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Point Slope Form Given two Points

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Tom from Every Step Calculus EveryStepPhysics.com we’re going to do a point slope form problem with two points that are given. And index8() to get to my menu. Choose point-slope equation and we’re going to choose number 1, 2 points are given. Press alpha before you enter anything, Use alpha line in these lines here when you enter any problem in my…enter the information in a problem that i’ve programmed. alpha 8 for x, 1 point alpha minus three for the y value. For the other point alpha five and alpha minus nine. I always show you what you’ve entered, you can change it if you want I say it’s okay. And we find the slope M by using this formula here. The answer is two and we find b line crosses that V is the line that crosses the y-axis y minus M times X 1 is minus 19 so the equation of a line y equals MX plus B it’s 2x plus minus 19 pretty neat huh everystepcalculus.com go to my site buy my programs and get through calculus somehow have a good one.

Point Slope Form Calculator

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Hello again, Tom from Every Step Calculus, Every Step Physics dot com. A point slope form calculator, that’s what we’re doing here. This calculator when you buy my programs and load them into you titanium here. It’s really my notes from many tests I got from all over the country. And when I find out a problem, I want to take the notes. Well, you take the notes you put them, great notes, and you put how to do the problem ok right? Then you always have it, that’s what my calculator does ok? Index(8) to get to my menu. Oh, by the way, there is a little square to the right of your screen here of this video, if you click on that you can subscribe to my channel and get more videos to you and you can see more videos. If you do that for me I certainly appreciate it. Index(8) to get to my menu you load that into the home screen of the calculator. And then we’re going to scroll to, I’m going to go with the curser up because it goes to the last, bottom of the alphabet work pumping, oil. And then we’re going to just just click up here to, to the P section and look for a point-slope equation. See R’s. There’s point-slope equation. Click on that when it says busy here it’s loading the program. Question of a tangent line, here’s where we learn y equals MX plus B that’s what we learned in algebra right? It’s the point-slope form equation. I always give you the equations because in a test you’re you’re scored by partial credit in the class curve so whatever you can throw down in your test paper even the equation is give you two points, and you’re looking for the most points okay? To get a good grade and get get out of there so now I gave you the choice here’s the two points right? So we’re gonna choose number one scroll to it or choose number one. We’re gonna enter the points we have to press alpha first, alpha five alpha 7, alpha six, alpha 8, now we show you what you’ve entered, you can change it if you want. I say it’s okay, and let’s do it the slope M is the slope with y1 minus y2 divided by x1 minus x2 pretty simple yeah, if you learned that and study it and remember it pay attention I guess you can remember this stuff. So here’s the y1 which is 7 and then 8 and divided by, divided by 5 minus 6 then equals minus 1 divided by minus 1 equals 1 that’s the slope. Now slope just as a kind of an information thing that’s this is slope is what the first derivative gives you okay now if they gave you a 1 here notice there’s a 1 over 1 and you have rise over run so to graph this slope. When you do the derivative of it you’re gonna go up 1 on the y-axis and 1 over on the x-axis and draw a line from that point down through the origin of the graph and that’s the slope that you’ve just figured when you do a derivative so keep that in mind okay? And then we’re gonna find the line crosses y-axis so we’re gonna take y1 equals M x1 7 minus 1 5 7 minus 5 equals 2. Equation of a line is y equals MX plus B equals 1 X plus 2 that’s C now if you graph that which I will show you right now that’s what you found okay? That’s the line that goes through that point okay? Let’s do a new, let’s do the other problem here here’s number two a point and the slope they got the point and the slope here okay? So let’s do that one number two alpha – 2, alpha 3 and to the slope alpha – one. I know it is there say it’s okay we could have changed it if we made a mistake and here’s the calculations plus three you can go slow here and learn it if you want to learn it but best things to buy my program just do it. And here’s y equals MX plus one now let’s look at that slope. That’s the other direction almost hits the origin of the graph but not quite okay? So anyways pretty neat huh everystepcalculus.com go to my site buy my programs and pass calculus plus I’ll give you the all the help you need you can call me and if you send me tests I’ll make sure it does the test for you but your practice tests or whatever I’m your private tutor here that it will help you all I can I’m interested in in fooling that professor and getting out of calculus that’s my goal that was my goal when I program these in the first place and I got A’s in calculus okay so I’m a genius at calculus now that’s for sure after 25 years so buy my program that enjoy passing calculus, hey have a good one!

Line Integral Test Question 5

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Hello, Tom from Every Step Calculus Every Step Physics .com. This is a problem submitted by a student. How do you evaluate this line integral? A line integral the original formula is the integral of a function times d R which is the derivative of the RT function here and then C is the what the parameters are going through which in this case is a curve. Trying to find the integral of this curve here. index(8) to get to my menu. Scroll down here to line integrals because that’s what they are asking you to do in the problem in your test or whatever, homework. So we’re gonna go down here to line integrals, press Enter . And we’re gonna choose number six I and J because that’s what they give you here, we’re in I and J okay? So you have to get used to that and my programs a little bit. You have to think a little bit in college so I and J we’re gonna choose number six and then in calculus 3 this is an M this might be calculus 2 here because we’re only dealing in XY & IJ parameters not XYZ and ijk parameters. So if it’s calculus two then this is an M which which you know tells you that this whole function here is an M and this is an N function before the J ok? So enter the m function term before the I, here’s the term before the I okay? And we’re gonna put that in now you’re gonna press alpha, that’s the only thing you have to do in my program, just press alpha and then you can enter the two times x times y plus one here in the calculator here okay? And then, so we did that, and then the end function here and we’re going to enter that x-squared plus 1 by pressing alpha first and then putting that term into the calculator okay? Now I’ve already done that in the simulator to save time. So you can see that it matches what they’re talking about the calculator always puts the XY term before the consonants you know so switch them around a little bit but no big deal that makes sense to us. And we’re saying it’s okay because you can change it in case you made a mistake in entering it. Our line segments given no they’re not, a segment is XY point, to another XY point, or XYZ point to the number another XYZ point depending upon what they tell you in the problem but no line segments so we’re gonna say no. But they do mention I and J in the in the problem okay so here’s I and J where is it? Here’s I and J we’re gonna say yes to that and we’re gonna enter the term before the i hits which is T okay so we we’d enter alpha t normally alpha t. And then we’re gonna enter the term before the J well here it is over here now this is a tricky problem because sine of PI over 2 is really 1 okay so you can put that in here you can enter sine of PI over 2 times T squared just like it says here but the calculator well we’ll assume that you’re talking about sine of PI over 2 is really 1 so it’s 1 times T squared. So if you enter that it’s just gonna come up with T squared anyways okay so here’s the I and the J that they’ve entered here and now we’re going to put in the range range is zero less or equal than t and then less or equal than one okay? So the lower limit is zero you’d put alpha zero in here and then you put alpha one just in case they gave you something different here. But generally when you parameterize something in a problem like this it’s always 0 to 1 ok? So we say it’s okay so now we start to do the the derivatives etc here’s the X term which is T which is T right here and the derivative of that is one here’s Y as T squared the derivative is 2t DT here’s the original integral and we’re going to put the all these what we found for X&Y; into the problem here which we’ve done I have to use quotation mark so the calculator does that but you’re gonna put parentheses around here like for x put parentheses around for T squared okay and you’re in your paper. Then we of course we multiply it out and condense it and here’s the real which you’ve entered here. And we’re gonna integrate the problem which is here and there you could do that really easy couldn’t you? I couldn’t, but you could over the range of 0 & 1 so we integrated this problem now at t equals 1 we compute it through there and we get seven halves at t equals zero we computing its 1/2 upper – lower always in physics and calculus upper – low which we’ve determined. And the answer is 3 pretty neat huh? everystepcalculus.com go to my site buy my programs are only 40 dollars and you’ll pass calculus and you don’t have to study this stuff hours and hours and hours. I’m pretty good at line integrals right now because I’ve been doing it for three weeks 25 years actually but then cleaning it up here for three weeks for other problems and almost all the problems I can I can find so have a good one, I hope you pass calculus!

Line Integral Test Question 4

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Hello Tom from EveryStepCalculus.com, EveryStepPhysics.com. A line integral problem again. I’m going to do this until I get done with the formulas that I’ve found so that you can do all these on your tests or whatever might come up on your test or homework . My programs are one-of-a-kind and the only one that does it. I’m not the only one they could do it but I’m the only one that does do it because I was into calculus at one time and then of course into programming which I like. Anyways you can go to my site buy my program for $40 and they’ll have you pass calculus, simple as that. And calculus I have determined is a worthless math concept, it doesn’t solve anything as far as I’m concerned. It can’t solve anything because it always comes up with one single number. It doesn’t come up with formulas or wild things that you can solve cancer with, or help with physical fitness and which they all say it does you know. But they’re just completely uninformed and and they don’t do this calculus as much as I do it. But anyways, index(8) to get to my menu, here’s a problem right off a test. Get to my main menu we’re gonna scroll down here I’m just gonna hold down on the cursor here to scroll down to the L section verse as line integrals. And choose that because that’s the subject we’re talking about. And here’s line integrals and we’re gonna do this, notice that these are all the possible test entries for line integrals, in other words you got dx dy, dx dy dz, function (x,y) a function of (x,y,z), and i j, and i j k and x equals okay? That’s what’s gonna be on any test you might take one of those. You have to choose which one. Now this one would be X Y Z they’ve left off the Y in here somewhere but you’ve got a Z here so you’re gonna choose, you certainly couldn’t do X Y. Choose number for X Y you’d have to choose number 5 X Y Z. So I’m going to press the number and scroll down or press the number. I’m gonna enter the function, now you’re gonna press alpha and enter this function here which is x squared times Z okay? Alpha X squared times z. Now I’ve already entered this in the these all these variables I’m not going to do each variable because I just wanted to show you how do you enter the variables in here. And, um, but I’ve entered them in here for speed you know. Now was given here as points you see points here? You know there’s segments that’s for sure they’re going from this point to this point, that’s a segment. So we choose points in my next menu and we’re going to X Y Z to X Y Z here’s X Y Z to X Y Z okay? So we’re going to choose that. I can’t put X 1 and X 2 etc because of the width of this screen here so I can, I have to keep it within the constraints of this area here. So you have to get used to my programs a little bit, try them out. Press number 3 and we’re gonna go quickly through entering these now you’d press alpha and put each one of these in. alpha 0 alpha 6l so alpha minus 1 and I show you what you’ve entered you can change it if you want I say it’s okay. And we’re gonna parameterize it okay? So and this is the formula, don’t let any professor tell you any different this is 1 minus T times the X 1 plus X 2 times T okay? That’s it, now you get all kinds of professors that throw this stuff around and these are the tough things you don’t know what they’re talking about, what they do, and they think it’s simple they’ve been, teaching this crap for years so they just you know throw it around but we have to know the formula and to be able to to make it couldn’t stream which with every other possible thing you know. And here’s y and here’s Z the parametrization of this okay? And when you parameterize the limits are zero the range is 0 to 1. So x equals 40 the derivative of that is 4 y equals 6 minus 5 T times T the derivative is minus 5 etc okay? Do it exactly like this system. ds right over here there’s a DS that equals the magnitude of the of this segment and that’s the derivative of RT which is these here X prime of Y squared and Y prime of square of Z prime of squared ok? 4 minus 5 6 does everything for you write this stuff down don’t even study it it’s not worth studying it’s a waste of time. so it ranges from zero here’s the integral here we were substituting all of those in for x and y here’s X and then of course we have the DS here okay? We bring all constants outside of the integral in calculus okay? If there’s a constant in here we bring it out. We integrate it, of course you could all integrate this like nothing couldn’t you? Of course you could. And then at T equals 1 we enter the T and in for all the t’s in this what we just found in the screen before we come 56 divided by 3 at T equals 0 we enter the zeros for all the t’s now comes 0 we always upper – lower in physics and in calculus 56 and 3-0 answers and then we add the the square root of 77 in your 57 56 you know this answer here okay? Notice that a line integral, notice how stupid everything is because this doesn’t even represent anything. What is this? feet? meters? is it square cube roots what is it? You never know just a bunch of garbage and go to my site buy my programs and pass calculus, believe me it’s the greatest programs ever, have a good one!

Line Integral Test Question 3

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Hello Tom from everystepcalculus.com I do these programs for one reason for you to pass a test and and if you’re lucky do some homework, get some homework done. Mainly tests, I get tests from all over and I program from test problems. Test problems have to be relatively simple even though nothing’s simple in calculus because it’s 10-million puzzles like Sudoku or crossword puzzles. That’s all it is, it never solved anything at NASA, never solved anything in life worth anything in my opinion. After 25 years of studying this stuff any more than multiplying two numbers together can give you the answers for curing cancer or whatever. Look at this problem here we’re dealing with sine waves okay, here’s sine waves, nonsensical adding things together in the functions three-dimensional XY and Z and then all of a sudden somebody decides that you’re going to have a time, times you know to the fourth power or minus time to the fifth power of T you know none of this nonsense would ever happen in real life is it just puzzles to see if you can calculate them. Okay well I can calculate them in a program why would I ever try to learn this stuff and memorize it and put it to my fabulous memory if it was such if it’s worthless and I know it’s worthless I’m not gonna memorize that, I’m not gonna study it okay? That’s the reason I created these programs and you can take advantage of them too, you can go to my site buy my program for $40 and pass calculus as simple as that. You don’t have to study this crap okay so anyways index(8) to get to my menu let’s do this problem. This is the problem one I got off at Yahoo one of many now and one thing I do want to teach you but lining it was well I’m gonna press number 7 here because this is what we’re after here, no we’re not after that at all, we’re going after line integrals sorry. I was ahead of myself here I got so upset over calculus. now I’m an expert at line integrals now because I’ve studied it for this program this stuff for well around for the years but really pushed it hard for the last month here. so now line integrals and we’re going to go to line integrals here, press Enter. And this problem is I J and K okay so we’re gonna choose that number seven in the menu I J and K. Now these are all of the possibilities of line integrals that I have found. When they give you a problem, they will give you dxdy you’ll choose that dxdydz the function of under with x and y no Z, function with XY and Z or the function with I and J and I J and K. These are all tricks by professors to screw you up and see if you know everything that they teach they put a simple problem of line integral down the book and the blackboard and then all of a sudden teach this god-awful stuff with sines and cosines and etc how they do this. So anyway is that we’re gonna go to number seven press number seven to get to that in the program. And we’re going to enter the variables now this turns into calculus 3 2 m and and O ok? M and then O so here’s the M 1 know if we’re gonna enter this sine of here we’re gonna press alpha and we’re gonna press sign we’re gonna go to sine which is second and Y there’s sine we’re gonna put the X in there close up close off the parenthesis and press enter okay? Now the n function the same thing you press that you go to here and press alpha and put in the sine cetera et cetera in my program got to press alpha first remember that I’ve already loaded this stuff into the simulator here to make it easy and here’s here’s what they’ve entered right x times Z here’s x times Z here for the K and D are okay the D are shows you there’s an RT function and you’re going to do the derivative of the RT function which equals the dr here okay? So I say it’s okay when you enter these variables and was segments given? No segments is when you get when they give you points a segment going here like in a triangle segment going up and a segment going here etc two segments three signal whatever so there’s no segments given here so we say no okay now is I J and K given yes yes for C okay and so we’re gonna we’re gonna press YES on that and we’re going to enter the T term before the I here’s a here’s the I here’s a before T to the fourth power. Now again you press Alpha and put T to the fourth in there and I’ll do it for you here alpha T to the fourth, there it is right there okay? And we press ENTER now we’ve got that entered there now we’re doing for the other three but I’ve already entered them in here now okay so we have T to the four minus T to the five and then T for K notice this nonsense you know x times the power of the fourth times minus time to the 5th power you know how would that ever read till anything worthless worth anything in this lifetime okay? So I say okay if it’s all good we’re gonna enter the range, range with RT functions is always 0 and 1 so I’ve already entered that to 0 and 1 but if they gave you something different I suppose they could if it’s a puzzle and you you could enter that yourself okay? I say it’s ok, now we have x equals T the 4th here the derivative of that DX is 4 times T cubed okay DT etcetera etcetera you mark this on your paper systematically does this, these computations for you here’s the original integral it’s changed with DX dy and DZ here instead of I J and K because you have to multiply the problem with DX and dy and DZ which we just found in the previous screen we’re gonna substitute all these in for the X&Y; here we do it here no you’re gonna put your gonna put parentheses around these quotation marks because I’ve entered a in in for X and here’s the DX right here’s the Y and here’s the here’s the dy etc and it equal and it’s over the range of 0 and 1 here’s the integral okay? We’re gonna condense it make it clean it up here’s the real integral after we multiply things together we’re going to integrate, try integrating this stuff in a test huh? easy right I don’t think so over the range of 0 and 1 now at t equals 1 you have cosine we’re entering the one for all the t’s and here in the pretend that what we just found and here’s the answer – cosine 1 – sine 1 plus 1 over 6 ok? At t equals zero turns out to be minus one upper minus lower that’s always the case in physics and calculus when you do anything over a range it’s always upper – lower okay and here’s the answer and here’s – – one you probably put down plus one here which would be correct and plus one you know is course six six and then you’d have seven six here but here’s the answer – point two one five okay? Now go into my site buy my program and enjoy passing calculus have a good one!

Line Integral Test Question #2

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Hello tom from everystepcalculus.com we’re gonna do a calculus problem here is you see on the screen line integral this is an IJ and K. Once you see that this that’s what you’re going to do. It triggers you to for the menu to choose that okay anyways let’s do it index(8). I’ve already added the parameters and these variables into the into the simulator here so that it can save time when doing that index(8) to get to my main menu. We’re gonna scroll down here to the line integral or in the menu that’s at the alphabetical letter of L of course so we’ll scroll here down and then look for that. Here’s line integrals here we’re gonna press Enter and wait for when it says busy here it’s loading the program only does that the first time you loaded after that it’s very quick and we have IJ and K which is number 7 here you can scroll down like this or you can press the number 7 I like to press the numbers it gets there quicker and it loads the program for this type of problem. You’ll notice that the line integral this is the integral of the function for dr over the range of C and that equals (M)i+(N)j+(0)k here do you write all this on your paper exactly like it says here ok. We’re going to enter the function now normally you’d have to press alpha and then enter the minus 2x I here but I’ve already done that so and then before n2 asks you and before the O it asks you ok and it turns out to be here now here’s minus 2x I which is just like in the problem here plus y + 5 z k dr okay? You can change it if you want i say it’s okay. There’s line segments given no there’s no line segments given here is there okay so we’re gonna say no number 2 to his I J and K mentioned in the problem yes here’s here’s IJ and K here. RT there should be i’s this person didn’t put these in here right there should be i’s after each one I, J and K after these but yes and normally the problem will always have an IJ or K here so you add the I’ve already added them in here you can see it’s sine of T cosine of T i, J and K okay. I say it’s okay we’re gonna enter the range the range is given from zero to three times pi over two notice there’s no x in here this guy doesn’t know what he’s doing or girl whoever put this problem on the internet here and I’ve already added them so we’re going to go from 0 to 3 PI over 2. Say it’s okay. Now we parameterize the whole thing which we take x equals sine of T DX is the cosine of T y etcetera cosine of T dy is minus sine of T etc Z and DZ okay here’s the original integral minus 2 X 2 DX Y dy 5 times Z DZ the reason that dxdydz because we got to add that we’re gonna add that in the problem so we substitute all these into the problem here you notice here’s 2 times X there’s minus 2 times X and X is sine of T so then and then of course the DX is cosine of T and that’s the same thing for the other entrances here you’ll notice the quotation marks and your problem on your sheet you’d put you’d put parentheses around these quotation marks to indicate that you’re substituting it for x y&z; we clean it up here and condense it here’s what you get put this on your paper just as clean as that over the range of 0 and 3 times pi minus and divided by 2 we integrate okay here’s the integration you can do that in your head can’t you but I couldn’t I needed income calculator to do it here’s three times divided by two over that range so we’re adding the 3 PI over 2 to all the T’s in the problem the answer is 45 times PI squared over 8 at t equals zero we had zero for all the t’s in the problem and the answer is three-halves upper minus lower that’s always the way it is upper minus the lower and the answer is 54 you notice I’m pathetic line integrals are because you don’t even know what this is what is 54 mean what 54 square units 54 meters 54 feet what you know but calculus is stupid that way calculus is sudoko of math all right so anyways why would you do this in your brain or try to learn how to do this and memorize it for tests or anything else you can when you have a program like this you buy my programs for 40 bucks and 39 bucks I guess whatever it is and then you load them in your calculator and you can do a million of these including physics so make sure you think about physics too unless you want to just study this stuff for hours and hours and hours like I do I know a line integral is pretty good because I’ve been doing it for what two or three weeks now so now I know everything about line integrals and they’re all a waste of time in calculus but we need them to pass our homework and pass our tests okay you get out of that class move on with our lives so anyways go to my site buy my programs and enjoy passing a calculus okay have a good one

Line Integral Test Question 1

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Use the Fundamental Theorem of line integral to evaluate the integral (y^2-3*x^2)dx + (2*x*y+2)dy

:: Transcript ::

Hello Tom from EveryStepCalculus.com a
problem in calculus regarding line
integrals one of the most pathetic
sections of calculus completely worthless
just really puzzles all calculus is is
puzzles like sudoku puzzles
and some professor will say that it’s
important for something else and I’ve
never found that after 25 years. So
anyways let’s do this so you can pass
your test here’s a question that
some kid put in and I’m going to show
you how that works and my programs. index(8) to get to my menu I’m going to scroll
down here to line integrals that’s these
subject matter here’s the line integrals here
and we’re going to go number six here
dxdy because that’s what they’re asking
here D with DX and dy okay so then we’re
gonna choose that in the menu number six
I press the number or scroll to it I’ve
already entered the functions in here to
save time with the simulator here so
here’s what the function is the integral
DX dy you’d put that in yourself
pressing alpha first and then you
I have a choice of sitting it’s okay or
you could change it in case you made a
mistake but you’re gonna enter that
yourself by pressing alpha first
remember that E and there’s line
segments given yes they are and they’re
X Y and X 2 y 2 because they give you x
and y and then X 2 y 2 I say okay
and I put these in already one and minus one zero looks pretty good we’re going to say okay to that
parameterize it, change it with t-values and here’s the
form of the trick for that remember this
this is the big trick took me
quite awhile to figure this out from the
way professors teach things
they skip the easy stuff which is tough
for maybe you or me
and we do the y-value that’s what it is
anytime you’re parameterizing
a function the range becomes zero to one
okay remember that so now we have the X
and now we’re gonna do the derivative of
that which is minus three here’s the y
value that we got and we’re gonna do the
derivative of that which is minus one
original integral is this okay now we’re
going to substitute all this in here
putting the one in just like you says I
have quotation marks here which this
simulator does but you’re gonna put your
gonna put parentheses around this on
your paper okay because remember they
want to explain a step it’s that when
you get points for each step you do
and so you’re gonna add all these
anytime you see quotation marks you’re
gonna put a left or right parenthesis
okay to clean it up and then we
do the actual math of that which is this
okay three we’re gonna integrate it
of course you could do that without this
help right no problem just
anybody can do integral is really easy I
don’t think so and we’re gonna at T
equals one which is the upper limit
we’re going to enter the one for all of
all the steps etc and we’re gonna enter the zero at T
equals zero we’re gonna enter the zero
for all the exes okay that equals zero
the other one was five so take the upper
limit minus the lower limit five minus
zero and here’s the answer five units
okay now we don’t even know what units
are I mean what are we doing with line
intervals let’s see we’re finding the
distance or something around this thing
is so Pythagorean theorem anyways so but
what do we care we just need to pass the
test so go to my site buy my program
they’re only forty dollars you get three
hundred and over three hundred and
seventy programs in your calculator to
answer most of test problems in calculus
and you’re going to get six or seven or
eight right versus the person sitting
next to you and that’s all you need
because of the class curve to pass that
class or maybe get even an A so think
about that okay the price is pretty
cheap compared to what you’re paying for
college nowadays so think about that go
to my site, buy my programs, and pass
calculus hey have a good one

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

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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)

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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)

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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, x*e^(2*x)

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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

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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 5th. 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)

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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

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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]

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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]

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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

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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]

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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.

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

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

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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.