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 Related rates problems again regarding a cone or a cone shape, let me read it here one test problem, water is withdrawn from a conical reservoir 8 feet in diameter and 10 feet deep at the constant rate of 5 cubic feet per minute, how fast is the water level falling when the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.
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.
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.
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
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
Let √(x) = sin(u)
Differentiate both sides
= 1/(2*√(x))dx = cos(u)du
dx = cos(u)du / 1/(2*√(x))
= cos(u)du / 1/[2*sin(u)]
Invert and multiply
∫√(x)dx / √(1-x)
= ∫sin(u ) * cos(u) * 2*sin(u) / √(1-sin(u)^2) du
= ∫ 2*sin(u)^2 * cos(u) / √(1-sin(u)^2) du
= ∫ 2*sin(u)^2 * cos(u) / √(cos(u)^2) du
= ∫2*sin(u)^2 * cos(u) / cos(u)du
= ∫2 * sin(u)^2 du
= ∫2*[1 – cos(2u)]/2du
= ∫ [1 – cos(2u)]du
= ∫(1)du – ∫cos(2u)du
= u – (1/2) * sin(2u) + C
u – (1/2)(2) * sin(u) * cos(u) + C
u – sin(u) * cos(u) + C
= sin-¹(√(x)) – √(x) * √(1-x)”
Green’s Theorem, Triangle, x^2y^2+4xy^3
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!
A man 5 feet tall walks at a rate 5 feet per second away from a light that is 16 feet above the ground. When he is 8 feet from the base of the light, find the rate at which the tip of his shadow is moving.
Hello again everyone, this is Tom from everystepcalculus.com and everystepphysics.com. We’re gonna do a triple integral from Calculus 3 right now. this is an example of Patrick JMT, my favorite instructor on the Internet and youtube. So I’m gonna show you how it works on my program. I don’t know anybody can do that problem, he can do it because he’s a genius, but for us students, etcetera, how do we do it? So let’s get started. Index 8 to get to my menu. I’m gonna scroll up because I can go to the bottom of the menu then, instead of going down quicker to the T section and we’re gonna choose triple integral. And we’re gonna enter our function, you have to press alpha before you enter anything into these entry lines here in my programs, okay. Alpha X times sin of y. I always show you what you’ve entered you can change it if you want. And we’re gonna use the order of integration, which is dx, dz, dy, which is in the example. You have the other choices in case that’s given on a test also. region cue enter these. And we’re gonna enter the region, q. We’re gonna enter these limits. This is alpha 0 for the x one. Alpha square root of 4 minus Z squared. Made a mistake so I gotta go back. Choose number 2. Alpha 0. Alpha square root of 4 minus z So here’s what you write on your paper That’s better, say it’s okay. Next one for the y is alpha 0, alpha pi. That looks okay. and alpha 0 for the z. alpha 2. That’s okay. So here’s what you write on your paper, the way you write a triple integral with dx, dz, dy order of integration. Here’s the function in here. We’re gonna do the dx first, and you put this over here with these lines, when you’re doing the range over this integration here. And here’s the integral of the first here. Of the first function. And if x equals the upper range… I show quotation marks here but you put parenthesis in there because you’re substituting this amount for any X in the integral, and it equals this, minus sign etc. And then we do the lower integral. X equals zero, and there’s 0 plugged in, you put parentheses around this instead of quotation marks, okay. And here’s the answer, upper range minus the lower range equals this right here. So that becomes the new integration function and I show you that here. Dz, dy is left, okay? So now we integrate that, come up with this, minus sin z, z squared, et cetera and over this range here, 0, 2. Add z equals 2. Here’s the answer here, at z equals 0 plus these in for all the z’s in the problem. And the answer is this, the upper range minus the lower range is 8 sin y divided by 3. We’re gonna use that for the integration function with the range of 0 and pi. At y equals pi minus 8 cosine, here’s the 8/3. Y equals zero, you plug that in here, you get minus 8/3. Upper range minus the lower range, notice the minus times the minus, you can’t remember that stuff a lot of times. Turns out to be, the volume is 16/3. Okay. New problem. So go to my site, subscribe so you can see other videos I might make. Or you can go to the menu on my main site and go scroll down to what you need to learn. And see my program works, because it sure teaches you quicker than a book or anything else. Okay, so have a good one.