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You are here: Home / Integrals / Integrals / Line Integral Test Question 3

Line Integral Test Question 3

January 31, 2019 by Tommy Leave a Comment

:: Transcript ::

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!

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