When calculus fails Related rates question

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The discussion centers on the limitations of calculus in predicting area changes using related rates, specifically when applying the derivative to the area of a square. The user illustrates a scenario where increasing the side length of a square from 2 to 3 results in an incorrect area prediction using the first-order differential approximation. The correct approach involves integrating the differential 2sds over the interval from 2 to 3, which accounts for all infinitesimal changes in area, leading to the accurate result of 9 instead of the erroneous 4.

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Hi I was just curious as to why calculus does not seem to work in the situation of area calculation. I came across this doing some HW and noticed that the booked worked out a seemingly similar question in two separate ways

First was predicitng the chance in the length of the hypotenuse. Using the pythagorean theorem take the derivative and your done. Plug in values and one has a relation that will predict the value of the hypotenuse given the orginal sides and their respective changes.

Now I had a similar problem with an area questions where I was trying to determine the new area given a defined change in the sides. Repeating as I have seen earliar. Take the derivative plug in values and ... wrong answer.

So now I am curious to know the explanation why... Take area of a square given by a=s^2
the derivative is da = 2sds (I was taking the derivative in respects to time and eliminated dt) So given a 2x2 square with area 4, I want to predict the area if I increase the side by 1. Given the previous result it tells me that the da would be 4. But the next perfect square is 9.
I tried for triangles and as expected not correct. I just wanted to know why for some related rates calculus works and sometimes it doesn't.

Thanks in advance
 
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Well, the issue is, what you are using is a first order differential approximation, which is just that, an approximation. The thing is, differentials are made for use with infinitesimal, or really small, values. Were your change in side length more like .0000000001, the approximation would be much more accurate. However, your value is 1, which in the world of infinitesimals, is extremely large. The first order approximation requires the portion of the graph that you are looking at(from s=2 to s=3)to appear to be essentially a straight line, hence it is a linear approximation.

However, if you were to integrate your differential 2sds with your bounds (from 2 to 3), and by doing so sum up all the infinitesimal changes in area along the way, you would get the correct answer.
 
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