Problem

(a) Sodium chlorate crystals are easy to grow in …

03:49

Problem 11 Hard Difficulty

(a) A company makes computer chips from square wafers of silicon. It wants to keep the side length of a wafer very close to 15 mm and it wants to know how the are $ A(x) $ of a wafer changes when the side length $ x $ changes. Find $ A'(15) $ and explain its meaning in this situation,

(b) Show that the rate of change of the area of a square with respect to its side length is half its perimeter. Try to explain geometrically why this is true by drawing a square whose side length $ x $ is increased by an amount $ \Delta x. $ How can you approximate the resulting change in area $ \Delta A $ if $ \Delta x $ is small?

Answer

a) $P(x)=4 x,$ so $A^{\prime}(x)=2 x=\frac{1}{2}(4 x)=\frac{1}{2} P(x)$
b) see answer

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Video Transcript

So we have a square computer chip and its dimensions are X by X, and we want to find its rate of change, the rate of change of the area at the time, or the moment when the side length is 15. So first of all, the area of the square would be X squared. And then we would differentiate that to get the derivative the rate of change and that would be two X. So then a prime of 15 would be to times 15. So that would be 30 and the units on that So the units on area would be square millimeters and the units on ah, side length would be millimeters. So what that's telling us is that at this moment there would be a 30 millimeter change, an area for every one millimeter change in the side link. We could take a look at an example. If the side length was 15 the area would be 15 squared, which is to 25. And if the area, if the side linked was 16 the area would be to 56. That would be 16 squared and those air off by what 31. So that's fairly close to the 30 that was described as the rate of change. Okay, so for part B, So we have our square with side lengths of X. We just found the rate of change was two X and the perimeter would be the sum of all the side lengths, and that would be for X, So the rate of change of the area is half the perimeter. Now what we can do to kind of analyze this geometrically is imagine that we have our square, but that it grows on each side by an increment of Delta X. So the area of the new square is X plus Delta X quantity squared, and the area of the old square was just X squared. So what's the change in area? The change in area is the area of the new square minus the area of the old square, and that would be X plus Delta X quantity squared minus X squared. Let's go ahead and work this out. Algebraic Lee. We're going to have to use the foil process and we get X squared plus two x Delta X plus Delta X squared and then we have minus X squared from subtracting the area. The old one notice that we can cancel the X squared in the minus X squared. And so what we have is two x Delta X plus Delta X squared, which is the change in area. So suppose we want to know the change in area over the change in X. Suppose we want to know Delta X Delta A over Delta X. So let's go ahead and factor a Delta X out of Delta A and we have Delta X Times a quantity two x plus Delta X over Delta X notice that we can cancel the Delta X on the top of the bottom. And what we're saying is that roughly the changing area with respect to X is two X plus Delta X, And if Delta X is very small, then it's approximately two times x half the perimeter. So if you were interested in approximating a change in area for a very small Delta X, you could approximate it by just taking two times X or half the perimeter

Heather Z.

Oregon State University

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Problem

(a) Sodium chlorate crystals are easy to grow in …

Problem 11 Hard Difficulty

Answer

a) $P(x)=4 x,$ so $A^{\prime}(x)=2 x=\frac{1}{2}(4 x)=\frac{1}{2} P(x)$
b) see answer

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a) $P(x)=4 x,$ so $A^{\prime}(x)=2 x=\frac{1}{2}(4 x)=\frac{1}{2} P(x)$b) see answer

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Grace H.

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Kristen K.

a) $P(x)=4 x,$ so $A^{\prime}(x)=2 x=\frac{1}{2}(4 x)=\frac{1}{2} P(x)$
b) see answer