 Welcome back to our lecture series math 1210 calculus one for students at Southern Utah University as usual I'll be your professor today. Dr. Andrew missile nine in lecture eight. We are gonna start chapter two Which is about the idea of limits for which we really are now gonna start Calculus everything we've done previously was just a review of pre calculus materials that wax on wax off that The karate kid had to go through now We're actually going to start applying these techniques into this setting which we call calculus now to motivate that We really have to start talking about the idea of a limit and while many Calculus textbooks and classes try to start motivating this with tangent lines and velocity that does make sense We're gonna take a slightly different start to our discussion of limits in this course And we're actually going to introduce the idea of a limit using the vehicle of error and tolerance And so to explain what one means by error and tolerance I'm going to borrow homework exercise actually from James Stewart's calculus textbook as of the eighth edition This should be exercise two four twelve. So imagine a crystal growth furnace is used in research to determine How best to manufacture crystals used in electronic components for the space shuttle this does kind of date the problem a little bit Sorry, they don't have space shuttles the way used to but you didn't Beside the point right for proper growth of the crystal the temperature must be controlled accurately by adjusting the input power Suppose that the relationship is given by the following which you can see here on the screen T of w where t is the temperature in of the oven of the furnace in Celsius I know a chemist just tried somewhere it should be in Kelvin, but we're gonna measure in Celsius So the formula t of w is given as 1 or 0.1 w squared plus 2.155 w plus 20 Where w right here is measuring the power input into the furnace Measured in watts. Alright, so we have this relationship between the temperature of the furnace and the wattage of the furnace So this is how a typical electrical oven works that maybe we send electrical current through some type of heating element or other Capacity, you don't need all the details of that But basically we insert more power when we want it to get hotter and then we lower the power when we wanted to get it cooler This is an important thing to remember when we have a function relationship that with a function relationship You essentially have two variables in play. There's the variable w here Which in our situation is our input variable Right, this is the variable sometimes It's called the direct variable because this is the variable that we actually have direct control over we control How much power is entering the system on the other hand t here is what we could call the output variable Or we can call it the indirect variable and I really like that name there because with the indirect variable We don't actually have control on what it is We have control over w and then from w this function relationship Then causes t to change because after all variable the word means in able to vary Right, so t will vary based upon w. We actually have control over w. We don't have control over t not direct control So we as the operator of this furnace We have our direct variable w and the indirect variable is t here But keep that in mind as we continue on with this problem So imagine how much power is needed to maintain the temperature at 200 degrees Celsius Let's suppose that the engineers who work in this lab here know that 200 degrees Celsius is the optimal Baking temperature for growing these crystals so that we know but we can't just wave a wand and make the furnace 200 to 200 degrees because again, we don't have direct control over the temperature We have control over the wattage, so it's important to know the variable We have control of how do we control it to cause the variable which we don't control But we want to influence. How do we choose the wattage to influence the temperature to the desired temperature? All right. Well given that we have this function relationship I'm going to put a big green box around it so we can remember it because we have this function relationship here It if we want the temperature to be 200 degrees Essentially, we're trying to solve the equation t of w equals 200 degrees now that's an equation. We really can't solve by itself But if we make the relationship, oh T of w is this quadratic polynomial Then we have to solve the equation point one w squared plus 2.15 w plus 20 equals 200 This is an algebraic equation which we potentially could solve Like if we were to subtract 200 from both sides of the equation you get 20 minus 200 which would be negative 180 We then have this quadratic equation right here in standard form for which from a previous Algebra or pre-calculus class. We then have an inkling on what to do We could try solving it by factoring by completing the square given the decimals going on here I think probably that the best approach is just going to be use the quadratic formula So sing along with me pop goes the weasel. I'm not actually going to sing it here But we see that based upon the formula right here W equals negative e plus or minus the square root of b squared minus 4 a c all over 2a All right, you can do that to the tune pop goes the wheels all right now for us I didn't do plus or minus because notice as this number right here is negative and the square roots always can be positive by definition If I take a negative minus a negative I'm going to get a negative wattage which that doesn't make sense for this problem, right? This is why we often have to analyze things like domain What are the acceptable values for my variable w a negative wattage wouldn't make any sense in this context So we're we can we can restrict our attention just to the positive case Now that with the quadratic formula if we just focus on the positive side Then we know that this just becomes a number crunch at this moment We're going to get 2.155 squared is approximately 4.6 4403 If you take negative 4 times well actually if you take point one times 180 that gives you 18 4 times 18 is gonna be 72. It's a double negative. So makes it a positive So we get this number 4.6 4403 plus 72 if you add together the terms of the radicans you would get 76.644 etc. Take its square root. That's about eight point seven five four six six And then subtract from that negative 2.155 we get six point five nine nine six six divide that point by point two Because remember the numerator was two point Two times point one so you get point two. So we divide that out. You're gonna get thirty two point nine nine eight three Which is approximately thirty three watts. All right, so by solving this algebraic problem We see that 33 watts would be how much power we want to enter into the system to get the ideal temperature of 200 degrees, but 33 watts isn't actually the answer. This is an estimate of the answer. We round into the nearest watt If if this formula remember the green box from a moment ago if this formula is to be taken without any Criticism whatsoever like how reliable is the formula? We'll think of that as a black box or as a green box in this case if we just accept that this formula is absolutely true Then the correct answer is not actually thirty three watts It's thirty three thirty two point nine nine eight three But even that answer isn't actually precise because you'll notice along the way, right? 20 or two point one five five squared. Is that exactly? 4.6 4403 I would probably guess not That was rounded right four times point one times 180. That is actually 72 But the thing is this right here was probably rounded, right? So there's some type of rounding here. You add it to 72 then take the square root This number is rounded as well because after all we're an irrational numbers at this point In which case we add that together. That's legit, but they divided by point two There's a lot of rounding that we did along the way and then we finally rounded at the very end So with all of these rounding we get error into the system that is our calculations are not Exactly precise the precise answer would be an irrational number again That's not even talking about how erroneous the original model is if the if the model is accurate We still are gonna have some type of error So assuming thirty two point nine nine eight three is pretty good is rounding that that One hundredth, oh, you know two one thousandths of a watt. Does that make much of a difference? Well, it's not gonna be precisely two hundred degrees Celsius So one then sort of ask well, what's allowed then? Getting exact precision like perfect two hundred degrees Celsius. That's not gonna happen, right? Even if our instruments were so precise we could get that 33 32 point nine nine eight three watts each and every time, right? There's error in that number It's gonna fluctuate a little bit, right? And so I mean even though 200 is the optimal temperature to grow crystals It's very unreasonable to expect the temperature to be perfectly set at 200 degrees at all times during the growth of the crystal Since fluctuations in environmental temperature like the room the city that this laboratory is in Also, there's power issues from the power supply, right? That could cause marginal maybe even Microscopic defects in maintaining the temperature. So instead we should be concerned with how close to 200 degrees is close enough That is how much error is considered allowable And so let's say that our researchers have come to the conclusions through their data that plus or minus one degree Celsius is Allowable amount of error that is as long as it doesn't get above one degree or below one degree from 200 degrees Then the creep are then the crystals will grow in an optimal temperature So if the temperature allowed to vary from 200 degrees Celsius by up to plus or minus one degree Celsius What range of wattage is allowed for the input value? Well, if we go one degree above 200 that's 200 plus one sets 201 If we go one degree below 200 degrees, that'll be 200 minus one, which is 199 degrees so this right here is the Allowable fluctuation in the temperature. So this gives us an interval of Temperatures so our temperature can range from 199 to 201. So this is considered acceptable. Okay, this is allowable But how do we get these exact numbers? Well, if you take the function from the previous slide, yeah, you know the point one W squared plus two one five five w plus 20. That's t of w here We had to solve that equation again set t of w equal to 199 It's a track 199 from both sides put in the quadratic formula if you approximate your answer You'll get approximately thirty two point eight eight three nine which remember the the ideal temperature Excuse me the ideal wattage we did on the previous slide was thirty two point nine nine eight three watts Right. That was that was the ideal the optimal one So if we if we allow ourselves to go down to hundred and ninety nine degrees Then we can allow the wattage to go down to thirty two point eight eight three nine On the other hand if we want to figure out how high can we get in terms of wattage? Well t of w if you set that equal to 201 and solve the result in quadratic equation I won't go through the details of this. You'll get approximately thirty three point one one two four And so this then gives you another interval So in terms of the temperature the temperature is allowed to fluctuate between 199 degrees and 201 degrees which means that the wattage is allowed to fluctuate from thirty two point eight eight three nine watts up to thirty three point one one two four watts So telling the telling the operator to keep it at thirty three watts would be an allowable power level to maintain this level of This level of temperature and so this example is really motivating This idea that we're trying to build towards with the idea of a limit So when working with these functions we can control the range of the output by controlling the domain of the input Right, we didn't have direct control on the temperature But we did have direct control on the wattage right when you go home and bake cookies, right? You don't actually control the temperature you control assuming you have a conventional electric oven You have control over the power right when you turn the knob that doesn't make the temperature increase That makes the power increase which that has a consequence of changing the temperature, right? And so well from what we saw from this example here if though if we put the watts W in the interval thirty two point eight to a lap thirty three point eleven that will guarantee that the temperature is between 199 and two hundred and one Alright, and so this this is what we're mean by error And so now we're officially in a place where we can define what that means let the Greek letter Epsilon denote our numerical error Tolerance the Greek letter Epsilon which kind of looks like an E It's sort of like the Greek equivalent of the E for error here And so what is our error tolerance? It's the marginal amount of actual output Which is acceptable from the desired one, right? So when it came to when it came to the The example with the furnace our desired output, which we call L L's actually short for limit here We'll talk about that in the next video L was 200 degrees All right, but then we discovered that the error was allowed to be one degree Celsius Okay, and so it's important to it's important to know when you talk about error You can't have negative error or negative error doesn't make any sense Also, you can't let you can't put error to be zero because zero error mean perfection for which although that would be ideal That's not practical. We anticipate error in the system. So we're not gonna allow negative or zero error so oftentimes in Context of limits people often say things like let epsilon be greater than zero. We're saying that we have some You know some specific positive value for error, although in proofs we like to keep this generic for the setting of the exercise We just did the error was specifically one that was the allowed Error in that example and so if you then allow yourself to vary We took two hundred minus one and we took two hundred plus one We took that interval that gave us the interval of 199 and 201 if we do that Generically we're gonna take our desired output minus the error as the lower bound And then we're gonna take our desired output L plus the error and this didn't this interval gives us the so-called Margin of error. This is the interval for which we can fluctuate our Indirect variable or output variable and that would be considered acceptable Okay, there's so the margin of error is this allowable Fluctuations and you know as as students as many of you watching this video are probably students another good example of error Take your typical calculus students say in a class like this one this of course You know students typically want to get good grades in a calculus class, right? So if you want to get a passing grade in my calculus class, you'd have to get a 73 or above that would give you a C above above a C C or better is considered a passing grade But for some students a passing grade is not good enough Maybe a student wants to ace the class right in which case if you want to get an A in the class then your final Average needs to be at least 93 or above And so you know for the C student who just wants to pass their allowance would be 73 to a hundred percent But for the a student theirs would be 93 to a hundred percent, right? So that the a student has a much smaller margin of error in terms of their final grade that is they allow Less error in terms of calculating their final grade here. Alright, so that that's sort of like a funny little aside Okay, maybe it's not that funny, but you know, it's relevant to the typical view right now So continuing on with this idea of error, right this margin of error If we have some function, right? So you see a function f right here We have some function which we don't have direct control on the output variable Which for the sake of argument, let's call the output variable here y and what we do have direct control over is the input Variables call it x for the lack of a better name right here So let L be our target value. This is the number we want to output from the function So for our furnace example, this was 200 degrees now this desired value will coincide with some You know some portion of the function, right? If it's inside the range of the function It should be hitting the function somewhere But we do allow ourselves to go a little bit below L and a little bit above L by a Quantity which we call epsilon the error, right? So we allow to go above or below L So this little strip you see on the screen right here This is our margin of error as long as we're within these two Horizontal lines then we're considered acceptable. We can tolerate any fluctuation in the margin of error Okay, and so therefore what we're looking for is this expression right here f of x Which is the y-coordinate. So y equals f of x f of x as long as it's greater than L minus epsilon the lower bound But less than L plus epsilon which is the upper bound that's considered acceptable if f of x is between L Is it if f of x is between L plus or minus epsilon that is acceptable Which because of the symmetric bounds right here We can actually simplify this box in the following way if you take the absolute value of f of x minus L That needs to be less than epsilon the idea is if you take the absolute value We don't know if you're a little bit above or a little bit below L So that way you can get the we can get rid of the plus and minus epsilon. We can consolidate as a single statement All right, so now that a margin of error is given it is desirable to have an interval of inputs Which will guarantee that the outputs land within the margin of error This interval of input is what we will call the domain of allowance which will be denoted by a Minus delta comma a plus delta. Let me explain what those mean a little bit here So we have L our desired value have epsilon the allowable error and then we get this margin of error like so Well, this this value L is going to coincide with the function somewhere, right? So in order to hit L like what's a perfect bullseye what what do we aim to hit L? Well for this function We're calling that number a that if I take the number a and put it into my machine put inside my function It'll coincide it'll produce the value L and the reason we call it a in this context think of it It's like we're playing archery. We're aiming right If we have some target over here, right? There's a bullseye that we want to hit and we're shooting our arrow I know that these graphics are superb, right? I should be working for Pixar I think right now, but when you're if you're if you're an athlete in the sport of archery if you're an archer, right? What your desired target is the bullseye you want to hit the bullseye now the bullseye is not just a single point It's actually a circle anywhere in that circle is considered the full point That's considered a bullseye and so you then need to shoot your arrow into the bullseye now You can't just like run up and thrust your arrow into the target, right? That's not acceptable So you can't choose where you're going to hit what you can do is you can choose how you're going to aim, right? So the variables you would have control over as an archer is you can control the angle of inclination You can show control how far back you pull the string You can choose to shoot side to side based upon the wind There's a couple variables in play here, but these are the variables you have direct control over and then Where you hit that is where you have indirect control over that Controlling a true a good archer is one who can control the direct variables to get the indirect variable that they want That's what we're trying to do right here. So a here stands for aim that if we aim using the number a a Perfect aim would hit the value L perfectly over here But again, we allow some type of error So what fluctuation can we have on a to guarantee that will be in the margin of error? So like how far to the left of a can we go how far to the right of a can we go? And so this gives us this blue strip gives us the domain of allowance If we're in if if our input is within the domain of allowance That'll guarantee that the output is within the margin of error And so we want to choose delta. So how far to the left or right of a can we get? So we have our margin of error and our domain of allowance Let's consider the previous example and make sure we label all of these parts right here So our target value our target value like we said earlier was L to be 200 200 degrees That was the target our epsilon was one. We were allowed to go one below or above 200 and so this then gives us our margin of error our margin of error Sitting right here. So then the perfect value the perfect value that hits 200 degrees That was our number 32.9983 so that was the perfect value if you hit the wattage at that value, you'll get 200 degrees precisely and then we saw that our domain of Allowance here was that if you had 32.8839 watts you would You would hit 199 degrees And also if you had 33.1124 that would give you the 201 degrees. I'll just label those real quickly 199 degrees now when it comes to our picture that we have right here You'll notice that you go the same amount above as you below but because of a function's Curvature it could be that one side is more oblong than the other and So what we're trying to do what what what the type of homework questions? You'll see for this type of unit right here is you're trying to figure out what? Symmetric choice of Delta are you gonna choose and so what you're gonna do is you're gonna look for the piece on the left side The piece on the right side You're gonna subtract it from a and then pick the smaller value So the minimum the minimum is gonna be this value Delta that we're looking for here And so for our function if you take the a value 32.9983 and subtract it from the left bound 32.8839 you get the value point one one four four Okay, if you take the bigger value thirty three point one one two four and subtract that from a which is thirty two point nine nine eight three That'll give you point one one four one which you'll notice that this one was a little bit smaller point one one four one Versus point one one four four so it turns on the left you actually have a little bit more allowance than you do on the right But we're always gonna choose the restrictive we have to round to the smaller one here And so for this problem the Delta value is gonna be point one one four one and Then that gives us the domain of Allowance you're gonna take your a value and subtract from it and add to it Delta So thirty two point nine nine eight three subtract Delta gives you thirty two point eight eight four two You'll notice that's a little bit closer to a than we started with and then the other one will be Thirty three point one one two four, which is a plus Delta You