- [Facilitator] Let's start today with a problem. "A particle's position on a line is given by the following function, S of T is T to some number D." We don't know that number yet. "Minus 11, T cubed plus 27 T squared." Now D, all we know is it's positive, and, "is some parameter to be chosen." Now we have the following information, "At time T equals one, the acceleration of the particle is zero." So that's what we know, that's what we're given. What do we wanna do? "Determine another time when the acceleration of the particle is zero. And also determine all times T, when the particle is at rest." So, what does it mean to be at rest? Says not moving left or right. Well, okay, we'll think about what that means. So first off, this tells us information about the acceleration. We're given information about the function, which is the position. How do we go from position to acceleration? - [Students] (indistinct) - [Facilitator] Yeah, two derivatives are the second derivative. So we're gonna start out by saying, let's just take a derivative. S prime of T, well, I have T to a number, even though I don't know what that number is, it's still T to a number. So the number comes down and it becomes one less number in the exponent. How about the next term? What will be the dirt of minus 11 T cubed? 33 T squared and, whoops, 33 T squared, we subtract. Root of 27 T squared? 54 T, good, good, nice. And second Derivative S double prime of T. Well, the D is a constant, so that stays. Now how do you take the root of T to the power D minus one? The D minus one comes down, so it's derivative of something to a power. The power comes down, T to one less power. Derivative of 33 T squared? - [Students] 66T. - [Facilitator] 66T, and derivative 54 T? - [Students] 54. - [Facilitator] 54, so there's our second derivative. Now, what we wanna know is, is when the acceleration is zero. But the thing is we don't know what D is and that really seriously affects the answer. So we have to figure that out. So to help us figure this out, what we'll say, we'll say, well, let's use our information. "At times T equals one, the acceleration is zero." So, we'll have that one, sorry, zero is what we get when we plug in S double prime at one. So plug in one into the second derivative we get zero. So, if we plug in one, one to the power D minus two, what is one to the power D minus two? It's one, yep. One to any number is one. All right, then we have minus 66 times one plus 54. Now of course this can be simplified. D squared minus D, negative 66 plus 54 together gives us, negative 12. So we wanna know when does that equal zero? Well, what should we do? - [Students] Factor it. - [Facilitator] Factor, if it does factor, does it factor? How does it factor? - [Students] D minus four D plus three. - [Facilitator] D minus four D plus three. We're looking for two things that multiply together, give negative 12. The fact that it's negative means we're gonna be different signs. And when we combine 'em together, we get negative one. So negative four times three gives us the negative 12. And you'll see that the minus four D and the plus three D combine together to give us just a minus D. So that says, well D could be two possibilities. D could be four, or D could be negative three, but they're not really two possibilities because what else do we know? - [Students] D is greater than zero. - [Facilitator] Yeah, D is greater than zero. So we can knock that possibility out. So I say, aha, now the good news, we really have what S prime of T is. So I can replace everywhere where I see a D by four. So it's really four T cubed minus 33 T squared plus 54 T. And for S double prime of T we'll have four times three, which is gonna give us 12 T squared minus 66 T plus 54. So now, we've determined D using our information. So we're ready to start answering the questions. So, we want sometime when the acceleration is zero. So, we set the derivative second derivative equal to zero. Zero equals 12 T squared minus 66 T plus 54. We say great, we just have to factor that. And we're probably thinking, ah, great. Now we have to factor that? So, well actually, it's not so bad. What do they have in common? - [Students] Six. - [Facilitator] Six, so pull out what they have in common. Two T squared minus 11 T plus nine. But there's something else, we already know beyond a shadow of a doubt, to the core of our very soul, that this factors. And not only do we know that it factors, we know what one of the factors has to be. What's one of the factors and why? - [Students] T minus one. - [Facilitator] T minus one has to be a factor because we know when T equals one, the acceleration is already zero, which means that one is a root. So, we don't even have to think hard about one of the factors, which is good. That means we only have to think a little bit about the other one. If I need to get it to a two T squared and I have a T, what do I have to put here? - [Students] Two T. - [Facilitator] Two T. And then if I have to get to a plus nine, and here's a minus one, what has to be there? - [Students] Minus nine. - [Facilitator] Minus nine. Now of course we can double check, two T squared, then we have minus 9 T minus two T, which is minus 11 T, and then a plus nine. So, we already knew about the time T equals one, the other time's gonna come from here. When does two T equal nine? Well, that says T equals nine halves. So this is when acceleration is gonna be zero at our other time. All right, well, that's something, but the problem said we want more. This problem is greedy, it wants more. It says look, all the times when the particle is not moving, it's at rest. So let's rephrase that in terms of our notation here. In terms of S, or S prime, or S double prime, what is this last thing asking us for? It's talking about S prime and it we wants S prime to be? - [Students] Zero. - [Facilitator] Zero. So this is secret code. So, not moving means that our motion is zero. There's no change happening with our motion. So, all right, so we come up to our motion, four T cubed minus 33 T squared, plus 54 T that's equal to zero. Now that's a cubic that's a little bit harder. It's not terribly hard to get started. What's something we can do? - [Students] Pull the T out. - [Facilitator] Yeah, get the T, pull that out. T times four T squared minus 33 T plus 54 equals zero. Now, what would be amazing? It would be amazing if this factored, right? I'm gonna go on faith, I believe in miracles. All right, so let's think, how could it factor? Well, it could, think about ways to break four apart and four can break apart as four and one or as two and two. And 54, well, there's a couple ways to break 54 apart. One in 54, two in 27, three in 18. There's a lot of ways actually, six and nine, but I think it's gonna be a four T and a T, that's my guess. And let's see, 54, how does 54 factor? I think, so wait, what did you say? - [Students] Minus nine (indistinct). - [Facilitator] Minus nine, where, here? - [Students] Minus nine, (indistinct). - [Facilitator] Here? Minus nine, and you said? - [Students] Minus six. - [Facilitator] Minus six. Let's see, does that work? Four T squared minus 24 T minus nine, oh my gosh, minus 33 T and plus, oh, our belief has paid off. It factored. Now, if you didn't know how to do that, quadratic equation would work. So, and I will tell you that if you're getting nervous, would we put these kind of numbers on a test to factor? No. - [Students] Yes. - [Facilitator] Yes? You haven't had any of my tests yet. You don't know what they're like. So, no, we put the bigger numbers on the on the practice prompts. The funds on the test are fun. Okay, so we're ready to read off our answers. So, what do we get from that point? - [Students] Zero. - [Facilitator] Zero. Okay, how about from this one? - [Students] Nine four. - [Facilitator] Nine fours, 'cause you take the nine across, divide it by four, and how about this one? - [Students] Six. - [Facilitator] Six. So these are the three times when the particles is at rest. So there's three times when the particle just stops and says, "I need a break." At time zero, time nine fours, and at time six. All right, good, nice, fun stuff. All right, well, what's today? Well, we've already mentioned what today is. Today we're gonna add more rules to our list of derivatives. So, what rules are today? And the answer is, the trigonometric functions. So the trigonometric functions are really important and so, we're gonna start there. Now, we need a couple of of facts. We need to to know a few things that we've already talked about, we've already covered. And so what are they? Well, some basic facts about limits, because remember, if we're gonna take a derivative, we need to use a limit. We always use the limit definition of derivative to establish the rules. So, the limit as H goes is zero or sign of H over H is one. And we also showed the limit as H goes to zero of cosine of H minus one divided by H is zero. So these are things that we've already done before, things that we've seen in our past. All right, we're also gonna need some trig identities. These are trig identities that you don't need to have memorized, but we can use. If you want to see proofs of these trig identities. If you look at the trig review video on the website, we actually do have a short proof of these. So, sine of A plus B, that's sine of A times cosine of B, plus sine of B times cosine of A, and cosine of A plus B is cosine A, cosine B, minus sine A, sine B. On a side note, you can get the double angle identities from these. So for example, if A and B are both your angle theta and you get sine theta, cosine theta, plus sine theta, cosine theta, which is your two sine theta, cosine theta. And if here, A and B are theta, you get cosine theta, cosine theta, minus sine theta, sine theta, which is your cosine squared theta minus sine squared theta. So those doubling identities, you should know, you don't need to know these identities here. Now, we're gonna start by doing the sine function. Now what you see on the very bottom here is you see two curves. The top curve is the sine function and it starts at zero, and then it moves in what you could describe as a sinusoidal pattern, which is basically saying it looks like a sine curve, it goes up and down. Now what's on the bottom? Well, what's on the bottom is, we're trying to think about what should the derivative look like. So for instance, at zero, if you were to sketch a tangent line, it would look something roughly like that. And you'd say, okay, well what's, eyeballing what does it look like? What's the slope? - [Students] One. - [Facilitator] About one. So we say, okay, so in the derivative we'd hit about one. And then we'd say, okay, if we move along we'd see, okay, so, there's this part right here. Well, what's the slope of that? - [Students] Zero. - [Facilitator] Zero, So if you were to drop down to the derivative it should be zero. And you can see the slope is plus one, but then the slope is sort of shallowing as you move up. So the derivative is going down. Now, say we move further along, and so we'll go through a variety of colors here. Suppose I'm over here and I look at my slope. Well what is it? - [Students] Negative. - [Facilitator] It's negative, and it's not hard to tell, exactly, but who knows, maybe about negative one-ish, I don't know. But you can see I start a positive one and my slopes are going down. Now, suppose we move over, all right, now we're over here. What's happening, what's that slope? - [Students] Zero. - [Facilitator] Zero. So if you go down to your derivative, you'll see you get zero. So you can kind of see that by thinking about the shape of the curve, if you look at the shape of the curve, you can start to extract the shape of the function. If you want one more, 'cause we're only going out little bits and pieces over here, we're back to our start. So we're seeing that when the sign curve goes through one revolution, it looks like our derivative should start up one-ish, go down to maybe negative one-ish, and head back up to one-ish by the time we end. And then since the sine curve is repeating, this curve, whatever it is, should also be repeating, should repeat the same frequency. So we think to ourselves, huh, have we ever seen a curve that has this sort of behavior? Does that, 'cause remember this is the sine curve. What does this, does this look like anything? - [Students] The cosine. - [Facilitator] The cosine? Oh that would be amazing, wouldn't it? The derivative sine of, like that would be like, whoa. Like it was meant to be, like the math gods have smiled on us. Well the math gods have smiled on us. So the derivative of sine is cosine. All right, so, let's prove it. So we do what we always do when we wanna take the derivative of a function. We say okay, what's the rule? So if I wanna take the derivative of any function, so in our case it's sine of X, I say the definition for the limit, limit as H goes to zero, I plug in X plus H into my function. So sine of X plus H, and then I subtract off sine of X over H. Now we start thinking about what we wanna do. We wanna get to a point where we can use our rules for limits and we're not there yet. So we say, okay, what can we use to rewrite anything? Is there anything that we have that can help us rewrite our expression? - [Students] (indistinct) - [Facilitator] Yeah, we have that sign of A plus B. So, we apply it, so here A is X and B is H. And so we just apply our rule. So it becomes sine of X, cosine of H, plus sign of H, cosine of X, minus sine of X over H. And now what we're going to do is, we're gonna just do a little bit of bookkeeping, move stuff around. But the next step is nothing more than just rearranging. So, there are two terms that have sine of X in them upstairs. I pull that sine of X out, there's a cosine of H minus one, still over H. There's one term that has a cosine of X and that's a sine H over H. And we can see that these two pieces are pieces which are really nice. And as we take the limit H goes to zero, the sine of X and the cosine of X, they don't change. So we just need to understand what happens to those two parts. Well, they go to zero and one, and therefore we end up with cosine of X. All right, well that was nice. So the derivative of a sine is cosine, but you know we should do the flip side. How about the derivative of cosine, what should that be? Wouldn't it be cool if it were like sine? Wouldn't it be like, yeah? That would be really convenient and easy. Well, no, we can't have everything be convenient and easy. So, the derivative of cosine is negative sine. Okay, so what's the proof? The proof is the same. Start by saying we take the definition of the derivative, so we plug in X plus H, we apply the rule for cosine, and the rule for cosine is different. So, it's cosine X, cosine H, minus sine X, sine of H. Now, that's the only step that's different, everything else we do is the exact same. So for example, here we're gonna put, the cosine Xs are gonna come together, and that'll end up giving us a cosine of H minus one over H. The sine of X will have a sine H over H. and then you have that minus sign. As before the cosine H minus one over H goes to zero, the sine H over H goes to one, and so we get our result. Now, if you look at the curves, you can actually check and see that it makes sense that this is the case. So what's happening here is the top curve, now this top curve is cosine of X, the bottom curve is the derivative of cosine of X. And so, what you can check is, you eyeball a couple things. Okay, initially at the start we should be zero. Okay, we are, then if you look over here at the next point over a little bit, so this would be at pi halves, that's when the cosine is at zero. What's your slope? It's going down so it's negative. So that's why you have that sort of, it's not sine of X, it's that negative sine X because your slope is going in a slightly different way. And of course you can keep going. You go over a little bit here, this would be at pi, you see we have a zero slope, here would be at three pi halves. We have a positive slope, and so forth, and so on. All right, so the derivative sine, positive cosine, derivative cosine is negative sine. I wish there was a nice convenient way to remember this. Just practice and you'll get better. In a few weeks, as we get really close to the end of the class in Calc one, we're gonna talk about what are called anti-derivatives, and then we're really gonna throw off the confusion. Or I shouldn't say throw off, we're gonna layer it on. So, be careful, be careful. Okay, so let's just check a couple things, make sure we know how to do it. So we're gonna take some derivatives. So our original function is sine of X. So I wanna find these various pieces. So DY DX stands for the derivative. The derivative of sine? - [Students] Cosign. - [Facilitator] Cosign, all right. So, D two Y, DX squared, what does that mean? - [Students] Second derivative. - [Facilitator] The second derivative, and we're gonna end up with? - [Students] Negative sine. - [Facilitator] Negative sine, because we have it right here. Derivative of cosign is negative sine. So it says take the derivative of the derivative, so we come up to the previous sine. The D three Y DX cubed, which derivative is this? - [Students] Third derivative. - [Facilitator] Third derivative. What does that turn out to be? - [Students] Negative cosine. - [Facilitator] Negative cosine. And the way you get the third derivative is, you go to the second derivative and you say, what's the derivative there? In that case you say, look, the negative is a constant. So you take that constant times the derivative sine. All right, what about D four Y, what will that be? That's the fourth derivative that turns out to be? - [Students] Sine. - [Facilitator] Sine. D five Y, the fifth derivative or DX, the fifth? - [Students] Cosine. - [Facilitator] It's cosine. Now, we could keep going because eventually we need to get down to the 107th derivative. Now, there's not a lot of space here, so I can't imagine in the dot, dot, dot we're gonna be able to put in all the intermittent 101 derivatives that are missing. But, what do we see? Someone mentioned the word pattern, what's the pattern? - [Students] (indistinct). - [Facilitator] Yeah, notice it's going through length four. Sine, cosine, negative sine, negative cosine, sign again. So when you hit that fourth derivative, it's like you've reset. So as an example, the fifth derivative matches with the first derivative. So we could think and say ourselves. Oh, so every four derivatives we are restarting. So let's think about the number 107. Well, what's true about 107? Well, one thing that's true, it's 104 plus three and you're probably thinking, well that's true, it's also 105 plus two Steve. But what's special about 104? - [Students] (indistinct). - [Facilitator] Yeah, see this is a multiple of four. It turns out to be four times 26. So what happens is, we're gonna take a bunch of derivatives, but remember every time we go through a step of four, we're back to where we started. So we're gonna go through 26 cycles, and then three more. So, if we're back where we're started, so the 104th derivative will be here, back to sine, then we just keep going. 105th, 106th, 107th, negative cosine X. So in other words, it's the same as the third derivative. So we like the derivatives of sine, cosine 'cause they repeat themselves. One of the things I should say about sine and cosine is, these are very important functions in mathematics because sine and cosine are what we call periodic functions. They repeat themselves after a certain length of time. And periodic functions are very important because a lot of things that happen are modeled by periodic things. So for example, if you look at the sunrise and the sunset, that's periodic, it's very predictable. Tides coming in and out are periodic, lots of natural behavior happens to be something which falls on a somewhat periodic basis. So these are gonna be important functions, especially when you start getting into modeling real world phenomenon, and of course, differential equations, which is a great class. Well, let's do some more practice. All right, so we wanna take the derivative of sine of two X. Now I know some of you're like, "I know how to take that derivative." (chuckles) But we don't know the really cool way to do it yet. So let's think about how we could do it. What we do know is we know how to take derivative of sine and cosine. So I can say instead of taking the derivative of sine of two X, I'll take derivative of two sine X cosine X because these are equivalent expressions. So the derivative of sine of two X is equal to the derivative of two sine X, cosine of X. All right, the nice thing is, we do know how to take derivative of sine and cosine, and we know how to take derivative of a product of sine and cosine because we have a rule, which is the product rule. So, the product rule, it'll be two times the derivative of sine, which is? - [Students] Cosine. - [Facilitator] Cosine, times cosine. So, I took, the constant just stays as the constant, derivative of the first times the second, plus two. Now I leave the sign alone and then I times it by the derivative of cosine, which is? - [Students] Negative sine. - [Facilitator] Negative sine. All right, now that is the derivative. We could justifiably stop there, but I don't think we get the great aha moment. We're like, oh, I guess, okay. We want that great, oh, isn't that an interesting thing? Okay, let's get to that moment. What do these have in common? - [Students] Two. - [Facilitator] Yeah, they have a two in common. And then if we factor it out, cosine times cosine is cosine squared X. And now, there's that minus, minus sine squared of X. Now cosine squared X plus sine squared X we really like, it's super nice. Cosine squared X minus sine squared X is not too bad either. What does this turn out to be? - [Students] (indistinct). - [Facilitator] It's a double angle, that's cosine two X. So we've got two cosine, two X. So isn't this interesting? See, we get that the derivative of sine of X just by itself is cosine of X, and the derivative of sine of two X is two cosine two X, that's an interesting pattern. It kind of vaguely reminds us, we've actually did previously that we saw, hey look, the derivative of E to the X equals E of the X. And then we said the derivative of E to the two X, do you remember what that was? - [Students] Two E to the two X. - [Facilitator] Two E to the two X, oh, my goodness. Hmm, I suspect there's something going on here. I'm very suspicious, I wonder what it could be. Well, we'll figure that out, I think, in the near future. Find the derivative of X squared sine of X, plus two X cosine of X minus two, times sine of X. So, well, we're gonna have to be patient here. What's the derivative of X squared sine of X? What rule do we need? - [Students] Product rule. - [Facilitator] Product rule, okay, so let's just start. I like to do derivative of the first, first. So derivative of X squared? - [Students] Two X. - [Facilitator] Two X, and the sine X, what do we do? - [Students] We leave it alone. - [Facilitator] We leave the sine of X alone. It will get its chance, don't worry, every term gets its chance. Next? - [Students] X squared. - [Facilitator] X squared times? - [Students] Cosine. - [Facilitator] The derivative of sine, which is cosine. Okay, now we go to our second term. What rule do you need here? - [Students] Product. - [Facilitator] More product rule, that's cool, lots of product. We're gonna be very productive today. Okay, derivative of two X? - [Students] Two. - [Facilitator] Two. And the cosine X we leave alone, plus two X times the derivative of cosine X, which is? - [Students] Negative sine X. - [Facilitator] Negative sine X, all right? And the root of negative two, sine X. - [Students] Negative two. - [Facilitator] Negative two is a constant, so we just start by writing down negative two. Derivative of sine? - [Students] Cosine. - [Facilitator] Is cosine. All right, now, again, we could say we're done, but we suspect that something nice might be happening here. We've lived a good life, something beautiful might be about to happen. Do we see anything happening here? - [Students] Two X, sine of X. - [Facilitator] Okay, someone said the two X, sine of Xs, they actually cancel 'cause there's a minus attached to this one but not to that one. So those cancel, anything else? - [Students] The two cosine X. - [Facilitator] The two cosine X. Again, there's a minus attached to this one but not to the other one. And of course, the X squared cosine X is X squared cosine X. Well, that's remarkable. So we had something which is a very large expression. When we took its derivative, it collapsed down. So that's fun. Now, something to look forward to next semester when you're doing calculus part two, the return of everything. What you see in calculus part two is, instead of starting here and going there, you start here, and go there. So it's getting to go backwards, that's the fun stuff. Good times, good times. All right, well, we've done some trigonometric functions, are there any left? - [Students] (indistinct). - [Facilitator] Oh yeah, there's more. You know that's the nice thing about trigonometry. There's always another identity. You know, there's always another trig function. All right, okay, so, for the other trigonometric functions, we can use our fun facts. Okay, so we don't wanna keep going back to limits. Our goal is, use limits as much as possible. So we're gonna be in that mindset. I don't want to use limits if I can avoid it. The good news is, once you have sine and cosine, everything else as far as trigonometry is concerned is some combination of sine, cosine. So let's start with tangent, how can I write tangent in terms of sine and cosine? It's sine over cosine. So, if I wanna take the derivative of tangent, I'm gonna take the dirt of sine over cosine. What rule applies? - [Students] Quotient. - [Facilitator] Quotient rule. So, you apply the quotient rule, you take the bottom, which is cosine, times the derivative at the top, which is cosine, minus the top, which is sine, times the derivative at the bottom, which is negative sine, all over the bottom term squared. Now upstairs you'll see cosine squared X plus sine squared X, what does that become? - [Students] One. - [Facilitator] Becomes one. So you get one over cosine squared X, which is known as secant squared X. All right, now, since we're talking about tangent, let's do the cotangent. How do you find cotangent? - [Students] Cosine over sine. - [Facilitator] It's cosine over sine. For code tangent, very similar process. The only thing that happens is you end up getting a negative in the exponent, and downstairs it's a sine squared. So it's negative cosecant squared. Now, because we greatly prefer things without negatives, I will tell you that one of these that you should memorize is the tangent one. The cotangent, if you have a few extra brain cells, and you're young, you have a couple of extra brain cells left, you can memorize. But, if you must memorize only one, definitely memorize the tangent one. Okay, are we done with our trig functions? - [Students] No. - [Facilitator] No, can't be excited. What's a good one, what's another one? - [Students] Secant. - [Facilitator] Secant, okay, how do you write secant? - [Students] (indistinct). - [Facilitator] One over our cosine. Okay, do we have a rule for one of over cosine? Yeah, you can use the quotient rule again, or you can say the reciprocal rule, which is really a sort of a special case. So the quotient rule says, you take negative the derivative downstairs. So derivative of cosine is negative sine. So it's negative of negative sine, that's your numerator. Then you square your denominator. Now what you end up with this sine X over cosine squared X. We wanna rewrite that, so how can we do that? So, someone said, hey, there's like a tangent and a secant floating around. What you can do is you can say, well look, I have a single sine and I have two cosine. So I'm gonna pair up one sine with one cosine, that'll be sine over cosine which is tangent. Then I'm left with a one over cosine which would be a secant, and therefore the derivative of a secant is secant times tangent. Now the last one, you remember what the last trig function will be? Cosecant, it's same process and lo and behold you get minus cosecant cotangent. So those are your various trigonometric functions, or at least the other ones. Now you'll notice I wrote the tangent and secant down here and I put a box around them, because those are the ones you wanna remember. These are the ones that are gonna be used more often. Now, it's not because tangent is somehow an inherently better function. I honestly truly believe that it's primarily because, the derivative of tangent does not have a negative coming in where the derivative of cotangent does. That's the big difference. That's the big difference, all right? Hmm, okay, how are we doing on time? We probably have plenty of time. Plenty of time, plenty of time. All right, let's see, so, let's do some examples. Probably gonna regret this example but we'll see how we go. Find, and what does this notation here mean? - [Students] (indistinct). - [Facilitator] Third derivative with respect to theta, we're thinking of theta as our variable of tangent theta. All right, so we're taking the third derivative of the tangent. Well, if we're gonna take the third derivative, we should probably start by taking just plain old derivative. So, here I'll just call, F of theta be tangent of theta. So I'm after the third derivative. Do we know the first derivative? Derivative of tangent? - [Students] Secant squared. - [Facilitator] Secant squared of theta, 'cause we're gonna keep, our variable here is theta. Now, how do we take the second derivative? Well, what's another way to think of secant squared theta? - [Students] Secant times secant. - [Facilitator] Secant time secant. So, if we take it this way, well, derivative of secant is secant times tangent or tangent times secant, some order, times secant, plus secant times derivative of secant, which is secant times tangent. Which becomes, well, these are the exact same, so there's two of them. Each term has two secants, so that's a secant squared and then there's a tangent. So two secant squared theta, tangent theta. All right, now, we want to go do one more, why? 'Cause what derivative are we after? - [Students] The third. - [Facilitator] The third derivative, triple prime. All right, so, product rule. Two times derivative of secant squared. Do we know the derivative of secant squared? - [Students] (indistinct) - [Facilitator] Yeah, it's this thing. That's the derivative of secant squared, we just did it. I'm not redoing it, I'm not crazy. I recycle, I'm a green mathematician. Okay, so two times zero of secant squared times tangent, plus two secant squared times the derivative of tangent, derivative tangent? - [Students] (indistinct) secant squared. - [Facilitator] Is secant squared. So what you end up with, is you end up with four secant squared theta, tangent squared theta, plus two secant to the fourth theta. And there you go. All right, that wasn't so bad. Now let's do the other one. So, for the other one, we'll think of our function here as being secant theta, you don't have to go quite so far. What's the first derivative turn out to be? - [Students] Secant times tangent. - [Facilitator] Secant times tangent. Okay, so the second derivative, what rule do we need? - [Students] Product rule. - [Facilitator] Product rule 'cause we have two things multiplying. Derivative of secant? So yeah, secant times tangent. Then we're gonna multiply it by tangent. So derivative of the first times the second, plus secant times the derivative of tangent which is secant squared. So we end up with secant theta, tangent squared theta, plus secant cube theta. In general, it turns out that if you're taking derivatives of secants and tangents, you always end up with some combinations of secants and tangents. All right, find the tangent line to the curve Y equals six times tangent of X, minus root two secant of X, at X equals pi fours. All right this is a good question because it says, find a tangent line of a function. We can always do that, even if our function has trigonometric terms involved. So if I need a tangent line, what information do I need? Slope and? - [Students] Point. - [Facilitator] Point, remember we need both. We don't have a full point yet 'cause this is an X-coordinate, so I need a Y-coordinate. So let's get our point. So our point will be at pi fours, comma, and then we'll plug pi fours in. So six times tangent of pi fours, minus root two, times secant of pi fours. And you're probably thinking great, let's leave it like that. But I think we can go further. For example, tangent of pi fours, what does that turn out to be? - [Students] One. - [Facilitator] That turns out to be one. Secant of pi fours, do you know that one? - [Students] (indistinct). - [Facilitator] So, that one may not be quite as familiar, but let's think about secant, what is secant related to? - [Students] (indistinct). - [Facilitator] It's cosine, it's one over cosine. How about cosine of pi fours, you know that? Root two over two, you can also write, cosine of pi fours as one divided by square to two. So if you flip that, secant of pi fours is square root of two. - [Students] It's six minus two. - [Facilitator] So, pi four's six minus two. Pi fours comma four. That's a reasonable point. Now the slope, what will we need? - [Students] The derivative. - [Facilitator] We'll need the derivative. So let's take our derivative, Y prime, derivative of six times tangent of X? - [Students] Six secant squared X. - [Facilitator] Six secant squared X. Okay, minus root two, derivative of secant? - [Students] Secant times tangent. - [Facilitator] Secant time tangent. I should keep saying secant of X, tangent of X. All right, now is this our slope? - [Students] No. - [Facilitator] No, not yet, it's our derivative. Now our derivative won't be a slope until we evaluate. What our derivative is, is it's a function that gives our slope at any given point. But you have to evaluate to actually get the slope. So, we now plug in PI fours. So at pi fours, well, we're gonna get six times secant of pi fours squared, minus square root two, times secant of pi fours, tangent of pi fours. But the good news is, we know these numbers. Secant of pi fours is root two, route two squared will be two, six times two makes the first term, 12. And I have a root two from the secant of pi fours, a one from the tangent of pi fours, root two times root two times one gives two. So 12 minus two gives 10. So, there's our slope, our point pi fours four, our slope 10. So we're now ready to say our line is Y minus our Y- coordinate, which is four, is equal to our slope, 10 times X minus our X-coordinate pi fours. And that is a perfectly fine way to leave our answer. All right, so, I wanna stop after our next thing. So the motion of a spring, and there's something that you'll hear about in physics called Hooke's law. And Hooke's law says that if you have a spring, so, if you're not familiar with what a spring looks like, think about it as a slinky. So, here's a spring, and on the end there's some sort of weight attached to it. And what happens is, if you pull a spring, so if I take the spring and I were to pull it down the spring would try to pull it back up. So in other words, if this is its natural state, if I pull down the spring says, "No, no, go back up." Similarly, if I tried to push up, then the spring would push against and say, "Hey you need to go back down." So if you think about what's happening, you can think about it as the acceleration, which really means a force here. It depends on where you're at. So displacement is just a fancy way to say location. So if you pull down, it's trying to pull you back up. If you push up, it's trying to pull you back down. So the nice thing is that, a spring really wants to be balanced, it wants to get into an equilibrium. Well, so how do you actually go from an equation and say, "Hey, I'm looking for something where the second derivative is somehow some constant of the first, sorry, of the function itself." So S double prime equals minus KS. The answer is, you take differential equations 'cause they love these kinds of things. Like, you're gonna go crazy about this in week two of differential equations, you're gonna have lots of fun. Now for us, it's just suffice us to say you can solve it. And it turns out that the type of function that solves it are the sine function or the cosine function. And in general, you get that the solution looks like, A times sine of root K, times T, plus B. Now, it's very easy to verify if you take the derivative, we don't have the tools yet to to go all the way through, but it turns out to be true. Now, if you think about what's happening here, essentially what it says is that the second derivative is the opposite sign of location. So, if you come here, this is a really crowded picture, but here on the blue, that's the sine function. So you can think of it as you have a spring, and this is how the spring is behaving. It's going up and down, up and down, as you march forward in time. In the red, that's the first (indistinct), so that tells you velocity, but at the very bottom is the green, and that tells you acceleration. So, if you look at how the position relates to acceleration, you can see it has this property that they're exactly opposite, that they march opposite to each other, so that it has this beautiful property. So somehow this relationship that says, lo and behold, I'm exactly opposite to my acceleration, things happen. Well, something else can happen, which is you can have what's called the damped motion. So, there's this theory that says if I have a spring and I release it, it'll just keep going up and down, and up and down, and up and down. Because of course in physics you always start to say, look, there's nothing else acting on it. And it's a perfect world, there's no friction, there's no gravity, make lots of other assumptions. But oftentimes there's something called the damped motion. And what happens with the damped motion is that something is causing it to slow down. So what you have is, over time you get smaller and smaller variations, and things disappear. So, I claim this as an example of something which would have a damped motion. Now the sine of T, you should think of it as saying, that's causing it to go up and down. So sine of T looks roughly like this. When I multiply sine of T by something like, either negative T, what's happening is, I'm applying this E to the negative T. So I draw the E to the negative T, and I draw it on both sides. And what the sine does, is it goes in between these. And so you can see what happens is it's still going up and down, but it starts going up and down at lower and lower heights. So what happens to the acceleration? Well, so that's says take the derivatives. So in the interest of time, I'm gonna try to go through these derivative really quickly, but there's nothing very surprising here. So, you take the derivative by the product rule, derivative of first times the second, plus the first times the derivative of the second. And if you take the second derivative, well, you'll get E to the minus T, sine of T, then you're gonna get minus E to the minus T times cosine of T. Then here you're gonna get a plus, actually, a minus E to the minus T, times cosine T, and a minus E to minus T times sine of T. And these two cancel and you end up with a minus two E to the minus T cosine T. The important thing is, look at this, S double prime versus S, they're not exactly opposite each other, they're out of sync. And that's what's causing the the spring to slow down. So if you were to graph something, here's an example, you can see that the motion is the blue, the acceleration is the green, they're not in sync and that causes it to slow down. You'll do this a lot more when you take differential equations. All right, see you next time.