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I am attempting to follow Paul's calculus notes, but am having trouble, in particular at this page: http://tutorial.math.lamar.edu/Classes/CalcII/HydrostaticPressure.aspx

I get to the part with the "The height of this strip is $\Delta x$ and the width is $2a$. We can use similar triangles to determine a as follows,"

and I am completely lost, I do not follow at all what is happening. Is there a mistake? The calculation seems impossible. Is he writing things backwards? $\frac{3}{4}$ is a constant how is it equal to that weird statement? I don't get it, is that solving for $a$?

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    Do $y$ou know what *[similar](http://en.wikipedia.org/wiki/Similarity_%28geometry%29) triangles* mean? Do you know [Thales' theorem](http://en.wikipedia.org/wiki/Intercept_theorem)?2015-04-06

3 Answers 3

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$\hskip 1in$ triangle

In the above, the base and height of the larger right triangle are $A$ and $B$ respectively, and the base and height of the inside, smaller triangle is $a$ and $b$ respectively. The two triangles are similar, which is a term from geometry meaning the ratios between the sides of one triangle are equal to the ratios between the sides of the other triangle. That is, in the above, we have that

$\frac{a}{b}=\frac{A}{B}.$

(Equivalently, the inner triangle is the bigger one scaled down in size.) Hence we have:

$\hskip 1in$ similar2 $\huge\displaystyle \frac{3}{4}=\frac{a}{4-x_i^*}$

Keep in mind we've approximated the triangle by covering it with thin rectangular strips, and intend to calculate the hydrostatic force associated to each strip and then add them up. As we make the strips smaller and smaller our calculation will be a Riemann sum for an integral, and hence this integral will be our desired answer. In order to calculate the force for a strip, we first denote its dimensions as $\Delta x\times 2a$. Note that $a$ is not really a constant: it varies with choice $x_i^*$. In order to do the Riemann sum though we need to write the hydrostatic force (of each strip) purely in terms of $x_i^*$, which means we need to solve for $a$ in terms of $x_i^*$, and similarity is what allows us to do that.

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    I have an understanding of this part but I get lost after this. Thank you for the pictures with explanations though it really helps a lot.2012-06-13
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The big half-triangle has width $3$ and height $4$ while the similar small half-triangle has width $a$ and height $4-x_i^{^.}$

Since you have similar triangles, you can then say $\frac{3}{4}=\frac{a}{4-x_i^{^.}}$ and so $a=3 -\frac{3}{4}x_i^{^.}$

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    @Jordan, "I memorized the process and formulas" will fail you and is the reason you keep bringing up problems that are solved with only slightly different methods. If you don't try to understand why you'll just end up taking this course again and you'll never be able to use the concepts in it. I don't want to sound harsh, but this point has been made several times and you don't seem to take it to heart very well. These are not "one off tricks": you're getting$a$function out of$a$geometric problem. You need to try and see connections like that instead of assuming they aren't there.2012-06-13
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Note that the height of the triangle is $4$. Because the small triangle is similar (i.e. same angles) to the big one, we know the ratios of side lengths are the same.

On the big triangle, one side is $3$ and the adjacent is $4$. On the smaller one, the similar side has the length of $a$ and its adjacent is $4-x_i^*$. Thus $3$ is to $4$ as $a$ is to $4-x_i^*$. Thus

$\frac{3}{4}=\frac{a}{4-x_i^*}$

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    I understand that part but I have no clue what is going on for the rest of it. He just magic wands the part with the definition of an integral and throwns in random xs whereever, why is this>2012-06-12