Show Work Step by Step on your TI-89 Calculator Screen
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Hello, Tom from everystepcalculus.com,and everystepphysics.com. A problem in calculus dealing with critical numbers in critical points of a function. Let’s do it. Index 8 to get to my menu. Then your going to scroll down to critical point or numbers.I always have you start a graph on your paper. When you enter the function you have to press Alpha before you enter anything into these entry lines. Your going to press Alpha X cubed minus 3 times X. I always show you what you’ve entered. You can change it if you want. I say it’s okay. And we’re gonna choose number five, critical points.
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This video is on parametric equations. A parametric equation is, you’re adding a parameter of t time to every x y and z function, and that’s where we add the parameter and that’s the reason we call them parametric equations. And let’s do it. Turn the calculator on here. We’re going to get back to the home screen here. Clear the calculator we can go F1 eight and it clears that screen here. You press second alpha and put in the letters i n d e x, and then you push alpha and get into the number 8 and closed parenthesis to add this and get my formula for my menu. Press enter and we’re into my menu. And you can see all the things available in my menu for you to pass calculus and do your homework. Position vectors, product rule, projection of a and b, all those kinds of things you will be involved with in calculus one two or three. We’re going to do parametric equations now. That’s concerned with position vectors. If z is not given you enter zero in for z, then you can do the other two, x and y. So there’s the vector r t is generally an r t, is equal to this vector here, x t, y t and z t. So you have to press alpha to enter the functions in the entry lines here, so let’s do it, three times t let’s say plus four for x t, here’s y t, let’s enter have to push alpha, five times t and let’s do the z one, or let’s put alpha just for to make it simple and put z you can see that the z one is zero. Gives you a chance to change it if you’ve made a mistake, and we have all these things that we can do with this formula now with these functions in there. We can eliminate the parameter. Which eliminates the t and changes it back to an x function. Let’s do that quick, I’ll go through these quick so you can see. You solve for t, here’s the solution for t, and then you substitute t into every other x y and z, but ah here’s one point six seven times x minus three point nine and that eliminated the t parameter. Let’s go length of arc, you want to do that, fine, let’s go press four, notice it’s an integral over a and b, with the derivative of the r t formula. And rt we’re going to put in what we entered, I’ll go through it quick, put this all on your paper, write it down exactly as you see it, and we’re doing the square of each one, over the time of let’s say, you have to push alpha, let’s say from two to alpha six, shows you from two to six, here we’re doing this, write this on your paper and each individual one is gone, there it is, and here we substitute etcetera in there, and here we have approximately twenty three point four units. We can do speed, do you wanna do speed, let’s push number seven here and do speed, unit vectors or speed is the square root of these squared. Square root of nine, twenty five zero. 5.8 meters per second. Ah pretty neat huh? everystepcalculus.com, check it out. Go to my site and you’ll love these programs.
Transcripts
Hello again, everyone. This is Tom from everystepcalculus.com and everystepphysics.com. I’m going to do a Triple Integral for Calculus 3 right now. This is an example of a Patrick JMT, my favorite instructor on the internet, on YouTube. So I’m going to 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,etc. How do we do it? Let’s get started. Index 8 to get to my menu. I’m going to scroll up because I can go to the bottom of the menu then instead of going down quicker to go to the T’s section. And we’re going to choose Triple Integral. And we’re going to enter our function. You have to press Alpha before you enter anything into these entry lines here in 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 going to 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 test also. And we’re going to enter region q. Enter these limits. This is Alpha 0 for the x one Alpha square root of 4 minus z squared I made a mistake so I gotta go back. Choose number 2. Alpha 0, Alpha square root of 4 minus z squared. Close up the parentheses. That’s better. I say it’s okay. Next one for the y is Alpha 0. Alpha pi. That looks okay. and Alpha 0 for z. Alpha 2. That’s okay. So here’s what you write on your paper. The way you write it with triple integral with dx dz dy order of integration. Here’s the function in here. So you’re going to do the dx first and you put this over here with these lines. Showing you’r doing a range over this integration here. And here’s the integral of the first function okay. And if x equals the upper range. I show quotation marks here but you put you put parentheses in there. Because you’re substituting this amount for an X in the integral. And it equals this, minus sin, etc. And then we do the lower integral. X equals 0 and there’s 0 and you put parentheses around this instead of quotation marks, okay? And here’s the answer, you have the 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, etc. over this range here 0 2. Add z equals 2 Here’s the answer here. And z equals 0. Plug these in for all the Z’s in the problem. And the answer is this. 8, the upper range minus the lower range is 8 sin y divided by 3. Now we’re going to use that for the integration function. With the range of 0 and pi. At y equals pi minus 8 cosine is 8 thirds.
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A problem in calculus called Surface Area of a Revolution. So let’s do it. Index 8 to get to my menu. We’re going to go up to get to the bottom of the alphabet. And scroll up to Surface Area of Revolution.
We’re going to enter our, I’ll show you the actual formula here. Let’s write down your paper first. Press Alpha to enter anything into these entry lines here. Alpha x cubed over the range of alpa zero for a for b, Alpha 2. I show you what you’ve entered so you can change it if you want. I say it’s okay. The first derivative is 3x squared. That goes into the formula here. That’s part of the formula. Over 0 and 2 for the range. And we start doing the computations. Squaring things. This is U Substitution. U equals this. DU equals 36. You always take 36 and put it on the other side as a denominator. DU/36 equals that. And this x cubed equals the problem. X cubed dx. And then we substitute for U. In the problem here, DU 36 has to come out of the integral. So that 2 pi, there’s the 1 over 36 here and that computes to pi over 18, etc. We’re doing the integral of one half, of course you add two halves to that and Upper Range less than the Lower Range. And you get 203 squared units. Pretty neat, huh? everystepcalculus.com. Go to my site, subscribe to see more videos or go to the menu and look up what you need to learn about and pass your calculus. Have a good one.