Introduction to polar coordinates

[@oldgraybeard3659]

I'm a visual person, and almost numbers blind. It took me an hour per page to read this information out of my text book. In ten minutes, you went to the point. Your visuals were neat, and visually literate. After watching your video, I was able to jump into working the problems in my text book; with confidence and understanding. Thank you!

[@chrisodden]

Thanks so much for the feedback! I'm happy the video helped you, and good luck with your studies.

[@katherinekelm7439]

This is fantastic! Thank you for such clear explanations and wonderful visuals to match.

[@g.gardiner4517]

This is amazing! I was completely lost on how -r relates to the placement of P, but now it's all clear. Thanks so much!

[@resyarsunil5613]

Nice presentation nd teaching everything ...really nice...

[@swapnil72]

Crystal Clear Explanation ✌️

[@ShrutiIyer88]

Thank you so much! This has been super helpful!

[@anasghaffar7837]

I wish I had found this video, the first time!

[@MrZak-rf3vq]

Same

[@rkumaresh]

Polar coordinates are hard to understand and this video was simple to learn the basics. @8:37 when the arctan(2/5) is 0.381 which lies in the range of inverse tan i.e -pi/2 to pi/2 why would we have to go back and add pi to that.

[@enashameed471]

Thanks جزاك الله خيرا

[@maqdala]

This is an epic video. Thank you.

[@lux27.42]

waw... thank you so much !!! love your visual explain.

[@sumansen3687]

Really nice explanation 👌

[@chrisodden]

Thank you 🙂

[@dugunu34]

Very clear explaination

[@shamimhussain5918]

Using which application you write mathematical terms?

[@juromebey1922]

255 degrees should be 4pi/3 radians on your unit circle 1:13

[@garymartin9777]

Now an introduction to polar bear coordinates. Wherever there's a polar bear, my butt is vectoring with great magnitude pi radians relative to it.

[@justinli19901027]

well explained, thank you

[@Hud_Adnan]

Great instructions

[@swapnil72]

Helpful, a lot

[@catedoge3206]

amazing.

[@kevinscheengsbier6130]

Awesome

[@abcdef2069]

at 3:40 negative r didt they already define r is always postive? like r^2 = x^2 + y^2, then please explain this when r= sin (theta) * sin (theta) this totally violates (r, theta)= (-r, theta+pi), i never seen any mathematicians who could explain this. or can they please expand number system to explain negative r, like they expaneded number system to complex number to satisfy x^2 + 1 =0

[@chrisodden]

The key distinction is between r and r^2. It is true that r^2 must always be greater than or equal to zero; however, even when r is negative it is the case that r^2 will be positive. By the way, let's examine the conversion equation y = r sin(theta) and see it is compatible with the ambiguity we are discussing. Suppose we simultaneously turn r into -r and theta into theta + pi. Recall the trig identity sin(theta + pi) = - sin(theta). Now watch what happens when we recalculate the y coordinate with these new polar coordinates: y = ( -r )( sin (theta + pi) ) = -r ( - sin theta ) = r sin(theta). This proves that the y-coordinate is the same when we use these new polar coordinates. You can work out a similar calculation to show that the x-coordinate also remains the same. The conclusion must be that these new polar coordinates really do give you the same point in the plane.

[@abcdef2069]

...new polar coordinates really do give you the same point ... yes it works when you deal the points with rectangular coor and polar coor on the same sheet of paper. x=rcos(theta), y=rsin(theta) are more rectangular things in my eyes. maybe polar coors and polar functions are two different things. for r= sin (theta) * sin (theta), (r, theta)= (-r, theta+pi) doesnt work in my eyes. for r= sin (theta), you can not plot anything on the 3rd quadrant. the moment that theta hits the 3rd, it starts to plot on the 1st quad. i still do not understand the negative r(theta) of polar function.

[@chrisodden]

OK. When you are plotting a polar curve then there is less ambiguity. As you suggest, given a particular value of theta then you have a theoretical direction in which you “should” be pointing (i.e. first quadrant for 0

[@bismabisma9892]

Thanks 👍

[@_shivani_gupta]

Thanku so much.....😘😘😘

[@johncharles3907]

nice graphic, thanks

[@arnoldstephen5668]

Thanks

[@footage6402]

Whyis polarcoordinate useful ?rather than rectangular

[@shelbyolofson]

We use it in upper levels of calculus to rewrite complex equations that have x,y and lots of sines and cosines. We rewrite the equation with polar coordinates (a process call "reparameterizing the equation"), which simplifies the calculations of derivates, integrals, and such

[@iruranyiha6248]

NIce

[@hlrnoctopus8244]

Yo prefiero theta 0 a 2pi ) Ponto final!!!!

[@shelbyolofson]

tambien!

[@peony325]

Thank you for this T _ T