
Korean: 
여기에 있는 것은
t로 정의된 x와
t로 정의된 y이고
모든 t값들을 넣어
그래프로 도식화한다면
바로 이런 꽤나 멋있는
그래프를 얻게 됩니다
시도해보세요
t가 0일 때 x와 y가 몇인지 구하고
t가 1일 때 x와 y를 구하고
그리고 모든 다른 t들도 넣어보세요
그러고 나면 이 멋진
그래프를 얻게 됩니다
그러나 이 영상의 목표는
매개 함수로 나타낸 함수나 곡선의
멋있음을 느끼는 것이 아닙니다
대신 우리는 계산을 할거고
정확히 말하자면 도함수를
구해볼겁니다
우리는 x에 대한
y의 도함수를 찾으려 합니다
x에 대한 y의 도함수
t값이 －⅓일 때의 것으로요
t값이 －⅓일 때의 것으로요
t값이 －⅓일 때의 것으로요
만약 여러분이 그러고 싶다면 저는
멈추고 한 번 풀어볼 것을 추천합니다
이제 제가 여러분과
같이 풀어볼 겁니다
여러분이 이미 풀었거나 아니면
그냥 제가 풀어주길 바라건 간에요
그래서 핵심은

English: 
- [Voiceover] So what we have here is
x being defined in terms of t and
y being defined in terms of t,
and then if you were to plot
over all of the t values,
you'd get a pretty cool
plot, just like this.
So you try, t equals zero, and
figure out what x and y are,
t is equal to one, figure
out what x and y are,
and all of the other ts,
and then you get this
pretty cool-looking graph.
But the goal in this video
isn't just to appreciate
the coolness of graphs or curves,
defined by parametric equations.
But we actually want to do some calculus,
in particular, we wanna
find the derivative,
we wanna find the derivative
of y, with respect to x,
the derivative of y with respect to x,
when t,
when t is equal
to negative one third.
And if you are inspired, I encourage you
to pause and try to solve this.
And I am about to do it with you,
in case you already did,
or you just want me to.
(chuckles) All right, so the key is,

Thai: 
สิ่งที่เรามีตรงนี้คือ
x นิยามในรูปของ t และ
y นิยามในรูปของ t
แล้วถ้าคุณพลอตสำหรับค่า t ทุกค่า
คุณจะได้พลอตที่เจ๋งทีเดียว แบบนี้
คุณลอง t เท่ากับ 0 แล้วหาว่า x กับ y คืออะไร
t เท่ากับ 1 หาว่า x กับ y คืออะไร
แล้วก็ t อื่นๆ ทั้งหมด
แล้วคุณจะได้กราฟที่ดูเจ๋งดีอันนี้
แต่เป้าหมายในวิดีโอนี้ไม่ใช่เพื่อซาบซึ้ง
ความเจ๋งของกราฟหรือเส้นโค้ง
ที่นิยามจากสมการพาราเมทริก
แต่เราอยากทำแคลคูลัส
โดยเฉพาะ เราจะหาอนุพันธ์
เราหาอนุพันธ์ของ y เทียบกับ x
อนุพันธ์ของ y เทียบกับ x
เมื่อ t
เมื่อ t เท่ากับ
ลบ 1/3
และถ้าคุณรู้สึกอยากทำ ผมแนะนำให้คุณ
ลองหยุดแล้วแก้ปัญหานี้ดู
และผมจะทำไปกับคุณ
ในกรณีที่คุณทำแล้ว หรือคุณอยากให้ผมทำให้ดู
[หัวเราะ] เอาล่ะ ประเด็นคือ

Bulgarian: 
Това, което ни е дадено в тази задача,
е х, дефинирано като функция на t,
и у, дефинирано като функция на t.
Ако трябва да изобразиш 
всички стойности t,
то щеше да получиш доста приятно
 изображение, точно като ето това.
Ако заместиш t = 0, ще получиш
 на какво са равни х и у.
За t = 1., отново ще намериш 
на какво са равни х и у.
И така за всички останали 
стойности на t.
Така ще получиш тази приятно
 изглеждаща графика.
Целта на настоящия урок обаче
 не е само да оцени
колко е хубава графиката или кривите,
зададени чрез параметрични уравнения.
Всъщност искаме да използваме 
математически анализ,
за да намерим производната,
т.е. производната на у спрямо х.
Или производната на у спрямо х (dy/dx),
когато параметърът t
е равен на минус 1/3.
И ако имаш много ентусиазъм, 
те насърчавам
да спреш видеото и да се опиташ 
да решиш задачата самостоятелно.
Готов съм да я решим заедно,
в случай, че не можеш самостоятелно,
 или просто искаш аз да го направя.
Добре, ключовото в задачата е

Czech: 
Souřadnice x a y máme
vyjádřené pomocí proměnné t.
Když za t postupně dosadíme
všechna reálná čísla,
dostaneme tento
zajímavý graf.
Dosadíme například t rovná se 0
a spočítáme, čemu se rovná x a y,
potom spočítáme x a y pro t rovná se 1
a tak dále pro všechna t,
čímž dostaneme
tento zajímavý graf.
Cílem tohoto videa ale není jen obdivovat
křivky popsané parametrickými rovnicemi.
Chceme na ně také použít
diferenciální počet.
Konkrétně chceme spočítat
derivaci y podle x,
když se t rovná
minus (1 lomeno 3).
Pokud už máte nějaký nápad, tak si
zastavte video a zkuste to spočítat,
nebo to teď můžete
počítat spolu se mnou.

Korean: 
어떻게 x에 대한 도함수를
찾아낼 수 있는가 입니다
x와 y 둘 다 t로
정의되어 있을 때의
x에 대한 y의 도함수를요
이를 위해 반드시
알아둬야 하는 것은
x에 대한 y의 도함수가
t에 대한 y의 도함수가 분자이고
t에 대한 x의 도함수가 분모인
분수와 같아진다는 겁니다
도함수들을 숫자처럼 보면
실제로 수학적으로 성립하는 것을
볼 수 있죠
이런 방식으로 하는 것이
조금 엄격하지 않기는 합니다만
그래도 이렇게 하면
그래도 이렇게 하면
무언가에 대한 또 다른
무언가의 도함수는
t에 대한 y의 도함수가 분자이고
t에 대한 x의 도함수가
분모일 때와 같습니다
자 그럼 이게 우리를
어떻게 도와줄까요?
우리가 할 수 있는 것은
t에 대한 x의 도함수와
t에 대한 y의 도함수를
구할 수 있습니다
t에 대한 x의 도함수는
다음과 같습니다
안쪽에 대한 바깥쪽의 도함수는

Czech: 
Hlavní je přijít na to,
jak spočítat derivaci y podle x,
když máme oboje definované
pomocí jiné proměnné t.
Odpovědí je,
že derivace y podle x se rovná:
Derivace y podle t
lomeno derivace x podle t.
Kdybychom se na tato písmena d
dívali jako na pouhá čísla,
tak by nám po vykrácení dt
skutečně vyšlo to samé.
Není to sice formálně
korektní postup,
ale když se na
to díváme takhle,
tak je snadno vidět,
proč by tohle mohlo dávat smysl.
Derivace něčeho podle
něčeho jiného se rovná:
Derivace y podle t
lomeno derivace x podle t.
Jak nám
tohle pomůže?
Derivaci x podle t
i derivaci y podle t už umíme spočítat.
Derivace x podle t
se rovná...

Thai: 
คุณหาอนุพันธ์เทียบกับ x
อนุพันธ์ของ y เทียบกับ x ได้อย่างไร
เมื่อพวกมันนิยามในรูปของ t?
ประเด็นสำคัญคือว่า อนุพันธ์ของ y
เทียบกับ x, เทียบกับ x
จะเท่ากับ จะเท่ากับ
อนุพันธ์ของ y เทียบกับ t
ส่วนอนุพันธ์ของ x เทียบกับ t
ถ้าคุณมองดิฟเฟอเรนเชียลพวกนี้เป็นตัวเลข
อันนี้จะออกมาสวยงามในทางคณิตศาสตร์
ทีนี้ มันไม่รัดกุมเท่าไหร่
ถ้าคุณทำอย่างนั้น
แต่ ถ้าคุณคิดอย่างนั้น มันจะคิดได้ง่าย
ว่าทำไมมันถึงสมเหตุสมผล
อนุพันธ์ของอะไรสักอย่าง เทียบกับอีกอย่าง
เท่ากับอนุพันธ์ของ y เทียบกับ t
ส่วน x เทียบกับ t
เอาล่ะ มันช่วยเราอย่างไร?
เราหา
อนุพันธ์ของ x เทียบกับ t
และอนุพันธ์ของ y เทียบกับ t ได้
อนุพันธ์ของ x เทียบกับ t
จะเท่ากับ
ลองดู อนุพันธ์ของตัวนอก

Bulgarian: 
как ще намериш производната 
спрямо х,
т.е. производната dy/dx спрямо х,
когато и двете променливи 
са дефинирани като функция на t.
Основното е да разбереш, че 
производната на у спрямо х
ще бъде равна на
производната на у спрямо t,
върху производната на х спрямо t.
Ако можеше да видиш тези диференциали 
като числа,
това действително би изчерпало
въпроса математически.
Решението няма да е строго,
ако го направиш по този начин,
но това е лесен начин да си представиш
действителната логика на решението.
Производната на нещо спрямо нещо друго
е равна на производната на у спрямо t,
върху производната на х спрямо t.
Добре, а това как ни помага?
Е, може да намерим
производната на х спрямо t
и производната на у спрямо t.
Производната на х спрямо t
ще бъде равна на следното.

English: 
is well, how do you find the
derivative with respect to x,
derivative of y with respect to x,
when they're both defined in terms of t?
And the key realization
is the derivative of y
with respect to x, with respect to x,
is going to be equal to,
is going to be equal to,
the derivative of y with respect to t,
over the derivative of
x with respect to t.
If you were to view these
differentials as numbers,
well this would actually
work out mathematically.
Now, it gets a little bit non-rigorous,
when you start to do that,
but, if you though of it that,
it's an easy way of thinking
about why this actually might make sense.
The derivative of something
versus something else,
is equal to the derivative
of y with respect to t,
over x with respect to t.
All right, so how does that help us?
Well, we can figure out
the derivative of x with respect to t
and the derivative of y with respect to t.
The derivative of x with respect to t
is just going to be equal to,
let's see, the derivative of the outside,

Thai: 
เทียบกับตัวใน มันจะเท่ากับ 2 ไซน์
โอ๊ย อนุพันธ์ของไซน์คือโคไซน์
2 โคไซน์ของ 1 บวก 3t
คูณอนุพันธ์ของตัวในเทียบกับ t
มันจะเท่ากับ อนุพันธ์ของ 1 ก็แค่ 0
อนุพันธ์ของ 3t เทียบกับ t คือ 3
คูณ 3 นั่นคืออนุพันธ์ของ x เทียบกับ t
ผมแค่ใช้กฎลูกโซ่ตรงนี้
อนุพันธ์ของตัวนอก 2 ไซน์ของอะไรสักอย่าง
เทียบกับตัวใน
อนุพันธ์ของตัวนอกนี้
2 ไซน์ของอะไรสักอย่าง เทียบกับ 1 บวก 3t
คืออันนั้นตรงนั้น
และอนุพันธ์ของตัวในเทียบกับ t
ก็แค่ 3
ทีนี้ อนุพันธ์ของ y เทียบกับ t
นั้นค่อนข้างตรงไปตรงมา
อนุพันธ์ของ y เทียบกับ t
ผมแค่ใช้กฎยกกำลังตรงนี้
3 คูณ 2 ได้ 6, t กำลัง 3 ลบ 1
ได้ 6t กำลังสอง

Bulgarian: 
Нека да видим. Имаме производната
 на външната функция
спрямо вътрешната. Ще бъде равно
 на 2 по синус...
Опа! Производната на синус е косинус.
Получава се 2 по косинус от 1 
плюс 3 по t,
умножено по производната 
на вътрешната функция спрямо t.
Производната на 1 е равна на 0.
Производната на 3 по t спрямо t 
е равна на 3.
Тоест, по 3. И този израз е равен 
на производната на х спрямо t.
Просто приложих верижното правило.
Производната на външната функция 
2 по синус от нещо,
спрямо вътрешната.
Тоест, производната на функцията отвън
2 по синус от нещо спрямо 1 
плюс 3 по t,
което е този израз ето тук.
И след това производната на
 вътрешната функция спрямо t,
което е равно само на 3.
Сега следва производната на у спрямо  t,
която се получава малко по-лесно.
Производната на у спрямо t.
Просто ще приложим правилото 
за намиране производна на степен.
3 пъти по 2 е равно на 6, по
 t на степен 3 минус 1,
т.е. се получава 6 по t на квадрат.

English: 
with respect to the inside,
it's going to be two sine
whoops, the derivative of sign is cosine,
two cosine of one plus three t,
times the derivative of the
inside with respect to t.
So that's going to be,
derivative of one is just zero.
Derivative of three t with
respect to t is three.
So times three, that's the
derivative of x with respect to t
I just used the Chain Rule here.
Derivative of the outside
two sine of something,
with respect to the inside,
so derivative of this outside,
two sine of something with
respect to one plus three t,
is that right over there.
And the derivative of the
inside with respect to t,
is just our three.
Now, the derivative of y with respect to t
is a little bit more straight-forward.
Derivative of y with respect to t,
we just apply the Power Rule here,
three times two is six, t to
the three minus one power,
six t squared.

Czech: 
Derivace vnější funkce
podle té vnitřní je 2 krát sinus...
Ouha, derivace sinu
je přece kosinus.
...2 krát kosinus
v bodě (1 plus 3 krát t),
což teď musíme vynásobit derivací
vnitřní funkce podle t, která se rovná...
Derivace 1 se rovná 0
a derivace (3 krát t) podle t je 3.
Zde tedy musíme
vynásobit 3.
Takto vypadá
derivace x podle t.
Jde jen o derivaci
složené funkce.
Je to derivace vnější funkce, tedy 2 krát
sinus něčeho, podle vnitřní funkce...
Derivaci vnější funkce 2 krát sinus
něčeho podle (1 plus 3 krát t) máme tady
a derivace vnitřní funkce
podle t je toto číslo 3.
Spočítat derivaci y podle t
už je o trochu jednodušší.
K výpočtu této derivace stačí
použít vzorec pro derivaci mocniny.
3 krát 2 se rovná 6,
tohle krát t na (3 minus 1),
tedy 6 krát (t na druhou).

Korean: 
다음과 같아집니다
다음과 같아집니다
2cos(1+3t)
그리고 t에 대한
안쪽의 도함수를 곱합니다
그 값은, 1의 도함수는 0이고
3t의 t에 대한 도함수는
3입니다
따라서 3을 곱해주고 그것이
t에 대한 x의 도함수입니다
여기엔 단순히
연쇄 법칙이 사용되었습니다
안쪽에 대한 바깥쪽의
2sin(x)의 도함수
즉 이 (1+3t)에 대한 바깥쪽의
2sin(x)의 도함수가
바로 저기있는 저것이고
그리고 t에 대한 안쪽의 도함수는
바로 3이죠
t에 대한 y의 도함수는
좀 더 간단합니다
t에 대한 y의 도함수에는
그냥 다항식의 미분을 사용합니다
3×2는 6이고
t의 (3－1)제곱인
6t²입니다

Bulgarian: 
Тогава dy/dx ще бъде равно на 
6 по t на квадрат –
6 по t на квадрат – върху следното.
В този израз имаме 2 по 3,
така че се получава 6 по косинус
от 1 плюс 3 по t.
Шестиците се съкращават 
и остава само
t на квадрат върху
косинус от 1 плюс 3 по t.
Интересува ни моментът, когато
 t е равно на минус 1/3.
Когато t е равно на минус 1/3,
то производната ще бъде равна на
минус 1/3 на квадрат.
Минус 1/3 на квадрат,
върху
косинус от 1 плюс...
3 по минус 1/3 е равно на минус 1,
т.е. 1 плюс –1, и се получава
 косинус от 0.
А косинус от 0 е равно на 1.

Czech: 
Hledaná derivace se tedy rovná
6 krát (t na druhou), to celé lomeno...
Máme tu 2 krát 3, takže ve jmenovateli
bude 6 krát cos(1 plus 3 krát t).
6 můžeme pokrátit a zbyde nám t na druhou
lomeno cos(1 plus 3 krát t).
Zajímá nás hodnota derivace
pro t rovno minus (1 lomeno 3).
Když se t rovná minus (1 lomeno 3),
tak naše derivace bude:
(−1 lomeno 3) na druhou
lomeno kosinus v bodě (1 plus...
3 krát (−1 lomeno 3) je −1,
takže to je 1 plus −1,
tudíž tu
bude cos(0).

Thai: 
นี่ก็จะเท่ากับ 6t กำลังสอง
6t กำลังสอง ส่วน
เรามี 2 คูณ 3
เราจึงได้ 6 คูณโคไซน์
ของ 1 บวก 3t
แล้ว 6 ของเราตัดกัน และเราเหลือ
เราเหลือ t กำลังสองส่วน
โคไซน์ของ 1 บวก 3t
และถ้าเราสนใจเมื่อ t เท่ากับลบ 1/3
เมื่อ t เท่ากับลบ 1/3
มันจะเท่ากับ
มันจะเท่ากับลบ 1/3 กำลังสอง
ลบ 1/3 กำลังสอง
ส่วน
ส่วน
ส่วน
โคไซน์ของ 1 บวก 3 คูณ
ลบ 1/3 ได้ลบ 1
มันคือ 1 บวกลบ 1 มันก็คือโคไซน์ของ 0
และโคไซน์ของ 0 ก็แค่ 1

Korean: 
그러므로 이것은 6t²을
6t²을 분자로 하고
여기에 2×3이 있으므로
우리는 분모로
6cos(1＋3t)를 갖습니다
그리고 6들은 약분되고
남아있는 것은 t²이 분자로 남고
cos(1＋3t)가 분모로 남습니다
그리고 t가 －⅓일 때를 구해보면
t가 －⅓일 때에는
그 값은
그 값은 (－⅓)²이 분자이고
(－⅓)²이 분자이고
(－⅓)²이 분자이고
(－⅓)²이 분자이고
(－⅓)²이 분자이고
분모가 cos{1＋3×(－⅓)}인데
3×(－⅓)은 －1입니다
그러므로 이것은 1＋(－1)이고
즉 분모는 cos(0)입니다
그리고 cos(0)은 1이 됩니다

English: 
So this is going to be
equal to six t squared,
six t squared, over,
well, we have the two times the three,
so we have six times cosine
of one plus three t,
and then our sixes cancel
out, and we are left with,
we are left with t squared over
cosine of one plus three t.
And if we care when t is
equal to negative one third,
when t is equal to negative one third,
this is going to be equal to, well,
this is going to be equal to
negative one third squared.
Negative one third squared,
over,
over,
over
the cosine of one plus three times
negative one third is negative one.
So it's one plus negative
one, so it's a cosine of zero.
And the cosine of zero
is just going to be one.

Bulgarian: 
Следователно този израз се опростява
до плюс 1/9.
Нека да видим сега дали може да визуализираме 
какво се случва на графиката.
Нека да начертая една малка 
таблица ето тук.
Ще изобразя...
т.е. ще мисля за t, х и у.
t, х и у.
Когато t е равно на минус 1/3,
х ще бъде равно на
синус от 0, т.е. х ще бъде равно на 0.
А у ще бъде равно на
минус 2 върху 27.
Следователно разглеждаме 
точката (0; –2/27).
Това е ето тази точка тук.
Това е точката, в която 
искаме да намерим
наклона на допирателната.
Получихме, че наклонът 
е равен на 1/9.
Наклонът е равен на 1/9.
Възможен начин да мислиш за това е,
 ако се преместим

Czech: 
cos(0) se rovná 1, takže tohle
celé se rovná 1 lomeno 9.
Zkusme si nakreslit,
co se zde děje.
Udělám si k tomu
takovou tabulku.
V tabulce budou sloupečky
pro t, x a y.
Když se t rovná
minus (1 lomeno 3),
tak se
x rovná...
Tady bude sin(0),
takže x se rovná 0.
y se pak rovná
minus (2 lomeno 27).
Této hodnotě t tak odpovídá bod
[0; minus(2 lomeno 27)],
což je
tento bod.
Zajímá nás tedy
směrnice tečny v tomto bodě,
která nám tady dole
vyšla jako 1 lomeno 9.
Směrnice je 1 lomeno 9,
na což se můžeme dívat třeba tak,

Korean: 
그러므로 이 값은
양수 1/9입니다
이제 이 그래프에서 일어나는
일을 시각화할 수 있을지 봅시다
여기에 작은 표를 하나 그려봅시다
표를 구성하면서
t, x와 y에 대해 생각할 겁니다
t, x, y
t가 -⅓일 때
x는 바로
sin(0) 즉 x는 0이 됩니다
그리고 y는
－2/27이 됩니다
그러므로 우리가 얘기하는 것은
점 (0,－2/27)에 대한 것입니다
그건 바로 저기있는 저 점이죠
저 점이 바로 우리가 접선의 기울기를
찾으려 했던 점입니다
그 기울기는 1/9라고 나와있고
기울기가 1/9이라는 것을
알 수 있는 하나뿐인 방법은

Thai: 
อันนี้จึงเท่ากับบวก
บวก 1/9
ทีนี้ ลองดูว่าเรามองภาพสิ่งที่เกิดขึ้นตรงนี้ได้ไหม
ขอผมวาดตารางเล็กๆ ตรงนี้นะ
ผมจะพลอต
ผมจะคิดถึง t, x และ y
t, x และ y
เมื่อ t เท่ากับลบ 1/3
x ของเราจะเท่ากับ อันนี้จะเท่ากับ
ไซน์ของ 0 ค่า x ของเราจึงเท่ากับ 0
และ y ของเราจะเป็น 2 ส่วน
หรือลบ 2 ส่วน 27
เรากำลังพูดถึง เรากำลังพูดถึง
จุด (0, -2/27)
นั่นก็คือจุดนั้นตรงนั้น
นั่นคือจุดที่เราพยายามหา
ความชันของเส้นสัมผัส
มันกำลังบอกเราว่า ความชันคือ 1/9
ความชันคือ 1/9 ถ้าเราเลื่อน
วิธีคิดอย่างหนึ่งคือว่า ถ้าเราเลื่อน

English: 
So this is going to be equal to positive,
positive one ninth.
Now let's see if we can
visualize what's going on here.
So let me draw a little table here.
So, I'm gonna plot,
I'm gonna think about t, and x, and y.
So t, and x, and y.
So when t is equal to negative one third,
well our x is going to
be, this is going to be
sine of zero, so our
x is going to be zero,
and our why is going
to be, what, two over,
or negative two over 27.
So, we're talking about,
we're talking about
the point zero comma negative two over 27.
So that is that point right over there.
That's the point where
we're trying to find
the slope of the tangent line.
It's telling us that
that slope is one ninth,
that slope is one ninth, so if we move,
I guess one way to think
about it is if we move

Bulgarian: 
до едно, две, три, четири и половина,
а след това се преместим
 с половин деление нагоре.
Ако искахме да начертаем 
допирателната в тази точка,
би изглеждала по следния начин.
Като нещо такова.
По подобен начин.
Като нещо такова.
Нека да проверим. Имаме
 едно, две, три, четири
и половина. Това е което се е получило
и е начертано с висока точност.
Следователно това е, което 
току-що намерихме.
Намерихме, че наклонът 
на допирателната
точно в тази точка е равен на 1/9.
Мисля, че не само 
е приятно да го видим,
но и предполагам, че е полезно.

Czech: 
že když se posuneme
o 1, 2, 3, 4, 4 a půl doprava,
tak se musíme posunout
o půlku jednotky nahoru.
Tečna v tomto bodě
tak vypadá nějak takhle.
Opět si tu odměřím
1, 2, 3, 4, 4 a půl.
Nějak takto by
to mohlo vypadat.
Právě jsme tedy spočítali, že směrnice
tečny v tomto bodě je 1 lomeno 9.
Nemáme tak jenom
napohled hezkou křivku,
ale byli jsme o ní
také něco schopni zjistit.

Thai: 
ไป 4, 1, 2, 3, 4 1/2
เราจะขึ้นไปครึ่งหนึ่ง
ถ้าเราอยากลากเส้นสัมผัสตรงนั้น
มันจะเป็นแบบ
แบบนั้น
เป็นแบบ
แบบนั้น
ลองดูว่าเราได้ 1, 2, 3, 4,
1/2 นั่นคือสิ่งที่เราได้
อย่างนั้น เจ๋งดี
นั่นคือสิ่งที่เราเพิ่งหาไป
เราหาไปว่า ความชันของเส้นสัมผัส
ตรงจุดนั้น คือ 1/9
มันไม่ได้ดูง่ายนัก
แต่ผมว่ามันมีประโยชน์ทีเดียว

English: 
four, one, two, three, four and a half,
we're gonna move up half.
So, if I wanted to draw the
tangent line right there,
it would look something like,
something like that.
Something, something,
something like that.
Let's see if we've got,
one, two, three, four,
and a half, that's what we got,
just like that is pretty close.
So that's what we just figured out.
We figured out that the
slope of the tangent line,
right at that point is one ninth.
So, it's not only neat to look at,
but I guess somewhat useful.

Korean: 
4칸, 하나, 둘, 셋, 넷과
그리고 위로 반만큼 더 올라가서
제가 저기에 접선을 긋는다면
그것은 아마 무언가
이렇게 생긴 것일 겁니다
무언가, 무언가,
무언가 이렇게 생긴 것
하나, 둘, 셋, 넷
그리고 반을 가면
저것이 꽤 비슷하네요
그러므로 저게 우리가 구한 답입니다
우리는 접선의 기울기를 구했습니다
저 점에서 1/9이죠
이렇게 하니 보기 깔끔할 뿐만 아니라
꽤 유용하군요
