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Find equations of the normal plane and osculating plane of the curve at the given point.

$$x=2 \sin 3 t, y=t, z=2 \cos 3 t ; \quad(0, \pi,-2)$$

$$

x+6 y=6 \pi

$$

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Missouri State University

Campbell University

Harvey Mudd College

University of Nottingham

{'transcript': "So the first thing we're gonna do is find a normal plane, which is useful in multi variable calculus. And when we have parametric equations, Um, because when we have these three planes, we want to find normal planes in the same way we want to find normal vectors. So we don't have our of tea being equal to the sign of two t the negative co sign of two t and working. Then we would end up having that Our prime of tea is equal to to co sign to t to sign to t think for we noticed that the 0.12 pi will correspond Teoh t equals pi over two. So since we have t equals pirate to the normal vector, um, to the normal plane at 012 pi will be the same thing as our crime of pi over two not going to give us negative 20 four. So with that, we have that the equation. Um, we have our our vector right here, considering the fact that we have it at pi over two. We want to remember that the equation of a plane passing through a point ABC with the normal vector. LMN is going to be l times X minus a plus M times why minus b flux and times Z minus C equals zero. With this in mind, we have negative too zero for the M. So this is just gonna go away and then our end value is for so then since, uh, a is we know to be zero Uh, why doesn't really matter or be? Doesn't matter because it's we'll get zero for that term. That C is two pi. So we'll have to pie right here, then simplifying everything out. We end up getting our equation X minus two Z plus four pi equals zero. Um, now, finding the oscillating plane, um, we know that the normal vector to the escalating plane at that point is be of pi over two. Okay, so our t of teat is equal to our crime of tea over the magnitude of our prime of tea. So what, we'll end up getting, um, as a result of that is to this whole thing right here, we can just copy and paste it that's going to be divided by the magnitude. So we know that the magnitude is the square root of for cosign t four coastline square T plus four signs squared He plus 16 and these are to be clear to t and duty. We know that this right here when we combine these is just going to give us four. Um, so we'll end up getting as a result is four plus 16, which is 20. We can factor out before, so we get squared of five and out here would be a two. But then we have a bunch of choosing comin up here so we can just cancel these out, um, to simplify. So this is our answer right here. And then when we evaluate this at pi over two, we know that the cosine of two times power to will be a negative one. Yes, you know that this would just be a zero, and this will stay it too. So then what will end up getting as a result for that? Is this right here? Then when we differentiate t prime of tea, what we'll have is, uh, if we take this and then differentiate it, we'll get a negative too. Sign to t. And this is a vector. Keep in mind to co sign duty zero. And when we evaluate that, that's gonna be t prime of tea. When we evaluate that at high over to we end up getting zero a negative too, and a zero. And then the magnitude of that we know is going to end up giving us to over route five. So combining all these things the normal vector, the normal vector of pi over two is equal Thio the T prime of pirate or two over the magnitude of t prime over to. So we'll have this right here. Um g prime of pirate, too. This needs to be divided by on expiry five. So what we'll have is that will get, um, one over route five times the vector zero negative 20 And then that thing, that portion right there will be divided by our value for the magnitude of t prime evaluated pirate, too, which is just too over route five. Now assist in mind. Um, ultimately, we want to find be of pi over two, which is equal to t of pi over two cross and of pi over two. The tangent vector across the normal vector and what we'll end up getting as a result of that. Um, we see that. Be of pi over two. Well, give us 1/5. Oh, 20 Uh huh. So, based on that, we know that are escalating. Plane is going to be, um, going back to what we had before all the way up here. We bring this down here, we have that, uh, we'll get to route five X. Um, Plus, in this case, um, one route five over Z minus two pi one over route five. I am Z minus two pi. And when we do that and multiply everything, um, through, we see that we can multiply both sides by the square root of five, which will simplify this further. Um, ultimately giving us this right here. Onda. We can move the two pi over to the other side to guess that to act plus Z is equal to two pi. And that is, um, the equation of our normal plane. We already software. And the equation of the escalating plane is two x plus the equals two pi"}

California Baptist University