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Home » Video Blog » Archives for Tommy

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.

Filed Under: Integrals

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.

Filed Under: Maclauren Series

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.

Filed Under: Maclauren Series

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.        

Filed Under: Local Min and Max

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.

Filed Under: Related Rates

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.

Filed Under: Tangent Line

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.

Filed Under: Calculus Help

Difference Quotient, 7*√(x+13)

April 10, 2017 by Tommy Leave a Comment

Raw Transcript

Hello again, Tom from everystepcalculus.com. Difference quotient, regarding the square roots. And let’s do it. Index 8 you come to my menu here. I’m going to scroll down to what’s asking your test problem. Do the difference quotient to find the definition of derivative. So we scroll down here to difference quotient. Press ENTER and we under the program. Always write the formula on your paper first okay. Sometimes Delta X might be used in you from your professor for character H. So do that. Any time you see an H you’re going to put delta X okay. And then you’re going to have look at my program and change the character as you go through it. Enter the function here you have to press alpha first and enter my programs. And so the function we’re going to enter is 7 times second which is the time sign, which is the square root sign X + 13. X + 13 closed off at parenthesis. Press ENTER, here’s a few minute I will show you that so you can change in case you made a mistake it’s what we entered so that’s cool. It’s busy loading the program. And we’re going to rendering X + H for every one of these X’s in the function all over H. we get rid of the radicals by squaring them. Here’s the denominator with H. we’re going to multiply the same thing up here at the top which you can do equals 1. So you can multiply whatever and then whatever you think up on the bottom of the denominator and put it in the top. Right now we’re going to add the radicals and put in the top of the cube. Which gets rid of these okay. When you multiply radical times of radical. It becomes a not radical okay. You want to find out if we’ve gotten rid of the radicals here. No problem, you still have the radicals in the bottom of course. Expand the numerator, turns out to be this. Combined the numerator turns up by 7H. We divide the H out becomes 7 and then we have this remaining in the denominator okay. Now there’s an H here. So as H approaches 0 which is the definition of, or the rule when you’re dealing with limits. As H approaches in this case. It is approaching 0 so what an equals 0. The H comes out. And we have X + 13 plus square root of X + 13. So we get two times the square root X + 13 which is the answer is something denominator okay. So this is all basic algebra I’m in this very difficult who knows elder were this good. But I mean you can study it for weeks and know it. But I mean who wants to study it. Who needs to study it? It’s complete waste of time. So use my programs and pass calculus okay. so have a good one. everystepcalculus.com

Filed Under: Difference Quotient

Difference Quotient, √(x)

April 10, 2017 by Tommy Leave a Comment

Transcripts

Hello Tom, from everystepcalculus.com from everstepphysics.com. The difference quotient is what we’re going to do here. And I need to do one more because the square roots are kind of difficult because you have to get rid of the radical signs. And get of radical. You multiply times a radical and our radical squared is turns out to be whatever is inside the radical okay. So I’m going to show you kind of a very simple one here. So maybe you can learn this better or to understand it better. We’re going to scroll down here to difference quotient. Here’s the form you write that on your paper all the time we were to do that. Enter the function. And to press alpha first, alpha second. Time sign is a square root sign of X. And we’re going to this close up the prices. So in the front are going to add X + H, to every X in the function. I say it’s okay, it gives a chance to change it in case you may never stay put. And we’re loading the program here. Again the formula. So we’re going to put X + H here in the radical and we’re going to subtract the function. Visual function which is the funding divided by H. To get rid of radicals you need to multiply it. Somehow the radicals, times itself. Okay spire radical you get what’s inside the parentheses and the radical. So we’re going to here’s, the radicals here we’re going to have to put that in the denominator. Because you can’t just do it in the numerator you have to do the denominator also which actually equals 1. It’s multiplying x 1. So we look at what radicals we need to get rid of. We need to multiply that times that. And there is we’re doing that we’re putting in the denominator also okay. When you do that you get what’s inside the radical which is X + H – X okay. And we have H down here and we still have the radicals down here. We have to have that. In algebra expand it, you get this up in the numerator. We get H in the numerator and divide out the H, you get a 1. Still have an H here. As H approaches 0 when you’re taking the limit of a function. In this case that equals 0 so X + 0 is X. So you get two square root X on the bottom. Sorry this is not over here further but I have to make room in cases are extremely hard function and they take up the whole line down here okay. So you get the idea and so the answer is two times the root of X okay. This is the derivative the retribution function. Pretty neat everystepcalculus.com. Don’t waste time. Go to my programs buy them. There cheap for what you get here. I’ve studied years in this crap to help you pass calculus and of course you can buy my programs and pass calculus. So have a good one okay. everystepcalculus.com

Filed Under: Difference Quotient

Green’s Theorem, unit circle, 9y,3x

March 30, 2017 by Tommy Leave a Comment

0:00

Hello. Tom from everystepcalculus.com and everystepphysics.com.  Another Green’s theorem regarding a unit circle and its dividend. Index 8 get to my menu scroll down to Green’s theorem year what is the only reason that we’re studying this stuff is crap is to pass our calculus test passed her test that’s the only reason we have no interest in it whatsoever unit or going to work at NASA we have no interest in it keep that in mind raising our professors charity she disliked there are some basis for some goodness 40 are evergreen serum here problem from Yahoo gets have to press alpha beforehand or anything in these entry lines here is now four times why where the axe function and three times C looks pretty good reason to I say it’s ok they gave me MRP using MNE and here and it goes as the partials hebrew 3-6 unit circle convert to pull their situations circles Mexico said it takes our way it was caught sight of theater times are and we’re gonna do the first or detain Dr agreed defeated come up with minus six times over 250 compute that come up with minus 12 hi I mean agreed that Dr in a girl putting minor stroke pioneer over 10 subtract them and we come up with minus 12 high school career. Pretty neat, huh?  everystepcalculus.com and everystephysics.com.  Don’t do this step without my programs.  Have a good one.

Filed Under: Green's Theorem

Arc Length of a curve x=y^3/6+1/(2y)

March 12, 2017 by Tommy Leave a Comment

Transcripts
Hello everyone, Tom from everystepcalculus.com and everystepphysics.com. We’re talking about arc length today.  Got this on a new test that somebody sent me so I’m going to do a video of it right now. Pretty complicated nonsense in calculus but still there for calculus two, people. So let’s do it,  index 8, type in into the home screen there to get to my main menu. Your going to choose number 5, arc length and because the problem gives it an x, or a g a y.  I’m going to choose number 2. And we’re going to, I show you the formula here, here’s the formula for the arc length and the x function given is, you have to press alpha before
 you enter anything into my entry lines here.   Alpha y cubed divided by 6 plus one divided by 2 times y.  Looks good.  I always show you which of entered, you can change it if you want. I say it’s good.  I choose number 1, okay.  I always show you the derivative first; take the derivative of that. I give you the choice of which system they give you too. If they give you y, then we’re cool in this system here. We just enter what they give you. If they entered x then you have to compute the y values that were given the X values.  So we’re gonna do alpha 2 or, lower one is alpha 2, upper limit is alpha 3, or range, upper range. And I show you that also in case you made a mistake. I say  it’s ok.  We do our computations. This shows you the exact problem which is what you’re dealing with and what you write down in your paper, here.  And we’re going to square that as part of the functions, here.  You notice that calculus is so pathetic that they do one derivative and the rest of it is you have to do square roots, you have to do squares,  you have to do integrals. Well, I do that all for you. Here’s squared. Now we’re going to add one to what we’ve squared which equals this, here.   We’re going to take the square root of that whole mess right here.  And we’re going to take the integral of that which is this right here. And over this range 3 and 2. Y equals 2, substitute that in for the original equation here and you get thirteen twelves, add 3 substitute that in, you get 13 over 3, you take the upper minus the lower and you come up with thirteen over 4 or 3.25 units.  But pretty neat, huh?  everystepcalculus.com.  Go to my site and hope that I make other videos for you.  Have a good one.

Filed Under: Arc Length

Linear Approximation, 1 variable, f(x) = √(x), x=49. 4

March 2, 2017 by Tommy Leave a Comment

Transcripts
Hi, I’m Tom from everystepcalculus.com and everystepphysics.com. This is a linear approximation problem in calculus one generally because one variable and we’re going to read it here because it’s off of a test using your approximation to approximate square root of 49.4.  We’re going to let f of X equals the square root of x and then that can be written as the point-slope form of y equals MX plus B so we’re going to compute m and B.  Let’s do it,  Index 8 to get to my menu. We’re going to scroll down here to linear approximation,  in the L section. And we’re going to use one variable because there’s only x given.  So linear approximation, one variable,  I’m going to use the point-slope form equation for a tangent line to occur y equals MX plus B. We’re going to enter function,  you have to press alpha before you enter anything into these entry lines here.  So it’s alpa and second X will give us the square root sign put in the X close off the parentheses. Press Enter. I always show you what you’ve entered you can change it if you want. We’re going to enter the point but you want to evaluate this at. We’re going to press alpha again and I’m going to pull 49.4. Check that make sure it’s right again. I say it’s okay.  First thing we do is we enter the function and get it do the derivative of that which is the slope and it equals m.   Here’s the derivative of that. And we find that the closest square root number for that given number 49 and we add that into the derivative and so the slope is equal to one over 14.  Now we take that number 49 again and add it into the original function. We came up with y equals seven. So the point is 49 for x and 7 y,  okay. Y equals seven,  then we’re going to find B so we do the calculations here in algebra. And b turns out to be 7 halves . We found that already, ok. If you want to go further and they’re asked for this point slope form this is equal to y equals MX plus B slope times X to speak now we’re going to enter the original point where after in the same formula and so then y is equal to 7.029, okay?  And the XY is 49.4, 7.029.  Pretty neat,huh? everystepcalculus.com. Go to my site, buy my programs, and pass calculus. Have a good one.

Filed Under: Linear Approximation

How to solve: ∫(√(x))dx / √(1-x)

February 27, 2017 by Tommy Leave a Comment

Let √(x) = sin(u)

Differentiate both sides

= 1/(2*√(x))dx = cos(u)du

So:

dx = cos(u)du  / 1/(2*√(x))

= cos(u)du / 1/[2*sin(u)]

Invert and multiply

 = cos(u)du*2*sin(u)

So:

∫√(x)dx / √(1-x)

Substitute

=  ∫sin(u ) * cos(u) * 2*sin(u)  / √(1-sin(u)^2) du

=  ∫ 2*sin(u)^2 * cos(u)  / √(1-sin(u)^2) du

Identity

=  ∫ 2*sin(u)^2 * cos(u)  / √(cos(u)^2) du

=  ∫2*sin(u)^2 * cos(u) / cos(u)du

cos(u) cancels

=  ∫2 * sin(u)^2 du

Identity

=  ∫2*[1 – cos(2u)]/2du

=  ∫ [1 – cos(2u)]du

=  ∫(1)du – ∫cos(2u)du

=  u – (1/2) * sin(2u) + C

Identity

u – (1/2)(2) * sin(u) * cos(u) + C

u – sin(u) * cos(u) + C

Back substitute

Answer:

 = sin-¹(√(x)) – √(x) *  √(1-x)”

Filed Under: Integrals

Green’s Theorem, Triangle, 3x^2+y,3xy^2

November 1, 2016 by Tommy Leave a Comment

Transcript

Hello. Tom from everystepcalculus.com and everystepphysics. com. Another example of Green’s theorem and show you how that pathetic. Index 8 lyrics to get to my menu. Scroll down to Green’s theorem. Generally in a test, they give you, your’re supposed to use Green’s Theorem or something like that. So we choose Green’s Theorem. And we enter our x and y or i and j functions. This one is You have to press alpha before you enter anything in my entry lines here on the calculator ok? Alpha plus three times x squared plus y. three times for the j or y value. Three times x press alpha. Alpha three times x times y squared. Imagine the knucklehead professor that’s still refer to which taking up these functions here that would work. That’s what they are asking for you to do. Just find the area of a triangle. Which is of base times height. I give you a chance to change it. The m or the p function is 3 x squared plus y and the q or the n function is three x y squared. It uses different letters for which professor you’re talking to or which book you use. And the partial of m, i use m here. The partial of m with the partial of y is one and the partial of n with the partial of x is 3y squared. And we subtract those going to the formula. That’s 3 times y squared minus 1. And we’re going to choose, I’m going to go up and press the up cursor to go to the bottom of the menu to get to triangle. This is a Yahoo problem. Right triangle, okay? What do they give you? They either give you 3 points or the give you this y equals, y equals, x equals. This case, it’s the first one there. Choose that. Then I’m going to go alpha zero, it chooses the boundary if you’ve ever worked with Green’s Theorem, that’s the most difficult thing is to figure out what the boundaries are. And that’s 2nd alpha, alpha x 3rd one is the x value is alpha 1. I say it’s ok. And we do the green set up. This is what we found here. For the partials these regions d y dx.
So we integrate dy.And then we over the range that was given which is a x is 0. Comes out to be x cubed minus x.  And then we integrate dx. x cubed minus x with respect to x over the range of 1 0. So the answer is minus one quarter squared units. Pretty neat huh? everystepcalculus and everystepphysics.com. Don’t try to pass Calculus without my programs. Believe me, it’s much easier. Have a good one.

Published on Mar 6, 2016

Filed Under: Green's Theorem

Green’s Theorem, Triangle, x^2y^2+4xy^3

October 25, 2016 by Tommy Leave a Comment

Green’s Theorem, Triangle, x^2y^2+4xy^3

Transcripts

Hello everyone, Tom from everystepcalculus.com and everystepphysics.com. You can buy my programs to pass your physics and calculus classes step by step on all the problems which you need to show your work. Index 8 to get to my menu. We’re gonna do the Green’s Theorem, today, in this one. And they give us a triangle with vertices to figure out the double integral. So we’re going to choose index 8 to get to my menu and we’re going to scroll down here to Green’s Theorem it’s alphabetical. There’s Green’s theorem there. They give us the x or the i function from the line integral. You have to press Alpha before you enter anything in these entry here. Alpha x squared times y squared. And then we’re going to choose for the y the J function. Alpha 4 times x times y cubed. I always show you what you’ve entered. Here is the line integral, it really should really be called the curved integral. There are no integrals on lines. Straight things, everything needs in exponent to work. That’s the reason they use cosine and sine so much because it’s a sine wave which is a perfectly smooth curve and the use balls and spheres and etc, to do all their calculations, calculations as I see it. Calculus solves nothing in real life does nothing for anybody. Just a Soduko of math. Something that people do to waste time just like crossword puzzles in English. So anyways there’s a line integral of it and I’m going to say it’s ok to go with. P and q system or m and n system. I use the the n and m, depending on which professor you’re dealing with or the book. Compute the partials, partial of m with a partial of y is this. Partial of n with a partial of x divided by the partial of x is this . And we’re going to subtract the m from the n partials. And this right here. Ok and so now, we’re going the triangle, which they give us. We are going to choose a number to the points. They don’t give us y y x, they give us 3 points. First point is 0 0 , second point is 1 and 3, we’re going to choose number 3, scroll to it. We’re gonna add it. Alpha 1 alpha 3. And the next point is 0, 3 so we’re going to use 0 and then the question mark here, number 1
we’re going to enter number 3, alpha 3. And again, we look and see if we have entered correctly, and we have. Enter umber 1 or scroll to it. We’re going to set up the Green’s Theorem. Double integral 01 in three x Notice this, if you do the vertices and put it on a piece of paper notice that the line here is an angle such as this where certain parameters and this is the y function remember the y equals mx+b that you learned in Algebra? Well the mx is the slope and of course 3 over 1 is the slope of this line. So if you graphed it, you’d put 3x into the Y function on your graphing calculator and that would be the angle just like this. And then of course we have 3 over 0 and then 3 down the origin of the graph. So that’s the triangle we’re working with. And we’re going to integrate this right here with respect to Y and over this range here. And that equals this over the range. We take the 3x and put it in for all the y’s and we come up with 72 x to the 4 and then 0 of course is a 0. And upper minus lower is 72 x to the fourth. Now we’re going to integrate that with respect to x over this range here which equals 72x to the 5. Remember we add 1 to the exponent and divide by 5. That exponent in the problem. And over this range here. Remember Newton and Leibniz were working on exponents in the sixteen hundreds, that’s how they came up with calculus and of course logarithms, also. Logarithms is an exponent. And so we’re gonna do that range at X equals one. we substitute that for the x, come up with this, and the answer is 72 over 5 square units. 14.4 square units. Pretty neat, huh? everystepcalculus.com. go to my site buy my programs. It’s the best deal around. Much, much more economical and worth worth it then they even the calculus book. So enjoy my program and pass calculus or physics. Have a good one!

Filed Under: Green's Theorem

Equation of a plane

October 3, 2016 by Tommy Leave a Comment

Transcript

Hello again, everyone. This is Tom from everystepcalculus.com, everystepphysics.com. A problem with the equation of a plane that I found on Yahoo questions from a student. And I’m going to answer it in my programs here. Index 8 to my menu scroll down here to equation of a plane now and your calculator you can go second and which is quicker with your fingers and go screen by screen down to in the menu. There’s equation of a tangent plane is not what we’re looking for a look at our equation of a plane right there. There it is. And we’re given a point and a line in the problem. You have to press alpha before you enter anything into these entry lines, here. The point is 1 alpha 6 and alpha minus 4. These again all contrived problems, they never happen in real life ever. It’s just crunching numbers for no reason which is perfect for calculus. Calculus is like studying crossword puzzles to learn English. It’s a waste of time, in my opinion. But we have to pass your test we have to get through this course which you’re required to take so that’s the reason I program it for myself and I’m program for you. There’s the poin.t I say it’s ok I give you a chance to change in cases not. We’re gonna add a parametric line equation which is alpha 1+2 times t.
and the y value is alpha 2-3 times t. and the z value is alpha 3 – t. Again I show you what you’ve entered you can change. it looks pretty good to me. It’s a vector now. That’s what these little indications here are. It’s called a vector. I say it’s ok. So we do the calculations. We take T equal to zero, we come up with 1, 2, 3, here. At t equals 1 you put that in the problem. substitute one in for t’s. you’ll put parentheses in here, I have to put quotation marks because the calculation can’t go to let me do anything different and so then therefore we have 3-1 2. We have our 3 points for a plane ab&c. Here they are. And now we’re going to go find a vector from point c to b which is called vector d. I call it vector d, I call it vector d is equal to 1, 2, 3 minus the 3 minus 1 2. We do those calculations and we come up with this. Vector c to a, vector e is this. 1 6 minus 4, 3 minus 1, 2. We’re going to find the normal perpendicular well normal is perpendicular to the two vectors. Cross product, they call it normal but I mean I like to do it practical. And we’re gonna cross product d times e the two we just found which is these times this ok. Now, you try to do this from memory. Go ahead, try it; not me. It’s a waste of time. So anyway, here’s the the actual calculations. Matrix calculation multiplication and we’re doing these calculations. Make sure you write these on
your paper, now. and equals minus 25 14 8. That’s the normal vector to the two vectors we found. And so we do computations of the equation of a plane, etc Make sure you write all this in your paper, exactly like to see. And the answer is minus 25 X plus minus 14 y plus minus eight z plus 77 equals 0. pretty neat, huh? everystepcalculus.com. Go to my site, buy my programs, pass calculus. You’ll burn your calculus book but you’ll never throw away the calculator with my programs ever because you never know when your life who your partner is or your grandkids or somebody’s gonna do calculus eventually and you have all this in your calculator from then on. Have a good one.

Filed Under: Equation of a plane, Vectors

Eliminate the Parameter, (x,y), t^2+1, t^2-1, 0

September 29, 2016 by Tommy Leave a Comment

Hello, Tom from everystepcalculus.com and everystepphysics.com. Eliminating the parameter. This is an rt equations involving t and so let’s do it index 8 get to my menu. We’re already at eliminating the parameters we scroll down to that if you scroll down with the cursor here. rt equation. This came from a Yahoo problem, a kid couldn’t do it. And of course, the people that answered the question didn’t do it very clear and of course the only thing we’re interested in is passing a test and passing Calculus and getting out of there. We have no interest in calculus beyond that. Calculus is the Sodoku math, should be taught only to math majors. It’s like it’s like being required to do crossword puzzles in if you’re an English major or trying to learn the English language to work day after day and crossword puzzles. You’ll end up with nothing. But anyways we’re going to enter the function. The function is, you have to press alpha before you enter anything in these entry lines. Alpha T squared plus one and in this for the y t, function of t. Alpha t squared minus one. Notice that these are all contrived functions. For z, we’re going to have a alpha 0. Put 0 if they don’t give you z. Nothing happens like this in real life. Calculus solves nothing I say but it does solve completely nonsense nobody ever uses it nobody ever did use it. I say it’s okay. I always give you a chance to change functions if you made a mistake but you it’s a good and we’re going to choose number five. We can do all these things to, acceleration, angle, length of an arc
etc. Right now we’re going to choose number five. Press the number or you can scroll down there press a number and try to do this without a program be my guest. Solve for t It seems simple. Y equals t squared minus one. Answer is square root of y plus 1. We substitute for t into x and z equations. There was no z equation, it was zero. And so x equals y plus 2. Notice we we put in here, you should put parentheses around this for your professor but then x is equal y plus 2. and then this equation, you want to put y equals x minus 2. and z of course is 0. You can go back for more calculations or you can go back to the main menu. Pretty neat, huh? everystepcalclus.com Go to that come to my site, buy my programs, pass calculus, keep your head on straight. Don’t let these doing that these professors scam you too much however they do all the time so have a good one.

Filed Under: Parametric Equations

Eliminate the parameter, t+1, 0, t^2 1

September 29, 2016 by Tommy Leave a Comment

Transcript

Hello everyone. Tom from everystepcalculus.com and everystepphysics.com. Again we’re going to eliminate the parameter. More nonsense in calculus stuff that never happens. Calculus is Sodoku of math. Let’s do it. But we have to pass calculus, we have to get these tests passed etcetera etcetera so that’s the reason I just say my programs. Index 8 to get to my menu. I’m already at eliminate the parameter, you can scroll down to it when you get your menu. Also you can use second and the cursor here to go you know page by page down to the menu and we’re going to enter our r t functions. You have to press alpha before you enter anything in these entry lines here so the first one is alpha t plus 1. And of course alpha for y is zero and alpha for z is T squared minus one. All contrived variables, of course. Here they are you can check and see if it’s correct. If it’s not, you can change them. I say it’s ok. and we’re going to choose eliminating the perimeter, number five. Again, I show you what you’ve entered here. We’re not changing we’ve already checked in already. And we’re going to solve for t in the x equation, okay. So we take x and we solve for t and t equals x minus 1. pretty simple try solving harder questions even like this one. So we put this back into x X and we get x equals x. ok no big deal there. Z is t squared minus one. So you gonna put X minus one squared minus one. Now you doing the calculations yourself you you square this and subtract 1 to get this. Let me see you do that. I can’t do it wouldn’t do it without my program why would you do I would you waste your time on this nonsense. And the answer is x equals x, y equals 0, z equals x squared minus 2x. We’ve eliminated the parameter. Okay, great. everystepcalclus.com. Go to my site buy my programs, pass your calculus test. Tell your friends about this program. Have a good one.

Filed Under: Parametric Equations

Definite Integral, x^(1/3)+2x 1

September 27, 2016 by Tommy Leave a Comment

 x^(1/3)+2x 1

Transcripts

Hello, Tom from everystepcalculus.com and everystepphysics.com. I’m going to do a definite integral here. It was on Yahoo so I thought I might do it. Everybody got it wrong. Didn’t even really do the integral of it right so I’m going I show you how it’s done here. Index 8 to get to my menu. Then we’re going to school down here to definite integral. And the x is given. xy is given so it’s really double integral. Still is different though. It comes up with a certain area under the curve. So we’re going to enter the function. Alpha X to the 1 dived by 3 1/3 third plus 2 times x minus one.
I always show you what you’ve entered, you can change it if you want. I say its ok. We’re gonna enter the range. Alpha 1 to alpha 8. You have to press alpha before you enter anything in my entry lines my programs and we’re gonna say that’s ok. We’re going to do the integral which is this right here and then we do the calculations it x equals 8, 8 is substitute in all these in the integral equals 68. 1 is substituted for all the X’s. three-quarters, common denominator is 269 over 4 square units or 67.3 square units. Pretty neat, huh? everystepcalculus.com, go to my site, buy programs if you want to pass calculus.

Filed Under: Definite Integral

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

September 16, 2016 by Tommy Leave a Comment

x^3+6xy+y^2= 8

Raw Transcript
Hello again everybody. Tom from everystepcalculus.com and everystepphysics.com.
Calculus problem regarding a tangent line with a curve. Dealing with two variables. I 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 to save time. And you’ll scroll down to it and then you’ll enter the function Alpha, you have to press alpha before you enter anything into these entry lines. Alpha x cubed plus 6 times x times y plus y squared minus 8 equals zero. You always have to transpose everything on the other side of the equation here and bring it over the left side and make zero on the right or a number on the right either one would work. Now because you’ve entered y in here this is going to be an implicit differentiation problem. So here’s the formula here that you just entered. In the implicit system. And it looks pretty good. I say it’s ok. Differentiate both sides of the equation with respect to x. So here we go both sides. We’re going to do each one individually. This one here because xy, we need the product rule. So you can write the product rule down and do it like this or you can just keep clicking and you’ll finally get to the product rule answer to which is this right here and 6x dy dx and we keep going. y squared is equal to this because it’s a y dy dx also. And you combine the functions, dy dx here and then no dy dx just regular X Y and the implicit answer is this right here minus 3x squared minus 6x, etc. Division sign here, you know. Now we’re going to evaluate it at a point and the point is alpha 1 and alpha 1. So xy is 11. I say it’s okay. We enter those variables into the numerator and the denominator, come up with a slope of minus 98. Slope of the line and then we figure that y equals mx plus b which is here is the slope, m, and then you add the 1 for the x and 1 for the y and then figure out what b is here. Subtract this from 1, 1 of course 8 eighths and you subtract you get 17 eighths. Subtracting a minus so you get 17 eighths. Here’s the answer y equals mx plus b. Pretty neat, huh? everystepcalclus.com you can go, and I also give you that angle of the line is pointing, notice that it’s pointing in this direction not up but down to the right and if you want to do it and other points, it’ll take you back there. Pretty neat, huh? everystepcalclus.com Go to my site, buy my programs if you want to pass calculus and subscribe if you want to see more videos that I might make you. Have a good one!

Filed Under: Tangent Line

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