Small Angle Approximations (1 of 4: Proof)

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Eddie Woo

Күн бұрын

Пікірлер: 57

@maroctech761 5 жыл бұрын

Eddie you are the Teacher of teachers ! Thank you

@madshothd3452 4 жыл бұрын

he fucked up tho, cause 0 is sin0 which is tan0 throw approximations

@kkanden 5 жыл бұрын

*ENGINEERING INTENSIFIES*

@huntleymoore2486 3 ай бұрын

what an amazing explanation. this guy is a great teacher

@sukhmankahlon3496 Жыл бұрын

genuinely so amazing, I hope the best for this teacher

@llcstudios1846 5 жыл бұрын

For those who are interested, you can use the same setup to prove that theta is approx sin theta by considering that the area of the triangle approaches the area of the sector as theta approaches 0. So you have 1/2r^2theta = 1/2(r-x)h. We can cancel out the half and as theta -> 0, x->0, so we can set x = 0. Therefore, r^2theta = rh. So rtheta = h. But we know, h = rsintheta, therefore, rtheta = rsintheta which implies theta = sintheta for small angles of theta.

@MattColler 5 жыл бұрын

I normally love Eddie's videos, but this one gets it quite wrong. He fails to point out that θ (in radians) is equal to the arc-length divided by the radius (i.e. *not* an arbitrary way of measuring angle). And for small angles, this arc-length becomes very close to the opposite-side-length, h. Therefore, arc-length/r ≈ h/r ≈ h/(r-x) In other words, θ ≈ sinθ ≈ tanθ.

@joshuaronisjr 5 жыл бұрын

Right. Via his same argument, theta * 10 is also approaching 0, and then sin(theta) would be approximately equal to theta * 10, which it obviously isn't.

@Chris-hn4lp 3 жыл бұрын

@@joshuaronisjr This isn't his argument, this is an accepted mathematical proof thats used regularly. He was just demonstrating it. You're seeming to neglect that theta is considered to be infinitely small.

@joshuaronisjr 3 жыл бұрын

@@Chris-hn4lp When he says that sin(theta) = theta for small theta, he does so by saying that they both become infinitely small. Take the example of z=4x and y=x, where x is variable. As x comes infinitely close to 0 (and indeed, all the time - but that misses the point) we could say that z and y both approach 0. But we wouldn't say that z=y as x goes to 0. No, we would say that z=4y. The same thing with theta - just because theta and sin(theta) both go to 0, that doesn't mean that they can be considered equal for infinitely small theta (I'm not saying that they don't approximate one another - I'm just saying his argument isn't enough). As another example take z=e^(2x) and y=e^(x). As x goes to 0, z is approximately 2 times y, even though they both go to 0.

@JustMichaelMoreno 5 жыл бұрын

I’m watching this for fun.

@pratikraj3697 5 жыл бұрын

Sir, you are great... I want to meet you personally in India.

@skippychunkypeanutbutter 3 ай бұрын

goated teacher the way you teach is clear and captivating, so thanks for that! :))

@MS-sv1tr Жыл бұрын

Being a math teacher must be brutal. It's hard enough when the students actually want to learn, but when the first question they always ask is "when am I ever going to use this", you're fighting a losing battle. I almost think mandatory math education might be a bad idea. If classes were small and only had motivated and interested students, that could only be a good thing, imo

@praveenmaxy7200 5 жыл бұрын

I am looking 30yo sir I have learned all these at my school...But now I understand maths..U made me see maths

@abdullahsafarini8422 5 жыл бұрын

Great visualization of it. I also want to ask if you could explain later on why pie is the same for big and small circles.

@surrealisticinfinity2895 5 жыл бұрын

Pie is the same for all circles because that number is the relation between the circles diameter and the circumference of the same circle. In a perfect circle there will always be pi numbers of diameters around the circumference no matter how big or small it is.

@zuccx99 5 жыл бұрын

Another part of the fundamental theorem of engineering. That sin(theta)=theta.

@Chris-hn4lp 3 жыл бұрын

Yeah, I was struggling to understand why in vibrations we were equating sin theta to theta and this video made it perfectly clear.

@zurgno6781 5 жыл бұрын

Why does it look like you're doing the nae nae in the thumbnail 😂😂

@jasminvioletu 5 жыл бұрын

ahahahahhhahahahah omggg

@sergioh5515 5 жыл бұрын

I thought you were going to use maclaurin series for this one but this is good too

@agildashdamirli9812 Жыл бұрын

Perfect explanation . Thanks

@marsch000 5 жыл бұрын

May I have the link to part two of the teaching?

@em8136 4 жыл бұрын

Watching this in 4k is soo good

@afrozshaik7941 3 жыл бұрын

Crazy. Excellent teacher. Beyong words.

@vign1155 5 жыл бұрын

Can you please help me in taking the coordinates in matrix form??

@gldanoob3639 4 жыл бұрын

You haven't stated the proof completely yet. If sin x, tan x approaches 0 as x tends to 0 implies sin x, tan x could be approximated to the same thing for small x, then since x/100, 2x, x^100000000000000 all approaches 0 as x tends to 0, can't you also say that they are basically the same thing at x near 0? You have to also proof that, the rate of which the difference between sin x, tan x and x shrinks is much higher than that of which sin x shrinks, in other words, the limit of (sin x - x)/x is 0 as x tends to 0

@mohdishaq9250 3 жыл бұрын

Nice lecture

@thomaskim5394 5 жыл бұрын

You stated that sin theta approaches 0 as theta approaches 0. But that does not necessary mean sin theta approximately equals theta for small theta. You need more convincing proof for that. Again go back to the calculus for that part. Using the Taylor series of sin, sin theta is approximately equals theta for small theta.

@MattColler 5 жыл бұрын

You don't need calculus, just the fact that θ (in radians) is the arc-length around the circle divided by the radius. For small angles, the length 'h' in Eddie's diagram becomes comparable to the arc-length. So h/r = sin θ approximates arc-length/r = θ

@thomaskim5394 5 жыл бұрын

@@MattColler That is the point. Mr. Woo does not explain even though many calculus textbooks explain only using geometry.

@MattColler 5 жыл бұрын

Thomas Kim yes, I was meaning to agree with you that Mr Woo‘s explanation here is incomplete. My point is that you can explain it properly without venturing anywhere near calculus (which these students haven’t covered yet) - let alone Taylor series, which is well beyond that.

@llcstudios1846 5 жыл бұрын

@@thomaskim5394 Also by using calculus you defeat the point of small angle approx like this. This is needed to find the derivatives of sine and cosine which find the Taylor series for themselves.

@thomaskim5394 5 жыл бұрын

@@llcstudios1846 Read my other commitments about the Calculus textbook. Many Calculus textbooks have a great and complete proof without using Calculus. The proof in the video is not complete.

@amankumarsingh6744 5 жыл бұрын

Awesome sir

@gloriosatierra 5 жыл бұрын

I like how your videos abruptly end, like if a limit exist. 🌿

@Brilliant_Mysterious 5 жыл бұрын

Lol

@antonyzhilin 5 жыл бұрын

If the limits are the same, the functions are the same, wat? If there was ~= 0 at the end, I'd understand, but in-between relationship isn't implied

@Jack_Frost 5 жыл бұрын

If the limits as x approaches 0 of f(x) and g(x) are the same (let's call this limit a), then f(x) ~= g(x) as x approaches a. For example, take f(x)=x and g(x)=x^2. When x approaches 0, f(x)~=0 and g(x)~=0 , so we can say that f(x)~=g(x) when you get close enough to 0

@antonyzhilin 5 жыл бұрын

Jack Frost I see. I'm just used to equivalence meaning “same order”, i.e. f(x) / g(x) -> 1

@Jack_Frost 5 жыл бұрын

@@antonyzhilin if f(x) / g(x) -> 1, multiply by g(x) on both sides and you have f(x) -> g(x), so when you approach a certain value of x, the two functions are equal

@antonyzhilin 5 жыл бұрын

@@Jack_Frost Your definition is weaker. For example, take f = 2x, g = x. Then lim f(x) = lim g(x) = 0, but lim (f(x) / g(x)) =/= 1. So to prove sin(x) ~ tan(x) ~ x in one sense is significantly easier than the other

@MattColler 5 жыл бұрын

@@Jack_Frost You're quite wrong here. In fact, f(x)=x *does not* approximate g(x)=x^2 for small values of x. Try it out: f(0.01) = 0.01 but g(0.01) = 0.0001 - different by two orders of magnitude! Eddie Woo's derivation of the small-angle approximations is also wrong. It is insufficient to merely show that all three functions approach zero - you need to show that they do so with 1:1 proportionality. After all, the angle measured in degrees also approaches zero, but this is not a good approximation for sin θ.