Problem 36 Easy Difficulty

The curves with equations $x^{n}+y^{n}=1, n=4,6,8, \ldots$
are called fat circles. Graph the curves with $n=2,4,6,8,$
and 10 to see why. Set up an integral for the length $L_{2 k}$ of
the fat circle with $n=2 k .$ Without attempting to evaluate
this integral, state the value of $\lim _{k \rightarrow \infty} L_{2 k}$ .

Answer

$\lim _{k \rightarrow \infty} L_{2 k}=8$

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Calculus 2 / BC

Essential Calculus Early Transcendentals

Chapter 7

APPLICATIONS OF INTEGRATION

Section 4

Arc Length

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Video Transcript

All right. So for this problem we were analyzing curves form xCN plus yTN, It's equal to one um with even powers That are greater than two. So we have, Giving him the problem, we have 468 prototype and these are called fat circles. So we know the equation of a circle is just an equals two. So this is just read this red curve. So this graph contains um 55 plots from and equals two 4, 6, 8 and 10. And let me go ahead and mark each one. So we have 246 a 10. So as you can see um they're basically just very very rounded circles are um they're more they're larger than a circle, sort of enclosing the same sort of intervals from negative 1 to 1 in both X and Y. And it's pretty clear that as and increases as N goes up. Um This uh this curve basically approaches a square. Um so that would be very helpful for the end of this problem. So we want to do first is analyze specifically uh one N. Is equal to two K. Um giving us X. To the two K. That's why don't you? Two K. Is equal to one. And this will sort of be our case were given where we have even powers of even numbers as the powers affect someone. Okay so we want to go ahead and find the arc length for this uh this curve, this general curve. So we have first we need to know the arc length formula. Um so we have a lot of 2K. Is he equal to the integral for me to be square of one plus dy over dx squared? Yes, I will notice we have a very interesting figure. Um Just understanding how when we have even power, it's not going to be simple to just integrate over the whole sort or the whole figure at once. So you can split it up into a small portion since the symmetrical you can just work in the first quadrant and then the length we get of this sector here should just be equal to four times the entire. Should be just a quarter of the entire length that we can just go ahead and multiply by four to find the true the length of the entire figure. So this is just uh X between 01 and why? Also between zero and one. So it's equally simple to go ahead and solve it in both X and Y. So we're gonna go ahead and just solve it in with using our ex dependent as our independent variable. So if we go ahead and manipulate this, we first have to find a new tax so we can go ahead and salt for why? And we end up with this expression one minus 2 to 1 minus X. To the power All to the 1/2 K. Power. Right? So um we know for our clients we have to find a derivative of this. Um and I'm actually gonna do it over at the bottom just because it's a pretty long one. Okay, so we have to use power ruler chain rule. So we have on over to K one might as exit UKK 1/2 K. And that's attractive one in front of power. And then do the chain rule to give us uh minus two K times X to the two k minus one. Right then go ahead and make that a little nicer. Yeah. So now we can go ahead and knock out these two K factors and say both cancel, leaving us simply with dy over dX people to negative one s exit UK. So the one over to k minus one power All Times Exits UK -1. All right now we can go ahead and start plugging into our formula, we know that Alex UK Since we only do, since we're doing a quarter of the entire figured we can just do four times, let's go from 0 to 1 of one plus. Um So we're squaring dy dx square, so we have the I D X square here so we can go ahead and basically any negative uh uh negative one that appears in our question can basically just be ignored. So since these are going to be taking out once we go ahead and I swear it, so we can basically do one minus x. It okay to the one over to k minus one. Are sorry the 11 is up here Time Exit to K -1. All squared P. X. And we can simplify this. Um This to power basically just Turns this into a 1 -1 over K -2 and a 4K -2. But there's not really a need to do that since we're not actually going to be solving this expression, so we can go ahead and circle this in, it should be the length of um the whole figure, the whole fat circle and we can go ahead. And so for that we want to sell for the limits as K goes to infinity of altitude. Okay. No we don't want to actually solve this integral so we're gonna have to go ahead and use intuition using geometry. So like I said earlier, as n goes up to infinity, this curve gets closer and closer to a square and it's pretty simple to see that since we have goes from -1, the one we have side lengths up to so the arc length of square with silence of to adjust its perimeter, which is four times each side length. So our perimeter is just four times two or eight. So that just means that the the limits as he goes to infinity of hell to do K or L. Of two K is just equal to eight. Yeah.

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