sir still confuse towards the final answer. can you please do step by step till the final solution? my working outs are super straight forward but only the final solution which different from PI-3surd3/2.. please?
I'm just really confused why they double back on themselves. I'm just starting limacons and I dont know why they do that... If theta = pi then the equation for r should equal r = 1-2 =-1. But obviously, the curve does not pass r = -1. so like... whats going on?
Check out this video: ua-cam.com/video/MT1j6vdX2aI/v-deo.html It might help clear up some of your confusion. There are a couple of similar videos, too, that might help. This is my whole non-calculus based polar playlist: ua-cam.com/play/PLF35EF817EB13CA93.html The key to polar graphs is really understanding the rectangular graphs in my opinion.
Thanks for the reply. I'm definitely going to have to look into it. I'm not quite sure what you mean by 'facing' a particular direction from a number of your videos.
As far as "facing" goes, imagine (or actually do it) drawing the coordinate plane on the floor of a room. Start off standing at the origin and position yourself so you're looking down at the positive x-axis. While you're doing that you're facing the x-axis. If you rotate yourself so you're now looking at pi/4 (or 45 degrees) you're now facing into QI. If you're facing into QI and r is negative you'll walk backwards into QIII. That kind of thing.
Thanks again. Yeah, I didn't realise how different polar coordinates were to regular. It took a little bit of thought to work through it. Also, I realised I was confusing r and x. Realising x=rcos(theta) then x = [1 + 2cos(pi)] cos(pi) = 1. so, yeah x is definitely 1.
this was good but I was hoping for a more in-depth explanation of solving for the inner loop by hand :( that's what I'm currently stuck with on my calc 2 study guide i havent seen the sunlight in days
Hi! Maybe this video on graphing will be useful: ua-cam.com/video/MT1j6vdX2aI/v-deo.html If you graph the rectangular version of the polar equation (put r on the vertical and theta on the horizontal), the inner loop is when the graph "dips below the axis." That can help a lot with finding the bounds. Hope this helps! Good luck!
If you square the expression you will obtain 1 + 4cosx + 4cos^2x. To successfully integrate you must use the power reducing identity for cos^2x=.5(1+cos2x). There for you have 1 + 4cosx + 2(1 + cos2x). Distributing the two you obtain 3 + 4cosx + 2cos2x. Integrating will result in 3x + 4sinx + sin2x. Now evaluate at the proper interval!!
r=0 when the angle is 2pi/3 and thats start angle of inner loop.r=0 also for 4pi/3,thats the angle for end angle of the loop.,angle start to end counter-clock wise direction -they form the limits , in polar area integrals.If start angle is more, when sweeping the area in the counter clock wise, then subract 2pi from the angle to make it equivalent .It is a trick to make the larger angle smaller than end angle and then apply the integral limits..Hope you can understand.
Drawing the lines where the zeros where helped visualize the loop going back on itself. Simple but informative. Thank you sir.
Thank you so much! I understand it much better now!
sir still confuse towards the final answer. can you please do step by step till the final solution?
my working outs are super straight forward but only the final solution which different from PI-3surd3/2.. please?
I'm just really confused why they double back on themselves. I'm just starting limacons and I dont know why they do that...
If theta = pi then the equation for r should equal r = 1-2 =-1.
But obviously, the curve does not pass r = -1. so like... whats going on?
Check out this video: ua-cam.com/video/MT1j6vdX2aI/v-deo.html
It might help clear up some of your confusion. There are a couple of similar videos, too, that might help.
This is my whole non-calculus based polar playlist: ua-cam.com/play/PLF35EF817EB13CA93.html
The key to polar graphs is really understanding the rectangular graphs in my opinion.
Thanks for the reply. I'm definitely going to have to look into it. I'm not quite sure what you mean by 'facing' a particular direction from a number of your videos.
As far as "facing" goes, imagine (or actually do it) drawing the coordinate plane on the floor of a room. Start off standing at the origin and position yourself so you're looking down at the positive x-axis. While you're doing that you're facing the x-axis. If you rotate yourself so you're now looking at pi/4 (or 45 degrees) you're now facing into QI. If you're facing into QI and r is negative you'll walk backwards into QIII. That kind of thing.
Thanks again. Yeah, I didn't realise how different polar coordinates were to regular. It took a little bit of thought to work through it.
Also, I realised I was confusing r and x.
Realising x=rcos(theta) then x = [1 + 2cos(pi)] cos(pi) = 1. so, yeah x is definitely 1.
this was good but I was hoping for a more in-depth explanation of solving for the inner loop by hand :( that's what I'm currently stuck with on my calc 2 study guide i havent seen the sunlight in days
Hi! Maybe this video on graphing will be useful: ua-cam.com/video/MT1j6vdX2aI/v-deo.html
If you graph the rectangular version of the polar equation (put r on the vertical and theta on the horizontal), the inner loop is when the graph "dips below the axis." That can help a lot with finding the bounds. Hope this helps! Good luck!
If you square the expression you will obtain 1 + 4cosx + 4cos^2x. To successfully integrate you must use the power reducing identity for cos^2x=.5(1+cos2x). There for you have 1 + 4cosx + 2(1 + cos2x). Distributing the two you obtain 3 + 4cosx + 2cos2x. Integrating will result in 3x + 4sinx + sin2x. Now evaluate at the proper interval!!
Well done.
Thanks!
Really good video, thanks!
How do you know to set your upper limit at 2pi/3 ?
That's where the inner loop starts. See this video for how to figure that out: ua-cam.com/video/MT1j6vdX2aI/v-deo.html
r=0 when the angle is 2pi/3 and thats start angle of inner loop.r=0 also for 4pi/3,thats the angle for end angle of the loop.,angle start to end counter-clock wise direction -they form the limits , in polar area integrals.If start angle is more, when sweeping the area in the counter clock wise, then subract 2pi from the angle to make it equivalent .It is a trick to make the larger angle smaller than end angle and then apply the integral limits..Hope you can understand.
ty
thank you!
Thanks
thank you a lot