Problem

Compute $ \Delta y $ and $ dy $ for the given val…

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Problem 19 Medium Difficulty

Compute $ \Delta y $ and $ dy $ for the given values of $ x $ and $ dx = \Delta x. $ Then sketch a diagram like Figure 5 showing the line segments with lengths $ dx, dy, $ and $ \Delta y. $
$ y = x^2 - 4x, x = 3, \Delta x = 0.5 $

Answer

$\Delta y=1.25$ and $d y=1$

Discussion

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Video Transcript

All right. You're giving this function? Why? It's X squared minus four X. And X equals three. And delta X equals 30.5. And they want to calculate delta Y. And that and get this delta Y and the wife. But first is to delta Y. It means the actual real change in why as exchanges from 3- 3.5. All right. So, I want you to first figure out why at three And that would be 9 -12 which is -3. And then why at 3.5? And that would be 3.5 square calculator. Going here 3.5 times. Oops. 35 times yeah. 3.5 sq to a point 25 minus four times 3.5 14. Yeah -1.75. Yeah. Okay. So then the delta Y is Why of 3.5 -Y of three Which is -1.75 -3. 1.25. So if we looked at a picture of this as we moved from X equals 32 X equals 3.5. Why would move And would there would be a difference of 1.25 between them? Why would move from -3 -1.75. Now D. Y is the approximate change in case it was too hard to find your calculator or whatever approximate change in why as exchanges From 3 to 3.5. Can the formula from the Y is the derivative at X times dx Okay. DX is the same thing as Delta X. So it's .5. So we gotta find the derivative of that function F. Of X equals. I forgot what it was already. Um X squared minus four. X. So two x minus four. So you want to find it at three? How much was it at? Three? So two times three minus four which is six minus four, which is to and so D Y is two times .5, which is why? And So we got Delta Y is 1.25. The real change and our approximation is one. So if you didn't have to get really close then then that's good enough for approximation probably. Okay then it says draw a picture, like a picture somewhere else, I don't know. Um So you have this parabola X squared minus four X. So it's crossing the x axis at zero and 4 and it's got a its vertex is halfway in between there. Okay, I have it here somewhere. So if we look here, That's f. of three And then here's f. 3.5. Okay, and then the difference between the actual distance between them should be 1.25. And then if we draw this tangent line here, The distance between the uh and the point should be one

Linda H.

Oklahoma State University

Topics

Derivatives

Differentiation

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 10

Linear Approximation and Differentials

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