đź’¬ đź‘‹ Weâ€™re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

No Related Subtopics

Johns Hopkins University

Campbell University

Harvey Mudd College

Idaho State University

02:07

Fangjun Z.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

01:34

Scott N.

0:00

Felicia S.

Dungarsinh P.

Create your own quiz or take a quiz that has been automatically generated based on what you have been learning. Expose yourself to new questions and test your abilities with different levels of difficulty.

Create your own quiz

all right. It's now we want to find a flux integral or a flux line integral to be more specific. So we have the curve. Is the unit circle our favorite curve? So here's X and Y Yeah, the unit circle. Okay, And here's our vector field. And now remember what the flux in a girl we're actually doing this contour integral or just the line integral over the curves. See where our curve is. Just the unit circle X squared plus y squared equals one. And then I take my force or my my vector field F. And instead of taking the dot product with the tangential direction E. Now notice that this is positively oriented, so the tangential direction would just be along the curve. I'm taking the dot product with the normal direction. Now the normal direction points out of the curve like this. This is the normal direction, and of course, it's going to depend on the premature ization we get. We'll need to find a privatization of the unit circle, but that's really easy. And then the last thing is, we integrate with respect to these line elements, and of course, this will all fill in when we get the premature ization. So the parameter ization of the unit circle Well, hopefully by now. Do you remember this? Just go 70 70 and we'll just let t we want the entire unit circle. So that teeger from 0 to 2 pi. Okay, so the cool thing here is Okay, we want to know what the normal direction is, but if we look at, say are of tea, so this is our t. So this vector just points to a point on the unit circle. Notice that the normal direction points in the same direction is our of teeth, and this is actually a unit vector. So our t itself is a unit vector pointing in the normal direction, which is really going to help us out here. So I'm gonna do this. We're just gonna integrate from 0 to 2 pi f We're going to evaluate, so we wanna plug in x of T for X. So that's gonna be co sine squared tiu. And then why of tea for Why so sign t And then we take the dark part with normal direction. But the normal direction is just cosign t scientific. Now that doesn't always work out that nice. Sometimes we have to do a cross product. So, like a T cross K here, we can just observe that. Yeah. Wow. The normal direction is in the same direction as position vectors, and it's a unit vector, so we can just use that. So again, this is yeah of rt. This guy is a normal vector as a function of t, and we're just doing the dot product and we have the privatization. So everything is looking Caribbean. And so the other thing that we normally have here that I didn't include is we would really need to replace DS with the magnitude of the derivative notice. I didn't write anything or if you want, you can think I'm multiplying by one there because if I look at the derivative, the derivative is actually also a unit. So it's negative. Science t curse 19. If I find the magnitude of the specter, I get so here the kind of components that go in when I do my parent organization. And there we are good to go. Okay, So what is the integration? Become one integrating from 0 to 2 pi and if I do this Stop product. I get co sign que To plus science where t beauty. So I just want to evaluate center girl, and that's gonna give me the flux or, in other words, the amount of field that's going out of this curve. So you can think about, like, the ripple effect, how much is rippling out of this kind of founded? Yes, us. Well, there's one thing that we can notice. So if we look at Cosign Cube, we're integrating from 0 to 2 pi. And if you've been kind of paying attention, we've seen this a few times in some of the examples. This function has Justus much area above and below the X axis over one period, 0 to 2 pi. So this is actually going to give us zero. And if you don't believe me, it's a simple you substitution toe. Actually compute this, but you do get here. So this really is equal to integral 0 to 2 pi of sine squared to me. Me too. Plus So now Well, how do we integrate sine squared? Well, this is a trick identity kind of problem. So what we'll do is well right. This as Well, we have a factor of one half and we get one minus co signed to T. And that's just coming from this identity science where t is equal to one half times one minus coast on tt. Awesome. So again, this cosine two t dysfunction has period pie and we're integrating over two periods, so this integral is actually going to be zero. That's a very useful property, in case you were wondering. So all we're doing is just integrating 0 to 2 pi this part zero of one d t. But this is just one half times two pi, or, in other words, the total flux is hi.

44:11

04:14

07:28

04:12

10:03

26:52

05:28

06:38

07:48

05:33

28:40

06:44

10:25

08:33

09:29

11:59

07:49

12:44

03:43

04:53