FE Exam Math Review

Looks good from your side. All right, then we will just get started here and keep going. Um, so if there are issues with the stream, let me know. Put something in the chat. Uh, I will do my best to try to fix it and um hopefully uh we have fewer mistakes than the last time last go around. Last last semester I or last year I kind of did one of these on a on a whim. Um, I didn't really double check many of the solutions. So I apologize for errors there, but hopefully this semester we'll have fewer of those. So, this is the first in a series uh where we're going to be going through math and then I think next week we'll be on to statics uh structural design and a few different topics here. So, pretty pretty excited u to do this. We'll probably have between 12 and 14 videos when we're all set and done. Um each one will probably be about uh an hour long or so. And hopefully you found the um review problems as well that we're going to go over. if you've gone over through some of those, um, hopefully that helps as well. But, um, you know, if not, you can take a look in the description below. It'll tell you how to get access. So, we're just going to jump in. And what I do kind of with each of these problem sets is I start out by saying, okay, well, what is actually on the test? Um, so I went and took from the civil exam specs right from NCES. There's a link at the top of the page. What's on math? So there's analytic geometry, single variable calculus, uh vector operations, and um statistics. And this is actually a little paired down from what it used to be. I've talked to a couple people that have taken uh the test in more recent times more, you know, within the past year or so. They said they don't remember differential equations, so I I didn't put those on here. But we are going to go over some basic calculus and basic geometry, and hopefully uh we'll we'll get into that as well. engineering e economics. Matthew, I just saw your comment there. We will try to be hitting those uh later on in the semester. But in the beginning of the semester, I'm going to I'm going to focus on some of the more real core concepts because math is like one of those things that uh if you don't know how to do it, it makes the rest of uh the review harder. Statics another one of those. And then towards the end of the semester, I'll throw in like dynamics, I'll throw in the engineering, economics, um the ethics piece, but I'm going to put those towards the end as opposed to in the way that the NCS kind of has it in the spec right up at the beginning. But um you know, just to sort of focus where we're studying and um and help you with that. So what we're going to do is we're just going to jump in and try to solve some of these problems. So if you have the problems, feel free to uh you know, try to solve them. Um you know, don't spoil it. don't tell everybody what the answer is or maybe where I got something wrong. But um we're just going to start in with some basic kind of things and then um yeah, so this is this is what I'm trying to do. Some some just basic fundamentals here. And obviously I don't have a inside scoop on what NCS exactly is going to tell you to do or or tell you that's going to be on the exam. However, um I I I kind of have been looking at some of the basic concepts, the basic ideas, and talking with different people, look at having these um questions reviewed by some people that have recently taken the test, they said, "Yeah, this kind of covers covers the basics." So, as we get started, I mean, there's some basic formulas and even this first question, right? We're given a couple things. We're given a slope and we're given a point. So, right there, I mean, when you think of lines, right? And I mean my initial thought is y = mx +b. Um but when you see this also maybe you're not as familiar familiar with the point slope formula but y - y1 = m * x - x1. Right? So these are two formulas of a line. Right? So these are formulas for a line. Right? And honestly so much of what we do is just a formula. um you know of of a line and that's one of those things where you kind of have to know proportional relationships and and formulas here. I mean mx plus b isn't exactly proportional but it's a it's a um it's a direct relationship with a constant. Okay. So once we have that we can also put in here well we know our x we know our y we know our slope and we can just really plug in. So, what I'm going to do is I'm going to say y - y1 um y1 is going to be 3 = the slope here 12 over 5 * x - x1. So, x1 is going to be uh 5 as well. So, um you know, we can multiply this through and we're going to get you know y - 3 = what? Well, 12 fths x - 5 * our 125ths. The fives cancel out and we're going to get -2. So our formula is going to come down to 12 fths x um we add three to both sides min - 9. Okay. And now we get a formula and that doesn't actually answer our question because we want to know that what this passes through and it pass through all these points except for one of them. And it you know you could break this problem down a few different ways. Um there are some problem solving techniques and strategies right but if you understand slope you what you understand is that the slope here is going to be delta y over delta x right so the change in y is going to be a constant 12 right so we could look at this a couple ways I mean yeah I I showed this formula but um when we look at this point 5a 3 we're going to be going up 12 every time so from three we're going to go to We're going to go to 27. We're going to go down 12 every time. So 3 - 12 gets us to good grief. I I I I'm not going to trust myself on that one. Uh but 3 - 12 gets to - 9. And if we subtract another 12, what do we get to? We get to minus 21. So right away, I mean, we didn't need to do our our point slope formula here. I wanted to introduce it just with a thought that you know this this um this does work and you are able to to make it work. So um however what we see real quickly is is the the the outlier here is going to be our our B value and that's going to be this this this uh the answer to this question. That point does not lie um directly on the the line. The other thing that I wanted to do with this question was I just wanted to introduce uh some some a little bit of an idea of how to use the calculator. So, I'm going to bring up my calculator here. And one of the things that you can do here is in the calculator, I like to use the TI36X Pro. So, that's what I'm going to be using to kind of demonstrate some of this. Um, but in this, you know, TI36X Pro, you can do similar things in the Casio, but if you go to the table, you can actually put in a function. So, I've already put this in 12 fths x - 9. It's good to know your order of operations. I did not need parentheses here because it sees the 12 over 5 first. And if I hit enter a couple times, I can start at minus h I don't want to answer. I want minus 5. If I start at minus5, right? I can get well -5 and minus 21. I can get uh 0 and minus 9. Um also what you see here is if we come back to the you knowg -3 here it does not you know that we at6.2 so that point does not land on this line. Okay. So we could even keep coming here till five and make sure that we got our the equation of our line correct and you know 10 and 15. So big picture um there's multiple ways of solving this problem but I wanted to show you a couple things. one with a calculator, one with kind of those those point slope formulas, the y= mx plusb formula, um formulas of a line, those those equations are are pretty important when you're just starting to look into basic algebra, basic uh geometry and that sort of thing. So, any questions or uh if not, I think I'm just going to keep moving here and move on because as we keep moving here um we also go to uh equations of a circle. Okay, so equations of a circle. You might not remember the equation of a circle as well, and that's okay. Um, I do have the NCS reference handbook here. I'm just going to click on math and hope that it's there. Um, honestly, I've done this before, so I know it is there. But you can see, you can even see here, here's our equations for the state, the straight line, the y= mx plus b, the point slope formula. Um, we see some quadratics, and eventually I think we are going to get to a circle somewhere down here. Here we see polar coordinates. Again, some of these things, you know, there's so much here. Where do you focus your time? Um, you you do want to know the basics of polar coordinates. Um, you do want to know the basics of trig for sure. Uh, you know, these are some of those things where you you got to know them. And, um, as we come down here, we have different types of geometry. And eventually, uh, I believe we get to a circle. And on the test, you're not going to be want wanting to just scroll through this. I'm doing this a little bit now just because uh it makes it easier to kind of to to get a a handle on it. But, um, if I type in circle, right? I mean, this is where we have 18 matches. And on the test, you're going to want to know what to type in. So, the the more familiar you get with with this or with the the manual, the better. So, um, I'm starting with conic sections. if I'm going backwards here. And here we have a circle where we have a formula of a circle. So we also have a radius which is x - h and y - k. So this gives us uh basically everything that we need to solve this problem. So I'm going to come back and use those formulas here to solve this problem. So I'm going to know that my equation of a circle, right? My equation of a circle is um x - h in parentheses squared uh plus you know y - k again in parentheses squar and this is going to equal r 2 sorry the radius squar okay so the good news is we have h and k um we have h and k here I'll turn my calculator off so you can see the whole problem here a little bit better um and we have a point that's going to be somewhere on the circle. So, in other words, if if we have a coordinate plane here, you know, we're going to come over to 2 4 and that's going to be the center of our circle. Then, we're going to have a circle. And if we come over to like 58, that point's going to be on it. Okay? So that tells us our radius here is going to be the difference between kind of this this x and this y and the center which is h k. Okay. So if we want to start there we can do that. We can say well the radius equals well what's this going to be? It's going to be the square root of because we're doing pyagoreans theorem here. Um basically I'm going to you know the x and the y are bigger than the h and the k. So I'm just going to put those first, but we're going to say x - h, you know, squared plus what? y - k^ squ. So this is just Pythagoreans theorem. Um hopefully that's again one of those basic things that you don't have to go look up. Uh you've done enough times that it should be second nature for you. Um so so here we go. So um so as we go here, what do we have? We have uh this is x. So the square t of uh five, what's this? x is 5 minus h, which is 2, that's going to be squared plus y, which is 8 - k, which is 4. That's going to be squared. And hopefully you've you've seen this before, but this is going to be our 3^2 + 4^ 2 and r is going to equal five, right? So So this is our 345 triangle. or beloved beloved 345. Um, which I love to use in like statics and that sort of thing when I'm teaching that class. So, just one of those things that you should um keep an eye out for. Okay? Because often times the I I think the F is not really trying to make you screw up on like basics like this. It's they're they're trying to test you understand the fundamentals. I mean, honestly, you might just be tempted to say, "Okay, well, this is this looks like a a circle formula. I'm just going to pick it and move on because I got three minutes for a question and I don't want to waste the time on it. But let's take a step back and look at it. So, what we get here is x - h. So, x - 2 uh squared plus what do we got? We got y - k which is 4. That's squar = 5^2. So, this formula is really close except that our x and our y are switched. So, it's wrong. Okay, this formula is really close except we don't have negative signs. So that's wrong. So we get into this other form here and we're going to end up being one of those two. But when you get sometimes you you know you see a problem and it doesn't look exactly the way you expect. And that's one of the things you have to work watch out for on the FE as well. If it's not exactly like you expect, um you can still make it work. Okay, so what we're going to do here is we're just going to expand these. So x - 2, if that's squared, we're going to get x^2 - 4x + 4. And if we have y - 4 squared, we're going to end up with plus um y^2 uh - 8x or 8 y um + 16 = 25. Okay? And hopefully you're starting to see this, but if we subtract the four and subtract the 16 from both sides, that's subtracting 20. And what happens? And so the fours and the 16s go away. We end up with x^2 - 4x + y^2 - 8 y uh equals 5. Okay. And really quickly hopefully you start to see this as well, but x * x - 4 uh + y * y - 8 = 5. And you see the answer. Okay, the answer is going to be um right down here, which matches uh what we just came up with. So again, this question I threw in there for a couple things. I threw this one in there because it it gets into Pythagoreans theorem. It gets into circle formulas. It gets into just looking at what happens when your answer doesn't look exactly like you want it to. Right? Sometimes you have to do a little bit of manipulation in there with the algebra in order to get it to work. Okay, so that's where we're going to go um with this one. It's just a little bit different, but it it sometimes um sometimes it's easy as they'll give you this this exact formula here. Um sometimes you'll have to do a little bit of manipulation. So, no questions there. I'll just keep moving on. All right. So, this question again, I threw this one in there for a couple reasons. One, um because it shows you some basic uh some basic, you know, how to use or basic algebra really. I see I shouldn't say basic algebra, but a series of three three equations aren't like basic basic algebra. Series of two equations um you know, that's a little bit more uh basic level. Three gets a little bit more complicated. But Josh Wilson, I noticed you're on the line, man. Um, I I still credit you with with showing me how to use my calculator. So, I so appreciate this. But, you know, let's just let's let's start there, I guess, because if you have a TI36X Pro, um this is this is a beautiful thing because if we pull up the calculator, you could do a second um system solve. And this is great because we can just do a 3x3 linear system. Okay. Um that this is this is pretty amazing. Um, I put these in already. So, the coefficients four, four, one, and minus two, right? That matches um what's on the the left here with equation one, equation two, right? The first coefficient, x coefficient 2, 1, one, and then our uh our constant term on the right is eight. Um, on the left or on the equation three, one, two, two, and four. And the beautiful thing about this is once you go through, you can solve. So if you know how to use your calculator, you know, we can see the 4 - 6 and 6, you know, equations. So again, you can save a lot of time if you actually know how to use your calculator. And if you don't know how to use your calculator, what's another alternative? Well, another alternative is to look for um some things that are going to make your life easier. Like if you're doing this calculator is acceptable. Yes. So, so what you want to do is you want to look for for ways where you can do elimination or substitution. In this case, the first thing that jumps out to me is the coefficients on these two terms are um are are multiples of each other. Right? So, what I can do here is I can multiply this whole top equation by two. And I'm when I do that, let's just let's work this out by hand just real quick. So I get 4x you know plus um 2 y + 2 z = 16. Okay sorry my z's looks a little bit funny there. And now when I take these next two equations this series here I can subtract one from the other. So x you know - 4x I get - 3x 2 y - 2 y these cancel out 2 z - 2 z these cancel out. So I get equal to 4 - 16 or -2. Divide both sides by you know -3 and -12id -3 is 4. And the nice thing here again we get that answer. Um you might have two answers with like a four in here or something but this is the only answer that works. It's the only x value that works. And the interesting thing that I mean one of the things that I did here is you can't just pick this one and say well does it work for one equation because all of these work for one of the equations. Okay. So sometimes the answers are obvious, sometimes um they're not. But in this case, you know, you can use elimination, you can use substitution to solve for a series of equations. But yeah, I this calculator, the TI36X Pro, um, another reason I do the I I like to do the math session first is just to show you the power of a of knowing how to use your calculator on the FE because for a lot of these math questions, you can plug them in and actually solve them using your calculator without like having to do a lot of crazy crazy math. So, if you get either, you know, I'd recommend the TI36X Pro or the Casio uh is it the FX115? I don't know, one of you guys online probably has that right now, but um you know, either one of those uh should work. And if you really like HP, uh you can go with HP. But, you know, it is what it is. So, um definitely I' I'd suggest grabbing a good calculator and knowing how to use it. All right, we're going to keep going. And um as we keep going here, we're at inverses. I I mean inverses, exponential functions, um again, topics that I I think could show up. They're kind of some of those basic basic manipulation topics, right? An inverse, you just flip the x and the y and then you have to manipulate the equation. So this question gets thrown in there. Not so much do you need do you know how to use a log or what the definition of a log is but kind of the idea of do you know how to manipulate an equation again one that's maybe a little bit more complicated than the y= mx plus b or a circle formula but do you know how to manipulate it right so with this one where we're going to come down to is to get the inverse we're just going to switch the x and the y so x equals um 3 y and it would be so nice if that showed up you know over here but it doesn't um so what do we need to do. Um, so we need to go ahead and actually take a look at this. So let's just come back to our our reference handbook for a second. And if we um come to logs, there are some uh identities in here for for logs and and really you know the B to the C equals X if B to the C equals X. So the log of X with base B equals C. And if you haven't done logs in a while, it can be um confusing. Honestly, I I hadn't used logs in a long time, and I had to help one of my kids with logs in their algebra class, and it took me a minute to go back. So, I was like, "Hey, let's just let's let's go there for for now." So, let's let's go to this side. And one of the ways that I'll look at this is I kind of again like with algebra, you can kind of do anything to either side of the equation as long as you do the same thing. And that's kind of an overstatement a little bit, but um let's keep going here. Let's let's take logs of both sides. So if we take a log of x and we take a log of 3 to the y, right? What happens? Well, this is one of those identities. Sorry about that. Um this is one of those identities, right? So what we have here is uh if we come down to the bottom, right? What we see is um one of those identities log x to the c equals c log x. You see that? Um so we're going to come back and we're going to say well log 3 y = y log 3. Okay, we can both divide both sides by log three. And I'm going to rearrange this a little bit, but we get, you know, divide both sides by log three. We get y log x over log 3. And this is your definition right here of y equals log x with a base three. Okay, so this is again if you know how to manipulate some functions, this is going to help you on the fee. And if you can do that quickly, um that's going to help you as well. But let's just go back and see that definition uh over here. Um as we come back one page, we see that log of x you know base a log of b base a equals log of x base b. Okay. So you know whether it's knowing how to switch bases or um it's you know knowing that log of x y equals log x + y right so or log of x over y is like log x minus log y if you have to deal with with logs or natural logs um then then knowing some of those properties are going to be uh going to be helpful. All right. So again, um kind of representative of exponential functions, logs, natural logs, and those different pieces. Um are you going to get this exact problem? Probably not. But do you know what you know the natural log is versus the log? Right? Um the natural log just has a base of e whereas log of x typically has a base of 10. Okay? Until we change it to a base of three. Okay. But this gives us an answer here. It gives us uh which one was it? It was uh answer answer C. All right. So, we can keep keep going. Any questions? We good. I'm moving. It's it's late, you know, but we can we can do this and hopefully it helps. And as we get into a rhythm here, hopefully um you can do the same when you're taking the test and just get into a rhythm. Analytic geometry. Again, geometry gets into the algebra. It gets into the trig. It gets into some of these pieces where oh man, you get into identities and maybe you haven't used advanced trig in a while or and you forgot what a secant is or you forgot what a cosecant is. Um I would recommend going back and taking a look at especially the identity sheets on um in the reference manual. So as you come down here there's some there's some trig identities and you want to kind of get get a basic understanding of what these are. So the cosecant 1 over s the secant is 1 over cosine right and then there's some some identities in here like sin^ squ= 1 you know plus cosine^ square= 1 or tan^ 2 + 1 equ= secant^ squ so some of these like some of these identities you're not going to memorize all of them u but hopefully you should um what is analytic geometry that's a good question it's kind of covers like everything um in a sense it's like your algebra and into trig as well it deals with plain systems and it deals with like how you describe those plane systems. So, it can get into lines, circles, trig. Uh that's kind of the way that I I think of it. Um so, I threw trig in there for identities. We'll talk about trig when we get into calculus as well. Uh but but this is where you're given a question and this kind of gets to um some of the identities, right? You I mean you could try just plugging in numbers in your calculator and looking to see like does one of them work. Okay. Um, another thing that you could do when you get to identities like this is you could jump in and you could try to try to change some of these pieces so that you get everything in signs and cosiness. That's one approach. Um, but one approach might be here to come back and say okay well well how can I do that? And and really what we see here is tangent is defined as s over cosine. Okay. So if we come back and we say let's just start plugging in and see what we get. So cossine x plus sinx times we're going to substitute in we know that um tangent equals and it's so easy to forget and screw these up but what what is it sin over cosine? Okay so sin x over cosine x and if you screw it up the signs will cross out and it won't work that well. But um where do we go from here? Right? What you see is you get cosine x plus sin^ 2 x over cosine x. And you might be thinking like, man, I'm farther behind than I was when I started. But but don't give up. Don't lose hope because as we come back to our identities, and the nice thing is when you're on the FE, you have two screens. So you don't have to keep flipping back and forth like I'm doing here. But this this major identity is one of those real common ones. Sin^ square plus cosine^ square= 1. You might be saying, why do I care? Because if I know sin^ 2 x + cossine^ 2 x = 1. I know that cosine I'm sorry I know that let me write it this way. I know that sin^ 2 x equ= 1us uh minus the cosine^ 2 of x. So now I can take this value here and I can substitute it in over here. Okay. So I can say well the cosine of x plus 1 - the cosine of x^2 of x all over the cosine of x. And things are looking a little bit better here. So I get cosine x. I'm going to separate this this fraction here 1 over cosine x minus cosine^ squar of x over cosine x. You'll see that the cosine squar and the cosine um they cancel out. So we're left with cosine x plus 1 / cossine x minus cosine x is the equivalent you know the equivalent function here or um expression and we're left with one over cosine which if you remember and some of you probably need to go back and look at the um reference the reference handbook again but one over cosine is equal to the secant right so that becomes our answer make some sense But again, are you going to get this exact question? No. But could you get a trig expression where you have to find equivalency and do some substitutions and some manipulation? I I think that's pretty likely. Okay. So, um so being familiar with those identities, going back and looking at them again because maybe you haven't looked at them since calculus, um it's probably a good thing to do. All right. So, um, so Matt, I don't know, maybe this is more a calculus question, but even still, it describes the physical physical universe in a plane system. Okay, let's go back here. And the reason I threw this one in is again, one of the things that you need to know how to do, um, is you need to know how to use your calculator. And this this is so easy to mix up and so easy to mess up uh that I thought I'd throw it in because again when you're looking at how to use your calculator um you want to be able to solve it quickly and solve it correctly. Okay, it we can do this one of two ways. We can say well x oops we can say that x equals um you know x= pi over 2 radians is equivalent to what? Well, this is equivalent to 90 degrees. Do you know how to do that? I mean, basically we're taking this is times 180 degrees over pi, right? So, so we can get to degrees or we can use radians, but you need to pick which one to use and use it. you need to pick it and use it because um it's so easy like for example um to go into your calculator and um say okay well I can take the cosine of 90 and get a certain value and is that the right value? Well, it depends. This is where you need to look at your mode. What you'll see here is right now I'm in radians. Okay. Um, that's a problem in if if I'm using degrees for my angle. So, I can come back here and use degrees. Okay. One of the things that I want to point out though is when you switch back from, you know, mode radians to mode degrees, all my history, actually, my history isn't gone. Look, it's still there. Um, so that's good. So, I can go back and do cosine 90 again and get a different answer. Okay? Because now I'm back in degrees. So I know that the cosine of um the cosine of 90 degrees equals zero. Okay. So the question is you know it's not equivalent to which of the following. So each of these are going to equal zero except for one. So now I just need to go through and evaluate. So the tangent of 2 * 90° this is going to be the tangent of 180° right? So the tangent of 180 is going to be zero. Okay. So that one that one checks and it's it's not that one. Okay. We can come back. We can say well what about the secant of 2 * 90°. So what's that? Well, it's like the secant of 180° which is also like 1 over do you remember which one it was the cosine or the sign? Um it it's let's just come back and double check right what is it secant is one over cosine right so this is going to be one over cosine of 180 degrees which if you don't remember cosine of 180 is um is did I do that right cosine 180 is1 so one overgative one and what you'll see here is I made a mistake because it's so easy to lose some of these pieces the plus one I lost So, it's easy for me to take this and move it over a little bit um and add my one back in. But just be careful because it's so easy, especially when you're trying to go fast um on a test uh to to make some of these mistakes where you miss part of the problem and you make a mistake. But minus one plus one, this is going to also equal zero. Okay. uh this. So we had a we have b um we have c um 2 * the s of what do we have? 2 * the sign of 90°. And I'm going to put the square out here because that's kind of how we end up doing it in the calculator. Okay. Um you know minus one equals what? Well, if I come back and I type the s of 90, I get 1. So 2 * 1^ 2ar - 1 and all of a sudden we see something. Well 2 * 1 is 2 - 1 gives us to 1. So this is a strong contender um because it doesn't equal zero. But just to make sure we didn't screw something up um let's go to d. And the cosecant of what do we have of 90° minus one is equivalent to 1 over the s of 90° minus one. So if again I I I have the s of 90 I already have that here. Um the s of 90 is is equal to uh one. So 1 over 1 minus one equals zero. And we can cross out D as well and be certain that our answer is this one. Okay. So again, um knowing some of these these how to use your calculator is going to be important. Um knowing that mode, whether it's degrees or radians, pick one, but just be be consistent. I'd say be consistent with what you do because if you're flipping back and forth between radians and degrees and you're trying to go fast on the test and you see a problem like this and you say, "Okay, that's radians. I'm going to switch it to radians." And then you get to another problem later on and it's in degrees. Um, that could make it a little bit more challenging because you could get it wrong. Okay, so hopefully that works. Um, if you have questions, shoot them out there. Okay, let's keep going here. We're still in kind of this geometry section and I bring up right triangles and similarity here because uh I use these a ton when I teach trusses. So when I teach trusses I I the idea of similar triangles is is huge. So when you have a perpendicular line like you know like this like am that shows up here. Now all of a sudden we have a bunch of similar triangles. Okay. So I I can label some angles here, but if I label this angle alpha and this angle like beta. Okay, what we see is we have a whole bunch of similar triangles. So we have the big we have the big triangle, right? We have the big triangle here. That's our big triangle. And then we have two smaller similar triangles. So we have one similar triangle that looks like that. And another similar triangle that looks like this. And when we start looking at these similar triangles, we can relate them and come up with some um relationships here. So what I'm going to do is I'm just going to take a look. I'm going to take this first triangle out. I'm going to take, you know, this B um or A and B out and I'm going to look at I'm going to write some relationships here. So what I really want to find from this triangle is I want to find a couple things like can we get rid of Cal. Oh man. Um would would love to. Thanks Josh. Um so what we're going to do here is we're going to um we're going to go and what we're going to do is we're going to say let's take a look at at this. So um am right let's say we want to find this line am um am over what do we know? Well we know am over five right because we we have um this line over the hypotenuse is going to equal what? It's going to equal the long side of the big triangle over the hypotenuse. So, this is going to equal essentially what do we have? We have AC, I'm sorry, AC wrote it in the wrong spot. Um, divided by BC. And you might say, well, we never solved BC. But this is where again, if you know some of those basic triangles, and if you don't, you can use Pythagoreans theorem, but this is a 512 13 triangle. So, the hypotenuse is 13. So, we're going to get AM over 5 um equals what? AC which is uh AC is the equivalent of 12 over the hypotenuse of 13. So I'm just going to write that AM equals well if we multiply both sides by five we get 60 over 13. Similar triangles if you can see the similar triangles um it's going to help you a ton when you get into things like like trusses even in some of this stuff it's going to help. Okay let's go to the next one. Let's go to BM. So BM we're trying to solve for. So I'm going to say BM over the known BA or AB is going to be five is similar to what the short side of five divided by the hypotenuse of 13. Okay. So similar triangles again here. So we get BM um you know equals well now we have 5 * 5. So 25 over 13. And here let's go to our last triangle. I I think we're seeing an answer emerge here. But if we go to our last triangle, um what we can see is now we want to solve for I think this last one cm. So we're going to solve for cm over what do we know? Well, we know cm and we know um this value here of 12. So cm over 12. Okay. Equals what? It equals um the long side over the hypotenuse, right? So the long side is 12 over 13 and we get cm = 12 * 12 is 144 over 13. And what you see real quick is the 13s are all common. They go away and our solution is going to be the 60 25 144 ratio. Okay. So, so here we have um here we have uh similar triangles and something that's going to you know come up at some point in some way similarity I I believe so um something to be famili familiar with how to how to relate those whether it's on a problem like this or it's a truss um you're going to want to know how to relate one side of a triangle to another. All right, so let's keep moving here. This one I threw in. I had the idea of doing this one and then I actually went to try to do it in the reference handbook and was surprised because this I mean this is just like okay in the reference handbook as you're scrolling through the reference handbook. Let's go back to the reference handbook for a second. Um and what you see is in the math section there areas and volumes and we have parabas, we have um you know we have ellipses, we have circular segments and and I thought to myself, great, this looks good. We'll do a circular segment. Um you know, use something that's one of the more complicated ones. And all of a sudden, I realized, wait a minute, um this has R. It has FE has S over R, but we don't have a lot of this stuff. So in this problem, what we're asking for is an area, right? So we have a couple areas. I mean, we can have an area of like a trapezoid here. We could find that area. Um, we also have an area here of like this circular segment. Okay. Um, I had a student actually, I think Matt, I don't know if you're online, but um, you were doing this and you you just sort of said, I'm going to do it a little bit differently. I'm going to call this a parabolic section. And I think you got pretty close. Okay. But what I want to do is I want to go ahead and solve this as as the two areas that I I drew originally. So I want to show this as um a trapezoid and a circular sector. I mean it's just again there's going to be areas. Are they going to be exactly these two? I don't know. But can you go to the reference handbook, find the right area formula and use it? Right. So here we have some area formulas and these are dependent on their radius and d and fe and what you realize is all we have is this radius and that's about it. I mean, we we know this length here, this chord length, um, but we don't know a ton else. And that's where sometimes the reference handbook is a little frustrating because you look at this and you think like that doesn't give me enough. But maybe something clicks in your head and says, "Wait a minute, wait a minute, wait a minute." Um, in transportation, we did a lot with horizontal curves. We did a lot with uh other things along those lines. If I come down to transportation here and transportation's when it's after structures, right? Um, so if I come to transportation, if I get there eventually, actually it's after environmental. Sorry, I should have gotten there more quickly. I probably should have done a search for this. Um, there's transportation, right? So, we get into transportation. And in transportation, we have some more we have some more um equations. So here's vertical curves and here's horizontal curves as well which is interesting because we see in here um we see a formula r equals the length of the chord divid by 2 sin I here I was fe in the other one. So let's solve for that first and then kind of go back and work the other equation. Okay so when we come here um let me just label a couple things. We're going to find area one. We're going to find area two. Okay. And we're going to use a couple of these different formulas to do it. So area one, hopefully I'm just going to get that one out of the way because um area 1's just a trapezoid. So what's this? This is um height one. So 12 plus height 2, which is six all over two. So that's the average height times the width of 8. Um and that's going to give us one value. So 12 uh + 6 / 2 gives us 9 * 8 and that's 72 uh square m. So we know we're in the right ballpark. Now we're just seeing well how much does that little segment this little sector add? And this is you know maybe this isn't a threeinut problem but maybe it is. I don't know what we got from the transportation section was R equals what? It equal LC. So the length of the cord divided by 2 sin I / 2. Okay. So we get this and what do we see? Well, we know that we we can find the length of this chord. We know the radius. So we get uh essentially 11 m equals the length of the cord. And hopefully you see this, but again, I like these three, four, five triangles. We have a eight, six, 10 triangle. So the length of that chord is going to be 10 m / 2 sin i / 2. So we can we can manipulate this again. And when we do that, um I believe what we get here is well, let me just let me make it a little bit simpler here. we get um 2 sin I / 2 = um 10 / 11 or we could say sine of i / 2 = 5 over 11 we're dividing both sides by 2. Um we can take the inverse sign of 5 over 11 sin inverse of you know 5 over 11. Um this is going to equal I over 2. So from that I I believe what we get is I equ= um 54.07°. Anybody else get there? So this gets us this sector angle. So if we're if we're looking at a bigger picture here, this is some radius, right? And this angle here is going to be 54°. Okay. And now once we have that value, once we have that value, we can come back and go to our area section here. So if you come back to our math section and uh come down to our areas right we have a we have a area formula here that's r^ 2 v minus sin p /2. So let's come back here and we get the the the formula here um for that sector you know that for that that segment of the circle and we get a2 um equ= r 2* v minus sine of v over two and some of you if you have a calculator um you might be thinking like uh oh how do I how do I do this thanks man just a comment Um, you got it. 6 8 10 62 + 82 + 10^2 get or equals 10 squared. That's our Pythagorean theorem there. Um, so perfect. Um, I love you guys. You guys help out. So, so it's awesome. So, so if we plug some of this stuff in, you're you're you're going to want to know not only what the equation is, but how to use it. And this is where it gets a little dicey because we know the radius is 11. So, I'm just going to drop the units here for a second. Oh, man. And now what we have to worry about here is um is is honestly we have to worry about radians and degrees. This formula is in um this formula is in radians. Okay. So we need to we need to we need to put things in um essentially in radians here. So I'm going to take my 54.07. I'm going to multiply it times pi over 180. And I'm going to get I = what is that? 54.07 times um well actually I'm just going to divide it by 180. And I'm going to leave the pi in there. I'm going to get 0.3. Um so this bas basically 0.3 pi. Okay. Uh did I do that right? No. I think I did something wrong. Um what did I do wrong? Anybody got it? Um, oh, because this is this is is this I over two or is this I? So, I think I I don't know. Let me back up and and make sure I get got this right here. Um, let me back up and make sure I got this right. So, let me pull the calculator up and we will uh we'll take a look here. So, if I bring the calculator up um let's do the inverse sign. And to do the inverse sign here, I'm going to hit sign twice of 5 over 11, I get 27. So yeah, so if we multiply that by this two, we get 54.07. Okay, so that works. So um let's see what we get. So if we're looking at radians here, um uh I my notes are a little sketchy here. Um, this is bad, but you know, let's see what what let's see what happens. Um, we get I think I just dropped a pi, but we get I think we had the 0.3 pi up here. Okay. Um, 0.3 pi um minus the sign of, you know, this 54.07 or 0.3 pi. Okay. Um, all over two. Sorry, let me get rid of the calculator there. Um, so this is a little bit a little bit um interesting on the calculator when we get to it, but if we put it in here, right, we get 11^ squared, right, times um times 0.3 pi minus s of 3 pi. And you might be thinking, am I in radians or degrees? I don't know. That's a good question. Let's check. I'm going to divide that by two. I'm going to come back here, go to my mode. Um, I'm going to go to radians. And if I hit enter, I get like 8.07. And that is the correct value. So 8.07 um square meters. When you add this 72 plus the 8.07, you actually get this 80. So yeah. Yeah. The parabola equation is so close to this and it's so much easier. So that is a huge that exactly the parabola equation you have the 2BH over three 2BH over three right that gets you essentially to 80 as well. So what's what's our base? So 2 * the base 8 * the height 6 over3 um that's going to get us what? 2 um 2 * 8 * 6 over 3. It's going to get us 32 square m. So, if we're looking at um let me just do a different color here. But if we're looking at, you know, this this area here, that's you know, this is an approximation. It's it's not exact, but it's really pretty close. And it saved us having to go to transportation. Um it's it's it's not exact because um this is a circle. The the radius of 11 is a circular segment. Okay? It's it's it's not a parabolic segment. It's it's really close to a parabolic segment, but it's not a parabolic segment. So, you know, um if we look at this differently, um you know, have these other two sections here, like we could say, um essentially like 8 * 6 um equals what? That's 48 for this blue section, right? We could do this a couple different approaches. And honestly, on the test, your approximation is going to be okay. It's going to be close enough. But but this is where that circular segment sometimes you have to pull formulas from different sections of the book. Okay. So I don't know good questions. Um but the precise answer is to get the circular segment right this formula here is is for the circle or the circular you know circular you know segment or portion. Okay. And that's that's the precise way to do it. But again, you have to know whether you're using radians or degrees, and it gets a little bit a little bit tricky. So, um, know how to do it, and you guys should be good. But also use the formulas a couple times because you can look at the formulas and until you actually start plugging numbers in, you don't realize that that formula is in uh is in what's up, Justine? Um, that formula is in radians. Okay, so definitely check that out. All right, this question I actually took from my son who's in algebra and there's a way to solve this algebraically. But um I I thought it was interesting because when I saw it and he was working on algebra, I I thought like man why would they had this type of a question in algebra? This is a calculus question. Okay. And normally this type of calculus question you're doing something different. Normally when you have a parabola, you're looking to find like the minimum point, okay? Or a maximum. And if you remember how to do that, what you're going to do is you're going to set the derivative equal to zero and figure that one out. So in this one though, it's it's you're not finding minimum maximum. You're actually just changing the slope. So we want to find when that slope equals to what? To two. We want to find when that slope equals to two and essentially figure out this b value which is going to be, you know, minus b is going to be our y intercept. So that's what we're trying to do. So when we get started here, we're just going to take a derivative f, you know, frime of 1 x is just going to equal 2x - 4. So we know the slope. Again, we know the slope is going to be two at that point because this question is saying we want these line segments to be parallel or tangent at at a specific point and tangent means the slopes need to be um the same. So we know the slope of the line is going to be two. When does that happen? It happens when the first derivative um equals 2. So 2 = 2x - 4. So what does that mean? It means if we add 4 to both sides, we get 6 = 2x and x = 3. Okay. Once we know x, we can plug that back in. f of f not frime, but f of x f of 3 equals what? We have 3^ 2 - 4 * 3 + 2 and f of 3 = 3^2 is 9 - 12 is -3 + 2 is going to be I think -1. So all of a sudden we know where this point is. We know this point is what is it? 3 comma minus1. And now what we need to do is we just need to solve kind of this formula for b. So we know you know f2x in other words we have min -1 is our value when x= 3. So 2 * 3 + b. And if we do this out we get minus1. So this is f2 um minus1 um = 6 + b. We subtract six from both sides we get b = -7. So this is f_sub_2 at you know just for clarity here at um you know 3 comma minus1 and we just solved you know for b which means we have an answer here and we were able to do it. Okay, so you probably won't maybe you maybe you'll see this question maybe you won't but but the the question of maximum minimum you know when does the slope equal to or when does the slope of your parabola equals zero you've done all all along in calculus this one takes it and twists it a little bit right and a lot of times that's what the FE is going to do it's going to take a basic concept that you know and twist it just a little bit to see do you understand the principle or just how to apply straight things okay and this is same principle as finding a max or min it's Instead of finding a max or min when the slope equals zero on the parabola, you're just specifying when the slope is or what the slope is. Okay, so kind of a a cool question in that sense. Oh, more calculus. Um, I wish we could do derivatives on our TI36X Pro. You you can do some um you know, I don't know actually. Somebody tell me, can you actually do uh do derivatives like this? But yeah, see this is being recorded. You can come back. even if you can't. Let's go over how to do derivatives. And there's different derivatives. Um I'm just going to come back here for a second to the reference handbook. And if we come down to derivatives, we have some different formulas. And I just picked one of them. Okay. So this question is going to deal with the derivative of uv dx. And we get a formula u dv dx v du dx. So we'll go through this one. But you want to be familiar with these various formulas. you know, whether you're integrating, you know, sums or constants or um products of two functions or um potentially, you know, an exponential function. So, we're just going to come come back here and I I'll write this down, right? But the formula that we came up with or that's in um the reference handbook here is du v um dx equals what? It was um it was u dv dx plus v du uh dx. Okay, so this gets us a formula. So we need to define these. We need to find um the derivatives and that's where we need to go next. So u equals the cosine of 2x um v = e 4x and now we want to take derivatives. So, du uh dx equals Oh, man. Do you remember the derivative of a cosine? If not, you're in luck because it shows up here, right? Um the derivative of a cosine is a negative sign. So, if you don't remember it, it does show up. You just have to know where to find it in the reference handbook. Okay? And if you're not sure, you can find it. Um but this is 2x. So, what happens there? So, let's come back here. D of cosine u is minus sin u du dx. So, we have to also take the derivative of what's inside um the cosine. So, this is going to be minus sin 2x times the derivative of this 2x, which is going to be time 2. And I'm just going to throw that um back over on this side. So, - 2 * 2 sinx um dv um dx. If you can't remember derivatives of exponential functions, um what do we have? Do does that show up here? E to the u is e to the u du dx.x. So, it's very very similar. And we get this. We get we take the derivative of the inside part. we get 4 e 4x. And now we just kind of plug and chug a little bit here. Um we'll we'll take these different parts. We'll say cossine um 2x uh dv dx which is time 4 e 4x and then we're going to add v which is e 4x um time dv dx which is going to be min - 2 sin of 2x. And hopefully what you see is some common pieces here. So this e 4x is common. We can factor that out. e 4x uh what's left? We get 4 * cosine 2x plus what what do we have left? Well, actually I'm going to take that away because we have a minus sign here on the two. So - 2 sin 2x. And fortunately that matches something in here. I did my homework this time. Um so fortunately that matches something up here. and we get another answer. Okay, not too crazy, but you have to know how to do those derivatives. You got to know some of these um these functions and you get you get the answers. So, vlog I I did you factor out an extra two? Um you did, which is fine. Um that's actually a little bit more simplified. Uh but yeah, I mean we could have factored out um the a two from this four and from this two. Um, that would have been a little bit better, but I think you lost the two on the cosine 2x. So, take a look at that. All right. Okay. So, um, or you lost a two somewhere. I don't know. Yeah, take a look. Okay. So, let's keep going. If functions below intersect at two locations, create an enclosed area. How come A and B are the same? A and B, did I make them the same? because I screwed up. Thanks for the thanks for the the question, Margaret. Um I I uh probably should have made these different, but if you get a repeat answer, um it probably means um the question's bad and somebody didn't do enough QAQC. So, I did check these to make sure that the right answer was in there. Um I screwed up on the the wrong answer. So, I apologize for that. I will fix that. That'll be like version 1.03. Okay, so let's keep going. But thank you. um functions below intersect at two locations create an enclosed area bounded by these two functions is most nearly so this gets to a definite integral right basically what we have is we have two functions and you know if we if we look at a coordinate plane here um one thing that to notice real quick is they both have this plus one term so they both have the same yincept right we both we know that they're both going to intersect kind of at this plus one the x+1 is kind of an easy function right this is your f2 and what we see as well is we have this negative sign. So we know our parabola is negative. It's going to come back down. So it's going to go through here and back down. And um there's there's our parabola. There's f1. What we're trying to find here is this area bounded by the two curves. So essentially all that we're doing here is is we're going to take the integral, right? integrals find areas of um basically from 0 to x or I shouldn't really call that x. I should call it like 0 to a or something. Um but 0 to a 0 to a. My integral looks like a weird row. Um integral from 0 to a of f_sub_1 minus f_sub_2. And I'll put a dx out here kind of for completeness. Okay. But we need to do two things. We need to find a. Where do these intersect? And we need to then take the integral. Okay. So, um what we'll do is we'll start by first finding a. To find a, we're going to we're going to we're going to say, well, f1 has to equal f_sub_2. Okay. Um so, what does that look like? Um I'll just write it out. - x^2 + 4x + 1 has to equal x + 1. And right away, some things can cancel out. The ones can cancel out. um we get minus I'll put that min - x^2 over here. Actually, no, I'm just going to leave it this way. Um we get min - x^2. We can subtract an x um you know + 3x = 0. We can factor this. We can take an x out. So we get x * 3 - x = 0. So we get x has to equal what? It has to equal 0 or x has to equal uh 3. So this solved a essentially we know that this integral is going to go from zero um to three, right? This is kind of our a value over here. Um where where this point of intersection is. Okay, that's our that's our a value our three value here. And I probably should have labeled a closer to that, not the the um x intercept. Okay, so that means what we need to do here is we need to do this integral from 0 to 3 of our min - x^2 + 4x + 1 - x + 1 that needs to be in parenthesis. Um, and honestly we kind of already did this a little bit because on this side we did this and we know that this is the integral from 0 to 3 of - x^2 + 3x. Okay. So, you can do that integral out, but for for the sake of showing you something on a calculator, I trust you guys have all taken a calculus class. Um, I do want to show you this. This is kind of cool. Um, you can go to the the the second integral function in your calculator and actually do definite integrals. So, you can do it in this mode. Um, I kind of like when I'm doing definite integrals to go to math print. Um, if you go to mode and go all the way down to the bottom, go to math print and then to do the integral, it just shows up so much nicer. And I wrote this wrong. This is 0 to 3. Um, but let's let's enter this in 0 all the way up to three of what? Minus um x^2 + 3x. Okay. And and hit enter. You wait. It's it's moving. And you see the thing and we get 4.5. I mean, honestly, it's it's that easy. Sort of. Um, right. If you know how to use your calculator, it can be that easy. Which is nice. I mean, we could do this integral. But we can say well it's - x cub over 3 plus you know 3 x^2 / 2 from 0 to 3 the zero goes away. So this just works out to be what? - 3 cubed over 3 uh + 3 * 3^ 2ar over 2. And I think you should also get 4.5 here. Um if you do it that way as well. But I thought this would be a cool way to show you how to do a definite integral on a on a calculator. Um, and again, it's it's figuring out how to use your calculator to work for you on the exam. Um, because there's a lot more room for error, I I would say on the exam in something like this, um, than there is just on the calculator and putting it in on the calculator when you're doing a definite integral, I don't know, you know, use it. Um, single variable calculus. Again, we're doing calculus. You might have an indefinite integral where you don't have certain limits uh where you're going from 0 to to three, but you have to evaluate uh that integral as well. So let's take a look at this. Again, this kind of becomes a trig identity. Um you have to, you know, come back here and there are some indefinite integrals here, right? But as we come down, the one the reason I picked this one is because you don't see the integral of secant squar you see the integral of tangent squ. That's great. Um, but I don't see the integral of secant squar here in this form. And that's where again if you know a little bit of of some of the other identities, it's going to help you. Well, let's come back to the derivatives for a second because one of the things that we see in the derivatives is the derivative of the tangent is a secant squar. So the integral of seeant squar is going to be the tangent, right? So, so if we're looking at this, we can we can use some of the information maybe not directly but indirectly to solve our problem here, right? So, this is this is going to be useful. Another identity that we could use here uh where did it go? Um where are our identities? Another identity we could use here is tangent squ + 1 equals the secant squared. Okay. So, a couple identities here that are going to help us. So, so I'm going to use I'm going to build off of this one and come back to our problem here. So, we know that um we know that what we know that the identity that we just looked up was the tangent squared of theta equals or x in this case. Um I'm sorry, the tangent squar was it? It was the tangent squar + 1 equals the secant squar. So, sorry, wrong button there. Um so the tangent squar + 1 equals the secant squared of x. Okay. So we can we can rearrange this a little bit but we can get well the secant^2 x - 1 = the tangent of x right and all of a sudden um we can take this and we can substitute in we get the integral of the tangent squar of x which actually um does show up on our identities table. Okay, so a little bit of manipulation. We get the integral of tangent squar and we get tan x - x. So this equals um tan x - x and we'll throw in the constant there for completeness. Okay. So again, you might not be given the exact thing that's on the identity, but maybe with a little manipulation, you can get there. So if you if you if you get familiar, do some practice um with those um with those those those those identities, it should help. Okay, let's keep going. The next couple ones, I mean, I I'm going to show you how to do them kind of both ways, but vector operations are going to show up on the FE in some way. I've talked to people that have taken it, and they do show up in some way, whether it's a dot productduct, it's a cross productd, it's what that cross or dot product means. Um, so we're just going to throw these in um real quick. And actually, before we even do that, what I'm going to do is I'm just going to pull up my calculator and throw them in the calculator. So, you'll notice on the calculator, there's a vector. So I'm going to come to second vector. I already have these put in here. So I'm going to go to edit. You can see um if I come to vector u um I got 54 actually no this is a different different 54 and minus3. Okay 54 minus 3. And if I come back to second vector um I'm going to edit uh v and it's three and it's going to be what? It's going to be two um minus three and four. So this is kind of like I J K XYZ. Um and these are our our values here. So I can come back I go to my vector. I can check them you know just to make sure I got them in right. 54 minus 3 and uh you know if I come down and and get this um I get two minus three and four. Okay. So once I have that I mean I can just come in to my vector. I can do my math dotproduct. Uh I can say come back to a vector again, second vector. Um I can pick vector u and then I'll put a comma in here and again pick a vector and this is going to be vector v. And what does this do? It just gives me a dot product. Gives me a scalar. And the answer without doing um really any um any hand calculations um we get is minus4. It's a scalar. A dot productduct gives you a scalar. It gives you um one one value. I mean what the dotproduct is a b equals essentially ax * bx plus um a y * b y plus um you know a z + uh bz, right? So that's just going to be what? 5 * 2 + you know 4 * -3 + - -3 or -3 * 4 and we should get -14 here as well. Okay, but in the calculator you can do all this and hopefully do it um do it quickly. Okay, so that's where um you have to be um careful with what what you're doing. Um this is not a we don't get a vector from a dot product. we get a scalar. Okay, which is just the the one number. Um, and then what do we do down here? Um, for for cross productds, similar thing. I'll pull up the calculator. We're already it's the same vector. I tried to do that to make it a little bit easier here, but I can come to um second vector again. I go to math um crossroduct, right? So crossroduct of second vector. There's a lot of seconds going on in here. Um but this is comma um second vector and v and what do we get? We get a vector now that's 7 - 26 and - 23. So again what we see there is we see an answer um that that that works. I'll show you how to do this quick, but A cross B. Um, normally what I like to do this is just draw it out kind of that looks something um along these lines. Uh resolution's dying on the stream here. We'll see how that goes. Uh but basically, I'm going to write this out. 5 4 minus 3. You guys still have decent resolution on the your side? Um 2 minus 3 and four. Okay. Um so if we if we go there um what we're going to get is basically we're we're we're looking for um how to solve this and basically the i term we're going to um look at uh we're going to look at these components and we're going to take the determinant um for those those components. So, the determinant for those components is going to be 4 * 4 - 3 * -3. And what's that? That's going to be 16 - 9 is going to be 7. So, so the nice thing about doing some of these by hand is you can cross a couple of them out. Um, real quickly, um, we can come to like the J term. You see, you know, these two look the same, but let's go to the J term to see which which one. So, the J term, um, we're going to go kind of like like this. We're going to take we're going to go this way. So, we're going to take the determinant um that way. And essentially, if we look at this, it's going to be like -3 4 um and 5 2. We're going to take that determinant. And what's this going to be? It's going to be -3 * 2 - 5 * 4, which is also going to give us - 26. And we could do K as well. And that would just be the determinant of 5 2, you know, 4 minus 3 here. Um, but that would get us out to our the same answer as we got on the uh so as as we got here. So, um, so you're saying this is wrong? Uh-oh. So, hopefully the calculator is not wrong, too. Um, you know, I I know I I make mistakes at times, but um hopefully I I didn't screw something up. So, if there is something wrong, let me know. But, um, I think we entered it into the calculator, right? I think this is this is going okay. But um let me know. I don't know. Tell me what you guys think. Um but but if it is, just drop some comments and uh we'll uh we'll go from there. Okay. So, so let's keep going. Um so let's keep going. And what we'll do here is we're going to do vector summation. We're almost done. I mean like um if we add these two vectors um I can do same type of thing. I can I can add them in the calculator but with adding them it's it's pretty basic here. Um I can add you know a + b and I'm just going to add the terms. So 5 + 2 is 7 i you know 4 minus 3 is um plus j minus 3 + 4. I lost the k there. Is that what you're saying? Oh man, you're so right. I lost the k on these vectors. So there's my K hat. Um, so thank you, thank you, thank you, thank you. Um, I will fix that. I totally lost that and nobody else saw it. Um, including me. Um, but yeah, so there's the K, there's the K, there's the K. And then we we keep working and it makes all the difference in the world. So thank you. So minus 3 + 4 is going to be plus K. Okay, so as we go through this, um, what we're going to get here is B looks so right, but it's just so wrong. Um because what we've missed here was the unit vector. Oh man. So what's the unit vector? Well, if this is equal to some vector C, right? Here's vector A, here's vector B, the unit vector is going to be what? Um the unit vector is going to be um essentially C over the magnitude of C. So we have to go and find the magnitude of C. The magnitude of C is just going to equal um the square of 7^2 + 1^2 + 1^2 which is going to I think work out to be what? 7^2 + 1 2 + 1 2 um and if you take the square root of that um we're going to get the square root of I think that's 51 um is like 7.14. Okay. So this is in the reference manual. Yes. Um but basically we get 7 over 7.14 I + 1 over 7.14 J + 1 over 7.14 K. And I believe the right answer is somebody can check me on this um but I believe the right answer on this one is going to be the last one here. um d when we take 7 divided by 7.14 um we get 0.98. So yes, this answer here looks so right. Um but you just have to be careful with what um what the question's asking. Okay. And lastly here, I think I got my IG's and K's all right. Um so um hopefully didn't screw too many people up. But um where are we going with this one? What we're going to do here is we're going to take a look at A and B and we are going to say well what this is kind of a little bit more of a conceptual problem and this is where it helps to kind of have a little bit of an idea of what this means. So if A and B are perpendicular what does that mean? That means um what the dotproduct is zero. So if a so this is just something to note and these these are identities you'll find in the reference handbook. So if we come to um vectors in the reference handbook, did I miss them? Matrices vectors, right? Um this kind of does this tell us here? I don't know that it does. Um for some reason I thought that we we saw this in here, but maybe not. Okay, so here it is. If a b equals z, then either a or b is zero or a is perpendicular to b. If a cross b is zero, then a equals 0. b equals 0. A is parallel to B. Well, um, if we come here, the these are two-dimensional vectors. They're not that crazy. We can actually plot these, right? So, if I take a coordinate plane here and plot these, um, right? This isn't too crazy. I could come over two, up four, um, and this is like vector A. I could come minus one and up three, and this is, you know, vector B. So, now the question is, are they perpendicular? Well, no. Are they parallel? No. Are they um you know, this would mean a dob equals z. Um this would mean a cross b equals zero. So I'm just throwing these out there. Um a is the same length. This would mean the magnitude of a equals the magnitude of b. Um which I mean we could calculate that out to check. So let's check. The magnitude of a equals what? It equals the square of 2^2 + 4^2. The magnitude of b equals what? The square root of um you know 1 2 + 3 2. So what do we get? Um 2 ^2 + 4^2. Oh, let's just put on the let's put on the calculator here. Um so we get what the square t of 2^2 + 4^ 2ar uh sorry squared two of five um or 4.47 47. And if we do the same thing here, square of what? 1 + 9, that's the square root of 10. And if I hit my my um uh my decimal point, my decimal converter here, I get 3.16. So we know that these aren't the same length. Okay. And we can prove that a a is at an angle of 45 to b by using the definition of a a dot productduct really. So a um actually let me come back here. Um what we're going to say is a dob equals the magnitude of a * the magnitude of b * the cosine of theta. So if we if we look in the reference manual um we see that identity in here. I believe here's our dot productduct identity right a magnitude a magnitude of b cosine theta equals you know a dob. So that's going to help us a little bit here. But um we could we could dot these two um what do we get? We get um if we dot these two 2 * -1 um + 4 * 3 has to equal essentially 4.47 * 3.16 cossine theta. So could you have stopped at earlier? Yes, you could have stopped earlier because um the three other ones didn't work. Um but I just wanted to show you this kind of for completeness because what do we get? We two times um two * minus one is is minus2 uh so on this side we get 10 um equals essentially 4.47 * 3.16 cossine theta. If we get cossine theta all by itself um I'm going to turn my calculator off here. But what we get is we get cossine theta equals essentially if you do this math out it actually comes out to 22 over2 and that means theta is equal to 45°. Okay so kind of a long process here but I wanted to show you I wanted to talk use this to kind of talk about some of these properties um but also show you how to apply the dot productduct in um you know with the magnitudes here. So, um, something to consider, something to to think about here. All right, we're we're getting getting down here. Um, sample six concrete cylinders. Again, you can you can put these in. You can you can um you can add these up all by yourself. I mean, you could come up and you can, you know, take um you could find the mean, which is going to be your Xbar. You can find your standard deviation, um, which is going to be your SX. And you know mean is just um the sum of x over you know over the number of of of samples and the the standard deviation is going to be the sigma of x - xar all over you know n minus one. So th those formulas are in the reference handbook. You can go look them up. But this one just screams um don't waste your time. Go into the calculator. Um use how to use a table. Um actually use a data table here. So we can go into data. Um we can enter in some values. So I'm just going to enter these in. 3900 what? 4150 4450 4350. I missed 4275 and uh 4250. So that's all six terms. And then if I go to second um second data, I can get one variable statistics. Um data list one frequency one is fine. And I can hit enter a couple times, hit calculate, calculate, and we get X Xar. So if you know that Xar is the mean, you're good. You get the 42, you know, 42229, and then it's a question of what's SX. Is it 190 or 173? And in here, the standard deviation, if you know what that is in your calculator, um you're going to know that the right answer um is going to use Sx, which is the sample standard deviation, which is typically what we're going to use. Okay. So, another one of those ones where if you know how to use your calculator, it's going to going to be helpful. All right, so let's keep going. Just a couple questions left and then we'll call it a night here. But, um, question 18. Concrete mix produces normally distributed concrete with a mean strength of 4,400 PSI, standard deviation of 200 PSI. So, typically when you when you create concrete, you're going to have some, you know, uniformly distributed curve. you're going to have some um you know mean value here and you're going to have some standard deviation. And what you want with the concrete is you want it so that you you you you set the target mean high enough above so that you know 99% of the time you're going to be in the green zone. There might be one break out of 100 that gets low, okay, but not crazy low. It's going to get just a little bit low possibly, okay? But you want to set that target mean high. So the question is um what's the you know where is this? How does this work? And this again screams um this screams it doesn't just scream it tells us I'm not going to put that in red. I'm going to put this in yellow. Um but that that screams a normally distributed concrete. So what we can do is this this brings us back into kind of the statistics uh section of the reference handbook. So, if I just go in and type in normal distribution, right? Um, this is going to help me to find what I'm looking for. And it's going to help me. I mean, I can come down here. I can look all sorts of things up. Um, but really where this is going to go is it's gonna bring me back down to eventually um if I keep coming here um it'll bring me to like a Z test and and this is going to um this is going to give me some statistics here. I'm going to be able to find what I'm looking for and I'm going to be able to come down one more here and take a look. So essentially what I'm trying to do here is I'm trying to use one of these charts um to make uh what I'm looking for uh possible to find. Okay. So um let me turn my calculator off. Sorry about that. Um but basically yeah, as we come in here, I searched normal distribution and I'm going to be looking kind of at this one, right? We we looked back at our our distribution here. It's the opposite direction, but you know, we're just going to look at at this and we're going to be able to find it. So what we want to do is we want to look and say okay our Z variable is essentially going to equal mu - x um divided by um sigma minus this um I'm sorry sigma divided by the square root of n. Okay. So so here um you know if we come back uh to our our our reference handbook right this is kind of what we're looking at here. It's like our x bar minus mu. Um, and you can, you know, I've written it slightly differently, but um your mean minus your um uh I'm sorry, this the the hypothesized mean over the sample mean. Um let's just come back here and plug in and see where we get. So so basically what we're going to have here is this is we're going to have the 4,400 minus the value that we actually got 4,000 divided by our sigma, our standard deviation, our population. um the uh you know over the square root of 1. So what do we get? We essentially end up with 400 over 200 or 2. So that's our Z variable. What we want to do next is we want to come back to our tables here and say if we come down to a Z variable of two. Okay, that variable 2 um tells us that 97.7% of the time 9772 97.7% of the time we're going to be, you know, on this side of the curve, which is essentially what we're looking for. Um, we're looking for all that area of the curve. And if we look it up, um, we're going to we're going to find that here. Or I mean, we could do it this way as well. And we could do one minus this 0 or 0228. So these kind of go hand in hand. Um, but essentially, um, what that means is 98% of the time, okay, 98% of the time, um, we're going to be good. Okay. Um, hopefully that makes a little bit of sense. Um, if you have questions, pop them in the box. But, but this is kind of using that that Z table um to understand how to uh to find percentages of normal distributions. There are calculators online that you can do this with. But um hopefully this kind of helps as well. Okay, let's keep coming. Two more questions then we will call it a night and uh come back next week for some statics which is one of my favorites. Um but what we have here is we have points below can can be modeled using a best fit linear regression model. So what this means is you know again we have like an xy plane. We got some points here. Um, you know, I I'm not plotting those points exactly, but we can put a straight line through those. And basically, we're going to get some model that says kind of y equals um ax plus b, right? And what we want to do is find that. We can go through and find that by hand. Um, but why would you? I mean, that's what Excel is for. Okay? That's what your calculator is for. So, um, I'm going to bring up my calculator again, and I'm going to come back in here, and I'm going to go to, um, my data. I'm actually kind of come back here, uh, and, um, go to data. I'm sorry, second data. Um, and where am I? Um, actually, no, I'm sorry. I'm just going to go to data. I'm going to clear data. Um, where am I going? Sorry. Oh, there we go. Data. Clear list one. Sorry. Okay. So, if I hit data twice, I can clear list one and I can start over. So, now I'm just going to put in my X variables. Two, um, enter. Four, enter, five, enter. Um, eight, enter. I'm going to come up to list two. Um, I'm going to come in and I'm going to say two, enter, um, five, enter, you know, 12, enter, and 16, enter. Okay, so there are ways to do linear regression, but but why why um what we can do now is we can come up with our one variable statistics. Um actually not our one variable statistics, I'm sorry. Um now if I go to second quit um I can come to actually second quit. Let me get out of here. I can go to second stat, right? Statistics. Um where am I missing it? Um so oh yeah, sorry I just have to come down further. I can do linear regression ax plus b. So if I go to my second stat or second data, I can do my linear regression. And this tells me list one, list two is list one's my x, list two is my my um y data. I can keep going down, hit calculate, and I get a is 2.44 and b is um minus 2.84. So I get this equation of y equals turn the calculator off here, but y equals a was 2.44. 44 x um minus 2.84. So now we have a qu an equation um we can plug in a value of 10 to x. y = 2.44 * 10 - 2.84 and y = what's this? 24.4 uh minus 2.84. And if we do that on the calculator, 28 or 24.4 minus um 2.84 is going to give us about um well actually it gives us exactly um 21.56. This one's kind of in between here. Um but what we're going to get is um did I did I type something in wrong? Let me just double check here. Are you guys getting the same thing? I'm just double checking because in my notes I have a little bit different, but let's let's double check here. 2.44 minus 2.84. Um 2.44 * 10 - 2.84. I just want to make sure I give the right answer here. Um 21.56. That's what I'm getting here. Uh honestly had a slightly different answer in my notes, but I I think this is going to be um the correct answer at this point, unless you guys got something else. Um, okay. So, that's that's what we're going to go with. I'll double check it in Excel. Um, but, you know, it is it's all good. So, I mean, we could double check it by hand and do, but we'd have to find the mean. We'd have to subtract variances from the mean. And it's just this is one of those questions you don't want to waste time on doing that in the test. Honestly, and 99% of the time you're going to do linear regression. You're going to put all your points in Excel and do it in Excel. Um, actually, I have that Excel spreadsheet. I think I already did. So, um, where is it? Let me pull it up. Why not? Um, this is something where it's like it's already there. Let's just take a look at it. Um, so did I put the points in correctly? Uh, 2458 251216 um 2.44xus 2.84 and I got different numbers in here. So 2.44 um minus um minus uh 2.84 84 and um we get that value 21.6. Okay, good. So I'm I'm happy now. Um and uh it works. Okay, so linear regression and hopefully you got that. So no other questions. We're just going to keep going here. Statistics. Oh man, binomial distributions. And binomial distributions are um I I they seem like they should be so simple and sometimes I just feel like they're more complicated than they need to be. But binomial distributions are one of those things that I think will probably show up at some point. Maybe it won't. I don't know. Maybe if you've taken the test before, you can let me know what you think. But binomial distributions are kind of the standard distribution. It's like, okay, I've rolled a dice so many times. I in this case I have three marbles versus two marbles. And if I pick, you know, if I pick, how many of the times am I gonna get? This question basically says I have um I have five marbles in a bag. Three of them are blue. Um two of them are red. And if I pick five times, so I pick one, look at it, put it back. Pick another one, look at it, put it back, pick another one. You know, so I do that five times. The probability of obtaining at least four blue marbles is most nearly. So, so basically what I'm looking at, I I I'm trying to find what's the probability that I get four marbles that are blue. Um that are blue, you know, four blue marbles. I probably should have written it that way. Um I'm going to write it that way. So, four um blue marbles or five blue marbles. Okay, so that's the probability that we're trying to find. We know that the probability of getting a blue marble is three out of five. The probability of getting what's left a red marble is two out of five. So it's it's not just a matter of figuring out all the different uh the different like scenarios and figuring out well okay how many of them are blue and how many of them are red. I mean you could you could take a look at that but you get the wrong answer. So what we have to do is we kind of have to use a binomial distribution. The binomial distribution kind of says well the probability of some value of x is going to equal n factorial / x factorial * n - x um factorial * p of to x and q to the n minus x. I mean that's wow that's uh that's a lot. Okay, you're probably copying that down. Um, I'm just going to skip over here to the reference handbook for a second because I want to show you where that that shows up in the reference handbook. Um, this is, you know, we have binomials. Actually, let me just search it here because, um, I'd like to bring it up here. Um, this is cumulative binomial. I'm going to come back to that in a minute. Uh, honestly, this is the easiest way to do it. Uh, but if I come back to I think it's page 72. Let me just double check here. Is that right? No, it's not. Um I I was looking this up earlier, but um let's take a look. I'm just going to keep going to binomials. Here we have this binomial equation that I I kind of showed up or showed you. And we can go through and solve that binomial equation. But like I said, the easiest way to do this is going to sort of use this chart, right? This chart tells us a binomial distribution. Page 66 of the is this is this the other place? Um okay, here it is. Yeah. So, I just didn't search long enough. But similar equa, the same equation I think I wrote down here. Sometimes trying to figure out how to use this equation on a test is going to be a pain. Um, but this is where if I come back to I think page 80. Um, this is where the binomial probabilities is going to make might make life kind of a little bit quicker. So you can go through and you can, you know, plug and chug this equation a couple times and it'll work. It'll get you a value. Um, you can also, if you know that binomial distributions, what a binomial distribution is, um, you can use this table. And the thing is, I didn't tell you, I didn't use the word the key word, there's no binomial, um, in this problem. Binomial. There's no binomial in this problem. But you have to recognize it's kind of this either or, right? The binomial is kind of this either or. Either you roll a um an odd number or an even number. You flip a head, you flip a coin, you pick a red marble, you pick a blue marble, and there's a probability of success and a probability of failure. So that that's a binomial distribution. If you can recognize those um you can come back and actually make your life easier by using something like this. So what does this say? This is the probability that X is going to be less than or equal to X. So let's come back here and take a look for a second. We said, but we want four or more blue marbles, right? So, we want four or five blue marbles. That means we need one, two, or um actually how many red marbles does that mean? So, that means if if if ah there's too much to think about. Um we want to look at kind of reverses here a little bit. Instead of saying, well, what's probability four or more? We get to look at the probability of three or less and subtract it from one. Does that make sense? So if instead of getting four or more, our table gives us x or less. So let's find three or less and subtract it from one. So essentially what we're going to say is one minus the probability of three or fewer um blue. Does that make sense? Um so so that's the way we're going to use this. So we've already kind of looked at it a little bit, but what we're going to say here is we're going to say, well, we know the probability of blue is 6. We know we have five trials. That's our n. and we're looking at three um three oh you know three blues. So when I come over here I get 66 I get 6663. Okay the probability to get three or fewer blues is 666. So we're going to have 1 minus 0.6630 and that's going to be giving us our essentially 0 you know 34 something%. um 1 minus 6630 is going to be like 337. So it's using a table. I mean we could go through and um let me write this out just we've got another minute I we've got all night you know but um let's let me write this out. We could also say this is the table you know we just did the table way. We could also do the the math way. Um right the math way is going to be say well what's the probability that you get four blues? Right. the the way that would write out is is like this. We're going to use this. We're going to just solve this equation here for a second. So, let's solve this equation. The probability that we get four blues is going to say, well, we have five trials. So, five factorial divided by the probability or the the success, which is going to be four factorial times what? five trials minus uh minus um four successes factorial times the probability of success which is 0.6 to the 4. Okay. And then times probability of failure 0.4 to the 1. So when we do that out I think I get like almost 26%. And then if we do p5 we it's something similar. the the trials is on top. The success 5 factorial um five minus 5 factorial. And you might be looking at 5 - 5 that's zero factorial. Uh-oh. Did we do something wrong? Um no, because 0 factorial is what? It's it's it's one. Okay. 0 factorial is one. So this is 0.6 um to the five and 0.4 to the 0. Again, 0.4 to 0 is one. So, this is going to give us I lost my sheet um of notes, but this is going to give us about about 8%. So, when we add those together, we get 34%. So, you can check the math out here. Um when I did it, it it did work out, but you can do um the table way, you can do the math way, but if you can recognize a binomial distribution, that's one of the most basic distributions. you've been learning about it uh with rolling dice, with um you know, flipping cards in a deck, that type of thing. But it's kind of an either or. Either you get this or that. Um and if you can recognize that distribution and use the right graph, you know, that's one of the most basic, I'd say, fundamental types of statistics that you might get into on the FE. So, hey, um this is, you know, a long a long long long um probably one of the longer sessions that we'll do, okay? In the future, they'll probably be broken up into slightly smaller chunks. We probably won't go past two hours, but um I I appreciate you hanging on here. Um and I hope this is helpful to you. I mean, like honestly, I appreciate all the comments and the feedback as you've been going. It's neat to see people from all over the world um you know, from California to Calgary. Um I mean, coming in and even here in upstate New York, um to to work on these these problems together. So, I hope this helps whether you're watching live or um in the future, you know, you come back to it. But I will repost these problems with a couple of those corrections that um I noticed in there like I missed a couple of the K's and uh um some you know those those duplicate values. So when I repost things I might you know add a version here and tell you um what changes but um if you have any questions um definitely feel feel free to reach out to me and uh I will do my best to uh to get you some answers. Um but until next time hey um thanks so much. Next week we'll be on statics and strength. Actually, next week we'll be on statics. But until then, keep working hard, moving onward and upward. And for those of you that want to joke before you leave, um I I spent a summer in Guatemala doing some engineering work. It was really fun. Um in in in in Guatemala, um there's a there's a Mayan population. The Mayans were actually the Mayans were actually responsible for creating the number zero in like the third like century BC. Um, so before the 3rd century, the number zero and math didn't exist. Um, but you know what? Um, I said to them, "Thanks for nothing." Get it? It's too late. Hey, um, keep working hard. I'm going on. See you next time.

Transcript for:FE Exam Math Review

Transcript for:
FE Exam Math Review