A polar curve is just like x and y except now it's θ and r. x is the input, and y is the output(usually), but now θ is the input, and r is the output. So imagine if at every angle you get a length, and when you graph it, it's a polar curve.
@@5000jaap with x and y it is called a function, and a function isn't allowed to have two y-values with the same x-value. This is why an inverse of a function only has one half. For example: f(x) = x^2 (normal parabola) inverse of f(x) = sqrt(x) With the inverse, you only see positive y-values, because for example sqrt(4) = +/- 2. -2 is not allowed, because otherwise it wouldn't be a function anymore. So no, we can't make these kind of things with x / y.
@@TheBrainReal there isnt any way do make a curve with that graph, i mean, a way to pass a polar curve into a curve, and what is the definition and the propriets and etc of a polar curve (sorry my english)
I was struggling with this polar curve at a question that required the area of one pedal. I solved finding the intervals of the pedal and tried the with proportion, the method of the video. I didn't even know that, when a polar curva envolve a cosine of a odd number, you should integrate from 0 to pi, not from 0 to 2pi, now I can move on with my Calculus course. Thanks BPRP
Half the integral of r squared is easier to work with. And it has a very precise definitely which is super easy to visualize. You are just adding up the infinitely many isosceles triangles of side r and angle da (so whose area is r²sin(da), which in the limit reduces to just r²da). It is pretty much the equivalent of a Riemann Sum for a polar curve, pretty nice!
Normally one rotation is 2pi, but since the point is reflected when r is negative and cosine is an even function, the period is just pi. Meaning after a full rotation you've traced the curve twice.
@@AuroraNora3 If you carefully trace the curve with respect to θ, you can see that you will close the loop at θ = π. Note that r will be positive for the first half of the first petal, negative for the second petal, positive for the third petal, and negative for the first half of the last petal.
How I find the limits of integration is just setting r = 0, so in this case (i replaced theta with x) cos(3x) = 0 to find when cos3x = 0, you need the unit circle, so using the unit circle we know that cosx=0 when the angle is pi/2, and 3pi/2. But we are dealing with cos3x not cosx, so what we do is pi/2*1/3, and 3pi/2*1/3 which makes the cos3x = 0 true. So therefore the integration bounds are pi/6, and 3pi/6 (or pi/2). Setup Integral: 1/2*integral of (cos(3x))^2 on (pi/6, 1/2pi) = pi/12 (which is about 0.26180) I was stupidly confused as well because a lot of people skip steps, so I found this method the easiest personally
Actually, if you keep going, the solution is EXACTLY π/12 which i think is pretty cool! Solving that integral exactly was a little bit tricky, at least for me
Why not integrate from 0 to pi/6 and double the result to get the petal area? Then one of the "ends" will be easy because you are calculating at theta = 0.
x=π/2 to 5π/6 , polar plot cos(3x) www.wolframalpha.com/input/?i=x%3D%CF%80%2F2+to+5%CF%80%2F6+,+polar+plot+cos(3x) ^ to see that one petal if anyone is unsure Next remove the word polar to see the linear graph - it's one half of a cosine cycle. Go back to the original polar plot, now make the function cos(3x)^2 and notice that you still have one petal that looks very similar to the first one. You may not even notice the difference at first glance. Now remove the word polar - and instead of one half cycle of a cosine wave, you'll see one full cycle of a raised cosine wave. Sometimes pictures are worth a thousand words, hopefully this helps if you are new to polar plots and hopefully give you some insights. Thinking about area in a polar curve is confusing. However, you can see it more clearly under the linear equivalent of the area function, copy and paste this in - x=π/2 to 5π/6 , ∫ 0.5cos(3x)^2 dx *This is not some alternative to the lesson he gave here ok.* It's just hopefully some visual aids to help you gain insights if you are new to polar plotting, especially if you are going into engineering or applied physics. You're probably going to be seeing polar plots a lot in your career and you're going to want to remember that the cosine component is the real part and the sine component is the imaginary part for complex exponentials. (And if you don't see polar plots often in your career then there's a good chance you're not thinking about the phase angle (group delay) in what you're looking at. Sometimes that's ok, sometimes that's a fatal error.) If this doesn't make sense yet, that's ok. Just keep a note of this and come back to it once a year. If one of your professors doesn't cover this in detail, you'll thank me later. 😉☺️
Hey! I solved the integral of fourth root of tanx, but I think I complicated myself (it is somehow similar to the square root). Please make a video about it :) also, can you, please, think about a recursive formula for the n-th root of tanx when n is odd and when n is even. Thank you!
Can you provide a solution for minimum positive value of x in the equation (x div k )*(x mod k) = n where k and n is given. Ex. n = 4 and k = 6 then min x is 11. P.S. = Don't use brute force . If you see this comment plz reply. Thanks
That second lobe was 3pi/6 to 5pi/6, so the final lobe would be disjoint: 0 to pi/6 and 5pi/6 to 6pi/6. This was going to be a question, but then I figured it out. At first, I thought the final lobe would be 5pi/6 to 7pi/6. That bothered me, because I knew the entire thing was only 0 to pi, and 7pi/6 > pi.
Isn't it just pi/12? If it were cos(theta), it would be a circle with diameter 1, and going through it 3 times as quickly simply divides the area by 3.
Can you calculate the surface area and volume of the solid obtained by rotating the Bernoulli lemniscate (x^2+y^2)^2=2a^2(x^2−y^2) around the x (from x=0 to x=2)?
DannyG Yes , but when you make them end at a value of n and you divide them , and then you make n goes to infinity, the answer will converge In other words , you divide them by the “same” infinity, this will make it converge
You can write is as lim Σ[ (1/k)/(1+1/(2k+1)) ] (from k=1 to n). doing algebra you end up with Lim Σ[(2k+1)/(2k(k+1))] = lim Σ[ 1/2k + 1/2(k+1) ] (by partial fractions) = 1/2* lim Σ[ 1/k ] + 1/2*lim Σ [ 1/(k+1) ], Both of these diverge, so the initial sum diverges
Well, I first simplified the second sum to be (sum from k=1 to n of 1/(2k-1)) and then realized I didn't know where to go from there. So I plugged it into wolfram alpha :D It tells you that the answer is 2.
JHawk24 lool😂😂 But I don’t think the answer is 2 , I tried to calculate it with large n value on my calculator, the answer is about 1.8639 Maybe because I didn’t try large n value to reach 2 . Idk😂
@@omarifady if you read my other comment it proves the answer is 2, also if you graph it on demos as a function it goes up very very slowly but it does look like it is approaching 2.
I did this one in my head without directly manipulating the integral, but a lot of hand waving. First is of course to note that the three petals are identical, so for convenience, I would just look at the petal on the positive X-axis, that is, integrating from -pi/6 to +pi/6. If we fan out this petal to where it covers from -pi/2 to +pi/2, the curve becomes r=cos(theta), and has three times the area. At this point, recognize that r=cos(theta) is simply a circle, with a diameter of 1. The area of this is pi/4. The single petal before the stretch was in turn one third of this, or pi/12.
I am pretty confused with the equation oficina the curve as it would allow for negative values of r for angles between pi/3-2*pi/3, pi-4*pi/3 and 5*pi/3-2*pi... I mean it just feels wrong... Do not know, but may be it would be necessary to make r=|cos(3*theta)|?
@@blackpenredpen Okay I swear I just had a "Mandela Effect" moment. I just rewatched it and the parentheses that I thought were missing, weren't. So either you're a wizard at editing or my memory failed me. Or both, but in this case it's my error. Sorry and disregard!
@@marks9618 bro Here is the question The hypotenuse BC = a of a right-angled triangle ABC is divided into n equal segments where n is odd.The segment containing the midpoint of BC subtends angle alfa at A. Also h is the altitude of the triangle through A. Prove that tan(alfa) =[4nh/{a(n²-1)}]
Reverse songs Oh, so a is the hypotenuse length, not the angle. Have you tried coordinate geometry? Center A or B at the origin and use equations for median and altitude and slope-length?
can you make a video explain what is a polar curve ?
A polar curve is just like x and y except now it's θ and r. x is the input, and y is the output(usually), but now θ is the input, and r is the output. So imagine if at every angle you get a length, and when you graph it, it's a polar curve.
@@antimatter2376 but why the graph take this shapes and there is a way to make a equation like y/x with that same graphs ?
@@5000jaap with x and y it is called a function, and a function isn't allowed to have two y-values with the same x-value. This is why an inverse of a function only has one half. For example:
f(x) = x^2 (normal parabola)
inverse of f(x) = sqrt(x)
With the inverse, you only see positive y-values, because for example sqrt(4) = +/- 2. -2 is not allowed, because otherwise it wouldn't be a function anymore. So no, we can't make these kind of things with x / y.
@@TheBrainReal but, for example, i can make infiite curves with two or more y values that are not functions
@@TheBrainReal there isnt any way do make a curve with that graph, i mean, a way to pass a polar curve into a curve, and what is the definition and the propriets and etc of a polar curve (sorry my english)
I don't even understand most of this math but I find it interesting
I always had problems visualising polar coordinate systems
I was struggling with this polar curve at a question that required the area of one pedal. I solved finding the intervals of the pedal and tried the with proportion, the method of the video. I didn't even know that, when a polar curva envolve a cosine of a odd number, you should integrate from 0 to pi, not from 0 to 2pi, now I can move on with my Calculus course.
Thanks BPRP
Black pen yellow jacket
Fuck macron
you really know how to get to the student. thank you that was helpful a lot
Cab you make a video where you explain the whole of polar curve and converting from parametric to polar
welp I definitely got this wrong on my calc 2 final. makes so much more sense now.
I have a calc 2 final tomorrow. Any advice?
@@sentientartificialintelligence That guys prob dead
It makes more sense with a double integral
Half the integral of r squared is easier to work with. And it has a very precise definitely which is super easy to visualize. You are just adding up the infinitely many isosceles triangles of side r and angle da (so whose area is r²sin(da), which in the limit reduces to just r²da). It is pretty much the equivalent of a Riemann Sum for a polar curve, pretty nice!
Im really confused on how people find the limits of these things?! Why is it just 0 to pi and not 2pi if 2pi is one rev?! Help please!!!
Normally one rotation is 2pi, but since the point is reflected when r is negative and cosine is an even function, the period is just pi. Meaning after a full rotation you've traced the curve twice.
@@AuroraNora3 If you carefully trace the curve with respect to θ, you can see that you will close the loop at θ = π. Note that r will be positive for the first half of the first petal, negative for the second petal, positive for the third petal, and negative for the first half of the last petal.
How I find the limits of integration is just setting r = 0, so in this case (i replaced theta with x)
cos(3x) = 0
to find when cos3x = 0, you need the unit circle, so using the unit circle we know that cosx=0 when the angle is pi/2, and 3pi/2. But we are dealing with cos3x not cosx, so what we do is pi/2*1/3, and 3pi/2*1/3 which makes the cos3x = 0 true. So therefore the integration bounds are pi/6, and 3pi/6 (or pi/2).
Setup Integral:
1/2*integral of (cos(3x))^2 on (pi/6, 1/2pi)
= pi/12 (which is about 0.26180)
I was stupidly confused as well because a lot of people skip steps, so I found this method the easiest personally
Actually, if you keep going, the solution is EXACTLY π/12 which i think is pretty cool! Solving that integral exactly was a little bit tricky, at least for me
Why not integrate from 0 to pi/6 and double the result to get the petal area? Then one of the "ends" will be easy because you are calculating at theta = 0.
Can you make a video on sketching these polar curves?
I used to have a js script, for parametric graphs. Such a fan of those. They are lovely 😂
I like the tri-petal form of the curve at the beginning
Cool, but can you make a video doing it with double integral? =)
When I read the question I thought he would resolve it with 2 integrals
x=π/2 to 5π/6 , polar plot cos(3x)
www.wolframalpha.com/input/?i=x%3D%CF%80%2F2+to+5%CF%80%2F6+,+polar+plot+cos(3x)
^ to see that one petal if anyone is unsure
Next remove the word polar to see the linear graph - it's one half of a cosine cycle.
Go back to the original polar plot, now make the function cos(3x)^2 and notice that you still have one petal that looks very similar to the first one. You may not even notice the difference at first glance.
Now remove the word polar - and instead of one half cycle of a cosine wave, you'll see one full cycle of a raised cosine wave.
Sometimes pictures are worth a thousand words, hopefully this helps if you are new to polar plots and hopefully give you some insights.
Thinking about area in a polar curve is confusing. However, you can see it more clearly under the linear equivalent of the area function, copy and paste this in -
x=π/2 to 5π/6 , ∫ 0.5cos(3x)^2 dx
*This is not some alternative to the lesson he gave here ok.*
It's just hopefully some visual aids to help you gain insights if you are new to polar plotting, especially if you are going into engineering or applied physics. You're probably going to be seeing polar plots a lot in your career and you're going to want to remember that the cosine component is the real part and the sine component is the imaginary part for complex exponentials.
(And if you don't see polar plots often in your career then there's a good chance you're not thinking about the phase angle (group delay) in what you're looking at. Sometimes that's ok, sometimes that's a fatal error.)
If this doesn't make sense yet, that's ok. Just keep a note of this and come back to it once a year. If one of your professors doesn't cover this in detail, you'll thank me later. 😉☺️
Good one. Polar coordinates gets glossed over too often. Thx for the closer look.
Thank you so much! Awesome video
Could you please make a video on how to go from cartesian curves to polar curves ?
I don’t understand how to build a polar graphic
I have three requests:
1. Deeper explaining
2. More examples
3. Why pi and not 2pai?
i dont want to say that the information at is useless at 4:58 , honestly the most important fact here lmao. Luv your videos
There's an example of this problem on the multivariable calculus book of James Stewart
this really helped a lot!
With a limit of 0-pi
You get half of the area, in order find the area of the loop you have to multiply your first integral with 2/3
I did this same problem in double integrals in polar
Can you prove the polar area formula?
Hey!
I solved the integral of fourth root of tanx, but I think I complicated myself (it is somehow similar to the square root). Please make a video about it :) also, can you, please, think about a recursive formula for the n-th root of tanx when n is odd and when n is even. Thank you!
I do not know how to find the exact value for the moment but the result is π/12 according to CAS software
I love bprp
absolutelyMath
Thanks!
Can you provide a solution for minimum positive value of x in the equation (x div k )*(x mod k) = n where k and n is given.
Ex. n = 4 and k = 6 then min x is 11.
P.S. = Don't use brute force . If you see this comment plz reply. Thanks
Or you can just do integral from π/6 to 3π/6 of 1/2[cos(3ø)]^2dø and then multiply the result by 2
What wil be the volume if the whole graph is rotated along y axis ?
That second lobe was 3pi/6 to 5pi/6, so the final lobe would be disjoint: 0 to pi/6 and 5pi/6 to 6pi/6. This was going to be a question, but then I figured it out. At first, I thought the final lobe would be 5pi/6 to 7pi/6. That bothered me, because I knew the entire thing was only 0 to pi, and 7pi/6 > pi.
Muy buen material, a nadie en español le veo haciendo esto, la verdad me encanta.
Isn't it just pi/12? If it were cos(theta), it would be a circle with diameter 1, and going through it 3 times as quickly simply divides the area by 3.
Making a cos curve on cartesian plane halps me out extremely but I do not see more people using it. Is there a catch to it?
Can you make a video explaining how the Feynman's method of differentiation works (proof)
Hey Steve! I left a comment on Dec. 11 of the solution to your equation using Lambert's W-function.
Can you check it out?
4X7A 175E
Sorry, maybe I missed it. Which one?
It was this link. You mentioned at that point you had no clue how to solve the equation using the W-function.
ua-cam.com/video/ef-TSTg-2sI/v-deo.html
More general pls
Can you calculate the surface area and volume of the solid obtained by rotating the Bernoulli lemniscate
(x^2+y^2)^2=2a^2(x^2−y^2) around the x (from x=0 to x=2)?
I’ll be happy if you solve this for me
Lim(n goes to inf) of : (sum from k=1 to n of 1/k) over ( sum from k=0 to n of 1/(2k+1)
DannyG
Yes , but when you make them end at a value of n and you divide them , and then you make n goes to infinity, the answer will converge
In other words , you divide them by the “same” infinity, this will make it converge
You can write is as lim Σ[ (1/k)/(1+1/(2k+1)) ] (from k=1 to n). doing algebra you end up with
Lim Σ[(2k+1)/(2k(k+1))] = lim Σ[ 1/2k + 1/2(k+1) ] (by partial fractions)
= 1/2* lim Σ[ 1/k ] + 1/2*lim Σ [ 1/(k+1) ],
Both of these diverge, so the initial sum diverges
Well, I first simplified the second sum to be (sum from k=1 to n of 1/(2k-1)) and then realized I didn't know where to go from there. So I plugged it into wolfram alpha :D It tells you that the answer is 2.
JHawk24 lool😂😂
But I don’t think the answer is 2 , I tried to calculate it with large n value on my calculator, the answer is about 1.8639
Maybe because I didn’t try large n value to reach 2 . Idk😂
@@omarifady if you read my other comment it proves the answer is 2, also if you graph it on demos as a function it goes up very very slowly but it does look like it is approaching 2.
I did this one in my head without directly manipulating the integral, but a lot of hand waving.
First is of course to note that the three petals are identical, so for convenience, I would just look at the petal on the positive X-axis, that is, integrating from -pi/6 to +pi/6. If we fan out this petal to where it covers from -pi/2 to +pi/2, the curve becomes r=cos(theta), and has three times the area.
At this point, recognize that r=cos(theta) is simply a circle, with a diameter of 1. The area of this is pi/4. The single petal before the stretch was in turn one third of this, or pi/12.
Video on polar curves?
I am pretty confused with the equation oficina the curve as it would allow for negative values of r for angles between pi/3-2*pi/3, pi-4*pi/3 and 5*pi/3-2*pi...
I mean it just feels wrong... Do not know, but may be it would be necessary to make r=|cos(3*theta)|?
black and yellow , black and yellow, black and yellow, black and yellow
why you put (1/3) outside the integral
Can you prove this equation for me?
(sum(x^3), n=1 to x) = (sum(x), n=1 to x)^2
Thanks!
could u just have the bounds from pi/2 to pi? then just solve the integral from there?
Make more videos for 4theta
I had a lot of math in my student days, but didn't that much with polar systems, so i can't solve problems like that. I do always make mistakes :(
Great
え、こうやってぐにゅ～っと曲がってるの、
そのまま積分しちゃっていいのね
なんかその、x軸方向の極小のとこで2つに分けたりしなくていいのね
Cool
Stewart feelings ❤️
Got a challenge for you: integer prod(i^e)
I am cooked.
1 Pi, that doesn’t make sense...
This would’ve been nice before I took my final lol rip
精确结果是π/12吧
Well isn,t that difficult for a 12 class
Sorry to be "that guy", but twice you forgot close your outer parentheses before dϴ. :(
Bryan Shepard where?
@@blackpenredpen Okay I swear I just had a "Mandela Effect" moment. I just rewatched it and the parentheses that I thought were missing, weren't. So either you're a wizard at editing or my memory failed me. Or both, but in this case it's my error. Sorry and disregard!
Bryan Shepard ok, no worries : )
Sir i m in a big problem ,only u can help me now ( ur my last choice)
I stuck in a question of straight line
,Reply if u wanna help me😭🙏🙏🙏🙏🙏
Reverse songs I'll help! What's the question?
Bro its a iit jee problem
Tell me your whatsapp no i will send its photo
@@marks9618 bro
Here is the question
The hypotenuse BC = a of a right-angled triangle ABC is divided into n equal segments where n is odd.The segment containing the midpoint of BC subtends angle alfa at A. Also h is the altitude of the triangle through A. Prove that tan(alfa) =[4nh/{a(n²-1)}]
Reverse songs Oh, so a is the hypotenuse length, not the angle. Have you tried coordinate geometry? Center A or B at the origin and use equations for median and altitude and slope-length?
First
Can you make a video on sketching these polar curves?