Picture me Differentiating (visual calculus)
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- Опубліковано 3 лип 2024
- 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).
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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. 👍
Glad you like it! Thanks
Appreciate da Visuals as usual
👍😀
Thank u ❤️... Seriously you cleared my problems of the topic “derivative”... No one can beat the level of explanation you can give through visualisation..
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
Happy to help
amazingly done!!
Appreciate your comment!
Great! Thanks!
Glad you liked it!
Thanks for this. The geometric proof makes it really simple. Subscribed :)
Thanks!
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?
Glad you like this! Thanks
you are a gift to humanity
Well not sure about that. But glad you enjoy the math :)
These will be awesome in my classroom. Thanks for making it!
Glad you can use it!
Hypotenuse "perpendicular to the circle"? Should it rather be "tangent to the circle and thereby perpendicular to its radius"?
3:55
Yes. My bad. I definitely meant to say perpendicular to the radius of the circle.
Before youtube it seemed like nobody ever knew what I was thinking all the time
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.❤
Happy it helps.
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
Glad that helped!
The derivative of a function is exactly that 😅
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
Glad this helped!
You opened my eye ! 😍
😀👍
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]
Awesome. just awesome! What graphing utility is that? Wow!
Thanks! I use manim for these
at 4:21 you say, "This means that the triangular wedge region has an angle mapped out by α." What does that means?
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 :(
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.
@@MathVisualProofsooooo I see, thanks :)
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".
@@anderslvolljohansen1556 yes. Good catch. I meant perpendicular to the radius of the circle 🤦♂️thanks!
🙏
👍😀
wow
😀👍
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?
1 is the change in x and the other is the change in y. Together, they make the slope.
I fix the base of the triangle to be 1 so you can see the slope change.
Really glossed over the geometry around 4:00
Not sure what more I can say about similar triangles 🤷♂️
Not the visual proof , the *_ACTUAL_* proof 🙏
The creator of this video is named “Visual Proofs” for a reason.
False. A real proof is purely logical, not visual. Visual proofs are too unrigorous to be considered “actual proofs”.
he did walk through the actual proof using limit def of derivative in the video
@@sonicmaths8285 Tadashi Tokieda would disagree with you.
@@godfreypigott It doesn’t matter who disagrees. It is a fact that visual proofs are not considered “actual proofs”.
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.
What is wrong with the derivation depending on the definition of the function?
And he was clearly assuming that everyone knew that limit.
Well try to prove actually the integrals of functions like e^x and hyperbolic functions
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
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.
@@MathVisualProofs thx
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.
@@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
@@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.
And what is the practical application?
Sha you na na Japanese good bye
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😂😂😂
NO ONE CAN UNDERSTAND….
Because sine and cosine MEANS NOTHING…. Tangent is 180*
Before youtube it seemed like nobody ever knew what I was thinking all the time