov.

### 15.6. Grafy v polárnych súradniciach

Až do tohto okamihu sme všetky naše grafy vytvorili v známom obdĺžnikovom súradnicovom systéme. Teraz predstavíme nový súradnicový systém, ktorý je pre niektoré druhy grafov užitočnejší ako obdĺžnikové súradnice. Väčšina našich grafov bude aj naďalej v obdĺžnikových súradniciach, ale v niektorých prípadoch budú pohodlnejšie polárne súradnice.

Polárny súradnicový systém (obr. 15-33) pozostáva z polárnej osi prechádzajúcej bodom O, ktorý sa nazýva pól. Umiestnenie bodu P je dané jeho vzdialenosťou r od pólu, ktorá sa nazýva vektor polomeru, a uhlom θ, ktorý sa nazýva polárny uhol (niekedy sa nazýva vektorový uhol alebo referenčný uhol). Polárny uhol sa nazýva pozitívny, keď sa meria proti smeru hodinových ručičiek od polárnej osi, a negatívny, keď sa meria v smere hodinových ručičiek.

Polárne súradnice bodu P sú teda r a θ zvyčajne písané vo forme P (r, θ) alebo ako (čítať „r pod uhlom θ“).

TI-83/84 graf r = 8 sin θ, kruh. Na kalkulačkách TI je v ponuke polárny režim Pol. Výberom položky ZSquare upravíte mierky tak, aby kruhy boli kruhové.

## Prevod funkcií

. Toto je zobrazené na obrázku nižšie:

komplexné čísla

Aký je polárny tvar komplexného čísla z = 1 + 1 i? z = 1 + 1i? z = 1 + 1 i?

Tento bod môžeme nakresliť na komplexnej rovine:

Z obrázku vidíme, že trojuholník má dve rovnaké strany. Z toho vieme, že musí mať aj dva rovnaké uhly. Pretože jeden z uhlov má 90 90 9 0 stupňov, môžeme bezpečne konštatovať, že θ theta θ sa rovná 45 45 4 5 stupňov alebo π 4 frac < pi> <4> 4 π radiány. Potom môžeme nájsť ∣ z ∣ | z | ∣ z ∣ vypočítaním prepony trojuholníka pomocou Pythagorovej vety: x 2 + y 2 = z 2 ⟹ ∣ z ∣ = 1 2 + 1 2 = 2. x ^ 2 + y ^ 2 = z ^ 2 znamená | z | = sqrt <1 ^ 2 + 1 ^ 2> = sqrt <2>. x 2 + y 2 = z 2 ⟹ ∣ z ∣ = 1 2 + 1 2

​ = 2

. Naše komplexné číslo teda môžeme zapísať ako z = 2 (cos ⁡ π 4 + i sin ⁡ π 4). □ z = sqrt <2> vľavo ( cos frac < pi> <4> + i sin frac < pi> <4> vpravo). _ Štvorec z = 2

(Cos 4 π + i sin 4 π). □

V jednom zmysle by sa mohlo zdať čudné, že prvý spôsob, ako sa učíme reprezentovať polohu objektov v matematike, je použitie karteziánskych súradníc, keď táto metóda lokalizácie nie je najprirodzenejšia alebo najpohodlnejšia. Na začiatok musíte na opísanie všetkých bodov v rovine použiť záporné aj kladné čísla a musíte vytvoriť mriežku (osi vrtov), ​​ktorá sa použije ako referencia.

Keď sa dieťaťa spýtate, kde opustilo loptičku, povie „len tam“ a ukáže. Popisuje (aj keď veľmi zhruba) vzdialenosť „len“ a smer „tam“ (podporované bodom alebo (keď kývnete hlavou). Keď sa niekoho spýtate, kde je jeho mesto, často povie napríklad „asi 30 dolárov na míle severne od Londýna“. Opäť vzdialenosť a smer. Nestáva sa veľmi často, že niekto uvedie zemepisnú šírku a dĺžku svojho mesta!

Použitie vzdialenosti a smeru ako prostriedku na opísanie polohy je preto oveľa prirodzenejšie ako použitie dvoch vzdialeností na mriežke. Tento prostriedok na určenie polohy sa používa v polárnych súradniciach a ložiskách.

Polárne súradnice bodu popisujú jeho polohu z hľadiska vzdialenosti od pevného bodu (počiatku) a uhla meraného z pevného smeru, ktorý zaujímavo nie je „sever“ (alebo hore na stránke), ale „východ“ '' (doprava). To je v smere $Ox$ na karteziánskych osiach.

V rovine zvolíme pevný bod $O$, známy ako „pól“.

Potom zvolíme os $Ox$ cez pól a nazvime ju „polárna os“.

## Polárne zápletky

Pri vykresľovaní polárne obvykle považujeme (r ) za funkciu ( theta ). Pritom dostaneme krivky, ktoré sú veľa iné ako krivky dostaneme ako funkcie (x ) v obdĺžnikových súradniciach. Tu je niekoľko príkladov animovaných ako ( theta ) prechádza z (0 ) do (2 pi ) (nebudete zodpovední za ich znalosť):

Začnime jednoduchšie jednoduchým zvážením toho, čo sa stane, keď zakreslíme krivky zodpovedajúce rovniciam tvaru (r = text <[konštanta]> ) a ( theta = text <[konštanta]> ).

Zvážte rovnicu (r = 3 ). Chceme vykresliť všetky body v rovine s (r ) súradnicou 3. Pretože (r ) meria vzdialenosť od počiatku, znamená to, že chceme všetky body, ktoré sú od počiatku 3 jednotky: jeho kruh. Ďalej je (r = 3 ) zakreslené a animované tak, že ( theta ) prechádza z (0 ) do (2 pi ).

Čo ak nastavíme ( theta ) na konštantu, povedzme ( theta = pi / 6 )? Potom chceme všetky body tvaru ((r, pi / 6) ), kde (r ) môže mať ľubovoľnú hodnotu, kladnú aj zápornú. V tomto prípade dostaneme čiaru. Nižšie je ( theta = pi / 6 ) zakreslené a animované tak, že (r ) prechádza z (- 4 ) do (4 ).

Teraz sa pozrime na niektoré funkcie ( theta ). Zvážte napríklad (r = 2 cos theta ). Začnime vytvorením tabuľky niektorých hodnôt.

( theta ) r
(0) 2
( pi / 6 ) ( sqrt3 )
( pi / 4 ) ( sqrt2 )
( pi / 3 ) (1)
( pi / 2 ) (0)
(2 pi / 3 ) (-1)
(3 pi / 4 ) (- sqrt2 )
(5 pi / 6 ) (- sqrt3 )
( pi ) (-2)

Všimnite si, že sme už späť na pôvodnom mieste! Ak tieto body pozorne vykreslíme, dostaneme nasledujúci obrázok:

Tu som animoval graf pre (0 leq theta leq 2 pi ), takže obieha kruh dvakrát. To bude dôležité, keď hovoríme o integrácii funkcií do polárnych.

To, že ide určite o kruh, môžeme ukázať tak, že trochu manipulujeme s rovnicou (r = 2 cos theta ). Najskôr obe strany vynásobte (r ). Potom máme

Už skôr sme videli, že (r ^ 2 = x ^ 2 + y ^ 2 ) a (r cos theta = x ). Takže máme:

Toto je rovnica pre kruh s polomerom 1 so stredom (1,0).

Keď to trochu zovšeobecníme, zistíme, že ľubovoľná rovnica tvaru (r = a cos theta ) vytvorí kruh sústredený na ((a / 2, 0) ) s polomerom (a / 2 ) .

Podobná vec sa stane, keď vezmeme do úvahy rovnice tvaru (r = a sin theta ). Napríklad uvažujme (r = -2 sin theta ). Opakovanie našich vynásobiť (r ) a doplňte štvorec trik, máme

Tu je vykreslený a animovaný:

Opäť obchádzame kruh dvakrát nad (0 leq theta leq 2 pi ).

Jedna dôležitá polárna krivka, s ktorou ste sa pravdepodobne nikdy predtým nestretli, je Kardioidné. Zvážte rovnicu (r = 1 + cos theta ). Začnime nakreslením niekoľkých bodov:

( theta ) r
(0) 2
( pi / 6 ) (1+ sqrt3 / 2 )
( pi / 4 ) (1+ sqrt2 / 2 )
( pi / 3 ) (3/2)
( pi / 2 ) (1)
(2 pi / 3 ) (1/2)
(3 pi / 4 ) (1 sqrt2 / 2 )
(5 pi / 6 ) (1 sqrt3 / 2 )
( pi ) (0)

Pretože ( cos theta ) je párny (t.j. ( cos- theta = cos theta )), dostaneme zrkadlové obrazy týchto bodov nad osou (y). Tu je vykreslený a animovaný pre (0 leq theta leq 2 pi ).

Zovšeobecnenie: kardioidy majú formu (r = a (1+ cos theta) ) a (r = a (1+ sin theta) )

Aplikácia: Kardioid je bežný polárny vzor v mikrofónoch. Polárny obrazec mikrofónu popisuje, aký citlivý je na zvuk prichádzajúci z rôznych uhlov.

Polárne súradnice $(r, theta)$ a obdĺžnikové súradnice $(x, y)$ súvisia s rovnicami $begin x = r cos theta y = r sin theta end$

takže rovnicu $y = -1$ môžeme vyjadriť v polárnych súradniciach ako $podprsenka_y = -1$.

Predpokladám, že $y = -1$ je skratka pre bod $(x, y) = (0, -1)$ a že $R$ je polomer v polárnej súradnici $(R, theta)$ . Polárne a obdĺžnikové súradnice súvisia s identitou $begin x & amp = R cos ( theta) y & amp = R sin ( theta). koniec$ Nahradenie do systému, $R cos ( theta) = 0$ a $R sin ( theta) = -1$. Dva uhly v $[0, 2 pi]$, ktoré vyhovujú prvej rovnici, sú $pi / 2$ a $3 pi / 2$. Ak zvolíme $theta = 3 pi / 2$, $-1 = R sin ( pi / 2) = -R$ a naša polárna forma bude $(1, 3 pi / 2)$.

V skutočnosti existuje nekonečne veľa polárnych reprezentácií ľubovoľného bodu, ak je $k$ kladné celé číslo, $(1, -k pi / 2)$ a $(- 1, k pi / 2)$ sú platné reprezentácie bodu . Keď hovoríme o „polárnej podobe“ bodu v $Bbb R ^ 2$, zvyčajne vyžadujeme $theta v [0, 2 pi]$ a $R & gt 0$, čo má za následok jedinečné zastúpenie. Podľa konvencie je pôvod zapísaný $(0, 0)$ v polárnom tvare.

## 2 odpovede 2

Ukážeme si, ako sa robí výpočet plochy pomocou polárnych súradníc. Existuje potenciálna menšia komplikácia. Existujú dva rôzne konvencie o význame polárnej rovnice, keď $r & lt0$. Niektorí hovoria, že krivka potom nie je definovaná. Niektorí tvrdia, že je to definované, ale treba sa zamyslieť nad pôvodom. Pre našu konkrétnu krivku dávajú obidve konvencie rovnaký región, takže sa nemusíme obávať. Ale pri iných krivkách budete možno potrebovať vedieť, aké konvencie používa váš kurz.

Krivka $r = 6 sin theta$ a kruh $r = 1$ sa stretávajú, kde $sin theta = 1/6$. Nech $theta_0 = arcsin (1/6)$. Od $theta = theta_0$ do $theta = pi- theta_0$ je krivka $r = 6 sin theta$ "mimo" kruhu $r = 1$. Symetriou sa môžeme iba pozrieť na časť od $theta = theta_0$ do $theta = pi / 2$ a výslednú plochu zdvojnásobíme.

Oblasť od $theta = theta_0$ do $theta = pi / 2$, ohraničená našou krivkou $r = 6 sin theta$, je $int _ < theta_0> ^ < pi / 2> (1/2) r ^ 2 d theta.$ Od toho musíme odpočítať oblasť vo vnútri kruhu $r = 1$, čo je $int _ < theta_0> ^ < pi / 2> (1/2) , 1 ^ 2 , d theta.$ Odčítajte a vynásobte 2 , aby ste sa postarali o druhú polovicu regiónu. Požadovaná oblasť je preto $int _ < theta_0> ^ < pi / 2> (36 sin ^ 2 theta-1) , d theta.$

Teraz sa integrujte. Použitím identity $cos 2 theta = 1-2 sin ^ 2 theta$ zistíme, že $36 sin ^ 2 theta = 18-18 cos 2 theta$. Naša oblasť je teda $int _ < theta_0> ^ < pi / 2> (17-18 cos 2 theta) , d theta.$

Zvyšok je rutina. Budeme potrebovať $sin 2 theta_0$. Keď to spočítate, narazíte na 35 $, ktoré ste už stretli pri analýze obdĺžnikových súradníc. Konečná odpoveď nie je „pekná“, pretože$ arcsin (1/6) $nie je príjemné číslo. Škoda, že polárna krivka nebola$ r = 2 sin theta : \$!

Komentovať: Výsledok môžeme získať aj prepnutím na obdĺžnikové súradnice, ako ste to urobili vy. Prepínanie v tomto prípade bolo priame a ukázalo sa, že máme do činenia s kruhom. V mnohých prípadoch však môže byť verzia obdĺžnikových súradníc krivky uvedená v polárnych súradniciach veľmi nepríjemná. Je preto užitočné naučiť sa pracovať priamo v polárnych súradniciach.

V situáciách so silnou kruhovou symetriou prichádzajú polárne súradnice veľmi prirodzene. Napríklad gravitačná sila vyvíjaná Zemou na malý predmet je nasmerovaná do stredu Zeme. Pri štúdiu pohybov satelitov sú cestou polárne súradnice.

• Polárna mriežka je reprezentovaná ako séria sústredných kruhov vyžarujúcich z pólu alebo počiatku.
• Ak chcete vykresliť bod v tvare [latex] doľava (r, theta doprava), theta & gt0 [/ latex], posuňte sa proti smeru hodinových ručičiek od polárnej osi o uhol [latex] theta [/ latex ], a potom predĺžte smerovaný úsečku od pólu v dĺžke [latex] r [/ latex] v smere [latex] theta [/ latex]. Ak je [latex] theta [/ latex] záporný, posuňte sa v smere hodinových ručičiek a predĺžte nasmerovaný úsečkový segment o dĺžku [latex] r [/ latex] v smere [latex] theta [/ latex].
• Ak je [latex] r [/ latex] záporný, predĺžte nasmerovaný čiarový segment v opačnom smere ako [latex] theta [/ latex].
• Ekvivalentné polárne súradnice možno nájsť pomocou nasledujúcich vzorcov: [latex] vľavo (r, theta + 2 pi k vpravo) text vľavo (-r, theta + pi + 2 pi k right) [/ latex] a [latex] left (r, theta + 360 ^ circ k right) text left (-r, theta + 180 ^ circ + 360 ^ kruh k vpravo) [/ latex]
• Ak chcete previesť z polárnych súradníc na obdĺžnikové súradnice, použite vzorce [latex] x = r cos theta [/ latex] a [latex] y = r sin theta [/ latex].
• Ak chcete previesť z obdĺžnikových súradníc na polárne, použite jeden alebo viac vzorcov: [latex] cos theta = frac, sin theta = frac, tan theta = frac[/ latex] a [latex] r = sqrt <^<2>+^ <2>> [/ latex].
• Transformácia rovníc medzi polárnym a obdĺžnikovým tvarom znamená uskutočnenie vhodných substitúcií na základe dostupných vzorcov spolu s algebraickými manipuláciami.
• Použitím vhodných substitúcií je možné prepísať polárnu rovnicu na obdĺžnikovú a potom ju nakresliť v grafe v obdĺžnikovej rovine.

2. V čom sa líšia polárne osi od X& # 8211 a r-osy karteziánskej roviny?

3. Vysvetlite, ako sú znázornené polárne súradnice.

4. Ako sú body [latex] vľavo (3, frac < pi> <2> vpravo) [/ latex] a [latex] vľavo (-3, frac < pi> <2> správne) [/ latex] súvisiaci?

5. Vysvetlite, prečo sú body [latex] ľavý (-3, frac < pi> <2> pravý) [/ latex] a [latex] ľavý (3, - frac < pi> <2> right) [/ latex] sú rovnaké.

a) [latex] text <> -2 pi leq theta lt 0 text <,> r gt 0 [/ latex]
b) [latex] text <> 0 leq theta lt 2 pi text <,> r lt 0 [/ latex]
c) [latex] text <> 2 pi leq theta lt 4 pi text <,> r gt 0 [/ latex]

a) [latex] text <> -360 ^ circ leq theta lt 0 text <,> r gt 0 [/ latex]
b) [latex] text <> 0 leq theta lt 360 ^ circ text <,> r lt 0 [/ latex]
c) [latex] text <> 360 ^ circ leq theta lt 720 ^ circ text <,> r gt 0 [/ latex]

Pre nasledujúce cvičenia preveďte dané polárne súradnice na karteziánske súradnice pomocou [latex] r & gt0 [/ latex] a [latex] 0 le theta le 2 pi [/ latex]. Pri určovaní [latex] theta [/ latex] pre bod nezabudnite vziať do úvahy kvadrant, v ktorom sa daný bod nachádza.

15. [latex] vľavo (5, pi vpravo) [/ latex]

Pre nasledujúce cvičenia preveďte dané karteziánske súradnice na polárne pomocou [latex] r & gt0,0 le theta & lt2 pi [/ latex]. Nezabudnite vziať do úvahy kvadrant, v ktorom sa daný bod nachádza.

Pre nasledujúce cvičenia preveďte danú karteziánsku rovnicu na polárnu a riešte pre [latex] r [/ latex].

Pre nasledujúce cvičenia preveďte danú polárnu rovnicu na karteziánsku rovnicu. Ak je to možné, píšte do štandardného tvaru kužeľovitého tvaru a identifikujte znázornený kužeľovitý rez.

36. [latex] r = 3 sin theta [/ latex]

37. [latex] r = 4 cos theta [/ latex]

40. [latex] r = 2 s theta [/ latex]

41. [latex] r = 3 csc theta [/ latex]

43. [latex]^ <2> = 4 sec theta csc theta [/ latex]

48.

49.

50.

51.

52.

Pri nasledujúcich cvičeniach vykreslite body.

Pre nasledujúce cvičenia preveďte rovnicu z obdĺžnikového do polárneho tvaru a graf na polárnej osi.

Pri nasledujúcich cvičeniach preveďte rovnicu z polárneho na obdĺžnikový tvar a graf na obdĺžnikovej rovine.

## Súvisiace zdroje

Nasledujúci obsah je poskytovaný na základe licencie Creative Commons. Vaša podpora pomôže MIT OpenCourseWare naďalej bezplatne ponúkať vysoko kvalitné vzdelávacie zdroje. Ak chcete poskytnúť dar alebo zobraziť ďalšie materiály zo stoviek kurzov MIT, navštívte MIT OpenCourseWare na adrese ocw.mit.edu.

PROFESOR: Dnes budeme pokračovať v diskusii o parametrických krivkách. Musím vám povedať o dĺžke oblúka. A dovoľte mi pripomenúť, kde sme naposledy skončili. Toto sú parametrické krivky, pokračovanie. Minule sme hovorili o parametrickom znázornení kruhu. Alebo jedno z parametrických zobrazení kruhu. Ktorý tu bol tento. A najskôr sme si všimli, že to parametrizuje, ako hovoríme, kruh. Tým je splnená rovnica pre kruh. A sleduje sa to proti smeru hodinových ručičiek. Obrázok vyzerá takto. Tu je kruh. A začína to tu t = 0 a až sem sa dostane v čase t = pi / 2. Takže teraz s vami musím hovoriť o dĺžke oblúka. V tejto parametrickej podobe. A výsledky by mali byť rovnaké ako dĺžka oblúka okolo tohto kruhu. A začneme týmto základným rozdielovým vzťahom. ds ^ 2 je dx ^ 2 + dy ^ 2. A potom vezmem druhú odmocninu, vydelíme dt, takže rýchlosť zmeny vzhľadom na t s bude druhá odmocnina. No možno to napíšem bez rozdelenia. Stačí to napísať ako ds. Takže toto by bolo (dx / dt) ^ 2 + (dy / dt) ^ 2, dt.

Toto je teda to, čo formálne získate z tejto rovnice. Ak vezmete jeho druhé odmocniny a vydelíte dt na druhú v-- vo vnútri druhej odmocniny a vynásobíte dt mimo, takže sa zrušia. A toto je formálne spojenie medzi nimi. We'll be saying just a few more words in a few minutes about how to make sense of that rigorously. Alright so that's the set of formulas for the infinitesimal, the differential of arc length. And so to figure it out, I have to differentiate x with respect to t. And remember x is up here. It's defined by a cos t, so its derivative is -a sin t. And similarly, dy/dt = a cos t.

And so I can plug this in. And I get the arc length element, which is the square root of (-a sin t)^2 + (a cos t)^2, dt. Which just becomes the square root of a^2, dt, or a dt. Now, I was about to divide by t. Let me do that now. We can also write the rate of change of arc length with respect to t. And that's a, in this case. And this gets interpreted as the speed of the particle going around. So not only, let me trade these two guys, not only do we have the direction is counterclockwise, but we also have that the speed is, if you like, it's uniform. It's constant speed. And the rate is a. So that's ds/dt. Travelling around.

And that means that we can play around with the speed. And I just want to point out-- So the standard thing, what you'll have to get used to, and this is a standard presentation, you'll see this everywhere. In your physics classes and your other math classes, if you want to change the speed, so a new speed going around this would be, if I set up the equations this way. Now I'm tracing around the same circle. But the speed is going to turn out to be, if you figure it out, there'll be an extra factor of k. So it'll be ak. That's what we'll work out to be the speed. Provided k is positive and a is positive. So we're making these conventions. The constants that we're using are positive.

Now, that's the first and most basic example. The one that comes up constantly. Now, let me just make those comments about notation that I wanted to make. And we've been treating these squared differentials here for a little while and I just want to pay attention a little bit more carefully to these manipulations. And what's allowed and what's not. And what's justified and what's not. So the basis for this was this approximate calculation that we had, that (delta s)^2 was (delta x)^2 + (delta y)^2. This is how we justified the arc length formula before. And let me just show you that the formula that I have up here, this basic formula for arc length in the parametric form, follows just as the other one did. And now I'm going to do it slightly more rigorously.

I do the division really in disguise before I take the limit of the infinitesimal. So all I'm really doing is I'm doing this. Dividing through by this, and sorry this is still approximately equal. So I'm not dividing by something that's 0 or infinitesimal. I'm dividing by something nonzero. And here I have ((delta x)/(delta t))^2 + ((delta y)/(delta t))^2 And then in the limit, I have ds/dt is equal to the square root of this guy. Or, if you like, the square of it, so. So it's legal to divide by something that's almost 0 and then take the limit as we go to 0. This is really what derivatives are all about. That we get a limit here. As the denominator goes to 0. Because the numerator's going to 0 too. So that's the notation.

And now I want to warn you, maybe just a little bit, about misuses, if you like, of the notation. We don't do absolutely everything this way. This expression that came up with the squares, you should never write it as this. This, put it on the board but very quickly, never. Ok. Don't do that. We use these square differentials, but we don't do it with these ratios here. But there was another place which is slightly confusing. It looks very similar, where we did use the square of the differential in a denominator. And I just want to point out to you that it's different. It's not the same. And it is OK. And that was this one. This thing here. This is a second derivative, it's something else. And it's got a dt^2 in the denominator. So it looks rather similar. But what this represents is the quantity d/dt squared. And you can see the squares came in. And squared the two expressions. And then there's also an x over here.

So that's legal. Those are notations that we do use. And we can even calculate this. It has a perfectly good meaning. It's the same as the derivative with respect to t of the derivative of x, which we already know was minus sine-- sorry, a sin t, I guess. Not this example, but the previous one. Up here. So the derivative is this and so I can differentiate a second time. And I get -a cos t. So that's a perfectly legal operation. Everything in there makes sense. Just don't use that. There's another really unfortunate thing, right which is that the 2 creeps in funny places with sines. You have sine squared. It would be out here, it comes up here for some strange reason. This is just because typographers are lazy or somebody somewhere in the history of mathematical typography decided to let the 2 migrate. It would be like putting the 2 over here. There's inconsistency in mathematics, right. We're not perfect and people just develop these notations. So we have to live with them. The ones that people accept as conventions.

The next example that I want to give you is just slightly different. It'll be a non-constant speed parameterization. Here x = 2 sin t. And y is, say, cos t. And let's keep track of what this one does. Now, this is a skill which I'm going to ask you about quite a bit. And it's one of several skills. You'll have to connect this with some kind of rectangular equation. An equation for x and y. And we'll be doing a certain amount of this today. In another context. Right here, to see the pattern, we know that the relationship we're going to want to use is that sin^2 + cos^2 = 1. So in fact the right thing to do here is to take 1/4 x^2 + y^2. And that's going to turn out to be sin^2 t + cos^2 t. Which is 1. So there's the equation. Here's the rectangular equation for this parametric curve. And this describes an ellipse.

That's not the only information that we can get here. The other information that we can get is this qualitative information of where we start, where we're going, the direction. It starts out, I claim, at t = 0. That's when t = 0, this is (2 sin 0, cos 0), right? (2 sin 0, cos 0) is equal to the point (0, 1). So it starts up up here. At (0, 1). And then the next little place, so this is one thing that certainly you want to do. t = pi/2 is maybe the next easy point to plot. And that's going to be (2 sin(pi/2), cos(pi/2)). And that's just (2, 0). And so that's over here somewhere. This is (2, 0). And we know it travels along the ellipse. And we know the minor axis is 1, and the major axis is 2, so it's doing this.

So this is what happens at t = 0. This is where we are at t = pi/2. And it continues all the way around, etc. To the rest of the ellipse. This is the direction. So this one happens to be clockwise.

Alright, now let's keep track of its speed. Let's keep track of the speed, and also the arc length. So the speed is the square root of the derivatives here. That would be (2 cos t)^2 + (sin t)^2. And the arc length is what? Well, if we want to go all the way around, we need to know that that takes a total of 2 pi. So 0 to 2 pi. And then we have to integrate ds, which is this expression, or ds/dt, dt. So that's the square root of 4 cos^2 t + sin^2 t, dt.

The bad news, if you like, is that this is not an elementary integral. In other words, no matter how long you try to figure out how to antidifferentiate this expression, no matter how many substitutions you try, you will fail. That's the bad news. The good news is this is not an elementary integral. It's not an elementary integral. Which means that this is the answer to a question. Not something that you have to work on. So if somebody asks you for this arc length, you stop here. That's the answer, so it's actually better than it looks. And we'll try to-- I mean, I don't expect you to know already what all of the integrals are that are impossible. And which ones are hard and which ones are easy. So we'll try to coach you through when you face these things. It's not so easy to decide. I'll give you a few clues, but. Ok. So this is the arc length.

Now, I want to move on to the last thing that we did. Last type of thing that we did last time. Which is the surface area. And yeah, question.

PROFESSOR: The question, this is a good question. The question is, when you draw the ellipse, do you not take into account what t is. The answer is that this is in disguise. What's going on here is we have a trouble with plotting in the plane what's really happening. So in other words, it's kind of in trouble. So the point is that we have two functions of t, not one. x(t) and y(t). So one thing that I can do if I plot things in the plane. In other words, the main point to make here is that we're not talking about the situation y is a function of x. We're out of that realm now. We're somewhere in a different part of the universe in our thought. And you should drop this point of view. So this depiction is not y as a function of x. Well, that's obvious because there are two values here, as opposed to one. So we're in trouble with that. And we have that background parameter, and that's exactly why we're using it. This parameter t. So that we can depict the entire curve. And deal with it as one thing.

So since I can't really draw it, and since t is nowhere on the map, you should sort of imagine it as time, and there's some kind of trajectory which is travelling around. And then I just labelled a couple of the places. If somebody asked you to draw a picture of this, well, I'll tell you exactly where you need the picture in just one second, alright. It's going to come up right now in surface area. But otherwise, if nobody asks you to, you don't even have to put down t = 0 and t = pi / 2 here. Because nobody demanded it of you. Another question.

PROFESSOR: So, another very good question which is exactly connected to this picture. So how is it that we're going to use the picture, and how is it we're going to use the notion of the t. The question was, why is this from t = 0 to t = 2 pi? That does use the t information on this diagram. the point is, we do know that t starts here. This is pi / 2, this is pi, this is 3 pi / 2, and this is 2 pi. When you go all the way around once, it's going to come back to itself. These are periodic functions of period 2 pi. And they come back to themselves exactly at 2 pi. And so that's why we know in order to get around once, we need to go from 0 to 2 pi. And the same thing is going to come up with surface area right now. That's going to be the issue, is what range of t we're going to need when we compute the surface area.

PROFESSOR: In a question, what you might be asked is what's the rectangular equation for a parametric curve? So that would be 1/4 x^2 + y^2 = 1. And then you might be asked, plot it. Well, that would be a picture of the ellipse. OK, those are types of questions that are legal questions.

PROFESSOR: The question is, do I need to know any specific formulas? Any formulas that you know and remember will help you. They may be of limited use. I'm not going to ask you to memorize anything except, I guarantee you that the circle is going to come up. Not the ellipse, the circle will come up everywhere in your life. So at least at MIT, your life at MIT. We're very round here. Yeah, another question.

STUDENT: I'm just a tiny bit confused back to the basics. This is more a question from yesterday, I guess. But when you have your original ds^2 = dx^2 + dy^2, and then you integrate that to get arc length, how are you, the integral has dx's and dy's. So how are you just integrating with respect to dx?

PROFESSOR: OK, the question is how are we just integrating with respect to x? So this is a question which goes back to last time. And what is it with arc length. So. I'm going to have to answer that question in connection with what we did today. So this is a subtle question. But I want you to realize that this is actually an important conceptual step here. So shhh, everybody, listen.

If you're representing one-dimensional objects, which are curves, maybe, in space. Or in two dimensions. When you're keeping track of arc length, you're going to have to have an integral which is with respect to some variable. But that variable, you get to pick. And we're launching now into this variety of choices of variables with respect to which you can represent something. Now, there are some disadvantages on the circle to representing things with respect to the variable x. Because there are two points on the circle here. On the other hand, you actually can succeed with half the circle. So you can figure out the arc length that way. And then you can set it up as an integral dx. But you can also set it up as an integral with respect to any parameter you want. And the uniform parameter is perhaps the easiest one. This one is perhaps the easiest one.

And so now the thing that's strange about this perspective - and I'm going to make this point later in the lecture as well - is that the letters x and y-- As I say, you should drop this notion that y is a function of x. This is what we're throwing away at this point. What we're thinking of is, you can describe things in terms of any coordinate you want. You just have to say what each one is in terms of the others. And these x and y over here are where we are in the Cartesian coordinate system. They're not-- And in this case they're functions of some other variable. Some other variable. So they're each functions. So the letters x and y just changed on you. They mean something different. x is no longer the variable. It's the function. Správny?

You're going to have to get used to that. That's because we run out of letters. And we kind of want to use all of them the way we want. I'll say some more about that later.

So now I want to do this surface area example. I'm going to just take the surface area of the ellipsoid. The surface of the ellipsoid formed by revolving this previous example, which was Example 2. Around the y-axis. So we want to set up that surface area integral here for you. Now, I remind you that the area element looks like this. If you're revolving around the y-axis, that means you're going around this way and you have some curve. In this case it's this piece of an ellipse. If you sweep it around you're going to get what's called an ellipsoid. And there's a little chunk here, that you're wrapping around. And the important thing you need besides this ds, this arc length piece over here, is the distance to the axis. So that's this horizontal distance here. I'll draw it in another color. And that horizontal distance now has a name. And this is, again, the virtue of this coordinate system. The t is something else. This has a name. This distance has a name. This distance is called x.

And it even has a formula. Its formula is 2 sin t. In terms of t. So the full formula up for the integral here is, I have to take the circumference when I spin this thing around. And this little arc length element. So I have here 2 pi times 2 sin t. That's the x variable here. And then I have here ds, which is kind of a mess. So unfortunately I don't quite have room for it. Plánovať vopred. Square root of 4 cos^2 t + sin^2 t, is that what it was, dt. Alright, I guess I squeezed it in there. So that was the arc length, which I re-copied from this board above. That was the ds piece. It's this whole thing including the dt. That's the answer except for one thing. What else do we need? We don't just need the integrand, this is half of setting up an integral. The other half of setting up an integral is the limits. We need specific limits here. Otherwise we don't have a number that we can get out.

So we now have to think about what the limits are. And maybe somebody can see. It has something to do with this diagram of the ellipse over here. Can somebody guess what it is? 0 to pi. Well, that was quick. To je všetko. Because we go from the top to the bottom, but we don't want to continue around. We don't want to go from 0 to 2 pi, because that would be duplicating what we're going to get when we spin around. And we know that we start at 0. It's interesting because it descends when you change variables to think of it in terms of the y variable it's going the opposite way. But anyway, just one piece of this is what we want.

So that's this setup. And now I claim that this is actually a doable integral. However, it's long. I'm going to spare you, I'll just tell you how you would get started. You would use the substitution u = cos t. And then the du is going to be -sin t dt. But then, unfortunately, there's a lot more. There's another trig substitution with some other multiple of the cosine and so forth. So it goes on and on. If you want to check it yourself, you can. There's an inverse trig substitution which isn't compatible with this one. But it can be done. Calculated. In elementary terms. Yeah, another question.

PROFESSOR: So, if you get this on an exam, I'm going to have to coach you through it. Either I'm going to have to tell you don't evaluate it or, you're going to have to work really hard. Or here's the first step, and then the next step is, keep on going. Alebo niečo. I'll have to give you some cues. Because it's quite long. This is way too long for an exam, this particular one. Ok. It's not too long for a problem set. This is where I would leave you off if I were giving it to you on a problem set. Just to give you an idea of the order of magnitude. Whereas one of the ones that I did yesterday, I wouldn't even give you on a problem set, it was so long.

So now, our next job is to move on to polar coordinates. Now, polar coordinates involve the geometry of circles. As I said, we really love circles here. We're very round. Just as I love 0, the rest of the Institute loves circles. So we're going to do that right now.

What we're going to talk about now is polar coordinates. Which are set up in the following way. It's a way of describing the points in the plane. Here is a point in a plane, and here's what we think of as the usual x-y axes. And now this point is going to be described by a different pair of coordinates, different pair of numbers. Namely, the distance to the origin. And the second parameter here, second number here, is this angle theta. Which is the angle of ray from origin with the horizontal axis. So that's what it is in language. And you should put this in quotation marks, because it's not a perfect match. This is geometrically what you should always think of, but the technical details involve dealing directly with formulas.

The first formula is the formula for x. And this is the fundamental, these two are the fundamental ones. Namely, x = r cos theta. The second formula is the formula for y, which is r sin theta. So these are the unambiguous definitions of polar coordinates. This is it. And this is the thing from which all other almost correct statements almost follow. But this is the one you should trust always. This is the unambiguous statement.

So let me give you an example something that's close to being a good formula and is certainly useful in its way. Namely, you can think of r as being the square root of x^2 + y^2. That's easy enough to derive, it's the distance to the origin. That's pretty obvious. And the formula for theta, which you can also derive, which is that it's the inverse tangent of y y/x. However, let me just warn you that these formulas are slightly ambiguous. So somewhat ambiguous. In other words, you can't just apply them blindly. You actually have to look at a picture in order to get them right. In particular, r could be plus or minus here. And when you take the inverse tangent, there's an ambiguity between, it's the same as the inverse tangent of (-y)/(-x). So these minus signs are a plague on your existence. And you're not going to get a completely unambiguous answer out of these formulas without paying attention to the diagram. On the other hand, the formula up in the box there always works. So when people mean polar coordinates, they always mean that. And then they have conventions, which sometimes match things up with the formulas over on this next board.

Let me give you various examples here first. But maybe first I should I should draw the two coordinate systems. So the coordinate system that we're used to is the rectangular coordinate system. And maybe I'll draw it in orange and green here. So these are the coordinate lines y = 0, y = 1, y = 2. That's how the coordinate system works. And over here we have the rest of the coordinate system. And this is the way we're thinking of x and y now. We're no longer thinking of y as a function of x and x as a function of y, we're thinking of x as a label of a place in a plane. And y as a label of a place in a plane.

So here we have x = 0, x = 1, x = 2, etc. Here's x = -1. So forth. So that's what the rectangular coordinate system looks like. And now I should draw the other coordinate system that we have. Which is this guy here. Well, close enough. And these guys here. Kind of this bulls-eye or target operation. And this one is, say, theta = pi/2. This is theta = 0. This is theta = -pi/4. For instance, so I've just labeled for you three of the rays on this diagram. It's kind of like a radar screen. And then in pink, this is maybe r = 2, the radius 2. And inside is r = 1. So it's a different coordinate system for the plane. And again, the letter r represents measuring how far we are from the origin. The theta represents something about the angle, which ray we're on. And they're just two different variables. And this is a very different kind of coordinate system.

OK so, our main job is just to get used to this. For now. You will be using this a lot in 18.02. It's very useful in physics. And our job is just to get started with it. And so, let's try a few examples here. Tons of examples. We'll start out very slow. If you have (x, y) = (1, -1), that's a point in the plane. I can draw that point. It's down here, right? This is -1 and this is 1, and here's my point, (1, -1). I can figure out what the representative is of this in polar coordinates. So in polar coordinates, there are actually a bunch of choices here.

First of all, I'll tell you one choice. If I start with the angle horizontally, I wrap all the way around, that would be to this ray here-- Let's do it in green again. Alright, I labeled it actually as -pi/4, but another way of looking at it is that it's this angle here. So that would be r = square root of 2. Theta = 7pi/4. So that's one possibility of the angle and the distance. I know the distance is a square root of 2, that's not hard.

Another way of looking at it is the way which was suggested when I labeled this with a negative angle. And that would be r = square root of 2, theta = -pi/4. And these are both legal. These are perfectly legal representatives. And that's what I meant by saying that these representations over here are somewhat ambiguous. There's more than one answer to this question, of what the polar representation is.

A third possibility, which is even more dicey but also legal, is r equals minus square root of 2. Theta = 3pi/4. Now, what that corresponds to doing is going around to here. We're pointing out 3/4 pi direction. But then going negative square root of 2 distance. We're going backwards. So we're landing in the same place. So this is also legal. Áno.

PROFESSOR: The question is, don't the radiuses have to be positive because they represent a distance to the origin? The answer is I lied to you here. All of these things that I said are wrong, except for this. Which is the rule for what polar coordinates mean. So it's maybe plus or minus the distance, is what it is always. I try not to lie to you too much, but I do succeed. Now, let's do a little bit more practice here.

There are some easy examples, which I will run through very quickly. r = a, we already know this is a circle. And the 3 theta equals a constant is a ray. However, this involves an implicit assumption, which I want to point out to you. So this is Example 3. Theta's equal to a constant is a ray. But this implicitly assumes 0 <= r < infinity. If you really wanted to allow minus infinity < r < infinity in this example, you would get a line. Gives the whole line. It gives everything behind. So you go out on some ray, you go backwards on that ray and you get the whole line through the origin, both ways. If you allow r going to minus infinity as well.

So the typical conventions, so here are the typical conventions. And you will see people assume this without even telling you. So you need to watch out for it. The typical conventions are certainly this one, which is a nice thing to do. Pretty much all the time, although not all the time. Most of the time. And then you might have theta ranging from minus pi to pi, so in other words symmetric around 0. Or, another very popular choice is this one. Theta's >= 0 and strictly less than 2pi. So these are the two typical ranges in which all of these variables are chosen. But not always. You'll find that it's not consistent.

As I said, our job is to get used to this. And I need to work up to some slightly more complicated examples. Some of which I'll give you on next Tuesday. But let's do a few more. So, I guess this is Example 4. Example 4, I'm going to take y = 1. That's awfully simple in rectangular coordinates. But interestingly, you might conceivably want to deal with it in polar coordinates. If you do, so here's how you make the translation. But this translation is not so terrible. What you do is, you plug in y = r sin(theta). That's all you have to do. And so that's going to be equal to 1. And that's going to give us our polar equation. The polar equation is r = 1 / sin(theta). Je to tu. And let's draw a picture of it. So here's a picture of the line y = 1. And now we see that if we take our rays going out from here, they collide with the line at various lengths. So if you take an angle, theta, here there'll be a distance r corresponding to that and you'll hit this in exactly one spot. For each theta you'll have a different radius. And it's a variable radius. It's given by this formula here. And so to trace this line out, you actually have to realize that there's one more thing involved. Which is the possible range of theta. Again, when you're doing integrations you're going to need to know those limits of integration. So you're going to need to know this. The range here goes from theta = 0, that's sort of when it's out at infinity. That's when the denominator is 0 here. And it goes all the way to pi. Swing around just one half-turn. So the range here is 0 < theta < pi. Yeah, question.

PROFESSOR: The question is, is it typical to express r as a function of theta, or vice versa, or does it matter? The answer is that for the purposes of this course, we're almost always going to be writing things in this form. r as a function of theta. And you can do whatever you want. This turns out to be what we'll be doing in this course, exclusively. As you'll see when we get to other examples, it's the traditional sort of thing to do when you're thinking about observing a planet or something like that. You see the angle, and then you guess far away it is. But it's not necessary. The formulas are often easier this way. For the examples that we have. Because it's usually a trig function of theta. Whereas the other way, it would be an inverse trig function. So it's an uglier expression. As you can see. The real reason is that we choose this thing that's easier to deal with.

So now let me give you a slightly more complicated example of the same type. Where we use a shortcut. This is a standard example. And it comes up a lot. And so this is an off-center circle. A circle is really easy to describe, but not necessarily if the center is on the rim of the circle. So that's a different problem. And let's do this with a circle of radius a. So this is the point (a, 0) and this is (2a, 0). And actually, if you know these two numbers, you'll be able to remember the result of this calculation. Which you'll do about five or six times and then finally you'll memorize it during 18.02 when you will need it a lot. So this is a standard calculation here. So the starting place is the rectangular equation. And we're going to pass to the polar representation. The rectangular representation is (x-a)^2 + y^2 = a^2. So this is a circle centered at (a, 0) of radius a.

And now, if you like, the slow way of doing this would be to plug in x = r cos(theta), y = r sin(theta). The way I did in this first step. And that works perfectly well. But I'm going to do it more quickly than that. Because I can sort of see in advance how it's going to work. I'm just going to expand this out. And now I see the a^2's cancel. And not only that, but x^2 + y^2 = r^2. So this becomes r^2. That's x^2 + y^2 - 2ax = 0.

The r came from the fact that r^2 = x^2 + y^2. So I'm doing this the rapid way. You can do it by plugging in, as I said. r equals-- So now that I've simplified it, I am going to use that procedure. I'm going to plug in. So here I have r^2 - 2ar cos(theta) = 0. I just plugged in for x. As I said, I could have done that at the beginning. I just simplified first. And now, this is the same thing as r^2 = 2ar cos(theta). And we're almost done. There's a boring part of this equation, which is r = 0. And then there's, if I divide by r, there's the interesting part of the equation. Which is this. So this is or r = 0. Which is already included in that equation anyway.

So I'm allowed to divide by r because in the case of r = 0, this is represented anyway. Otázka.

PROFESSOR: r = 0 is just one case. That is, it's the union of these two. It's both. Both are possible. So r = 0 is one point on it. And this is all of it. So we can just ignore this. So now I want to say one more important thing. You need to understand the range of this. So wait a second and we're going to figure out the range here. The range is very important, because otherwise you'll never be able to integrate using this representation here. So this is the representation. But notice when theta = 0, we're out here at 2a. That's consistent, and that's actually how you remember this factor 2a here. Because if you remember this picture and where you land when theta = 0. So that's the theta = 0 part. But now as I tip up like this, you see that when we get to vertical, we're done. With the circle. It's gotten shorter and shorter and shorter, and at theta = pi/2, we're down at 0. Because that's cos(pi/2) = 0. So it swings up like this. And it gets up to pi/2. Similarly, we swing down like this. And then we're done. So the range is -pi/2 < theta < pi/2. Or, if you want to throw in the r = 0 case, you can throw in this, this is repeating, if you like, at the ends. So this is the range of this circle. And let's see. Next time we'll figure out area in polar coordinates.