Picture me Differentiating (visual calculus)

Рет қаралды 47,078

Күн бұрын

Correction: at 03:54, I should say “perpendicular to the radius of the circle at the point” for clarity.
In this video, we investigate the derivative of the sine function experimentally and analytically. The analytic (limit-definition) proof relies on two limits that are often computed geometrically, so this poses the question: is there a visual proof that the derivative of the sine function is cosine? We then show a wonderful visual proof that demonstrates that this derivative fact is true by using a similar triangle argument.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoffee.com/VisualPr.... Thanks!
This animation is based on a visual proof by Selvaratnam Sridharma from the September 1999 issue of The College Mathematics Journal (www.jstor.org/stable/2687673 - page 314-315).
#manim #math #mathvideo #mathshorts #calculus #triangles #animation #theorem #pww #proofwithoutwords #visualproof #proof #sinefunction #sums #pww #sine #proof #algebra #trigonometry #mathematics #mathvideo #mtbos #derivative #cosine #limit #limitdefinition
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Пікірлер: 139

@lazmotron 5 ай бұрын

This is a great concept for a mathematical You Tube channel, thanks for making this channel. Visualizing a mathematical concept is the best way to simultaneously show and understand it. Kudos. This is the proper way to visual medium to teach mathematics. 👍

@MathVisualProofs 5 ай бұрын

Glad you like it! Thanks

@alanthayer8797 6 ай бұрын

Appreciate da Visuals as usual

@MathVisualProofs 6 ай бұрын

👍😀

@GeneralFX 6 ай бұрын

Thank u ❤️... Seriously you cleared my problems of the topic “derivative”... No one can beat the level of explanation you can give through visualisation..

@GeneralFX 6 ай бұрын

Idk if u remember it or not 😂, but you suggested me 3blue1brown channel for this topic 's understanding a few weeks ago 😂... But ur explanation tops it up

@MathVisualProofs 6 ай бұрын

Happy to help

@calebhubbell2290 5 ай бұрын

Thanks for this. The geometric proof makes it really simple. Subscribed :)

@MathVisualProofs 5 ай бұрын

Thanks!

@__melker6202 6 ай бұрын

amazingly done!!

@MathVisualProofs 6 ай бұрын

Appreciate your comment!

@PlanetNumeracy 5 ай бұрын

These will be awesome in my classroom. Thanks for making it!

@MathVisualProofs 5 ай бұрын

Glad you can use it!

@danielc.martin1574 6 ай бұрын

Great! Thanks!

@MathVisualProofs 6 ай бұрын

Glad you liked it!

@yolamontalvan9502 5 ай бұрын

Thanks for mentioning the software used for you amazing videos. I teach math and I need a software to make videos to teach 7, 8, 9 years old math. Thank you.❤

@MathVisualProofs 5 ай бұрын

Happy it helps.

@user-hb1tx6de5t 5 ай бұрын

After all these 20 years of partial math related works... now this clip is mentoring that my conception of sin and its derivatives result was wrong! or never thought about it deeply.. Deeply thank you Mathematical visual proofs!! No time is too late for starting a math again for everyone, isn't it?

@MathVisualProofs 5 ай бұрын

Glad you like this! Thanks

@virushk 6 ай бұрын

you are a gift to humanity

@MathVisualProofs 6 ай бұрын

Well not sure about that. But glad you enjoy the math :)

@anderslvolljohansen1556 6 ай бұрын

Hypotenuse "perpendicular to the circle"? Should it rather be "tangent to the circle and thereby perpendicular to its radius"? 3:55

@MathVisualProofs 6 ай бұрын

Yes. My bad. I definitely meant to say perpendicular to the radius of the circle.

@math_travel 5 ай бұрын

In the first part of the video, you graphed the derivative of sin(x). After expressing the slope at each point in a graph, you asked whether the graph was cos(x). This was very impressive to me. This was because I had not thought about the differentiation of sin(x) in that way. thanks

@MathVisualProofs 5 ай бұрын

Glad that helped!

@wendolinmendoza517 5 ай бұрын

The derivative of a function is exactly that 😅

@hmedina79 5 ай бұрын

Awesome. just awesome! What graphing utility is that? Wow!

@MathVisualProofs 5 ай бұрын

Thanks! I use manim for these

@kyoq3204 6 ай бұрын

You opened my eye ! 😍

@MathVisualProofs 6 ай бұрын

😀👍

@blackholesun4942 3 ай бұрын

Thanks 👍👍. My understanding of cos and sin is greatly related to triangles. The limit proof did not feel as intuitive as the visual geometric one

@MathVisualProofs 3 ай бұрын

Glad this helped!

@christiansmith-of7dt 5 ай бұрын

Before youtube it seemed like nobody ever knew what I was thinking all the time

@godfreypigott 5 ай бұрын

The limits in the analytical solution turn out much simpler if you use f'(x) = lim (h->0) [f(x+h) - f(x-h)] / [2h]

@FundamSrijan 6 ай бұрын

Not the visual proof , the *_ACTUAL_* proof 🙏

@Manisphesto 5 ай бұрын

The creator of this video is named “Visual Proofs” for a reason.

@sonicmaths8285 5 ай бұрын

False. A real proof is purely logical, not visual. Visual proofs are too unrigorous to be considered “actual proofs”.

@hunterk1575 5 ай бұрын

he did walk through the actual proof using limit def of derivative in the video

@godfreypigott 5 ай бұрын

@@sonicmaths8285 Tadashi Tokieda would disagree with you.

@sonicmaths8285 5 ай бұрын

@@godfreypigott It doesn’t matter who disagrees. It is a fact that visual proofs are not considered “actual proofs”.

@ingx32 6 ай бұрын

I'm confused on one part... at 3:54, how are you getting that one of the non right angles has to be complementary to theta? I feel like there's some geometry knowledge I'm missing here :(

@MathVisualProofs 6 ай бұрын

Theta is complementary to alpha. And then in the picture the radius of circle is perpendicular to the hypotenuse of the triangle (tangent to circle). We already have a theta angle in the 90 degree angle for those two. The remaining angle, which is the triangle angle, must be complementary to angle - so it’s alpha.

@ingx32 6 ай бұрын

@@MathVisualProofsooooo I see, thanks :)

@anderslvolljohansen1556 6 ай бұрын

The narrator says "hypotenuse perpendicular to the circle", but I think he should rather say "tangent to the circle and thereby perpendicular to its radius".

@MathVisualProofs 6 ай бұрын

@@anderslvolljohansen1556 yes. Good catch. I meant perpendicular to the radius of the circle 🤦‍♂️thanks!

@user-ls9fw7jb2m 2 ай бұрын

at 4:21 you say, "This means that the triangular wedge region has an angle mapped out by α." What does that means?

@razday8490 6 ай бұрын

wow

@MathVisualProofs 6 ай бұрын

😀👍

@pradyumnanayak9844 5 ай бұрын

🙏

@MathVisualProofs 5 ай бұрын

👍😀

@freedivemd9366 6 ай бұрын

I don't understand the very first graph where there are two coordinates given. As the point moves, one variable changes, but the other varibale is always a constant "1". What is this constant "1". Where does it come from? What is it supposed to represent?

@asianhaydenxd 6 ай бұрын

1 is the change in x and the other is the change in y. Together, they make the slope.

@MathVisualProofs 6 ай бұрын

I fix the base of the triangle to be 1 so you can see the slope change.

@innovationsanonymous8841 2 ай бұрын

Really glossed over the geometry around 4:00

@MathVisualProofs 2 ай бұрын

Not sure what more I can say about similar triangles 🤷‍♂️

@airman122469 5 ай бұрын

Ummm. Hold on. Sin(0) = 0, so sin(h)/h as h goes to 0 is technically 0/0, and further limit rules must be applied to get to the correct result that the limit approaches 1. Other than that, sure. Welllllll… actually… there’s another nit: the derivative definition provided is predicated on a definition of cosine and sine themselves because the sum angle formula was used to get at the expansion used to come to the result of cos(x) But the visual derivation was nice.

@godfreypigott 5 ай бұрын

What is wrong with the derivation depending on the definition of the function? And he was clearly assuming that everyone knew that limit.

@purabimondal6270 5 ай бұрын

Well try to prove actually the integrals of functions like e^x and hyperbolic functions

@yyy76yyvhxxffb32 5 ай бұрын

Wait, why is sin(h)/h=1? h=0.000000000... So sin(h) should equal 0.000000... So 0.0000000.../0.0000000...=1 But if thats true then why in the hell does Coss(h)-1=0? Coss(h) should be equal to 0.999999999999999999... and then 0.999999...-1 should equal -0.000000000...=-h So why is it not Sin(x) times (-h/h) which would give -sin(x) Then -sin(x) +coss(x) Im not understanding please explain pleeeeeeeeease

@MathVisualProofs 5 ай бұрын

You have to really examine these two limits. You can use a calculator to get heuristic answers, but if you plug in, say, h=0.0000000001 into sin(h)/h, you will get a value close to 1. You can use a geometric argument to prove that the limit is 1.

@yyy76yyvhxxffb32 5 ай бұрын

@@MathVisualProofs thx

@pizzawhisker 5 ай бұрын

Both sin(h) and cos(h)-1 are close to 0 for small h but the latter is much closer. You can see this if you start at (1, 0) and go up slowly along the unit circle. The y coordinate increases at about the same rate as the arclength while the x coordinate almost doesnt change. This kinda describes the difference between the two limits.

@yyy76yyvhxxffb32 5 ай бұрын

@@pizzawhisker bro Lim sin(h)= aproximately zero h->0 Lim coss(h)=aproximately1 soo 0.(9) h-> 0 Coss(h)=0.9999... 0.99999.... -1 =-0.00000...=-h So -h/h=-1 So -sin(x) +coss(x) should be the final answer but im gonna go to explication class soo

@pizzawhisker 5 ай бұрын

@@yyy76yyvhxxffb32 you cant just use 0.0000... because not all infinitesimally small quantities are the same 0.0000../0.0000... can tend to 1 0 infinity or any other number. Its called an indeterminate form and can be solved by lhopitals rule in some cases.

@Peter_Riis_DK 5 ай бұрын

And what is the practical application?

@anilkumarsharma8901 5 ай бұрын

Derived dharawahik derived dhara dharaatal Deliver😂derivative dwara 😂😂 India🇮🇳 do this👌 types of🔺 mind😏 from millions of🔺 years ago😂😂😂 Sanskrit knowledge📚 called it's chavi😂😂😂