 Hi, I'm Zor. Welcome to a new Zor education Many times I was talking that the most important part of any course of mathematics is to solve problems Well, let's solve problems today The problems will be about derivatives and everything we know by by now about derivatives just six yet six problems Where different aspects of whatever we have learned should be applied Now this lecture is part of the advanced course of mathematics presented on unizor.com. I Suggested to watch this lecture from this website Because the lecture every lecture actually has very detailed notes, which can serve as a textbook basically and also the assigned in students can Participate in an educational process which involves enrollment and for this you need your supervisor or a parent And you can take exams as many times as you want just to gauge your Success in learning the material and the site is free by the way. All right, so a few problems The problem number one is Let's consider this function So my statement is about behavior of this function When x goes to infinity and basically I'm saying that e to the power of minus x and you know The graph is something like this goes to zero much faster than X to the power of n which graphs like this approximately so this goes to infinity and this goes to to zero and Obviously just looking at it is it's not obvious where their product actually would go But here is what I suggest. Let's rewrite it as and Use the L'Hopital's rule now the L'Hopital's rule states that if you have This ratio of two functions it behaves when x goes to some limit including infinity Exactly the same way as the ratio of their derivatives now Let's think about derivatives Well, the derivative of this would be what there's no equal sign derivative of numerator would be n x to the power of n and e to the power of x right This is sorry minus one This is derivative of this and e to the power of x is derivative of e to the power of x as we know, right? so The limit of this as x goes to infinity is equal to the limit of this as x goes to infinity But now we still have some kind of Indeterminate limit because this is infinity and this is infinity as x goes to infinity, right? Well, okay, but let's use the L'Hopital's rule again. The next would be n times n minus one x to the power of n minus two That's the derivative of numerator and Here we will still we will still have the same Okay, is it better? Well, not much. It's still infinity, right? But now let's do this process and times and what do we have in this case? Well after n times we will have n factorial obviously, right? n times n minus two minus n minus two times n minus three, etc and x would be basically to the power of zero which is one and in the Denominator we still have e to the power of x because this is the n's derivative of this, right? and Now this n is a constant, right? Now this we know for sure goes to zero so after Implementing this after using this L'Hopital's rule and times we basically have Reduced our problem of determining this limit to this from this to this and from this to this and this is easy and Result is zero so that actually signifies that a to the power of minus x Goes much faster to zero than x to any numeric power. I'm talking about obviously Natural number n otherwise. We would not have nice Constant one at the very end. All right, so that's my first so x to the power of n grows slower than e to the power of minus x Diminishing Now let's consider a different combination of function Okay, so x to the power of n We know is growing not as fast as e to the power of x or e to the power of minus x goes to zero Faster than x to the power of n. Now. How about logarithm? well, the logarithm is going to Infinity slower than x to any power and How can we prove it again? Let's use the Lopital's rule. So the limit of this as x goes to infinity is equal to Ratio of their derivatives derivative of logarithm x is 1x derivative of this is n power n x to the power n minus 1 and Now it's obvious that it if x goes to infinity Then this goes to zero which proves that x to the power of n is is growing faster than logarithm x or if you wish x to the power of minus n goes to zero faster than logarithm n goes to infinity So that's my second problem So as far as speed of growing logarithm is slower than power Function and powers slower than exponential function okay next is about Approximity calculation Using differential instead of increment of the function. Let's just think about this Now given that natural logarithm of ten is equal to two point three zero two five eight five Somebody gave me this number. All right, but I do want to calculate really is logarithm of ten point one So remember how we will be dealing with this f of x plus delta x Now this is ten and this is zero point one and F obviously is logarithm, right? Now remember this this is Delta f of x now Approximately if Delta x is really small Remember we were talking about that this is approximately Or Delta x doesn't really matter So this is something which we were talking about when we were discussing differential. So this is increment of the function and We were talking that since if this is if we divided by Delta x and Delta x goes to zero We will get a derivative right remember this f of x plus Delta x minus f of x divided by Delta x if Delta x Goes to zero it's equal to derivative at point x right so if x is relatively Delta x is it is relatively small Then we can say that Forget about this limit then This is approximately plus some kind of infinitesimal variable, right now we go multiply by Delta x and we will get this and to be exact this is plus Epsilon Delta x Right from this but epsilon is infinitesimal and Delta x is infinitesimal While this is not an infinitesimal. So this is an infinitesimal of a higher order than this and that's why we can do this Approximation instead of equal we can put approximately and Get rid of this now obviously this needs to be Somehow some kind of justified because Delta x can go to zero but question is how close to zero which is etc So it all depends obviously However, what I'm asking in this problem problem is well forget about all these nuances Just consider that point one is as small enough and if it's a small enough how to use this formula for Differential of the function. This is differential of the function. This is increment of the function. So replace increment of the function So how much this grows relatively to this? replace it with Differential of the function now Why is it better? Well, it's significantly better because this is very easy to find out What is the derivative of logarithm? It's 1 over x So I know that this is 1 over x where x is equal to 10 now this is 0.1 So I can actually very easy calculate my value as Function of x which is given plus this piece 2.30 2585 plus This is one tenth Right one over x at x is equal to ten times another zero point one. So it's zero point zero one and the result is two point three one two five eight five Okay So that's approximate value of this That's what our calculations show Now how good is it? Well, let me give you an exact number which I got from the calculator So this is approximation now exact value would be two point three one two five three five so we have this fifth digit after the decimal point Wrong, which is not bad. So up to four digits after decimal. We are okay. All right well now you can ask why should actually do all these Mathematical manipulations if I can just go to a calculator and and get and get this number Well, it has educational purposes. All right, just to show that you can use differential as an approximation To increment of the function. So the problem is not practical problem is educational. All right, so let's just settle on that next problem Also educational I Would like to make this more precise and For this I would like to use the Taylor series. Do you remember the Taylor series? We can actually represent So this is Taylor series So if you don't remember it just go to the corresponding lecture Where I explained what actually this is all about and how to Drive this basically So my point is the following now. This is a series which means it's converging Which means we can cut some kind of a tail and we will get an approximate value of this thing Right, so what I'm asking you the next problem is cut this tail So leave only up to second derivative and Do exactly the same problem calculate? Approximately what exactly the result would be if I'm using this where x 0 is obviously 10 and X is obviously 10.1 x minus x 0 is obviously 0.1 and Now you can calculate it now the first derivative now this is Lagerism of 10 which is given right so it's 2.3 0 2 5 8 5 Now this is exactly what we had the first time right the derivative, which is 0.1 Because it's 1 over 10 One over x is derivative of logarithm and x is equal to 10 right so it's one-tenths and this is also Difference between these two so it's also one-tenth so it's times Plus one one hundredth right so zero point zero one Okay, second derivative. It's from one over x. It's minus one over x square right So minus one over x square x is equal to ten so it's minus zero point zero one times This is also one-tenths square one hundredth Divided by Two so it's zero point zero zero five right So we subtract Sorry we yes, I have to sorry. I have to multiply Sorry multiplication This is from the derivative and this is from this member right and if I multiply it will be minus 2.0 0 0 0 5 right to Zero and then I have to move it to the right by two positions. So that's what would be All right, so it's minus zero point one two three four five Okay, let's see what happens Well, it's two point three zero two Now it's three one two if I add this five eight five minus zero point zero zero Zero zero five zero So it's five three five two one three point two And this is by the way equals to whatever I got on calculator. So up to Six positions after the decimal point my calculations just based on two Members of the tail or series is already Corresponds to the calculator Right, so one position gives me four digits after decimal point correctly Two members of the Taylor series give me six, right? Now without this one, that's the previous problem basically and With the second Derivative that gives me even more precision All right, that was my Other problem, let's go so we had the problem related to Lopitas rule we had the problem related to differential and Taylor series and Now we will talk about Differential along the curve So my I have two problems about this my first Curve is a circle and let's imagine that we have a particle which moves On this circle of radius R So it moves on this circle and this is how it moves I basically give the angle as a function of time So as the time goes whatever this function is Well, it's obviously in in regions. That's how it goes. All right That's if it's increasing well, it might actually increase it might decrease. It's a function, right? So a function can go anywhere it wants it has its own mind on its own, right? So and the particle along this Circle goes along goes according to this particular function. My question is What is its Absolute speed as it goes along this curve and what are the components of this speed? which are projections of the vector of Velocity of this particular Particle on x and y axis Okay, so First of all, we have to find out what are the coordinates of our Curve and that gives us the movement as the function of time, right? So X coordinate of the point is equal to R times cosine Of the angle and angle is a function of time So angle is changing it's a function of time and that's why my x coordinate is and y coordinate It's obviously sine of this now If we are talking about the speed remember what speed actually is well, you have to take a Small piece of the curve of the length of the curve differential along the Along this curve and divide by the time Now obviously this is function of t, right? And you remember that actually it's equal to this It's a very easy thing to To derive this formula It was derived basically when I was talking about Differential along the curve. So this is the absolute speed of Our particle as it moves along the Along this circle Okay, so let's find it out equals to This is equal to R square Now the derivative of the x is r minus sine of phi of t Sine of phi of t and derivative of the t and I didn't put the minus sign because I'm still Raising it to the power of 2 so it's square. So it doesn't matter minus or plus here, right? now similarly here R square cosine square of 5 t times derivative of the phi Square and This is all under the square root and Obviously it is equal to r square and rail and and derivative of the phi square can be Factored out. I will have sine square plus cosine square, which is 1 and the square root would be r and absolute value of derivative of the phi So that's my absolute speed of this particle. So if I have the function Angle as a function of time then absolute speed as it goes along the circle Any direction any way that would be My absolute speed now whenever we are talking about Velocity we are talking about a vector which has certain direction and this length and Direction and lengths obviously are very much related to individual components the projection of this vector of velocity on the x-axis and the V-axis and the y-axis and These projections are obviously here so x Derivative of x of t that would be my Component Along the x-axis Which is equal to minus r sine 5g phi of t and the Y component of my velocity is derivative of the y coordinate and It's our cosine So these are my two components Now is there anything else which I'm missing aha. Yeah So again This would be equal to r and absolute value of the derivative This absolute value of the derivative of the phi of the angle. It's called angular speed by the way so if you have a function Which represents angle as a function of the time derivative of this Angle as a function of the time is angular speed it's exactly similar to if you have a function which is distance as a Function of time then derivative of the distance is the speed Now obviously It's different at different moments in time as the t goes phi is changing and The speed which is derivative of the phi at point t also might not be exactly the same however, if If my angle is proportional to time, which means my movement Actually is is going without changing of the speed Then the derivative would be equal to see right always regardless of the time. It would be constant Angular speed. So that's when we're talking about the movement of the particle Without Changing of the angular speed So that's it, right so that's my Basically the problem where I wanted to talk about the differential along the along the curve ds of t and The next problem also related to curve and differential along this curve Okay, now we are talking about a projectile which we are launching at certain angle to horizon It has certain velocity Which has absolute value and angle, all right, so what's given is an angle and initial Absolute value of the speed we are launching our projector and we basically have to do exactly the same in the previous problem find out vertical and horizontal components and Absolute speed at any moment t Well, let's first Restrict our movement up to the Top point when our projectile is is rising now it's rising obviously But the gravity pulls it down Now we know that there is a constant g Which is equal to 9.8 meters per second square Which is by how much per second my speed would be Decreasing if we go up or increasing if we go down, right my initial speed The has both horizontal and vertical components, right? So we can always represent this velocity as sum of two vectors V is equal to Vx plus Vy Right, that's the rule of parallelogram now it's much easier to approach this Problem from this perspective because we know for sure that Vx the x component doesn't really change with the time because there is no force which is Horizontal the acting but we obviously assume there is no air resistance or something like this However, vertical component is affected by the gravity. So if my initial Value of the vertical component is V times Sine Phi These are constant and phi is a constant. These are initial value So that's my initial value of the Vx. So it's a Vy. Sorry. It's Vy at moment zero Now as the time goes by Every unit of time we are changing my we are changing our speed by g We are reducing speed by g because the gravity pulls it down so it's slowing down and How much time actually? It will it will do it will do up until the point when my speed vertical speed will reach zero, right? That's the total time. That's the time of rising but any time Any time this the speed Would be equal to initial speed minus Constant g times t because every unit of time we are reducing by g so t unit of time we're reducing by g times t so that will be my Law of motion if you wish or dependency of the vertical component of the time so we have actually determined the vertical component and Horizontal component now as you see My horizontal component does not change. It's the same always because there are no forces which are Help or restrict its movement it just goes with this Horizontal speed from the moment zero to any moment time and it's equal to whatever it was initially Which is V times cosine of angle five, right? so that's actually our two components of The speed at any time up to the moment when it reaches the top So what by the way is the time when it reaches the top? That's when these two are equal to each other and the V y is equal to zero so the time to the top is equal to V sine phi divided by g g V sine phi divided by g g right To make it equal to zero So that's the time To the top now up until this time This is the law my horizontal my vertical speed is changing and this is the law my horizontal speed is changing and What's my absolute speed is well again the absolute speed? square root of V x square plus V Y square and It's equal to Okay, it would be square root of V square cosine square phi plus V sine Phi minus g t square under the Square root So this is a function which describes my absolute speed now these are two functions which describe Components of the velocity of the vector which is here now just out of some kind of I always Like to be sure that the formula is correct correct or something like this There is some checking which I would like actually to implement here Let's think about it at the very Moment when it reaches the top again my horizontal speed is still the same right as it was before which is V cosine phi Now my vertical speed is zero which means my total absolute speed should be the same as horizontal speed So if I will substitute this moment t Into this formula for absolute speed. I must have the result equal to be cosine phi, right? Well, let's just check it out Always try to analyze if you have reasonable result alright, so Let's substitute instead of t we will substitute this particular oh my my fault by by t here if this is equal then t is resolved as V sine phi the light divided by g alright, so what do I have I have v square cosine square phi plus v square sine square phi minus 2 v sine phi gt Plus gt square square and this is square Now if instead of t I substitute this I will get here v square sine square phi divided by g square right now instead of this v sine phi divided by g right So obviously g and g g square g square This is v Square and sine square and this is v square and sine square. This is minus 2. This is 1 and This is together would be v square cosine Square phi plus v square sine square phi minus 2 v square sine square phi plus v square sine square phi and this is all under the root and This is Canceling out and as a result I have v cosine As we should So at the top point my formula gives exactly what it would be expected to give all right, so This is just a second problem about The movement along the curve differential the speed and stuff like this Okay, I do suggest you to read all the notes for this lecture on unizor.com Other than that, that's it. Thank you very much and good luck