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The Gateway Arch in St. Louis was designed by Eero Saarinen and was constructed using the equation

$ y = 211.49 - 20.96 \cosh 0.03291765x $

for the central curve of the arch, where $ x $ and $ y $ are measured in meters and $ \mid x \mid \le 91.20. $

(a) Graph the central curve.

(b) What is the height of the arch at its center?

(c) At what points is the height $ 100 m? $

(d) What is the slope of the arch at the points in part (c)?

A)$|x|<91.2$$\\$B)$190.53 \mathrm{~m}$$\\$C)$x \approx \pm 71.56 \mathrm{~m}$$\\$D)Slope $=\pm 3.6048296591$

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Matt P.

September 9, 2020

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 2 3 x2, y = 5 3 ? x2; about the x-axis

Campbell University

Harvey Mudd College

University of Nottingham

Idaho State University

All right, guys, We're going to start with Chapter three Fiction eleven, Number forty eight and going to start with Part A, which is asking us to find to graph the curve which is represented by this equation right here. And this equation is representing the St Louis Arch. So if we think about an arch, we're thinking of an upside down parabola. That's the shape paragraph should, um, should follow. So the first step to find this crap, we know the general straight. We're gonna find the intercepts for it. And so if you guys are comfortable and you guys can go ahead and work on finding intercepts to find the intercept sugar to set X and Y equal to zero Okay, so we're going to start with wedding. Why be equal to zero? So we have Why equal zero right like that. You're going to take our original equation, which is this, and we're going to change this water here two zero with worry for zero. We're going to now solve for X. So we're going to bring the two eleven to the left hand side and divide over that negative at the same time to get everything back to positive here, Frank. Zero three to nine. Great. Now that we have this, we can divide over the twenty point nine six, which gives us ten point o nine. We hold to the co sign. Hyperbolic. Now we have to get rid of that co sign hyperbolic function on the right hand side. To do this, we're going to use the co sign hyperbolic inverse function that can be found on your calculator. If you have a t I eighty four, you can go to the second zero. Which brings you to your catalogue school down to the seas. Find the coastline hyperbolic function, and you should end up with about that number equal to point zero three to nine. Seventeen sixty five x on the final step here is to divide over the point zero three number to get X all by itself. Which means we are done solving for X intercepts and we have X equal to now. A parabolas should have two intercepts plus or minus ninety one point two. Then we're going to keep these in mind for the graph that we're going to do. Okay, That's the first part of finding the intercepts the next part said of Y equals zero. We're gonna let X equal to zero. I am. We're going to solve the same equation, but this time for why? So that we can find our Why intercepts. So here's our original equation and we're going to take this X right here, and we're going to do times zero, not one thing. I would like you guys to keep in mind. Is this identity I'm going to types of? This is an identity. You started writing the coastline, hyperbolic zero is equal to one, and this does follow coastline. Hyperbolic, unsigned, hyper evolved. You follow most of the same function most of the same properties as the regular Trig co sign and sine function. So again we're solving this. That's come back around here minus twenty point nine six. Code sign. Hyperbolic zero a number times zero zero. And we know from our identity branch because I have a broth of zero is just one twenty, forty nine, six times one still twenty nine point six. This leaves us with our final answer for the Y intercept. So this leaves us with why you will, too. One ninety point five three So why is equal to accept one ninety point five three? Now that we have these intercepts, we can go ahead and get our graph finished. Crate. So here are intercepts. This is part A continued. So we're still working on our graph here. Co ordinate plane tank. We have the tip of the arch, which is going to be our y intercept. And we have our two ex and her steps down here. We know our top one. Where's one ninety point five three friendlies are y in ourselves That's going to go. It's this point right here. We also have plus and minus are ex intercepts. So we have ninety one point two and negative ninety one point two. We're screwed, but the positive access label here. But just keep in mind that the negative one is right here. Okay? Now that we have that, we can go ahead and connect it. Make it look a cz arch Likas possible. Okay, some like that. So we have an art shape. We have our ex intercepts and we have our y intercepts. So that finishes up part a moving on to part B. We're asked to find the fight at the center. Right? So the height of the centre of the arts. So if we just recall quickly what are graft look like? This is just gonna be a rough sketch of our arch. Something like that here. So the centre this is supposed to be symmetric across the y axis. So the center is at the tip or the maximum point of the arch. We figured out that the maximum was at why equals one ninety point five three and that it's going to be the answer for this part. Now the way to prove this if you don't want to just use the graph is by recalling Go ahead and do that. Over here, you're calling that the maximum of a function. So the max of a function is when f primal wreck. So when the derivative is equal to zero, so this is how you can find it. So to do this, we're going to by the derivative. Second is when let me make us a little bit clear the max of a function you're going to set f crime of the ex, hold to zero and solve, and that's going to give us the X value grab will help us find our Y value for the max of the function. Okay, so we go ahead and do that, and we're going to start by finding the derivative. So here, we need to find the derivative. You're very original immigration. Now, something I want to make clear here when I am talking about these derivatives. This is when I am calling first the first term and in blue. This is what I'm referring to as the second charm when I'm about to go into the derivative here. Okay, so we're finding the derivative of our original function. I've prime of X. This twenty one point four nine just zero. We're not gonna worry about it. Derivative of the first that's in green is zero. So we're going to move on to the second part of the chain rule which says first, I'm derivative of the second, which is in blue road of of co sign. Hyperbolic is signed hyperbolic. Keep flea inside the same here and multiply by the derivative of the inside. Richard zero point zero three to nine. Which is this right here. So now we consume Qualify this to help us find her Max, breathe f prime. Our legs is equal to when you do that. You should get a negative point six eight nine nine sign hyperbolic zero point zero three to nine seventeen. Sixty five x. Okay, we're sending this equal to zero. So we're going to change this zero and we're going to solve this. You divide this over, you're gonna have zero. You find the sign hyperbolic in Bors. I'm going to quickly go through these. Stop. You end up with exit hole zero. No one replied. X equals to zero back into our original functions. Up here, you end up with what we had in part a which showed us that our intercept was at one ninety point five three, which is also the height at the centre. So that's how you would do part B now for part. See, for this approach, we are asked to find out what points the height is one hundred meter. So to do this, our approach here, if we're gonna plug in one hundred for why and solve for X. So this is the steps that we're going to use. And again if you feel comfortable now and go ahead and write on your own crap that you have this little hint here. So plug in one hundred for why and solve for X. Here we go. We have our originally equation right here. We're going to take this originally equation, and we're gonna replace one hundred right there, and we're going to solve this Similar to how we did in part A. So we're going to bring the two eleven to the other side, which will make this negative. But because we have this negative in front of the twenty, those negatives will cancel out. So I'm going to go ahead and make both sides positive. Twenty point nine six co sign hyperbolic zero point zero three to nine. And again, we're solving for X. In this case, we're going to divide over the twenty point nine six and continue to solve again. We're going to have to use that co sign hyperbolic inverse function. We could do that on your calculator. Co sign hyperbolic in bursts five point three one nine. You're going to divide over, finished solving the equation, and you should end up with X equal to and again. We're going to use the plus and minus because we know what intercepts on two sides in it is symmetric X equals plus or minus with Cem rounding seventy one point five six. So again, you're plugging in that one hundred for why? Solving for X to get seventy one point five six. That concludes part C for part D. We're going to use that intercept from sea and were asked to find the slope. So the steps for finding the slope ears what our approach is going to be we're gonna have step one. We're gonna find the director, find the derivative stuff to plug in the value for X. I were interested in this case we're using angst equals seventy one point five six from part C. And the question is what telling us. We're interested in that points. It's telling us to find the slope at the point that we found. Okay, so there we have it. Step one and step two and we'LL go ahead and get started. We have step one, find the derivative. I did find the derivative in part B. So we're going Teo, quickly grow through this. I'm going to talk through it. For the most part, here's our original function right here. And to find the derivative we have that the derivative is equal to. We know this is zero green have the first times the derivative of the second. So we change the codes, I hyperbolic nine X and then we take times the derivative of the inside. Then we're left with this, which can be simplified. Two, When you do this multiplication, you're gonna end up with negative point six eight nine nine sign hyperbolic no zero point zero three to nine seventeen, sixty five francs. So this is our derivative. And again, we got that in part B as well. And now we're going to go on to step to, to finish solving this part of the question and step two toes is that we need to plug in the value for X that were interested in. So we're going to take We just found in stuff one, which was this. So here's our derivative, and now we're going to plug in that xie that we need. So we're going to plug in and I'm going to do it and read. We're going to have this and times our value for X is seventy one point five six. We're gonna have a plus and minus here again. So I'm going to do the positive. Seventy one point five six when you do the negative, that should get you the same answer once. Gonna have a negative sign. One will have a positive sign. That's fine. You should have a positive and negative slope. And once you solve this, so now you're using regular sign hyperbolic. This again can be done in a t I eighty four, second zero. To get to your catalogue, scroll down to the sign. Hyperbolic. And once you do this, you should be left with negative point six eight nine nine. Sign hyperbolic of two point one for six, eh? I'm rounding here. When I did the math on my calculator, I tend not around. So your answer might be off by just a little. If you are rounding and this again sign hyperbolic, done in your calculator gives us a slope equal to plus or minus three point six, three point five, three point six depending on how you're grounded. So there you have it. That's party. Find the derivative and then plug in the value of X that you're interested in against sign hyperbolic cosa Hyperbolic can be found on your calculator. And that wraps up question number

University of Missouri - Columbia