 Hi, I'm Zor. Welcome to a user education. Today we will talk about derivatives of parametrically defined functions. In particular, second derivative actually, because the first derivative has already been addressed. Now, this lecture is part of the advanced mathematics course for teenagers and high school students. It's presented on unizor.com. I suggest you to watch the lecture from this website. Not only it contains very detailed notes for each lecture, it's also for registered students. It provides the certain functionality, like guidance, including exams, and the site is free. So let's talk about parametrically defined functions and second derivative. Now, first of all, let's assume that our function is defined parametrically. Now, the typical example of this is where t is time and x and y are coordinates of the point which is moving on the plane. Basically, making some kind of a trajectory, which can actually be something like this too. Okay, let's wipe it out. This art. Now, what is the mechanical, if you wish, meaning of the first derivative? Well, if you have a trajectory, well, as you know, this is delta y, this is delta x. And as these two points are getting closer and closer, the ratio is becoming the tangent of the tangential line at this particular point. So you know what the first derivative of y by x actually is. So we're talking about y first derivative by x as a function of t. It's a tangent of the tangential line at this particular point, which is characterized by parameter t. Okay, now what's the second derivative? Well, it's a little bit more difficult to talk about mechanical sense, but probably you can think about the speed of the changing of this tangent. So the steeper, for instance, the trajectory changes its direction. Well, the higher would be the second derivative, right? Okay. Now, so our purpose is to find the second derivative, but let's start from the first one. It's a little repetition of that particular problem which we have already discussed once. The first derivative is basically a limit of delta y divided by delta x, right? And what we do in this particular case, well, this limit is as delta x goes to 0, but we can as well say it's delta t goes to 0 because x and y are presumed smooth functions. And if delta t goes to 0, then both increment of these functions also go to 0. And we can obviously replace it with delta y divided by delta t divided by delta x divided by delta t, right? And we can put delta t converges to 0, which is the same thing, right? And that is limit of the ratio, again, presuming our functions are smooth enough and the bottom is not equal to 0. Then its limit of ratio is equal to ratio of limits. And you will get obviously the first derivative by t and the first derivative by t by t here. So that's what it is and that's the expression for the first derivative of y by x. Now, I'm interested in the second derivative. Now, what is the second derivative of anything by anything? It's a derivative of the first. So we have to derive, find the derivative, whatever, of this by x, which means we have to give some kind of an increment to x or increment to t, if you wish. And so we will have what? We will have incremented value of this ratio divided by incremented value of x, right? Incremented ratio, incremented of this ratio divided by incremented x and have a limit as delta x goes to 0. So that's what we have to find. This is our second derivative, right? Now, again, what I will do, I will change it this way. I will divide this by delta t and this by delta t. Same exactly approach as for the first derivative. And I will take limit of ratio as ratio of limits, right? So we will have limit of this divided limit of this, which is yt divided by xt derivative by t because the limit of this is derivative of this, right? Divided by and limit of this is derivative of t. So all we have to do now is take a derivative of the ratio and we know how to do that. If you have a ratio, derivative is what? G squared fg minus gf, right? So f is the derivative of y and g is derivative of x. So that's what. Now, in the bottom, I will have derivative of x as it is. Now, when I do derivative of this ratio, now in the bottom goes to g squared, which is derivative of x squared. Now, but I already have one derivative of x, so it would be in the power of 3, right? Cube, one of this and the square of this. And numerator will have derivative of this, which is the second derivative now. The first derivative is yt, y by t, and derivative of this would be second derivative by t times first derivative of t minus, again, derivative of this one of the x prime, which is x double prime by t times y prime. And this is the formula, basically. So this is my second derivative by x as a function of t. That's what it is. Okay. Now, knowing this, we will just apply it to a couple of problems, real life problems. And we will see if we can have enough patience and skills to go through some challenging calculations. And I wish I would do it correctly. All right. So let's just remember this formula. Just do it. It's, everything is function of t, so it's easier if I will just omit this t parenthesis. So that's my final formula. All right. Example number one. x is equal arc tangent t, and y is equal logarithm of 1 plus t squared. All right. So we need the first and we need the second derivative, right? Of both of them. So the first derivative is equal to, what's the derivative of arc tangent? Obviously, I don't remember. Let's derive it. So if x of t is equal to arc tangent of t, by definition of arc tangent, it's tangent of x of t is equal to t, right? I take the tangent of both sides. What is arc tangent? It's the argument. If we take the tangent of this, we will get the argument of arc tangent. That's from definition of arc tangent. Now, we will take the derivative of both sides. Well, it's easier on the right. The derivative of t is 1. The derivative of the left, obviously, I don't remember. All I remember is derivative of sine is a cosine. Derivative of cosine is minus sine. So it's sine over cosine of x of t. And we have to take derivative by t, right? All right. Now, this is a ratio and again, I remember that formula about ratio. So first, we have to take derivative of this ratio, which is what? It's a cosine square of x of t. And here I will have derivative of sine, which is cosine of x of t times derivative of x, right? And times the cosine of x of t. Minus. So this is my f prime times g. Now I have to have g prime times x. The derivative of cosine is minus sine. So minus, it will be plus. This is sine of x of t times x of t. And times the first one, the f, sine of x of t. Now obviously, I will have here cosine square sine square multiplied by x prime. So this is x prime of t divided by cosine square of x of t. All right. Finally, I've got my derivative. So that would be, okay, now I don't know what cosine of x of t is. However, I know that x of t is arc tangent. What's very useful in this case is to express one over cosine square as one plus tangent square of x of t. Why? Because tangent square is sine square over cosine square. And if we will go to the common denominator, I will have cosine square plus sine square divided by cosine square. And in the numerator, I will have one. So one over cosine square is the same as one plus tangent square. Why did I do that? Because I know what is a tangent of x of t. It's t. It's easier, right? And now I can replace the whole thing with tangent square of this, which is t square. So now I've got it. So this is my, and this is equal to derivative of this, which is one. From this, I can derive that this is one over t square. All right. I probably should have remembered my table of derivatives, but I don't, so I derived it. Sorry about spending the time, but it's a good exercise anyway. But those people with memory, which is as bad as mine, there is always the way to derive something. And actually I encourage you to do that. Okay. Now, how about the second derivative? Now, the second derivative is the derivative of the first derivative, right? Now, this is one over something. Well, I remember that if it's one over something, it's minus one over that something square, right? So it's minus one over one plus t square square times derivative of the inner function, which is derivative of one plus t square, which is 2t, right? Just a plain 2t. Finally, we got it. Okay. Now, why the first derivative is equal to derivative of logarithm is one over whatever is under logarithm, which is this, times derivative of the inner function, which is 2t. Now, second derivative is derivative of this guy. So, again, that's f over g. So I have g square square. Okay. Now, in the top, I have a derivative of this, which is 2 times this minus derivative of this, which is 2t times this, another 2t. So it's minus 4t square. Or I can simplify it. It's 2 minus 2t square, right? 2 minus 2t square. All right. Fine. So we have all the components for this formula. And let's just use it and see what happens. Okay. In the denominator, I have this thing to the third degree. So that would be 1 plus t square to the third degree. No, that's wrong. That should be on the top, because this is one over and this is denominator, so it will go to numerator. So it will go to a numerator. Okay. Now, and this thing is... Okay. Second derivative, which is 2 minus 2t square divided by 1 plus t square square times xt, which is this, and it will be cube minus. Second derivative of x, which is this, and this is minus, so it will be plus. 2t divided by 1 plus t square square times the first derivative of y, which is this. So it will be again the third degree here. And here 2t and 2t will be 4t square. Now, lo and behold, this thing, this thing and this thing are cancelled out. And what do I have? 2 minus 2 plus 4, so it's 2 plus 2t square. That's my second derivative of y by x, and of this calculations. Next problem. x is equal to cosine of 2t, y is equal to sine square of t. Okay. Same thing. First derivative equals 2. The derivative of cosine is minus sine of 2t times derivative of the inner function 2t, which is 2. Now, the second derivative is from here, minus 2 from a sine is the cosine of 2t times derivative of the inner function, which is also 2, so it's 4. Okay. And let me change this 2 in the back into 2 in front, minus 2. Okay. Now, first derivative of y by t is equal to 2 sine of t times cosine of t, right? First, I took the derivative from power function, which is 2, which is 2 times whatever is under power, with a power minus 1, which is 1, 2 minus 1 is 1. And then inner function is cosine. Now, the second derivative is, well, this is second derivative of x by t. Now, the second derivative of this is, well, it's a product. Actually, it might be a little bit easier if I will convert it into sine of 2t, right? Remember trigonometry, 2 sine by cosine is a sine of double angle. Now, the derivative of this is cosine of 2t times derivative of inner function, which is 2. Okay. I hope it's right. Now, let's go to my second derivative by x. Well, I wiped out this formula, but I do have it here. So it's the first derivative of x in the power of 3. So this goes to a third power. So it's minus 8 sine cube of 2t. Now, in the numerator, I have second derivative times first minus first times second. Okay. So this times this is 2 minus 2. It's minus 4 sine times cosine of 2t. Minus this times this minus 4 and minus it's a plus 4 cosine 2t times sine 2t. Cosine times sine. Wow. These two cancel out, so the whole thing is equal to zero. Isn't that wonderful? What does it mean, actually? Well, if the second derivative is equal to zero, it means that the first derivative is constant, right? And if the first derivative is constant, then the function must be linear function, right? Remember? So if you have a linear function, my first derivative is a, which is constant. And my second derivative, the derivative of a constant is zero. So it resembles, right? So it looks like we should have something like this between y and x. Well, let's just think about it. It's why Paul told this nonsense. And start from the beginning and see what exactly our y and x actually are. Look at it this way. What is cosine of 2t? Well, I do remember this, right? But at the same time, since cosine is, cosine square is minus 1 minus sine square, so it's 1 minus sine square and sine square, so it's 2 sine square of t, right? So you see, that's my y. No, sorry, that's not my y. That's my cosine of 2t. So in this case, sine of square of t is equal to, this goes here, this goes here, 1 minus cosine of 2t divided by 2, right? So this is equal to 1 minus cosine of 2t divided by 2, which is 1 minus x divided by 2. So as you see, we do have a linear dependency between y and x. And we have noticed this by taking the second derivative of the y by x. I mean, obviously, we could have noticed it in the very beginning and derived this type of relationship between y and x just looking at this. But maybe we were not smart enough and didn't realize it, so we decided to go by some kind of a paved road and use the formula and got 0 as a second derivative and from 0 as a second derivative, we actually concluded that the functions must be linearly dependent. And they are. And what's the graph of this? Well, don't forget that y is supposed to be from 1 to 0 and x is a cosine, it's from minus 1 to plus 1 and at x is equal to 0, it's 1 half. If x is equal to 1, that's 0. And if x is equal to minus 1, that's 1. So this is my graph. Only this piece of the straight line actually is the graph. So you see, that's a very important difference. If I will write the function y is equal to 1 minus x divided by 2, this function is defined for any x. But defined parametrically, it's not defined for any x because the x is a cosine that cannot be outside of the minus 1 and 1 interval. So we are basically reducing the area, but still the straightness of the line is retained. Line is still straight because the second derivative is always 0 and the first derivative is constant. Okay, that's my second problem. Next. Alright, x is equal to cosine of 2t and y is equal to t square. Okay, my first derivative is minus sine and times 2, so it's minus 2 sine of 2t. And my second derivative is, it would be 4 actually, right? Minus 2 cosine and 2, so it's minus 4 cosine of 2t. Okay, that's my second derivative. Now my first derivative is 2t and my second derivative is just 2. Alright, so let's again do this thing. Denominator is cube of this, which is minus 8 sine of 2t cube. Okay, in the numerator, I have second derivative of this times this. So it's minus 4 minus second derivative of this times this. So minus it would be plus 8t cosine of 2t. Okay, basically that's it. I mean we can reduce it by 4 and that would be my answer. That's it. Okay, these are just, you know, nice exercise to check if you have enough patience to go through all these calculations. I wish you do. That's it for today. Thank you very much and good luck.