- #1

esmeco

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Any help on this is really appreciated!Thanks in advance!

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- Thread starter esmeco
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- #1

esmeco

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Any help on this is really appreciated!Thanks in advance!

- #2

mathman

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ds

In practice, you would use:

ds=(1+(dy/dx)

- #3

HallsofIvy

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[tex]ds= \sqrt{\left(\frac{dx}{dt}\right)+ \left(\frac{dy}{dt}\right)^2}ds[/tex]

If the perimeter is given by two or more different formula (as a region defined to be between two curves) you will need to integrate each separately and then add.

- #4

esmeco

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Sorry for my "ignorence" but, what does de ds means?Is it the integral?

- #5

arildno

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We add all those tiny lengths up to get the entire curve's length.

- #6

Gib Z

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For the area, we cut up the region into tiny rectangles yes? well the integral basically makes it perfectly accurate by making lots and lots of rectanles. For that to happen, the base has to be smaller. When its really small, we call it dx. a small change in x. a small change in height would then be dy.

if we took those small changes in x and y, and applied pythagoras to them to get the straight line that connects them. That imitates the perimeter abit. Then we integrate it to sum all of the small ones. :)

- #7

esmeco

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ds=sqrt(2^2 + 2^2)=sqrt8

Is this correct?But what do I do now with the hipothenus value?

- #8

Gib Z

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if [tex]y=\sqrt{x+2}[/tex]. Try drawing the graph. Then zoom in. It gets abit straighter yea? Then in even more, it'll look even more straight. Since a tangent straight line then becomes a good approximation of that part, we make a straight line that hugs the curve. To do that, we take the lines vertical height, and horizontal height. From those 2, and Pythagoras Theorem, we get the length of the line that hugs the curve. The tangent will vary in different places, and this method of doing this is very long, taxing and sort of in accurate. With Integral calculus, we can do this process perfectly accurately and very quickly.

But you haven't done Differential Calculus yet, so your at least a good few months off this.

- #9

Gib Z

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[tex]\int^b_a \sqrt{1+ (y')^2} dx[/tex]. Thats the perimeter of the function y, from b to a. I tried use prime notation rather than Leibniz, but that's hard to do with Integrals lol.

From your other forums, it seems you haven't encountered integrals, but rather only primitives.

- #10

HallsofIvy

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