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 5th. 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.
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.
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.