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Show that the circle of radius R in C centered at w € € may be parameterized by ~(t) Reit + w. 0 < t < 2t.Use partto conclude thatdz = 2ti_Similar...

Question

Show that the circle of radius R in C centered at w € € may be parameterized by ~(t) Reit + w. 0 < t < 2t.Use partto conclude thatdz = 2ti_Similarly; show thatdz = 0.

Show that the circle of radius R in C centered at w € € may be parameterized by ~(t) Reit + w. 0 < t < 2t. Use part to conclude that dz = 2ti_ Similarly; show that dz = 0.



Answers

Show that the graph of
$$ \mathbf{r}=\sin t \mathbf{i}+2 \cos t \mathbf{j}+\sqrt{3} \sin t \mathbf{k} $$ is a circle, and find its center and radius. [Hint: Show that the curve lies on both a sphere and a plane. $]$

So for this problem we have a family of circles described by the inclusion export. That's why squared is equal to r squared in what what he wanted to do is to show that all the normal lines to this struggle for this family of circles will actually pass through the center of the circle. And just to emphasize this circle has the center at the origin. Okay, so how are we going to prove that? The first thing I wanted to do is to find out the slope of tangent and therefore the normal is would and I can do that by differentiating the equation of the circle. So we'll end up having to fcX just do it. Why is able to see you know? And simply dy already is negative X over Y. And they know that this expression here is the slope of the tangent line and that being said since the normal is perpendicular to the tangent line. The queen of the slope of the normal line is just a negative reciprocal of this one. So that should be Y over X. Okay. Now, if I will create an equation equation of foreman equation of normal line that is passing perpendicular to a tangent At X. one comma wise of one will end up having this equation. Why? So why- Ways of 1? Is you go to the slope of the normal line Times X- Except one. Why minus Y. Sub one is equal to Y over X. Uh if this is the tangent line then the slopes be wise, one over except one. Let me just put emphasize that here. So at. Okay, therefore At an arbitrary point except one. Wise 1. The slope of the normal line is simply about the voice of one over except one. Okay. And would have that slope here continuing we end up having this equation why my slice of one is equal to Y sub one over X. Of one times X minus. Wait up one D. How can I prove that this normal line will pass through the center. Then we can substitute The Centers Coordinate which is 00. And if this coordinate will satisfy the given equation of the line right here, then it is indeed passing through the center. So if once altitude would have zero the white coat, the zero for the value of Y and zero for the value of X. Will have finally, twice a one over X, one times zero- twice a woman. This expression becomes zero. We'll end up having negative voice of one here. And negative voice of one here. Which Manchester equation That being said, we were able to prove that this um family of circle, we will always have a normal pass into its center

Here, we want to show that in a circular disk of radius r. A sector with a central angle of fada radiance has an arc one half r squared Theta. We want to take data between zero and two pi and we know that the ark of the area of the circle is going to be pi r squared. So if we have a circular disk, we know that it's going to be one half r squared theta because we already know that are set to gives us the actual arc length. And then if we look back to the fact that the entire circle is too high are in circumference. If the circles area is high R squared and that means that it's going to be two pi provided by two or one half two pi. Where to pie is our angle measurements, So we have one half fay to r squared.

So here we have a circle and a point P. Exterior to the circle. We have P. T. Is the line segment um tangent to see at T. And we want the line to go through P. And through the center. We want to show that PM times PN equals P. T. Squared. So in this case we see that PM will be from the point exterior to the circle to the point where as the lines intersect so at PM. And then if we multiply that times P. N. Which is also the point at which um it's also a point where to see intersects because see is going to the circle. So M. And N are somewhere along the path. But Petey is tangent to see at the point T. And since he is tangent. Um and see is the center, we see that Pete M. Times P. N. Is going to be equal to the P. T. Value. And that is going to be a p. T squared. That's your final answer.

Okay, so first thing we're gonna do is we're going to convert some of these items too. Rectangular coordinates. So I can rewrite co sign as, um X divided by r and I can rewrite sign as why divided by our, um, still equals r. And then I wanna I'm just gonna multiply this entire equation by our so multiply each term by our I'll get r squared equals a X plus b. Why? Because the ours here will cancel now R squared is the same thing as X squared. Plus y squared equals a X plus b y. Okay, let's get Let's keep moving. Let's get everything toe one side's track day X and subtract b y from both sides. So I'll have X squared minus a x plus. Why squared minus B Y equals zero. Now, this is starting to look like I can complete square to complete the square. What I'm gonna do is I'm going to take this term divided by two and then square and added to both sides so this will end up being a over two squared eight. The fourth distribute that to to the numerator and denominator over four. And I'll add that to both sides. I'll do the same thing with the B term. I'll take be divided by two. Square it. It'll actually be negative. B and this will be positive. Be squared over four and I'll add that to both sides. When I do this, I'll get X squared minus a X over two plus eight to the fourth over. Four plus y squared minus B y over to plus eight to the fourth over. Four equals zero Now, actually, it'll still be just a and B y. Now, when I factor these thistle factor as X minus a over two squared similar to what we had when we divided it by two. Plus why minus b over to Sorry, This should be also be be to the fourth and this will be squared equals And I had the added to these this side Sorry, I missed that step eight to the fourth over four plus B to the fourth of before. The reason why this is on the right side is I just added this. So whatever I do, the one side I have to do to the other saying with the beat of the fourth and when I combine these, I'll just have, um, aid to the fourth plus speed of the fourth, all divided by four. And now this is kind of in the form of a circle. So my center will be at these values My center. My ex value will be a over to my Why value will be a be over to in my radius. This is gonna be the same thing as R squared. So I'm gonna take the square root of the numerator and denominator. My radius will be square root eight to the fourth plus B to the fourth e Don't know where the fourth came from. Should be squared. We should all be squared. This will be eat. So it was my mistake A square plus B squared over the square to four is too


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