Integration again..

jackdaw

Sorry i thought putting a new post would lighten this for me

in attached doc , i don't understand how he makes the first connection marked in red.

Michael Collins

In the line you're talking about he's going from "a differential" (i.e. an approximation to "y" - with error PQR) to the exact value or "a derivative", as the strip gets infinitesimally narrow i.e. the region PQR disappears and has no area...does that make sense? Or would you like me to elaborate on any part?

jackdaw

Michael Collins wrote: »

In the line you're talking about he's going from "a differential" (i.e. an approximation to "y" - with error PQR) to the exact value or "a derivative", as the strip gets infinitesimally narrow i.e. the region PQR disappears and has no area...does that make sense? Or would you like me to elaborate on any part?

so how does ∂Ax/∂x = dAx/dx ??

how are these 2 areas related , I can understand how as ∂x -> 0 the area ∂Ax -> 0, but Ax ?? how is this related ??

Michael Collins

Ok so you do see how ∂Ax ≈ y. ∂x? Here the ∂Ax is considered to be a very small but finite amount. Now as you say, as ∂x -> 0 then ∂Ax -> 0. So in the limit we can replace these finite amounts by infinitesimal amounts namely, dAx = y. dx (or dAx/dx = y). Note that we have replaced the ≈ sign with the = sign, since our equation is now exact.

Maybe I should explain the bit in italics above.

So let the area of the first shaded bit be A(x) and the area of the overall shaded region be A(x+∂x). So now we can say that the area of the second shaded region is exactly given by

A(x+∂x)-A(x)

i.e. just subtracting the two areas yeh?

Now this can be approximated by the rectangular area, given by

y. ∂x

so

A(x+∂x)-A(x) ≈ y. ∂x

And the smaller ∂x gets, the more accurate the approximatin will be. Rearranging we get

(A(x+∂x)-A(x))/∂x ≈ y

Taking the limit as ∂x -> 0, this becomes

lim (A(x+∂x)-A(x))/∂x = y (i.e. it is now exact)

but the first is just the definition of a derviative...so we get

dA/dx = y

as before.

Fremen

Hmm, I guess I'll have a shot at explaining it.

If you have a rectangle, and you divide its area by its width, you get its height. Our problem here is that we have a rectangle with the triangle PQR stuck on top of it. If we could find some way to make that triangle "disappear", then we could just divide the area, in this case ∂A, by the width, ∂x and get Y, the height.

We can make the triangle PQR as small as we want by making ∂x progressively smaller. However, if we want to make it so PQR has zero area, we need to make ∂x be zero units long. If you were to try to print this in a book, it would be a rectangle so thin that it just looks like a vertical line. In a very real sense, it is just a vertical line, namely the height, y.

The notation is changed from ∂A/∂x to dA/dx to reflect the fact that suddenly you're dividing something with zero area (∂A) by something with zero length (∂x).

Note that this explanation isn't strictly rigorous. You never actually divide zero by zero, but as both ∂x and ∂A get closer to zero PQR gets smaller, and thus giving you a better estimate of y. Really, the point is that you can get your estimate "as close as you like" to y by making ∂x smaller.

Now could someone please take a look at my rook tour problem?

berengar

Fremen wrote: »

Hmm, I guess I'll have a shot at explaining it.

If you have a rectangle, and you divide its area by its width, you get its height. Our problem here is that we have a rectangle with the triangle PQR stuck on top of it. If we could find some way to make that triangle "disappear", then we could just divide the area, in this case ∂A, by the width, ∂x and get Y, the height.

We can make the triangle PQR as small as we want by making ∂x progressively smaller. However, if we want to make it so PQR has zero area, we need to make ∂x be zero units long. If you were to try to print this in a book, it would be a rectangle so thin that it just looks like a vertical line. In a very real sense, it is just a vertical line, namely the height, y.

The notation is changed from ∂A/∂x to dA/dx to reflect the fact that suddenly you're dividing something with zero area (∂A) by something with zero length (∂x).

Note that this explanation isn't strictly rigorous. You never actually divide zero by zero, but as both ∂x and ∂A get closer to zero PQR gets smaller, and thus giving you a better estimate of y. Really, the point is that you can get your estimate "as close as you like" to y by making ∂x smaller.

Now could someone please take a look at my rook tour problem?

Thanks!!