Polar coordinate limit

In this video I use polar coordinates to calculate a limit in several variables. This is usually very hard to do, but polar coordinates simplify our work tremendously.

[@marcioamaral7511]

More multivariable calc please, these videos are for sure great!

[@plaustrarius]

@0:57 I think you mean y=rsin(theta) on the whiteboard but you change it in the next line and when you are speaking so everything works out haha @3:57 when you multiple each term by r^2, you can make a tighter restriction on the lower bound because it was previously zero it's still zero so you don't even have to evaluate a limit for the lower bound.

[@Aaron-lp3zt]

I appreciate how happy you are doing this limit. I was stressed about my exam, but after your video it clicked.

[@plaustrarius]

The squeeze theorem is so important!!! I wish I understood this concept earlier, it somehow tangentially helped me understand L'Hospital's Rule, along with looking at the series expansions. Thank you for the upload!

[@GabrielPohl]

I'm definetly gonna use this in my test, my teacher didn't see that coming! hahahahahahah

[@Smoothcurveup52]

Wonderful explaination sir The way of teaching is really awesome sir Full respect from India 🙏

[@TheDiederikdehaan]

Is the squeeze theorem necessary in this case though? Don't cos & sin have positive powers and therefore always larger than zero? I'm not sure - just observing.... Thank you Dr Peyam - love your videos!

[@moskthinks9801]

The answer for the limit is the average/current amount of dislikes you have generally/on this vid.

[@drpeyam]

❤️

[@tolgakabacasamala]

Could you please consider on the limit of the function (y*x^3)/(y^2+x^6) at (0,0). I think polar coordinate approach does not work in this case. After polar coordinate switch, I calculate the limit as 0. On the other hand, as aproaching to (0,0) along the path y=x^3, the limit is 1/2. Is there anything wrong that I calculated? My reasoning is "polar coordinate is just representing the linear aproach to (0,0), because we are keeping theta as constant and r is approaching to 0. Then, if limit does not related to theta, this means all linear paths to (0,0) give same limit. But this approach can not represent curvilinear path approaches." At this point you may ask "why don't we use y=mx switch, instead of polar coordinates" y=mx represent all linear paths except x=0. By this point of view, I am not agree with your spiral approach explaining as well. What do you think about my reasoning?

[@drpeyam]

No you did it correctly. Since you find 2 different limits, the limit does not exist

[@rodjaknenad6984]

@@drpeyam But in the video your statement at 4:08 implies that this method tests all paths, but it apparently doesnt. If this does not test a curved path such as the one that the commenter here gave, how can we test all of the paths?

[@xbz24]

nice vid dr, just what i needed, huge greetings from chili

[@michaelempeigne3519]

correction to your parameterization of x and y at beginning : x = r cos theta and y = r sin theta

[@real0067]

Sei il re di tutti i re!❤

[@katiegalgano3351]

Awesome video!

[@irok1]

Thanks for watching at the beginning of the video? That's great, almost as great as the video

[@amandaliu7409]

very good

[@armandoski-g]

Hey Dr. Peyam, couldn't we use a similar argument to say that if theta is close to k*pi , rsin(theta) tends to zero and so y tending to zero is equivalent to theta tending to kpi? Shouldn't we divide the limits into the two cases; when r tends to zero and also the case theta tends to kpi (in the y case, the x case would be theta tending to kpi/2)?

[@mathadventuress]

Dr peyam in my class my professor had us evaluate it like a substitution Like we do y=MX and y=kx² If the both go to the same value, then the limit exists otherwise it doesn't So...can we do this same thing that way as well? And factor and cancel out any terms?

[@BananaHead223]

This helped me out a lot! Thank you!

[@leonardromano1491]

I believe it is also possible to let x(n)=x0+a/n and y(n)=y0+b/n for arbitrary a and b and then let n--> infinity. If the result depends on your choice of a and b then the function is not continuous in (x0,y0). This works because you can have arbitrary paths depending on how you choose a and b.

[@AQWorldsOmmiiT]

Hey Dr. Peyam! At 3:14, there is a mistake in multiplying the inequalities by "r".

[@tofu8676]

thaha i struggled so hard with the idea that you have to prove such limits without assumptions on the path you take in calc class ^^

[@derekeby8654]

You messed that up quite a bit to get the correct result :)

[@elliewolcott6254]

Thank you so much!!

[@wesenfikre4821]

It's really helpful thanks Dr 😊👍

[@ompandey6345]

Nicely teach i am maths teacher from India

[@fazelbanoei6212]

haha thanks , you give wholesome vibes

[@vasundhrabisht8941]

Hi I tried to evaluate a limit using this. The function is x²y / (x^4 + y^2). And using this my answer came out to be zero but this limit doesn't exist. How do we know where to apply polar coordinates to evaluate limits and where not?

[@OtherTheDave]

I used to just say that “y = mx” (b was 0, since the lines won’t intersect there if there’s an offset), and then see if the limit depends on m.

[@dalek1099]

I don't think that works as you must consider all the possible ways x and y could go to zero not just straight lines. Peyam's proof works as it doesn't even matter whether theta varies on each path due to the squeeze theorem.

[@OtherTheDave]

dalek1099 Does that matter? It always got the right answer in school, but that might’ve been luck.

[@dalekonezeroninenine4588]

Consider the limit as x and y go to 0 of 2x^2y/(x^4+y^2) if you let y=mx then this expression becomes 2mx^3/(x^4+m^2x^2)>2mx^3/(m^2x^2)=2x/m which has limit of zero. However, if you let y=x^2 then this expression becomes 2x^4/(2x^4)=1. You would be alright if you said y=m(x)x and then showed the limit was zero like theta in Peyam's video has to be a function of x and y.

[@TechNewsAndReview101]

Thank you SO MUCH!

[@someone229]

A new video from Dr π times m! like before I watch 🙈

[@marcioamaral7511]

My professor uses rho as radius and phi as the angle for polar coordinates he likes weird notation

[@PackSciences]

It's not weird, rho as radius is a valid international notation. Phi as an angle is, however, not very common, as it can be misinterpreted as spherical coordinates. Also I like weird notations too, it widens your culture of mathematics.

[@marcioamaral7511]

I didn't said that it wasn't valid though Saying weird was just an expression, I like the notation a lot actually We are at the current moment talking about multiple integrals and for the Jacobian he used Lagrange's notation for derivatives trying to keep it minimalistic He's from Vietnam

[@tofu8676]

at our university we always write r and phi.

[@PackSciences]

Tofu which uni is it?

[@tofu8676]

PackSciences JKU in Linz, Austria most of the time we use (r,phi) for radial and (r,phi,theta) for spherical

[@ashokkumarsah9258]

Thanks Sir🙏 it's really fruitful 🙏🙏

[@ashokkumarsah9258]

From India

[@aneeshsrinivas9088]

do general changes of coordinates work for these limits?

[@colehonecker6436]

you are a gem :)

[@Luis-lw8fr]

Thank you😄

[@Arup497]

I found the center of c onic sections by following methodology: Step-1:let f(x,y)=0 be the equation of the conic section Step-2:then I calculated the partial derivative of f(x,y) with respect to x and y respectively. Step-3:Then,l equated those partial derivatives to zero and got 2 simultaneous equations Step-4:At last l solve those equations for (x,y) which was later found to be the coodinates of the conic section. I want to know the actual truth. So,l need your Noble help

[@user-jr9xc4fo3t]

thanks

[@shubhumchatterjee]

Sir, one question is that will y= r sin(theta)?

[@shubhumchatterjee]

@@paladise thanks bro

[@abduhalloqovabdusamad3849]

Thanks

[@sreerajnairg]

It’s not a supreme method to find if a function has a limit or not. You can use this to prove limit does not exist, yeah, but not to prove that limit exists or not. There might be other curve through which we may get a different limiting value.

[@fashionlover620]

Please do video on advanced calculus delta epsilon limit proof for function of two variable limit

[@danielkirilov8065]

Nicely done! Squeeeze.

[@harelkariv1477]

Since x and y are practically "dummy variable" and they both go to 0, can't we assume x=y?

[@drpeyam]

No you have to show it using all directions, x = y is the diagonal direction

[@duncanw9901]

can you do a video on Riemann's proof of the Fourier transform stuff? I tried to read through that on my own and it didn't work. :)

[@drpeyam]

What do you mean?

[@srpenguinbr]

As both variables approach the same value, I used the limit as x gos to 0 and switched y to x. I also got 0, is that a coincidence? Can I always do that?

[@srpenguinbr]

@@paladise got it. Later I took Calc 2 and learned about the infinite number of paths I could take.

[@_Jasleenkaur]

if limit exists or doesn't exist, we use polar coordinates. Why don't we check over other paths?

[@drpeyam]

Polar coordinates is over all paths, it’s a different way of saying (x,y)

[@federicopagano6590]

I knew this method and i loved it at first sight but i always was in doubt if this was legal to do now i can confirm that it is. But whats bothering me is if this is true (i dont doubt at all its perfectly done) why the hell in all courses keep on doing deltha epsilon to proove something wich it is much more easy to do in polar coordinates. If this "proove" that the limit is zero why is so popular the tradtitional hardwork? That question i really cannot answear the fact that we cannot deny is that in all courses do for this kind of limit the epsilon theta and thats the reason why i always was affraid to do this limit in polar coordinates i always think that there is maybe something hide that u cannot do. The statistics use a 99% epsilon delta in all courses the question is whyyyyy???? If this is perfectly done i feel like the statistics tell me dont ever think to do it this way but i dont kjow why , it seems to me perfectly done

[@rizkyagungshahputra215]

please do more about multivariable limit

[@danyaaaa31]

Amazing

[@federicopagano6590]

It looks great but I always wondered why on earth considering y=mx at a fixed m approaching x=0 doesn't seem to be enough to conclude anything but at a fixed theta approaching r=0 works ,it must be an extra fact behind because in both cases we fix one variable and then approach the remaining one to zero but in one case works and in the other doesn't. And i think ..the reason is because in y=mx we cannot conclude anything but in polar we can ...lol we already know that,... the question is why

[@flynnflanfck]

i think its because when you test y=mx, you only check along all straight lines. Therefore, you can't make any claims for other paths. However, using polar coordinates sets no constraints on the nature of your path

[@TheRedfire21]

i want to note that it is not always true you can find a limit using polar coordinates, a classic example is f(x,y)=((x^2)y)/((x^4)+y^2) edt; typos

[@mathadventuress]

Wow neat

[@meselmasal6602]

I think, to be able to use this property, we need the fact that f( rcos(theta)), rsin(theta))=F(r).G(theta)

[@MrRyanroberson1]

Lets see if I can manage. The limit is (x4-y4+4y4)/(x2+y2) = (x2-y2)+4y4/(x2+y2), so we get 4*Lim(y4/(x2+y2)), and we can do the same backwards to find: (4x4-3x4+3y4)/(x2+y2), and we get 4*Lim(x4/(x2+y2)), to be expected. Maybe now we can go to 4*Lim(x2/(1+y2/x2)), and see that for all y2/x2 not approaching -1, that it approaches zero.

[@drpeyam]

Watch the video...

[@cycklist]

Zero times r squared is not minus r squared :)

[@OtherTheDave]

Portsmouth FC Unless r = 0

[@PackSciences]

Portsmouth is refering to the fact that if he applies -r transform (before taking the limit operator), it should have been 0 as the bottom inequality, and not -r^2

[@drpeyam]

Portsmouth FC Haha, I originally meant to say -1 < cos^4 < 1 and multiply it by r^2

[@aboutmath2995]

thats just a greek Γ not r..

[@anissawilkinson9647]

Thank you daddy

[@davide467]

We also Need more ode

[@drpeyam]

Ode are coming :)

[@PackSciences]

What exactly did you smoke before this video? It's a short one indeed, but at first you messed up by writing r=cos(theta), but fortunately, you noticed it Then you wrote "y = r cos(theta)" while the line "x = r cos(theta)" was 2 cm above. I hoped you would have figured out it was sine when you replaced at 1:19 At 3:27, bottom boundary should be zero, if you apply *r^2 to all sides of the inequality.

[@drpeyam]

:(

[@PackSciences]

They are minor typos, but this video is dense in terms of error per time of video compared to your previous videos.

[@marcioamaral7511]

Try to be at least polite correcting other people We all make mistakes ,and more respect to Dr Peyam who takes his time to make these

[@blackpenredpen]

Márcio Amaral agree!!!

[@japotillor]

Even the best mathematicians make mistakes...contrary to popular belief were not robots.

[@nahombeyene7694]

Bro is excited

[@federicopagano6590]

mmmm let's see if we take z=x.y^2/(x^2+y^4) at (0;0) and we approach with x=0 then it goes to 0 but if we take x=y^2 the limit goes to 1/2 so the limit does not exist. However if we plug x=r.cost and y=r.sin (t) we end up the whole thing going to zero!! So....the polar method fails here there must be some hidden conditions I must to check before using this I think it's not so easy to plug polar coordinates and make r goes to 0(which I confess I did the same for several years until I stumbled with this example today!!!)

[@drpeyam]

I don’t think the polar method shows that it goes to 0, since you have r^3 / (r^2 + r^4) which doesn’t necessarily go to 0

[@Sam-hc4sd]

Ok I thought he was some random Indian KZbinr!!

[@drpeyam]

I’m Persian

[@cezarvintea7240]

The video is surely great but why does it look like you kidnapped some kids, tied them up and now you are explaining calculus to them?

[@drpeyam]

😂😂😂

[@exploringwithsd6253]

Lim (x^2-y^2)/(x^2+y^2),x,y tense to 0,0 =?????

[@drpeyam]

Same thing

[@MarcoMate87]

1:00 LOL

[@EIYEI]

Y=r sin(theta)

[@stefanj.8542]

Ba nebunule, ai grija ca acolo e sin la y, nu cos

[@yousseftaha8464]

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