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TimelimFind the arclength ofy = 2x 4 on 0 < = < 4 PreviewGet help: Video...

Question

TimelimFind the arclength ofy = 2x 4 on 0 < = < 4 PreviewGet help: Video

Timelim Find the arclength ofy = 2x 4 on 0 < = < 4 Preview Get help: Video



Answers

Find the exact length of the curve.

$ y = \ln (\sec x) $ , $ 0 \le x \le \frac{\pi}{4} $

First will want to solve for y so immediately we can see that we'll need to divide by 36. Getting why squared equals one over 36 times X squared minus four. You then we can square root both sides to get why equals 16 times X squared minus for to the three seconds now Because we are given that why is greater than or equal to zero we will only need the positive of the square roots. We won't have to worry about the negative one. You can go to our next page here and well, you will need to take the derivative of why so we can see that 16 times. Three over two times X squared minus four to the 1/2 now and then times the derivative of X squared minus four, which is two X Now the twos cancel. In this case, three over six is going to get this 1/2. So why prime equals 1/2 Times X? Because we have to remember our X over here. Times X squared minus four to the 1/2. So going over to our next page, we have our derivative. Now we can plug this in to our equation. Um, we were given the interval two three so we can plug in to for a on three or b radical one plus 1/2 x times X squared minus four to the 1/2 squared D X. Now let's go back for a second. Um, we will square this. So this is 1/4 x squared times X squared, minus four and then squaring. This eliminates the, um, 1/2 power here. So if we go over here, we end up getting 1/4 extra the fourth minus four x squared. Now we'll go to our next page, and we can factor out 1/4 and then we can move it outside and reorder terms the bit. And now we have X the fourth minus X squared, plus her minus four x squared plus four d x. And we can see that this can be also written as a radical X squared minus two squared D X. And now, um oh, I've already written that here. Oh, no. Okay, pardon me. So in this part will simplify um, rather integrate. So x squared will be x cubed. 1/3 minus two x now um, still with our boundaries. Um, 223 Um, And then we'll come to the next page where I've already written this out for us. So this comes down to 1/2 9 minus six, minus eight over three, minus four. And then, uh, simplifying about more. We have nine minus six equals three minus eight over three. Um, you have plus four. Distributing this negative equals 1/2 seven minus eight over three. Um, we'll continue to simplify. So this way we can write 21 over three for seven blindness, So this will give us 13 over three times 1/2 which equals 13 over six, which is our final answer.

He is clear. So when you married here. So we have our to be equal to co sign to the fourth power of data over for we see that one figure is completed from zero to four. Point included, huh? We're gonna find for D R. Over a defeat. No, we got negative. Four times 1/4 co sign cubed data over four times. Sign data over for this is equal to negative co sign cubed data over for sign of Dana over four. So next we're going to use the polar art link formula when we get L. A to be equal to from 0 to 4 pi square root of co sign the fourth power of data over four square plus negative co sign Cubed of Veda over four. Sign beat up over four square data. This is equal to me. Integrate from 0 to 4 pi square root of co sign to the eighth power of data or her four class coastline to the sixth power of data over for signs square no state of over four de seda me integrate from 0 to 4 pi square root of co sign to the sixth power of feta over four co sign square Seita over four. Close. Find square State over four. Dean fatum we integrate from 0 to 4 pi square root of co sign to the sixth power data over 40 fina since co sign squared plus signs Where is equal to one. Let me go 10 to 4 pi co sign cubed of data over for which is equal to integrate whose sign of data over four times CO signed square co sign square can also be equal to one minus signs square most data over for So this becomes equal to from 0 to 4 pi co sign of Dana over for minus coast Sign No staying up over four sign square folks data over four D data which is equal to four times sign of pain over four minus 4/3 sign cube of data over four from 0 to 4 pi and this is equal to zero, which is wrong, since the link cannot be equal to zero. So if we see here, we're gonna notice that from 0 to 2 pi the answer is positive. But from two pi to four pi, it's negative. So they cancel out. That's why I become zero. So to fix it, we're just gonna integrate from zero to two pi and double our answer, and we got 16 over three.

And this problem media to find the arc length of a curve using polar coordinates. So the first thing that we should do is let's just review the function that were given. Were given R equals co sign to the fourth of pi, pardon me of fate over four. So what we can see from the graph, um I believe that there's a figure in your textbook the region which we're finding this arc length is from when zero is less than or equal to fade up Which is less than or equal to four pi. So what we can do now is we can take the derivative with respect to theta. We'll get drd data equals negative. Cosine cubed a fatal over four times the sine of fate over four. So what we can do now is we can formulate the integral to find the arc length. So l which is our arc length is going to be equal to the integral from 0 to 4 pi of the square root of co signed to the fourth. Fade over four squared plus our derivative negative cosine cubed of fate over four Sine of theta over four squared in D. Theta. So now from here on out we're simply going to be simplifying what we have in this square root to make this integral so much easier. So we'll take the integral from 0 to 4 pi of the co sign pardoning of the square root of co signed to the six, fade over four times coastline square to fade over four plus Sine squared of fate over four in D. Theta. This inside the parentheses is a true identity. So this is going to simplify greatly. We'll get the integral from 0 to 4 pi of the square root of co sign to the sixth of fate over four D. Theta Because here we just got one because of our trick identity. So now we can simplify the square root. We'll get this is equal to the integral from 0-4 pi of cosine cubed Of data over four d. Theta. So now we can just use more trig identities will get the integral from 0 to 4 pi of the coastline of fate over four times one minus sine squared of fate over four and D. Data. And now we can just anti differentiate because we've simplified this much further down we'll get this is equal to two times four signed Fate over four minus 4/3 Sine cubed to fade over four. And we'll evaluate this from zero to pi and this is going to be equal to two times 12 minus 4/3 when we plug in. And simplify we get 16/3. So that is the arc length of the polar curve that we were given. I hope that this made sense. If you have any questions, please feel free to comment. And I hope that this applies to your current studies in calculus.

In this question, we want to find the intervals of T where the graph is concave upwards and when the graph is concave downwards. So before we do that we need to know a few things. We need to know that for compatibility, we need to look at the second order differentiation. But first we need to know the first order, which is the I D. X. And there is Dy over G. T divided by the X. Over PCI. For the second order, it will be D over the T. Of the dy dx divided by dx DT. Now, if we are looking at concave upwards mhm it will be like this. It means it's a minimum points. So our Second order deprivation is Square and zero. So when we set this condition will find the in the world T. That was satisfied. This. When his concave down what's in this we're looking at maximum point, our second order differentiation is actually negative. So I will find the interval of T where this condition happens. So let's find our dX DT and Dy DT first. So dX DT Okay, Since four is multiplied to the co sign for can be set aside of course 90. When you're differentiate with respect to you, I'll just get minus 70. Mhm. So it would just be minus fel sai inti for Dy DT since two is multiplied to the sign, it can be kept aside Now 70 when a differentiate respected T or just get co sign T. So our dy dx Yes. Mhm. Dy DT which is to call 70 Divided by the STT, which is -470. So this gives us -2. Put engine key. Let's look at our second order. Okay, Our second order says that we have two different shape. We respect to t the dy dx and there is minus half contingent E divided by the dx DT over here, which is this minus 4 70. Okay. All right now differentiating minus half contingent T the minus half. Okay look the minus can be canceled with this and the half can be brought up. So let me bring out the half. Now I'm only differentiating contingent E. That gives me minus who's second square T over 4 70. No um Half and the four here they would interact and I will get 1/8 With the minus sign of -1/8 here. Now 1/70 will become Corsican so cozy conjoined because he can square here I'll get Corsican Cube t. Now at this point it is a lot easier. We were to just grab the second order differentiation. Now if you were to grab it you will realize it looks like this. Okay. All right. Yeah, there is a simple pie here And then another isn't to two pi and this is like this and like this. Okay, this is great finger. Second order differentiation. Yeah. So this is the this one is the 2nd Order Differentiation Graph. Now we're only clearly looking at the weather is possible negative which region. So for concave we are looking at the second or the differentiation being positive. So this is the path they were looking at. So our Now this is tear not excellent. No, because it's in terms of T. Were graphing in terms of T. So our T. Is between pie and two pi for concave down. We'll complete the words. We're looking at the point where our second order differentiation graph is negative and this is the point here, right? This is the negative. So our t. is between zero and high and we're done.


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