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

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

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

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]

Transcript

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

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

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

Transcript

Hello! Tom from everystepcalculus.com and everystephysics.com. We are going to do a problem in calculus U substitution, and we are going to do a function with cosine in it. Index 8 to get to my menu. Press enter. I already had U substitution but you’ll scroll with this cursor here down or up or whatever to get to the [unclear 00:33] for the problem that you have at hand. So, U substitution is this problem and we are going to enter the function, we have to press alpha first before you enter anything in these entry lines here. So, we are going to press alpha, and then we want cosine, secant z here which is cosine, and then secant x which is log of x, close up the parentheses, and divided by x, is the function. So, it will show you what you have enter. You can change it if you want, else say it’s okay. Notice that the trick to use substitution is that whatever inside parenthesis here, you take the derivative of that and it’s got to be able to be matched to the outside somehow, okay? So, we are going to rewrite this because here we have 1/x here dx, so we are going to kind of isolate that so it’s already cleared for you. Cos(ln(x))*(1/x)dx, and then u = ln(x) du = 1/x

You have to memorize that. The integral of 1/x is ln(x) etc. so the opposite derivative of ln(x) is 1/x. So, now we are going to do the integral of cos u, du equal sin(u) + c. When you do the integral of cos u it’s sin u. so, now we have the answer of sin[ln(x)] + c, as the answer. Now, we are going to do another problem here, and we are going to press alpha, we are going to have secant. Cosine of second log of x, and without the x, divided by that. Okay? It shows you what you have entered, say it’s okay. And notice that this is not a U substitution problem but an integration by parts problem because notice that the derivatives of the parenthesis here, ln(x) is equal to 1/x but there’s nothing on the outside that equals 1/x. So, if that’s the case then you cannot integrate it by U substitution, you have to go to integration by parts. And that’s just pathetic here with the long, long here of getting the answer, more of Sudoku of math. So we have,

u = cos[ln(x)],

du = -sin[ln(x)],

## Mean Value Theorem, 2√(x), [4,9]

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.

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

Transcript

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

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

Transcript

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

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

Transcript

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

## Maclaurin Series, x*e^(-x)

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

## Maclaurin Series, 1/(1-x)

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

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

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

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

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

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

Transcript

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

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

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.

## Difference Quotient, 7*√(x+13)

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

## Difference Quotient, √(x)

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

## Green’s Theorem, unit circle, 9y,3x

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

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