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Derive Equations 1 for the case $ \pi/2 < \theta < \pi $.

The case $\frac{x}{2}<\theta<\pi$ is illustrated. $C$ has coordinates $(r \theta, r)$ as in Example 7and $Q$ has coordinates $(r \theta, r+r \cos (\pi-\theta))=(r \theta, r(1-\cos \theta))$ $[\operatorname{since} \cos (\pi-\alpha)=\cos \pi \cos \alpha+\sin \pi \sin \alpha=-\cos \alpha], \operatorname{so~} P$ hascoordinates $(r \theta-r \sin (\pi-\theta), r(1-\cos \theta))=(r(\theta-\sin \theta), r(1-\cos \theta))$$[\operatorname{since} \sin (\pi-\alpha)=\sin \pi \cos \alpha-\cos \pi \sin \alpha=\sin \alpha] .$ Again we have theparametric equations $x=r(\theta-\sin \theta), y=r(1-\cos \theta)$

01:58

Wen Z.

Calculus 2 / BC

Chapter 10

Parametric Equations and Polar Coordinates

Section 1

Curves Defined by Parametric Equations

Parametric Equations

Polar Coordinates

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So what we're trying to do here is, um derive equations one when we have a fatal value between pi over two and pie and we know that the circle rotated by an angle. Fada has a center of our theta. Uh huh. And the y coordinate of the circle will be our because the radius is our Andi it's in contact with the X axis at faded equals zero. We know that the x coordinate well, ultimately just be zero. So the circle is always in touch with the X axis, so we know that the distance moved by some point p will always be the X coordinate of the center. So one way that we can show that is that the X coordinate of the center will be, uh, the arc length rotated, which is exactly this right here our time stadium. So since uh, we'll call a are for arc length, that's gonna be the same thing as Arteta are, um and we will let Alfa be Seda. So Alfa Three equal to Seda minus pi over two. Um so when we do that, um, we can have a circle. And when we have this circle, what will end up showing is that if we draw circles, say, um X squared plus y squared. What's one? What I see is that we have say, the center right here and then some high pot nous right There are the radius and we see that it creates this right triangle right here. So if we have, say, um some value p and then I value em on the X axis and then a value down here somewhere N then we see that why is equal to P M plus a men where m n we know is going to equal the radius. So with that in mind, we now have that y equals P M plus r. And then since we have a right triangle, this is the center. The right triangle we have is P M. C. So it looks like this. We know that the sine of the angle measure Alfa which we already discussed the sign of Alfa is going to be equal Thio PM, which is the opposite side right here over the high pot news which we know is art. Then, um, we would have that PM is equal to our sign A so p m equals R signed outside that it's And in this case we know that Alfa is equal to high over to my next data, which ultimately just means that this is going to be a negative our coastline data. So when we substitute this in equation one, we end up getting that. Why equals ar minus are closing data which will be weaken do distributive property to get our times one minus Carson data. Similarly, we can, uh, do this again and with a different right triangle we get that the co sign of Alfa is equal to M C over arm then similar to before we get the M. C is equal to our coastline Alfa which is ultimately equal Thio are co sign of high over to minus data That's just equal. Thio are signed data with all that mind When we substitute this in to the equation, the second equation will get that X is equal to our data minus are signed data So acts is equal to our times Fada minus Yeah, and that's ultimately what we wanted to show and what we wanted to derive. So our work is done

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