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Find the curvature of the functiony = ln(2x)at x1.5.PreviewFind the radius of curvature at x 1.5. If the radius is infinite, enter '00' for &_PreviewF...

Question

Find the curvature of the functiony = ln(2x)at x1.5.PreviewFind the radius of curvature at x 1.5. If the radius is infinite, enter '00' for &_PreviewFind the center of curvature at € 1.5 . If the center of curvature does not exist; enter DNE"PreviewNote: For very large or very small numbers you will probably wish to use scientific notation to enter your answers_

Find the curvature of the function y = ln(2x) at x 1.5. Preview Find the radius of curvature at x 1.5. If the radius is infinite, enter '00' for &_ Preview Find the center of curvature at € 1.5 . If the center of curvature does not exist; enter DNE" Preview Note: For very large or very small numbers you will probably wish to use scientific notation to enter your answers_



Answers

Find the curvature and radius of curvature of the plane curve at the given value of $x .$ $$y=\frac{3}{4} \sqrt{16-x^{2}}, \quad x=0$$

Well, Cameron Y equals the hyperbolic tangent of X. So I I wrote a parameter ization of that function um with X equals T. And Y was then hyperbolic tangent of T. Again, there's various ways of doing this. We're told to look at the point where X equals natural look of two so that Y equals um 3 50. We can find the normal vector. Okay, so again I explained that all in previous problems, again, you can find the tangent and then across that with the out of plane vector to find the perpendicular vector, that's the normal. Um And that that winds up being this kind of ugly mess here. And they can you might basically get the sign of the tangent here. Right, Hyperbolic tangent. Again it's kind of ugly but you can figure it out. Um If you what you get might look different but should be mathematically equivalent. And so obviously when when you plug in numbers we should all get the same thing. So when X is the natural log of two, we get this value for the unit normal vector. So it's it's again kind. I didn't I should have probably calculated numerically what it is but it's you can see you can see that it is in the positive, it's in this direction here, slightly shifted downward, a little more than shifted a little more down than to the to the right Because -16 we have 16 -25. Now we can find the curvature and again if we do that again various ways of doing that then we can find the radius of coverage er and plug in this point and we can't. This really weird. 881 to the 3/2 over 12,000. So oddly enough that's what that's what we get for that. Um I didn't I should have calculated an american value for that. Um It's actually, let me see what is that was figure out what that is. Um And so 881 to the 2/3 power provided by 12,000 and in America equivalent of that. Hold on a 2nd. Is that right? Oh my God. Oh I put through the two times that she has. Yeah, so that looks a little bit. So that equals actually that equals two 18 I put 2/3 in the computer. That's why it was giving me a really weird answer which is so once we know the direction of the normal and then the radius curvature we can find the center of the circle of curvature by taking are not and adding to it. So we have are not plus row not and not gives us this point here. And that winds up being this again, I didn't calculate numeric value for it but it looks like it's around. And so this would be 12 two and a little over a little less than -1. And so once we find that point, we can then draw our circle. You can clearly see that again visually. You can you know this should all come out and make sense visually when you draw these things that you have this, the coverage of the circle and the coverage of this curve are the same at this point here. And so that's if you get a circle that looks like this, then something's not right? Or likewise its National Archives, you get an get the thing over here, so you get a circle like this and that's just the opposite because this is the this is also a normal vector, right? But this is the normal vector that we want, the one that is pointing towards the centre of curvature. There's a characters like this one, this 1. Um So again this is all you know when you when you make the plot you should you know everything should get confirmed that all these things are correct. Once you make this plot and if something looks weird then obviously you made a math mistake somewhere, which is very easy because these calculations get very ugly.

We were given why was squared facts. I analyzed this problem by parameter rising this this curve dysfunction here. Um So I said X equals T. And then are the position vector? Is T. I plus where to T. J. And then we're looking at the .11. So that's the point of interest in this case. Now again we can do this basically just using this if we wanted to. But I like the parameter, use the privatization. So we can find the unit tangent factory by taking the derivative of this and multiplying dividing by its magnitude. And then we can find the the normal vector by taking the cross product of the outer plane vector in a vector with the italian veteran tangent vectors this way. Um let's see here and then I crossed, it actually has the tension across with. Um So it's this one, right? So tent, the normal factor should always point towards the center of where the curvature is. So if you think about the circle here, it should point towards the center of the circle that ever here. So it always should point like towards this way. Now if the curvature is like this, the curvature is negatively should be pointing down, dad. Um So this is what we get for the for the unit normal vector, which is, you know, again ugly, but that's what we get. And we can obviously plug in teeth was one and that then simplifies down to get the normal vector at that point is one over square to five in the X direction, minus two over square to five in the Y direction. So this way, okay, so you can see that, don't you draw it? You say you can look at it and say, okay, that looks normal. So you can kind of draw the tangent in and see what the normal is, which is perpendicular to it. At that point we can get the curvature in a few different ways, but the curvature winds up being too Divided by five kinds of square to five, interestingly enough, Isn't that the same? Oh no, this was four times. Mhm. So so then the radius of curvature is five times scored five for 2. So we get the center of this circle of curvature by taking are not this vector and adding to it the radius of curvature times the unit normal vector at this point, which gets us down to here and that winds up being a nice value of 3.5 in the eye And then -4 in the J. So right out here. And so then drawing a circle, you can see that clearly the curvature match at this point here. So this circle has a radius of curvature the same as this curve does right at this point. Yeah.

It was given the function Y equals the cube root of X. So that looks like this. Now I parameter. Rised it with tea. So I said X equals T. Called X equal to T. Article X. And then Y equals just keep a lot of T. So this is the curve that parameter rises. This is the director of the parameter rises. This curve here. And so we're looking for again, I should call these are not and this should be maybe called R C. If I put a knot there too. So um again we can solve this very right of ways. I found that unit normal record just basically kind of in the standard way. That is actually in the next section of the book. Mm for the next Well, I guess it's after you probably should have read that section already. So you can find the unit to unit normal vector. Um If it's a plan or curve, it's easier because you can take the friend tangent and then find the normal because it's perpendicular to it, which is just taking the uh taking the tangent across with the unit normal vector. So uh is ugly and I didn't go through all the algebra to get this and what you get might be looked a little different, but it should be mathematically equivalent to this. I'm not sure if I simplified it down all the way, but again simplified it as much as I could. So you guys get this nasty factor out here with lots of, You know, 4/3 powers. Keep 1/3 powers, 2/3 powers. But anyway, we're looking at the .11. So that's a nice easy point. So peak was one there, so we just plug in one everywhere and then everything can simplifies down And we get one over square to 10. Yeah, becomes that's what this has us. And then this guy comes just boy, this is always one and then this just becomes -3. Has that become -3? I'm going to have a sign around there that may see here. Where is this problem? That one not that That one this 1. Um Actually I missed a minus sign. Yeah, this should be a minus sign up here, copy it down. All right. So then we get, you know, this is our, you know, normal vector at that point. So basically, you know, this direction here and so we can find the center here by taking are not and adding to it the well, I guess I need the the curvature. We can find again in a few different ways. Given this that it's a function. We can use the term in the book. What is that camera? Yeah. Right. So basically the absolute value of Y double prime all over um one plus Why Prime Square to the 3/2. And so we can plug that and plug this into here and then set it equal to set X equal to one and Get the the curvature and that's going to be 3/5 times square to 10. So the radius of curvature is one over that. And so to get this point here we go um go to this point and then add the radius of curvature times the normal vector the unit normal records in the direction that we want to go and the radius is the magnitude. So we're down here and so that we can draw our circle for the center here and a radius of this value here and again, we can see that clearly at this point here, they just grazed one another because the curvature of this circle here is the same as the curvature of the curve at that point. And so the center here winds up being at -3 At 8/3, or what is that? To 202/3. So here And then at -4. So this all boils down to, you know, friendly nice values here because we're plugging into equals one. And so lots of this stuff simplifies T equals zero, would simplify a lot more because then we just have, well we have an inflection point there, so it's a little, it actually gets a little tricky there. Um so you know, the tangent would be this way, so the normal would be this way. But again, it's hard if you're it's really you need to take a limit kind of because it's an inflection point there, I'm so um again we would find that we just have the tangent and the normal vector would be in the I. Direction. But again, we need to take a limit to see that

In this question we can Amanda family on the curvature. Okay, it was equal to the absolute white proper Bram, dividing by the one plus Y. Prime Square everything that were on the 302. And then the Raiders are equal to one over the curvature. In this question we even the Y. You go to be about three X. And the X equal to zero. Now the first step we need to fight the 1st and 2nd derivative of the function. So the white prime. We go to treat each of the three X. And then the white prem and zero. It will be equal to the three, the second degree with a with an echo to night. Each of the three X. They found the wind up from under zero echo 29. And if we apply the formula the curvature K. A week ago to Absolutely. I'm the wonderful prime will be nine. The Environment one plus Y. Prime will be three square everything about three out of two. So we have nine hour 10. Well 302. And they found the Rangers Echo two. We flip it over will be turned about 300-1.9. And this will be the answer


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