Tommy, Author at Every Step Calculus

U Substitution, Completing the square, 1/√(2x-x^2)

January 30, 2018 by Tommy Leave a Comment

Transcript

Hello, Tom from everystepcalculus.com and everystepphysics.com, I’m going to show you how to do this integration problem that a student asked for help with Yahoo and so I’ll show you how my programs work on this. Index 8 to get to my menu I’m already at U substitution, I’ll show you why it’s used substitution in a little bit. We’re gonna enter the function you have to press alpha first before you enter things in these entry lines here in my program. So, I’m going to press alpha the problem is one / what put the left parenthesis in for the divisor and second and then we get the square root sign of 2 times X – X² and then cap it off with the 2 parentheses. Press Enter and looks pretty good here we’re gonna rewrite it, bringing the X² – 2X in the proper form because we’re going to complete the square. Something you’re probably an expert on but not me so I wanted to show you how to do that and we’re going to complete the square by taking the half of the middle term here squaring it. This is – 1, – 1² and anytime in mathematics when you add something to a formula or function you have to take away the same thing to make it easy. You can’t change things just for no reason so we subtract the – 1² here also. Okay, this sets up the identity the integral of one divided by the square root of a squared minus U², DU is equal to the arc sine of U over A plus C. So, U is X – 2 and A = 1 here’s the U² or use substitution area. So, pathetic calculus you know and so we add the U and the A and we come up with the answer here. Arc sine of X – 2 plus C isn’t this wonderful.

Now, notice that the any time you do the arc sine you’re finding the angle in trigonometry okay. So, now you’ve found the angle and you still have to add some arbitrary C there’s ten million answers to this problem unless somebody comes up with a letter C okay. So, again when you find these answers notice how nonsensical they are and how useless they are the answers which calculus does totally it suggested hundreds of little puzzles that professors have dreamed up to solve here and they solve nothing. They just come up with answers and we get so involved when we’re trying to pass tests that we get involved with this stuff and say oh that’s good. We just got the answer here and the answer is generally is complicated or more complicated than the original function but that’s calculus okay. Now, you can go to my site buy my programs for $40, best $40 you’ll ever spend and you already can study this stuff for hours like I have to find the steps to do it. Find the best way of doing these problems and you know you can buy my programs and pass calculus and cut down on your study time and everything else okay. So, think about that but have a good one okay.

Finding Arc Length, & Unit Tangent Vector, given a position function r(t)

January 21, 2018 by Tommy Leave a Comment

Transcript

Hello, Tom for everystepcalculus.com and everystepphysics.com this is a another nonsense problem by calculus said to me by a student. I would imagine it never appear out of test but I’m going to do it anyways regarding an arc. Arc length and the unit tangent vector okay so, let’s do it index 8 to get to my menu we’re going to scroll down here to arc length in RTC. The RT here that’s a position vector, a vector has magnitude and direction in this case its tangent. So, therefore whatever curve this turns out to be on a graphing which I don’t know how to graph it. But somebody made this function here some mathematician or person made this function and they’re trying to convince us that this solves stuff in life and I say it doesn’t I say it comes up with a number which will show you this number right now and here it comes up with units. As if that was important to anything and it’s also one numbers like taking six times seven coming up with forty-two and saying hey there you see it’s that’s the orbit around Neptune, Neptune or something like that you know some wild statement.

So, anyways let’s do it arc length and RT, we’re going to choose number 3 RT I have program the X and Y values also and I’ve already entered these to save time okay. All these functions but you’re gonna press alpha and put every one of these variables in here okay exactly like you see it, make sure you put time sign between the T etc. But I’ve already added them so here they are two times T times cosine not this right here I guess what the original function this one here okay and – T etc. So, I always show you, you can change it if you want I say it’s okay and we’re gonna do number four arc length okay. You can do all these things with what you’ve entered here. We’re gonna do and here’s the formula you write this stuff down in your paper for your test or homework or whatever. L is length of an arc it’s really a definite integral and that it uses the magnitude of the derivative of the RT function to come up with this stuff here and I show you all this you just keep writing in your paper putting it down you look like a genius. Now X of T is equal to remember I is X, J is Y, K is Z okay those are the parameters that they give us. So, the derivative of X of T this one right here is this right here two times cosine of T I’m not going to go through each one of it because it takes too much time for this video and here’s why T the next one – right here – 2 times T sine of T. Derivative of that is this right here of course you knew that didn’t you, you didn’t pull that stuff out of the air just like anybody can. Here Z of T here’s this right here and that turns out to the derivative is 2 times the square root of 2 times T okay.

Now, we’re going to find the magnitude we have to square all this stuff and here we have the integral here over A and B that range and we’re going to put in the range okay. So, I haven’t program that’s where we’re going to go alpha 0 for the value of T in the upper range is alpha second I. Okay, so I show you this is the integral we’re doing with regard to T I just put that in there in case you made a mistake on the range. So, X prime of T is equal to this and then we’re going to so then X prime of T² is equal to this right here. Okay, the same thing with YT here’s the square of it and here’s the square of the Z parameter K. So, now we’re gonna do it over the range we got this that all in here we’re still doing the integral of this all stuff okay. You notice we got the square root of this here okay square root of this remember when you do an integral of the square root you change it to half, your exponent half right here. So, here’s what we’re still completing and of course the integral then is 2 plus 1 times the absolute depth T plus 1 okay and at T is 0 the problem as you put zero in for all the T’s and the answer for the integral. We get one and for pi we get PI plus 1² upper – lower always when you’re doing these this with a range a definite integral equals here right here so this is the answer pi squared plus 2pi.

Now, this is the answer that they get here on the right, right they giving you the answer okay so it’s 16.513 units. Now, what does that mean I mean you know it doesn’t mean anything to me doesn’t mean anything to you. It’s completely irrelevant you’re not gonna use this to go into space with let me tell you okay. So, this is the length of this unit tangent mention vector okay now we’re going to do more calculation we’re going to go back here number one and we’re going to scroll down here too unit tangent vector. You know tangible is given I’m against teaching this stuff but I’ll give you a little hint of what you’re talking about here and of course the length is one in a unit tangent vector okay. You can read that in your spare time and here’s the formula unit tangent vector is written just like this and all the calculus books derivative RT is over the magnitude of it and I show you what you’re doing here. Here’s the derivatives of each one of those pi J and K and the magnitude how you come up with that okay and so we’re going to do the derivative of this that equals this is what we found before. I drew the magnitude right here of course this is the numerator and this is the denominator.

We’re trying to and here’s the answer now you can see the answer that they have two times cosine of t minus two T sine of T okay, that’s this one here this one here over here for the J is minus two T cosine T two times sine of T okay. It’s all over four times T plus one squared now they took the square root outside the radical sign but no problem go to my site buy my program there only forty bucks and you’ll be able to do all this stuff forever you’ll notice when I. This is the greatest notebook ever these in the calculator here because my brain tells me why would I ever study something hard and not write it down somewhere where he could refer to it in the future. Well, I programmed it so in the calculator so that I can do actually do the problem no matter what variables they get. So, think about buying my programs and you’ll pass calculus you’ll get six or seven problems right in any test compared to the guy sitting next to you or the girl sitting next to you and because it’s all scored on partial credit in the class curve that’s all you need to pass the class and I never wanted to get a I did get A’s and in calculus but I never wanted to I’ve got a D I don’t care. Just get me out of there, I don’t ever let me touch this stuff again okay. Hey have a good one

U Substitution with square root involved, x√(x^2+4) or x(x^2+4)^(1/2)

January 15, 2018 by Tommy Leave a Comment

Transcript
Hello, Tom from everystepcalculus.com and everystepphysics.com again. This is a problem that was sent in by a student to Yahoo integrate X times the quantity X squared + 4 to the ½ power. So, let me show you how we do that in my programs okay. Index 8 to get to my menu we’re gonna go on a scroll up to get to U substitution the reason you know it’s U substitution because you take the derivative of what’s inside the parentheses here. Which is 2x and that is almost matching the outside of the problem except for the two you know and so then that’s someone my say that might be U substitution. So, then you go to use substitution in my programs, here we go there and we’re gonna enter the function you have to press alpha first. Alpha X times the quantity X2 plus four close off the parenthesis to the left parentheses, 1 divide by 2 power and you noticed the calculator and I would do anybody in a mathematician would you changed anything to the half power to a square root situation. Until you get into integrating it then you change it back okay so, this is the problem here I wish to give you a chance to change in case you made a mistake but that’s the correct one and we’re waiting for the program to load. So, here’s the problem we’re gonna rewrite it with the X next to the DX okay I like to do that because it keeps you organized as far as what you’re doing with the next step which is the most important one. Remember this one forever this is the one very simple U equals at the inside of the parenthesis okay, the derivative of that is 2X is DX. We take the 2 with algebra and divide it than the other side by ignore size by two with XDX notice this is the same is what we rewrote the problem with okay. So, you can forget about that portion of it now, we have integral of this right here so that’s the same thing as the integral of the square root of U, DU divided by two okay constants come out of the integral course and you do this write this stuff down exactly as what you’re looking at here. Excuse me that is a firetruck coming by and here’s the answer right here. Okay, pretty neat huh everystepcalculus.com go to my site buy my programs are only $40.00 nothing like you’d spend in a bar or a pizza house or out to dinner and yet you have this stuff in your calculator forever able to do all these problems, hundreds of problems in my; in the calculator okay. This is the greatest notebook because when I research a problem why? Why would I just put it on a piece of paper and throw it out you know I program it put it in here I have it for the rest of my life and you will too if you buy my program so keep that in mind and you’ll be able to pass calculus because you’ll get six or seven problems exactly right in any test of calculus compared to the guys or girls sitting next to you and of course the class is scored on partial credit and the class curve. So, that’s the reason that you don’t need to get a hundred percent on every test to get an A or pass the class or test. Hey, have a good one.

Integration by parts, integrate by parts of, xe^(2x)

January 13, 2018 by Tommy Leave a Comment

Transcript

Hello, Tom from everystepcalculus.com and everystepphysics.com. We’re gonna do a problem in calculus regarding integration by parts. Okay, I’m gonna do a series of these because that’s what I’m working on right now or cleaning it up whatever. Answering more problems for people this is once answer was given by a student for Yahoo, integrate Y equals X times E to the 2X. How do I do this does E stay the same I mean these are common questions are you gonna integrate that okay. Well, because it’s E to the X you know that you know you have to learn something in calculus you’re gonna say that’s integration by parts anytime you’re this called a transcendental functioning. So, you have to integrate that with the integration by parts you know so, I’m gonna show you how to do that index 8 to get to my menu. I’m already adding integrate by parts I’ll pull it up here so you can see it integrate by parts. Okay, so that’s what you choose and we’re gonna press enter and we’re gonna choose E to the X here okay. You’re going to scroll to that I’m going to scroll up to that you can press the number in front of it if you want I mean that’s quicker. I’ll do that I’m going to press number two we’re going to integrate some form of E to the X there is E to the X okay.

Now, I’ve already entered the problem here so for quickness and so the simulator doesn’t screw up on me but I’m gonna show you I’m gonna press number two here and go back for you to do it you press alpha first before you enter anything in here. Alpha you’d go X and then X and then this yellow button here and then the X which is the E portion of it. E to the two times X and close off the parenthesis on the right and here’s what you’d have right here okay and we’re gonna say it’s one okay. So, here’s what we do here’s the problem we go U equals x which is out here, D U is equal to a derivative of that which is 1DX okay and then DV is E to the 2X and we’re trying to get the V answer. Which is the integral of the DV with respect to X which turns out to be E to the 2x divided by 2 okay. Write this on your paper just exactly like this okay, the formula for that for integration by parts is UV minus the integral of VDU and so we’re going to add what we just found here. U is X ok times V which is E to the 2x divided by 2 minus the integral of E to the 2x my divided by 2 times the derivative of that which is D U which is 1DX okay and then we clean it up I multiply it together clean up this and here’s what we end up there okay.

So, now we’re gonna do this integral here and when we do the integral it becomes minus E to the 2x divided by 4 plus C. Anytime you do an integration it’s got plus C unless it’s a definite integral where you have a certain range to it and so then where’d we have to distribute the minus sign here. Which I do so it’s plus a minus E to the 2x divided by four now the trick here factor out E to the 2x, if we factor out E to the 2x we get 2x minus 1 times e to the 2x divided by 4 plus C. Which is the answer pretty neat huh go to my site buy my programs and enjoy passing calculus they’re only 40 bucks, well worth the money you’ll have it for the rest of your life in your calculator. You can do calculus like I can do it for the rest of your life where you couldn’t no matter how many books you have in your bookshelf or notes from your professor remember these are null notes from me. In other words once I do a problem and figure out how the step by steps go these are my notes. This is my notes and so you can bring this into any test and you’re gonna pass that test with calculus okay. So. think about that no big deal. Hey, go to my site buy my program and pass calculus. Have a good one.

Linear Approximation, 1 var, f(x) = x^5, x = 3

September 28, 2017 by Tommy Leave a Comment

Hello again, Tom here, we’re gonna do a problem with linear approximation with one variable, generally in calculus 1, and let’s do it. Index 8 of course to get to my menu from, to get to the main menu, I’m already at a linear approximation with one variable here to save time on the video. Press enter, linear approximation, 1 variable, and we’re finding the equation of a tangent line, and then working finding b in that, finding, also finding m which is the slope. Usual problem in calculus, and so we have to press enter before we enter anything in here and the function is alpha x to the 5^th. It’ll show you what you’ve entered, you can change it if you want, I say it’s okay, they give it as the point being alpha 3. and of course here’s the original function, x to the 5, the derivative of that is 5 x to the 4, that’ the slope of a line, remember that. We’re gonna find the slope of a tangent line, to the point 3, and we do that by taking the derivative and then substituting 3 for 4 in here, we come up with 405, the slope is equal to 405. now remember the slope, there’s a 1 over here of 405 remember the slope is rise over run, right? So you got really the rise is 405 and the run is, denominator is 1, so if you went up 405 on the x axes, and over 1 on the, on the x axes, draw a line and that’s really the origin of the graph, that’s what you’ve found here the slope, okay? Now we take the 3, we put it into the original function, we get the y value at that point, which is 243. and so here’s the point we’re at, we’re looking at 3, and y is 243. and so then we do the computations here, y=243, we do algebra stuff here, 405 and 3 for x to find b, turns out to be b is -972. so the equation of a tangent line is 405x + -972. now they give us what we’re supposed to approximate at which is alpha 3.6. so we’re gonna add 3.6 into this equation here, and we come up with y=486. So the x y coordinates 3.6 and the y is 486. pretty neat how everystepcalculus.com go to my site, buy my programs, and enjoy passing calculus. Have a good one.

Linear Approximation, 2 var, fx = √(20 x^2 7y^2)

September 26, 2017 by Tommy Leave a Comment

Hello, Tom from everystepcalculus.com and everystepphysics.com. This is a linear approximation problem in calculus 2, we’ve got two variables, x and y, and let me show you how it’s done. Index 8 to get to my menu at my programs here, I’m already at linear approximation, but you’d scroll to that if you, and we’re gonna choose two variables, here’s the formula, you have the original function plus the partial differentiation for x and partial differentiation for y, and then dx and dy, okay? So this is the formula you’re gonna use. You’re gonna enter the function, you have to press L for first before you enter anything in these entry lines, and the function is alpha. 20, I have to go slow because of the simulator it’ll mess up, minus x, squared, whoops, didn’t need second, squared, minus 7 times y squared, minus 7 times y squared. Oh, that’s a square root so we’ve gotta, we’ll put the parentheses in here, and then I’ll go all the way back to the front, put the square root sign in, the alpha x, see square root of, yeah, it’s correct, it’ll give you a chance to change it in case it’s incorrect, and they’re gonna give us the they want you to approximate the x variables and the y variables at, here’s your alpha first, the x is 1.95, 1.95, and the y value is 1.08, so you alpha 1.08, and again that’ll show you what you’ve entered, you can change it, say it’s okay, and then I choose the points [inaudible] you’re supposed to choose the points closest to it, and equals a and b, okay? So dx is equal to x-a is equal to the x value minus a, -.05, and dy is equal to y -b, 1.08-1 = .08. now we do the derivatives [inaudible] to x, with respect to x, turns out to be this here, with respect to y turns out to be here, and that a and b 2 and 1 you add that into the, you know, the derivative with respect to x, and the function here, and you come up with -.6667. here the same thing for y, for 2 and 1 comes up to -2.3330 or 333. and then we do the original function with 2 and 1, comes up with, here’s the visual function, comes out to 3. so, here’s the formula again, f + fx dx + fy times dy, you add these variables, here’s the answer 2.8467, that’s the answer linear approximation of that, of the original function. Pretty neat how every step calculus got [inaudible] buy my programs and help you do all this stuff for your homework and your tests. Calculus is nonsense so treat it that way, try not to learn too much about it, just pass the tests and get out of there. you have a good one.

Linear Approximation, 1 variable, f(x) = 3*√(x), x=27

September 25, 2017 by Tommy Leave a Comment

Hello again, Tom from everystepcalculus.com everystepphysics.com this is a video on linear approximation for calculus. It’s a one variable situation, so probably calculus 1, and I’m gonna read it to you. Use linear approximation, in other words the tangent line to approximate 3 times square root of 27 as follows. A and f(x) equals 3 square root of x the equation the tangent line of f(x) at x=27. If you’ve written in the form y=mx+b, find a and b. then get m and b. so, let’s do it. Index 8 to get to my menu, and we’re gonna scroll down here to linear approximation in the L section, and we’ve got to choose one variable because there’s only x given, okay? 1 variable. The point slope form is y=mx+b, it’s a standard tangent line equation, and we’re gonna enter what’s given. Function is given, and that would be if you press L [inaudible] if there’s any lines here and they give it as alpha 3 times e square root of x. and I’ll show you what you’ve got here, you can change it if you made a mistake. Now I’ve gotta enter the point that we’re given, which is alpha 27. I say that’s okay also. So the first thing you do, you write the function here, you take the derivative of a function equals three, I’m gonna find the slope. Okay, so here’s the derivative of 3 times the square root of x, and they, they’re supposed to decide the closest square root to that number, I do that for you which is 25. and so we enter 25 into the derivative, and so the m is the slope = 3/10. now we take 25 and put it into the original function here, comes up with y=15. So here’s the point, 25 for x, 15 for y. at y=15 then we solve for b using this calculations here, just straight ahead algebra, and equals this right down here, mx+b okay? Now we plug in the point to evaluate, 27, and with that we come up with these calculations here which is 78/5, so, the answer’s 27 and 78/5. so m is 3/10 and b is 15/2, that would be the answer to this problem, okay? Go to my site if you like these programs, buy them, and enjoy passing calculus, okay? Have a good one.

Mean Value Theorem, y=x^3+x 1, [0,2]

September 8, 2017 by Tommy Leave a Comment

Transcript
Hello! Tom from everystepcalculus.com and everystepphysics.com. Mean value theorem problem for calculus II. Index 8 to get to my menu. I’m going to scroll down here to mean value theorem and we are going to enter our function. You have to press alpha first. The function is alpha x^3 +x-1, [0,2] is the given limit here so alpha(0), alpha(2). It will show you what you’ve entered, you can change it if you want, else say it’s okay. It’ll show you the formula. I know we have f, Notice we have f(b)/2 – f(a) and (b-a), and we have f’(c). So, f(a) is equal to this. We put zero I for all x’s in the function, come up with -1. Put 2, the upper limit into the function, come up with 9. We subtract b-a, here is b-a, was 2, and we put it back into the formula: [9-(-1)]/2. Turn out to be 5. Now we are going to take our function, take the first derivative of it, 3x^2+1, change it to c, x’s to c: 3c^2+1=5. We are going to solve for c which equals this here. And the answer is two [2(3)^(1/2)]/3, the approximation is 1.1547. Go to my site, buy my programs and have a great time passing calculus otherwise suffer because it’s such a worthless course, lot of study but no reward except to get through calculus and move on with your life. So worthless math concept or subject and it goes along with crossword puzzles or Sudoku, [unclear 03:06] maybe. Same thing. After 25 years of studying this this is what I believe. So anyways, have a good one!

Mean Value, 400x+3x^2, [0,3]

September 7, 2017 by Tommy Leave a Comment

Transcript
Hello! Tom from everystepcalculus.com and everystepphysics.com. This is a mean value problem in calculus II not a mean value theorem problem, mean value. It finds the average value or the mean value under a curve using an integral. Integrals are areas under the curves, and this finds, actually finds nonsense in calculus, [unclear 0:31] calculus. Sorry. 25 years of studying this I find no value in it, practical value. So, index 8 to get to my menu. We have to pass our calculus class though. Okay. I’m already at mean value so I’m going to choose that. These programs are in your titanium calculator that you have in your purse or your schoolbag, and you’ll be able to solve all these problems in my menu forever. Compare to your calculus book, you can throw away your calculus books or your study materials from your professors, absolutely worthless but this calculator has all your calculus computations forever as long as you have the calculator for your children, your grandchildren, your partners or your spouses or sisters or brothers, whoever in the future. And not for a very much price either, let me tell you. Finding a mean value, let’s do it. Here’s the formula here. Compare to the root mean square formula which has the square root, all these are no square root signs. I’m going to enter the function, you have to press alpha first. The function is 400x+3x^2, the lower limit is alpha(0), the upper value is alpha(3). And it will show you which values you’ve entered, you can change in case you made a mistake, else say it’s okay. And we start the process of computations of these. Okay, here’s the formula. I put the integral sign down here, this is the mid of the [02:49] because it will off the screen if I did it any other way. So, remember that. No big deal. It still has a dx with respect to x here (3*x^2+400*x)*dx. We just subtracted the b-a here and get 3, and we still have the integral over the limits of 0 and 3 right here. We are going to do the integration now. Here’s the integration x^3+200x^2, and we are going to compute that, these x values with the upper limit and lower limit. It actually equals 3. You write these on your paper, you look like you are a genius, right? 1827, okay, now equals to 0. You write this on your paper equals 0. And the upper minus the lower always in physics, in calculus, upper minus lower. And the answer is, after we’ve put it back into the formula, the answer is 609. Hey, have a good one! Go to my site buy my programs and enjoy passing calculus.
Have a good one! Bye, bye.

U substitution, cos(ln(x))/x and cos(ln(x)), integration by parts

September 6, 2017 by Tommy Leave a Comment

Transcript
Hello! Tom from everystepcalculus.com and everystephysics.com. We are going to do a problem in calculus U substitution, and we are going to do a function with cosine in it. Index 8 to get to my menu. Press enter. I already had U substitution but you’ll scroll with this cursor here down or up or whatever to get to the [unclear 00:33] for the problem that you have at hand. So, U substitution is this problem and we are going to enter the function, we have to press alpha first before you enter anything in these entry lines here. So, we are going to press alpha, and then we want cosine, secant z here which is cosine, and then secant x which is log of x, close up the parentheses, and divided by x, is the function. So, it will show you what you have enter. You can change it if you want, else say it’s okay. Notice that the trick to use substitution is that whatever inside parenthesis here, you take the derivative of that and it’s got to be able to be matched to the outside somehow, okay? So, we are going to rewrite this because here we have 1/x here dx, so we are going to kind of isolate that so it’s already cleared for you. Cos(ln(x))*(1/x)dx, and then u = ln(x) du = 1/x
You have to memorize that. The integral of 1/x is ln(x) etc. so the opposite derivative of ln(x) is 1/x. So, now we are going to do the integral of cos u, du equal sin(u) + c. When you do the integral of cos u it’s sin u. so, now we have the answer of sin[ln(x)] + c, as the answer. Now, we are going to do another problem here, and we are going to press alpha, we are going to have secant. Cosine of second log of x, and without the x, divided by that. Okay? It shows you what you have entered, say it’s okay. And notice that this is not a U substitution problem but an integration by parts problem because notice that the derivatives of the parenthesis here, ln(x) is equal to 1/x but there’s nothing on the outside that equals 1/x. So, if that’s the case then you cannot integrate it by U substitution, you have to go to integration by parts. And that’s just pathetic here with the long, long here of getting the answer, more of Sudoku of math. So we have,
u = cos[ln(x)],
du = -sin[ln(x)],

Mean Value Theorem, 2√(x), [4,9]

September 6, 2017 by Tommy Leave a Comment

Long Division (x^4-9x^3+25x^2-8*x-2)/(x^2-2)

September 1, 2017 by Tommy Leave a Comment

Transcript
Hello, Tom from everystepcalculus.com and everystepphysics.com. Long division problem regarding functions you know in calculus if it comes up in calculus whatever, index 8 to get to my menu your going to change long division and your going to enter your function, you have to press alpha before you enter anything in these functions here I’ve already entered the problem so I’m going to press cause it’s a long problem and it takes time save time and you notice this is called the dividend (x^4-9*x^3+25*x^2-8*x-2) and this is called the divisor (x^2-2) you set it up just like a regular numerical long division problem, this off to the left if the screen was wider I could do that for you but I can’t so I have to do it this way and remember your dividing x ^2 into x ^4 first getting the answer of x ^2 and your multiplying x ^2*x^2 which is x^4 and then your going to multiple x^2*-2 which gets -2x^2and your going to change the signs because your subtracting from this so I’ll show you how to do that but remember that this is a divisor we’re operating dividing this into several things that come along here okay so it always give you a chance to change incase you’ve made a mistake but I’ll say it’s okay now here we’ve divided the divisor x^2 into x^4 and you get x^2 and you multiple that times it and then change the sign -x^4 we multiple x^2*-2 and we get -2x^2 and we change the sign to make it +2x^2 okay and then we add them, you notice now that we have 25x^2 up here we’ll if the screen was wider we would have -8x over here and -2 over here but I have to bring it down to show you that it’s still there okay but we are going to deal with this one and this one and this plus this is 27 and this of course becomes 0 so we have 0+(27*x^2) and we bring down the (-26*x)+(-2) okay we are going to divide x^2 into 27*x^2 and we are going to get 27 and we change the sign, write this all on your paper exactly as you see it and you have to keep the coulomb’s of x^2 and x and x^3 all together okay now 27*-2 because a minus -18*x okay -18x-8x is 26x-26x and we still have to bring down the -2 okay and we of course multiple that to come up with a minus which negates these two here and then when we multiple the minus 2 times the minus 27 we get -54 and that change its sign and become 54 okay and 54-2 here becomes 52 so we get 52 we got -2x now once in long division the divisor cannot be divided into this anymore this all becomes as one then you divide that by the divisor so we have [(-26*x)+(52)]/(x^2-2) here is the answer (x^2)+(-9*x)+(27)+[(-26*x)+(52)]/(x^2-2) complicated you have to do it on paper and you will see where it’s much clearer I have trouble explaining it here because of such a small screen and it’s all crammed into one but all the answer is correct and you will get all the problems are correct. Pretty neat hah everystepcalculus.com go to my site buy my programs and pass your calculus, have a good one.

Long Division, (x^4+3x^3-4x^2+5*x+7)/(x^3-3)

June 8, 2017 by Tommy Leave a Comment

Transcript

Hello everyone this is Tom from everystepcalculus.com and everystepphysics.com. This video is about long division of functions of a function I’m going to make others also, lets get started index 8 to get to my menu I’m already at long division to save time as you scroll down the menu and we are going to enter our problem you have to press alpha before you enter anything in these entry lines in my programs okay and the problem is alpha (x^4+3*x^3-4*x^2+5*x+7)/(x^3-3) press enter now it’s showing you what you have entered and you can change it if you want now notice that this is the divisor (x^3-3)and this is the is the dividend (x^4+3*x^3-4*x^2+5*x+7) here and there is not really any words for this line here the divisor line whatever it’s not hieratical sign if we had a larger screen than the titanium i would have the all just as a regular long division that we use in regular numbers, numeric long division but this is functions okay so remember this x^3-3 her now we are going to divide that first into x^4 it’s loading here when it says busy and once you’ve loaded it its very fast from then own in your programs so notice that equals x and x times x^3 is x ^4 and we have to change the sign to subtracting so subtract this an then x times a minus 3 is 3x -3x and then we change the sign to 3x okay now again if the screen was longer I would have 5x plus 7 right up here and you want to keep your answers for the, this is called a quotient up here this is a quotient and you want to keep these In there coulomb’s like this is in the x to the 4 Coulomb’s so minus this and it’s going to be zero and this is in the x coulomb’s 3x so you want to make sure this is still there don’t forget that yet and the next screen that equals zero and we pull down the 3x^3 and the minus x squared and then we add the X’s together which is 8x+7 so now x^3 into 3 x^3 which is 3 which we should you in quotient above and curses 3x 3 and when we subtract it we get zero of course here zero we pull down the -4*x^2 and the 8*x above and then we are adding this you notice that the 3 * the minus 3 and the divisor is minus 9 and we change the sign to make it a plus 9 so we get 16 now any time this is below the powers of devisers you are all done with long division in this okay so this is the answer, I’ll show you the answer here, here is the answer (x)+(3)+ [(-4*x^2)+(16)]/(x^3-3). Pretty neat hah everystepcalculus.com go to my site buy my programs and pass your calculus curse, do your homework etc. Have a good one.

Long Division, (x^5+x^2)/(x^2-1)

June 2, 2017 by Tommy Leave a Comment

Transcript

Hello everybody, this is Tom from everystepcalculus.com and everystepphysics.com. Doing a long division of functions problems with my programs showing you how that works, index 8 to get to my menu we are already at long division you will scroll down to that when you want to do it or whatever problem you want to tackle with my programs and we are going to enter the function and you have to press alpha before you enter anything in these entry lines here, alpha (x^5+x^2)÷(x^2-1) you notice now that your it up just like a regular division long division problem and numeric numbers her is your dividends (x^4-x^3+^2) and this is the divisor (x-1) there is no name particularly for the this line here and this line is not hieratical it’s a divisor line, so we are going to set it up like this and we are going to divide x into x 4 first and then whatever answer which is called the quotient your going to multiple that times both of these times here I’ll show you that in the next, busy means it’s loading the problem and it only happens slow like that when the first time you load it otherwise it’s very fast so x and the x4 is x^3 okay and x ^3 times x is x ^4 and you are going to subtract that right down here and this is equal to zero and then you are going to have x ^3*-1 which is a minus x^3 and you change the sign because you are subtracting so that’s a positive x ^3 and these will also cancel so what is left is x^2 here and you pull that down and all the rest have cancelled and since x ^3 can’t go into x ^2 you are all done with the problem and therefore this is pulled down here and then they’re divide by the divisor and your answer is (x^3)+(0)+(x^2)/(x-1). Pretty neat huh everystepcalculus.com go on my site buy my programs and enjoy passing calculus, have a good one.

Maclaurin Series, x*e^(-x)

May 31, 2017 by Tommy Leave a Comment

Hello, Tom from everystepcalculuseverystepphysic.com. Maclaurin series, let me show you how to do that in my programs, index 8 to get to the menu the main menu I’m going to scroll down to Maclaurin series and you can go on your calendar second and get through the alphabet quick where you want to go to I’m just going to scroll down here because the simulator sometimes screws up if you do that, so I’m to choose Maclaurin series, Maclaurin series is when your center is 0 (a) is 0 and Taylor series you have great than 0, here is the formula for Maclaurin series [f^n(0)*x^n]/ n! You should always write formulas on your paper you will get couple extra points and remember calculous is always good and partial credits on the class curves and you can get more than the person sitting next to you then you pass that class and that’s all we are interested in passing it not learning nothing, the Taylor series formula, we are going to enter the function we have to press alpha before you enter anything in these entry lines here so we are going to press alpha the function is alpha x* yellow button on then equal sign for e, (e) is a minus x and then we are going to put 0 in for the Maclaurin because that’s what we are doing finding the Maclaurin series and the we are going to have to press alpha again, alpha 0 and it’s going to show you what you have entered and you can change it if you want that’s if it isn’t right, we write this formula on our paper right? We are going to do that, we do the first 5 derivatives of that function which is here then we take those derivatives and we set it as to equal to zero, enter zero in for x and we come up with this an we take those answer’s and divide it by the factorials of 1,2,3,4,5 which turn out to be these answer’s 1,-1, 1/2, -1/6, 1/24 and we enter those answer’s into the formula here (x-0) to the 1,2,3,4,5 powers and here is the answer f(x)=0 don’t forget that 0 plus x plus minus x squared plus x cubed over 2 plus minus x to the 4/6 etc.etc. pretty neat huh everywhere is.com don’t try to pass calculous without these programs go to my site buy them enjoy passing calculous, okay calculous should only be taught to math majors not to us but since they do that we need to do anything we can to pass it and get out of there hey have a good one.

Maclaurin Series, 1/(1-x)

May 30, 2017 by Tommy Leave a Comment

Hello again everybody Tom from everystepcalculus.com and everystepphysics.com. Maclaurin series and other functions that I want to do comes up on the Yahoo answers problems students having trouble with it so let’s deal with these programs. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter you notice the difference between Taylor series and Maclaurin is if you use zero for the (a) which is the center point to do the series and this is the formula from Maclaurin series and here is the Taylor series formula now we are going to enter the function and we have to press alpha before we enter anything in these entry lines here and the function is alpha 1/(1-x) and because it’s Maclaurin we are going to put 0 in for (a) here, alpha 0, we could put the -1 here but that’s not what I entered let’s leave it and see what’s happening here I say it’s okay then, again write the Maclaurin series on your paper and you will get credit couple points you know because you partial credits in class curves so you don’t need to be a genius at this stuff just get a few more points than the turkey sitting next to you, and we do the derivatives here the first 5 derivatives equal this 120/(×-1)^6 and set that and put 0 in for x and we come up with 120, we take those answer’s and we divide it by the factorials of 1,2,3,4,5 and come up with all 1’s put all 1’s in here and multiplying it times x-0 to the powers of theses here and the answer is 1+x+x^2+x^3+x^4+x^5 that’s the answer. Go to my site buy my programs and enjoy passing calculous it’s a tough course and you have a good one.

Local Max & Min, 6x^3-9x^2+8, Graphing by Hand

May 30, 2017 by Tommy Leave a Comment

Hello everybody this is Tom from everystepcalculuseverystepphysic.com. We are going to do a maximum and minimum problem in graphing functions index eight to get to my menu we’re going to scroll down here to local max and min there is it there you can do all these things after you put the function in I always tell you to start on graph paper because when you have a test you are supposed to find these things through derivatives etc. then graph it by hand on a peace of paper, so you find the points and continue to graph it so we are going to enter the functions we have to press alpha before you enter anything in these entry lines here, alpha function 6*x^3-9*x^2+8 it’s going to show you what you have entered and you can change it if you want, say it’s okay and then we are going to scroll down here to local max and min and we are going to wait for it to load, this is busy when it’s loading the program, it only does that when your doing it for the first time you load it and the rest of the time it is very quick and first we find the first derivative and then we say it is equal to zero and factor it and then find the x values x= 0 or x=1 theses are critical numbers alone the x-axis they are not critical points, critical point is when you put theses into toe original functions then solve for y and you get the y value and that’s a point and we’ve done that here and (x,y) = (0,8), (x,y) = (1,5) these are the points, very important and a lot of professors don’t make that clear and then to find whether it’s a max or min you put the critical numbers into the second derivatives and if the answer is negative then you know you have a maximum and if the answer is a positive then it’s minimum, you notice that they are opposite positive minimum, negative maximum, so we find the second derivative here it is here put x=0 in for that we get -18, -18 means it’s a maximum now notice this is a positive and therefore it is a minimum, now I graph this for you so you can see exactly what is happening and you can see here at 0,8 we have a maximum at 1,5 we have a minimum. Pretty neat huh, everystepcalculus.com go on my site buy my programs pass calculus, have a good one bye bye.

Related Rates, Cone, Finding change in height dh/dt

April 25, 2017 by Tommy Leave a Comment

Hello again Tom from everystepcalculus.com everystepphysics.com. Related rates problems again regarding a cone or a cone shape, let me read it here one test problem, water is withdrawn from a conical reservoir 8 feet in diameter and 10 feet deep at the constant rate of 5 cubic feet per minute, how fast is the water level falling when the depth of the water in the reservoir is 6 feet? Let’s do it, index 8 to get to my menu then we are going to scroll, related rates is what we are looking for here and then we are going to choose cones number 4 now I haven’t choose the numbers here because I want the screen to be wide enough for you to see the problem properly so I’m just going to scroll down here to cones and what’s given is of course the volume, any time they give you cubic something its volume and then they give you the radius and the height so that’s what we are going to choose and we are finding dh/dt the changing height at a certain level etc. so choose that and have to pressing enter before you press anything in here entry lines alpha 5 feet per minute and its increasing its filling and it shows you what you have enter and you can change it if you want or say okay, the height is given as 10 alpha 10 say it’s okay and the radius is alpha 4 feet I say it’s okay, now we are trying to get rid of; we have r and h and the formula here this is the volume of a cone and we are trying to get so we can get one or the other and this is similar triangle; I don’t know what that means h/r=(10)/(4)=5/2 and so then r is what we use in algebra r=h/5/2 and so now we can substitute that in the formula here for r squared here it is here squared (h) we get (pie)(h) squared/ (75/4) (h) = (h)cubed * (pie)/ (75/4), we are going to differentiate with respect to the volume, time, and height so dV/dt=(h) squared * (pie)/ (25/4)(dh/dt) and they give is a certain height in the problem not radius so we are going to choose that and we are going to alpha 6 is what they give us, say it’s okay so now we are enter that in the (h) squared here (pie)/ (25/4)(dh/dt) =(18.1)(dh/dt) and we use the algebra (5.)/ (18.1) is dh/dt = .27631 (ft)/mn. Pretty neat ah, every step calculus.com go on my sight buy my programs you are going to enjoy them and you’ll have them for life in your calculator, you will throw your book away but you will never throw away your calculator with my programs because you can do it for your grandkids, your partner, your spouse and who ever in the future, somebody is going to run in calculus again in the future.

Equation of a tangent line, x^3+6xy+y^2= 8

April 21, 2017 by Tommy Leave a Comment

Transcript

Hello again everybody this is Tom from every step calculus.com every step physics.com. Calculus problem regarding a tangent line to a curve getting with two variables, I’m going to show you how that works on my programs, index 8 to get to my menu I’m already at equation of a tangent line here I’m going to enter the function you have to press alpha before you press anything in these entry lines, alpha x^3+6xy+y^2-8=0 you always have to transpose everything on the other side of the equation here and bring it over on the left side then make 0 to the right or a number to the right ether one would work, now because you enter y in here is going to be an implicit differentiation problem so here is the formula here that you just entered in the implicit system, it looks pretty good then click ok, differentiation both sides of equation with respect to x so here we go both sides and do each sides individually this one here because xy we need the product rule, you can write the product rule down and do it like this or you can just keep clicking and you will finally get to the product rule answer which is this right here which is (6+x) (dy/dx) and we keep going (2^y) = to the because it is a dy/dx also and you combine the functions dy/dx here and then no dy/dx just regular xy and the implicit answer is right here (-3*x ^2-6*y) divided by (6*x+2*y) now we are going to evaluate it at a point and the point is alpha 1 and alpha 1 so xy is 1 1 say its ok and we enter those variables into the numerator and the denominator come up with a slope of -9/8 slope of the line and then we figure the y=mx+b which is the slope m and the we add the one for the x and the one for the y and then figure out what b is here then you subtract this from one and one is of course 8/8 and when you subtract it you get 17/8 your subtracting a minus so you get 17/8 and here is the answer y=mx+b pretty neat, every step you go and I also give you the angle of the line the way it is pointing, if you notice the way it is pointing in this direction not up but down to the right and if you want to do it on a other points it will take you back there. Pretty neat ah, every step calculus.com go on my sight buy my programs if you want to pass calculus and subscribe if you want to see more videos that I might make. Hey have a good one.

Domain & Range, abs(x- 6), |x-6|, absolute function, calculus help

April 17, 2017 by Tommy Leave a Comment

Hello again everybody this is Tom from every step calculus.com everystepphysics.com. Problems in calculus 1 regarding domain and range and regarding absolute value of a function, index 8 to get to my menu we press enter we are already at domain it gives you a chance to if in case they give you a chance to test picture to find the domain and range you can do that with my programs by saying no because we want to enter our own function and the function is absolute and we enter it this way abs parentheses we have to press alpha now when we put the parentheses in, alpha parentheses the problem is x-6 close off the parentheses and it will show you what your entering and you can change it if you want and we need to find the value of x which is, we set the absolute value which is greater and equal to zero because that is the requirement for a under radical sign you cannot have a negative sign and then I show you the graph of it and then I show you the domain the domain is all real’s minus infinite plus infinite range includes this bracket here where the parentheses which means this includes 6 and it includes 6 to a positive infinity. Pretty neat ah, every step calculus.com go on my site buy my programs and have a good time passing calculus which is a very difficult subject as you probably already know and useless to us in all of our life.