Calculus 2 Lecture 10.4: Using Polar Coordinates and Polar Equations

Calculus 2 Lecture 10.4: Using Polar Coordinates and Polar Equations

[@keldonchase4492]

Day 1: Introduction to Polar Coordinates: 0:00 First Example (Graphing Polar Coordinates, positive r length): 7:32 Second Example (Graphing Polar Coordinates, positive r length): 10:36 First Example (Graphing Polar Coordinates, negative r length): 13:10 Second Example (Graphing Polar Coordinates, negative r length): 15:16 Brief Review of Graphing Polar Coordinates: 16:33 Introduction to Translating Between Polar Coordinates and Rectangular Coordinates: 17:10 First Example (Translating From Polar to Rectangular Coordinates): 20:28 Second Example (Translating From Polar to Rectangular Coordinates): 24:03 First Example (Translating From Rectangular to Polar Coordinates): 27:35 Second Example (Translating From Rectangular to Polar Coordinates): 33:41 First Example (Converting a Polar Equation into a Rectangular Equation): 39:41 Day 2: Introduction to Symmetry: 46:52 First Example (Checking for Symmetry): 48:42 First Example (Graphing a Polar Graph Symmetric about Polar Axis): 52:54 Second Example (Graphing a Polar Graph Symmetric about Polar Axis and Theta = pi/2) : 1:03:22 Third Example (Graphing a Rectangular Graph Symmetric about the X-Axis and the Y-Axis): 1:26:22 Day 3: Introduction to Finding Tangent Lines of Polar Graphs: 1:37:35 First Example (Finding the Slope of a Tangent Line): 1:46:47 Finding Horizontal and Vertical Tangent Lines to Polar Graphs: 1:59:16

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Keldon Chase Thank you!

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Thanks

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Many thank -last minute crammer

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[@danieljones6132]

at 1:11:00 I believe the correct trig identity is cos(a-b)= cos(a)cos(b)+sin(a)sin(b), not minus. But the answer is the same. I love these videos and Professor Leonard has given me such a better understanding of math than my previous classes. Thank you sir!

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[@ishtiaqhussain7766]

At 1:08:45 I suppose it’s meant to be a + sign rather than negative (check the identity for cos(a-b)). But it doesn’t make a difference in the result. Lmk if I am mistaken. Prof Leonard! I admire your dedication to teaching. If I ever lecture/teach, I will try to do in your way. Edit: It has been noted in comments by others I realise.

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YEAH, I was about to comment about that small error. fortunately, it doesn't affect the result.

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Thank you for this comment, I thought I was going crazy.

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[@ProfessorLeonard]

kinkehpanda Good Catch!! I meant + but I must have missed that. Thank you. For everyone else, make a note.

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1:09:20 the cosine identity he used should have a + not a -

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[@ProfessorLeonard]

I'm so glad you find them helpful!! I'm also very pleased that you can stay engaged and are learning. Keep it up!

[@MasonMcKinney-jx7jp]

These videos are a miracle from heaven, allowing me to self-study for AP Cal BC in just 3 months as a homeschooler. Anyway, I feel bad by pointing out a silly mistake because of how perfect Prof Leonard always is. But I'll do it for everyone watching in 2024 and beyond. At 1:10:00-1:11:15 he used a subtraction cos identity incorrectly. Leonard accidentally put a minus in between the identity's solution as opposed to a plus. Because cos(a-b) results in cosacosb+sinasinb. Hope this helps!

[@echo-channel77]

cos(A - B) should be [ cosA cosB + sinA sinB ] but the calc still works out in the demo as 1, i.e. 1-0 => 1+0. Awesome videos!

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[@marjavanderwind4251]

Hi @Professor Leonard I think I see a mistake at 1:10:06 because the cosine addition formula for cos(x - y) is with a plus-sign in the middle instead of a minus sign. In this case it doesn't matter because of the sin(pi) becoming zero, but it might be relevant for you guys.

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Oh wait this thing was said by others before.

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[@IlhamAzlanSyah]

1:09:14 In the cosine identities, the sign of cos becomes negative, doesn't it? But it doesn't matter because sin 2 pi equals to zero.

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[@georgesadler7830]

Professor Leonard, thank you for an excellent Introduction to Polar Coordinates and Polar Equations in Calculus Two. This material is a solid review of Precalculus and is very helpful in Multivariable Calculus. Graphing Polar Equations is not difficult; however, I found the examples a bit confusing. Problem solving will increase my knowledge and deep understanding of the material. This is an error free video/lecture on KZbin.

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[@saeedsaleh6873]

Hi Professor Leonard, I'm a big fan of you, I've just noticed a little mistake with the subtraction identity for cosine in 1:09:00 , it should be 2[Cos(2pi)Cos(2theta) + Sin(2pi)Sin(2theta)], luckily that mistake didn't change or effect the answer.

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The greatest professor of all time

[@animeshmajumdar66]

At 1:09:36 i think there should be a '+' sign in the identity and not ' -' 🙂🙂🙂

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[@fraze1228]

Hi Prof Leonard. Love your lecture. I think that the symmetry about the pole should be f(theta) = - f(theta) or f(pi + theta) = f(theta) instead. Just to check. Thanks a lot.

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[@tybargky3554]

Professor, I thought the subtraction Identity for cos(A-B)= cosAcosB+sinAsineB? correct me if i'm wrong. Thank you

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[@adnanmohamed6517]

@Professor Leonard I think there is a mistake in the video. Look the board @48:50 (or in the minute 48). Point 3 of symmetry says: If f(theta + pi) = - f(theta) then symmetric about the origin. But I think the negative sign should not be there in the R.H.S. The correct equation in my opinion is: f(theta + pi) = - f(theta) (without the negative)

[@fromfilmtocode]

Best video on the matter I found as a newbie.

[@ucbuixuan9940]

At 1:08:31, can you just think of 2π as a full circle and get rid of it? Hence: 2cos(2π-2θ) = 2cos(-2θ) = 2cos(2θ)

[@jaxonrholden]

In metrology when converting between coordinate systems, I use theta = 2*arctan[y/(x+r)] which automatically gives the angle in the correct quadrant without adding pi.

[@vk6297]

plz explain @professor leonard the symmetry rule is not working for r=theta sin(theta)

[@malevip]

@ProfessorLeonard, You have a mistake at 1:10:34 cos(A-B) = cosA cosB + sinA sinB not minus..

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[@abdullahmohammad5613]

An easy way to approach the right graph of the polar segment when it's negative is to add/subtract 180 Deg for the given angle :)

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[@kqnrqdtqqtttel1778]

1:08:32 Just note that cos has a period of 2pi, so adding 2pi to the angle -2theta and taking the cos gives the same result as simply taking the cos of -2theta, or the cos of 2theta.

[@BaySelvi]

Identity function is written wrong at(1:09:46) It must be written as 2[cos(2π)cos(2θ)+sin(2π)sin(2θ)] However it gives the same result because sin(2π) is 0 and it makes the sine part of the identity 0 in both cases whether its sign is minus(-) or not

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Thank you professor Leonard.

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[@kinkehpanda]

The trigonometric identity at 1:10:10 is slightly incorrect. It should be [cos2pi cos2thetha + sin2pi sin2thetha]. It didn't affect the outcome this time because the right term was zero. NVM it was mentioned in the comments already.

[@greense65]

But you added the timestamp, which was missing from the earlier comment identifying this small error. Thanks.

[@satyamverma1910]

Thanks mate!! I was wondering the same

[@yasminasabra8875]

yes true

[@jeramibellofatto817]

2Cos2(2pi - 2theta) = 2[Cos(2pi)Cos(2theta) + Sin(2pi)Sin(2theta)] In this case the second term cancels out but would matter for other problems. 1:10:17

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[@Salamanca-joro]

Tip for you guys , If you get theta as pi over something or in pi and you want to convert it to degrees use this : Multiply it by 180/pi For example this example 26:08 To find x we use this formula X=rcostheta Since theta was pi over 4 convert it to degrees Pi/4 times 180/pi Is 45 So x=-sqrt2 times cos 45 Which equals -1 And you can do plug 45 for sin as well to find y and you will get same answer -1 Try to do it with the other examples and you will see how good this method is

[@sidsingh9387]

When you increased the "size" of the origin so the poorly drawn curves pass through the point - I do that every single time too! :>

[@sandeepgrover395]

very good lecture. Thanks a ton sir! :)

[@Inspired_Cat]

Just to inform one minor error (1:08:46)... The result was still correct since it was zero but the subtracting identity of cos is positive at the center sign. Not negative.

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Is this a High school class or a college class you have when you film these calc 2 videos?

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College, but it seems like high school

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[@faiqfarhan2309]

May I Know Where Can I Download The Polar Coordinate Graph Just Like The One On This Video?

[@amre3237]

+Faiq Farhan Just google polar coordinate graph paper fam

[@ChrisChoi123]

the last thing you evaluated, the slope at pi/6, could be simplified to -1.

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thank you

[@anthonysantiago6017]

hey professor, your sum and difference identity was actually incorrect but you got the right answer. its cos cos sin sin then change sign for cos.

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Sqrt(2)=1.414 but you said 1.7 at 1:16:20 anyway love your videos, they are really helpful. Your efforts are appreciated. 😃

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35.00 why not in clockwise ..ı dont understand

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This motivates me to workout my biceps

[@Festus2022]

I think Leonard made a mistake at 1:55:20 when he "factors out a negative". I think the numerator should be -cos2theta MINUS cosTHETA. This changes the result.

[@Festus2022]

Correction--> Dr. Leonard was correct. I am wrong!!

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seriously awesome teacher

[@wildchild1161]

@ Professor Leonard @1:08:59 your identity is wrong , Cos(a-b) = Cos(a)Cos(b)+Sin(a)Sin(b) . It doesn't affect the out come though because your sin value ends up being 0

[@ProfessorLeonard]

Yep, I seem to remember watching this back a while ago and thinking the same thing. Good catch!