 Hi, I'm Zor. Welcome to Inusor Education. I would like to solve a few problems related to derivatives. Now, as usually, I do recommend you to first go to the website, unizor.com. This is the subject called calculus, and this is the topic called derivatives, problem number two. And try to solve these problems just by yourself. I think it's very important. And basically, as I said many times, the most important purpose of the whole course is to solve problems. And not because these problems will be of any practical value for you in your real life or your real profession, but because the solving problems really develops the analytical abilities of your mind. And that's the purpose. And to introduce you to the problems, I have to put some theoretical lectures in. So, the theory for the problems, the problems for development, and that's where it's very important for you to try to solve these problems just by yourself. And let me now go to these problems. Now, these are, well, at least two of the three, three problems for today. Now, two of the three have really kind of physical substance in them. And it does require certain knowledge of mechanics, which I will try to go into a little bit of details, but it's preferably that you know this stuff, actually. Anyway, so let me start from the first one. And the first one is about cannonballs. So, let's assume you are launching the cannonballs. And that would be your trajectory of the projectile, right? So, my question is, this is an angle which you should. So, the question is, which angle gives you the longest distance where your projectile will land? Now, obviously, if the angle is too small, it will be like this. And that will be shorter. If the angle is too big, then it will be like this. Also, too short. So, at a certain angle, and the angle is actually 45 degrees, P over 4, but we have to prove it, the distance will be the longest. So, the question is, what angle gives you the longest distance? So, let's just talk about this particular problem right now. And we will solve it, and the solution will be actually 45 degrees. So, how can we solve it? Well, again, if this is how it goes, how to determine this distance? Well, actually, it's kind of simple. You are shooting with some kind of a launch speed, a linear launch speed, which is the characteristic of the cannon itself, right? So, it's a constant. Whatever it is. Then, if you have this angle, phi, this particular speed can be composed from two components. This is the vector, right? Remember? So, this is a vertical vector, and this is the horizontal vector. So, my actual linear speed in vector form is a combination, a sum of two vectors. One of them is vertical, another is horizontal. Okay. Now, it's the horizontal component which gives you the distance, right? So, if you know, well, if this is the constant, now, if this v is some kind of a constant which is basically defined by the characteristic of the cannon itself, then these two components are also constants. Obviously, the size of the absolute value of the vertical is equal to vertical if equal to the real one times sine of phi, right? And the absolute value, the length of the vector horizontal is equal to absolute value of our linear speed of the launching times cosine of phi. Now, it's the horizontal speed which gives you the distance if you know the time, right? So, this distance L is equal to the horizontal times time. Well, if I don't put the vector here, it's basically this. It's the length of the vector, absolute value. So, that's what gives you the distance, but we don't know the time, right? Well, the time we can actually determine. So, what is the time? Well, look at this behavior of this particular cannonball. If you consider only its vertical component, it goes up as much as it can, right? Until the gravity actually stops it and then it goes down. Now, so the force which is actually acting upon our cannon, the vertical force, the force of gravity is actually characterized by this constant which is the gravity acceleration. Now, what does it mean? Which is approximately equals to 9.8 meters per second square. What does it mean? It means that your speed, if it goes down, free-falling, I mean, if it will go down, it will increase by 9.8 meters per second per second. So, each second, your speed measured in meters per second will increase by 9.8. Or if you go down, if you shoot something vertically up, it will decrease every second by 9.8 meters per second, right? So, if my initial speed is this vertical component of the speed V, which is V with an index, sub-index B, then the time it goes up is equal to seconds, right? If each second, this thing is diminished by G, then this is the number of seconds it reaches zero. So, that's my maximum whenever it occurs. So, after this time, it will reach my projectile will reach the top point and then it starts falling down. It will increase by G the speed from zero back to VV, basically, but going vertically, which means the time will double. So, the whole time will be this. So, this is my T in this equation. So, my length is equal to V horizontal, which is V times cosine phi times T, which is times 2 V vertical, which is again V and sine of phi divided by G. So, this is the constant, this is the constant, this is the constant. So, all we have to do is to find out where is the maximum of the function f of x is equal to cosine phi times sine phi. That's what we have to find out. That's the maximum of this, whenever this function is reaching its maximum. And by the way, phi, obviously, is from zero to pi over 2, right? It's from this to this. So, on this interval, we have to find the maximum of this function. That's basically the mathematical interpretation of the physical problem. Okay, now, if I were smart enough, I would notice that this is basically 1 half of the sine of 2 phi. Remember this formula for sine of 2 phi? It's 2 sine times cosine. Well, this is 1 half and that's why it's equal to this one. And in this case, I know that the sine has a maximum of 1, right? And it's reaching its maximum function sine, which reaches its maximum at pi over 2, right? So, 2 phi at the maximum point is pi over 2, so the phi is pi over 4, as we expected, right? Now, what if I'm not that smart? So, I just don't recognize the formula, right? I forgot the formula, whatever it is. So, I'll do basically direct search for the maximum of this function. How? By taking a derivative and making an equality that the derivative is equal to zero, right? So, what's the derivative? It's a product of 2 function. So, it's the first function times derivative of the second, which is cosine square, plus product of the first, I mean derivative of the first, which is minus sine, times the second one. So, it's minus sine square of x. Well, of phi. I use phi here. And this is phi as well. It doesn't really matter. All right. So, let me just equalize with zero this, which means cosine square of phi is equal to sine square of phi. Now, obviously, I should not put equal here. It's probably better to be, because if it's zero, it just doesn't go anywhere. And if it's pi over two, it goes vertically up. So, we can definitely exclude, which means cosine of phi is not equal to zero. Everything is fine. I divide by cosine square. I will have one equals to tangent square of phi, which means tangent phi is equal to one or minus one. But there is no minus here, because the phi is in the first quarter. So, we have phi is equal to pi over four, because that's the angle in the first quarter where the tangent is equal to one. So, we got this differently, using a straightforward approach of taking a derivative and equating it with zero. Well, question is, is it really maximum or minimum? I mean, we have really analyzed this, right? Well, if we are really very much interested, we can just take the second derivative and make sure that it's negative in this particular point. Well, basically, that's it for the first problem. Okay. Next. Next is not as physical as this one. Given an equation. Okay. We don't have to solve it. It's a cubical equation, although we can probably. But let's not attempt this. That's not what I'm asking. I'm basically asking how many zero points, how many solutions this particular equation have. Okay? Now, how can I determine how many solutions have a particular equation? Here is what I'm suggesting as a methodology. First of all, let's determine intervals where it's increasing and decreasing. Obviously, these intervals are separated by points where the first derivative is equal to zero, right? Because on the left of the zero, it can be, let's say, positive. On the right, it can be positive or negative. I mean, between the zeroes, it should not change the sign. It should be either negative or positive, right? Between two zeroes of the derivative, this is, and this is two points where derivative is equal to zero. Between these points, from minus infinity to this point, the derivative has some sign, positive or negative. Well, in this case, it's positive, so the function is increasing. In this case, derivative is always negative. At this point zero, at this point zero, and here it's again positive. So that's what I'm trying to find out. Which are the points where this particular function has change of directions from increasing to decreasing? Well, for this reason, we have to take the first derivative and equate it with zero. The derivative is three x square minus three equals to zero x square minus one is equal to zero, x is equal to plus minus one, right? That's the solutions. All right, now, graphically, it would be something like this. So this is y, this is minus one, this is one, something like this. So probably this curve looks something like this, probably. But in any case, minus one and plus one are two points where my derivative is equal to zero. Now, let's check what are the values of the function between these points. Okay, at minus infinity here, function is negative, right? If this is minus infinity, obviously it's negative. x cube is significantly faster growing to the negative side than minus three x in the positive side. As we reach point minus one, the value of the function is minus one plus three, two plus four, six. Look at the very important thing. At minus infinity, function is negative. I mean, when I'm saying minus infinity, it means to the left of the minus one, right? Significantly to the left, big, big, big left as much as possible. Now, but at minus one, at this point, it's positive. What does it mean if the function is monotonically increasing from something negative to something positive and it's monotonic? It means it crosses the x-axis just once. So we have one solution to this equation at this particular interval from minus infinity to one. Next point is one. At point one, I have one minus three, which is minus two, plus four, it's two. Okay, here is important thing. From point minus one to point one, from here to here, function is monotonically decreasing and it's still positive in both cases. Now, if the function is monotonically decreasing from six to two and it's monotonic, it cannot cross the x-axis because if it were crossing the x-axis, if there was a solution to this equation between these and this and the function cannot be monotonically decreasing from six to two, it goes to six to zero and then up to two. So there is no monotonic behavior, right? So there are no solutions in this particular interval. And next, at plus infinity, so significantly to the right of this when the function is increasing monotonically, obviously this is also positive. From positive to positive, again, no solution. So I have only one solution at interval from minus infinity to minus one and that's the answer. So by analyzing the nulls of derivative and thereby analyzing the intervals of monotonic behavior, we can actually find out on each interval comparing the values of this function whether there is or there is no zero point of this function on this interval. And in this case, we have only one interval out of three where function has different signs at both ends. So that's my second problem. Now, the third problem is also physical, mechanical, if you wish. It does require a little bit of knowledge, but I will give you all the information you need to solve this problem. Let's imagine that you have the rough road and you have certain object which you are pulling with a rope. So this is you and you are pulling this particular object this way with constant speed. Let's also assume that I said it's a rough road which means there is a friction. Now, the friction is characterized by coefficient of friction and the force of friction is equal to the weight of the subject times coefficient of the friction. The heavier the object, the greater the resistance of its movement, the friction, and they are proportional. If the object is twice as heavy, the friction force will be twice as big. So this is my friction force. Now, where is the friction force directed? Well, obviously this way. Now, I am pulling forward, but I'm not pulling completely horizontally. I'm pulling at some angle. Let's wipe me out and call it F. This is my force which I am pulling along the rope at certain angle. And the horizontal component of my force, let's call it F horizontal, should be equal to the resistance to the friction force to be able to move horizontal with the same speed. I'm not accelerating anything which means these two forces must be equal. So the force of resistance is equal to the force of my horizontal component of my force pulling forward. And these forces are supposed to be equal to the object to go the same speed without any acceleration or deceleration. What is my task? My task is to find an angle phi which minimizes my force of pulling, F. So I would like to minimize my force. Now, why is this important? Why different angles give me different efforts which I have to exhort? For a very simple reason because there is a vertical component. Vertical component, basically you can consider it as reducing the weight, P. So if my initial weight of this object was P, which means that's basically the force of gravity which the Earth pulls the object down. My vertical component of the first pulls it up. And it's the difference between these two forces, between the weight and vertical component of my force which actually is multiplied by coefficient of friction, which should give me exactly the force of friction which must be by absolute value equals to my horizontal component of my force. So this is the equation actually which must be held. And this equation basically defines my force F as a function of angle phi. So how is this actually looking like this? Vertical component is my absolute value of my force times sine, times k. It's supposed to be equal to horizontal component which is my F times cosine. So this is the equation which defines my function F as a function of angle phi. And this is the way how I would express it as F as a function of phi to be able to minimize my F, my absolute value of the force which I'm pulling forward. Okay, so let's just simplify it a little bit. So it's pi times k minus F times k times sine phi equals F times cosine phi from which F times cosine phi plus k sine phi is equal to pk. So my function looks like F equals p times k divided by cosine phi plus k sine phi. Now what we have to do, we have to minimize F, which means we have to find such an angle phi for which my F, which is absolute value of my force which I'm pulling the rope, would be minimal. Alright, as usually in this case, no tricks, just take the first derivative. Now this is a compound function, so my first derivative is equal to, well pk is actually just a constant. Then it's 1 over something, 1 over g of phi, right? So it's minus 1 divided by square of this times the derivative of the inner function, which is this, which is minus sine phi plus pi k cosine phi. Well, obviously we assume that this is not equal to 0, so we don't have any problems with this. So the only way how it can be equal to 0 is if these are, this is 0. So minus sine phi is equal to k cosine phi, or k is equal to, I mean plus, sorry, plus equals to 0. So k is equal to what? Sine goes there, tangent phi. And phi is equal to arc tangent of k. So that's the final answer. If you know the coefficient of friction, then you take the arc tangent of this coefficient to get the angle which gives you the minimum effort. In particular, is k is equal to 0, which means there is no friction, phi is equal to 0 as well. Which means what? Which means you have to pull horizontally. That's the best way, right? And as k is increasing, your angle should also increase to compensate the friction. The greater the coefficient of friction, the greater the angle should be. Now, if you are a person, let's say, you basically have this as a fixed height, right? So you always pull it, let's say, on your shoulder, right? You're pulling it on your shoulder. Now if you fix this one to increase this angle, you have to shorten this distance. Which means you have to basically shorten the rope. So if you are experiencing certain friction, your rope should be shorter so the angle will be greater. If, however, friction is small, then you're still here and you're still pulling on your shoulder. Take a longer rope and it will be a smaller angle. So the greater the friction is, the shorter the rope you should use if you always pull it on your shoulder. That's a practical advice you wish. Now, I'm sure you will not experience this type of problems in real practical life. Besides, you probably don't know what the coefficient of friction is, it's just only approximation anyway. But in any case, it's a nice physical, mathematical kind of problem. Alright, I do suggest you to go again to this website, to the calculus derivatives and problems number 2 and solve these problems just by yourself again. And obviously, see if you have exactly the same results. Other than that, everything is fine. That's it. Thank you very much and good luck.