Dear Mr. Avering,
Get a grip. Eve is not worth mooning over; she’s an evil,
manipulative itch that you can’t scratch. She destroyed the old
dipstick, and she lied about the shape of your tank. And I’m
going to prove it. I’m going to show you that your tank is not
hemispherical. And I’m going to show you that building a new
dipstick -- for your parabolic tank -- is as easy as pie. Or
rather, it’s as easy as 50-pi-H2; that’s the volume (in
cubic feet) of
material in your tank when it’s H feet deep.
I’ve drawn a picture below. It shows a superimposed pair of
cross-sections of two tanks: a hemispherical tank and a parabolic
tank. You can see that the main difference is that the hemisphere
is always wider, except at the very top and the very bottom.
Because that’s the only difference, it might be helpful to set up two
formulas to describe the widths. It’s easiest to do the math if I
let w stand for the width from the center (that is, measured out from
the dipstick to the wall of the tank). I’ll measure both w
(width) and h (height of the material) in feet.
Figure
1: In this figure, we
see that a parabolic tank is thinner
than a hemispherical tank.
For a hemisphere, width and height are related by that old circle
formula, w2sph + (h-100)2 = 1002.
For a parabolic tank, width and height are related by a different
formula: 100*h = w2par .
The first looks like the “(x-x0)2 + (y-y0)2
= r2” circle you learned in
geometry; the second one is sort of like the “y=x2” that you
might
remember as parabolas from your pre-calculus. The “100”s in each
of these formulas were included to stretch the or move the graphs so
that they match your tank at the bottom and at the top. At the
bottom, both width and height are 0; at the top, both width and height
are 100 feet. The following table shows you that both
these formulas line up with the measurements at the top and bottom.
|
hemisphere |
parabola |
formula relating w & h
|
w2sph
+
(h-100)2 = 1002
|
100*h = w2par |
top (w=100, h=100)
|
1002sph
+
(100-100)2 = 1002
|
100*100 = 1002par |
| bottom (w = 0, h = 0) |
02sph
+
(0-100)2 = 1002 |
100*0 = 02par |
Table: Both the hemisphere and the
parabola pass through the
correct points at
the
bottom and top of the tank: (0,0) and (100,100).
That’s why Eve was able to fool you for a little while: on the
parts you could measure with a long ruler, both of these shapes could
be right. But you helped a lot by measuring both the volume and
the depth of how much material you put in the tank so far, and that’s
how we’ll show that Eve was doing you wrong!
We’re going to compute a formula for volume by adding up thin layers of
material, one layer on top of another, like your grandpa’s favorite
16-layer cake. To do this, I have to assume that each new, thin
layer is a perfect circle -- meaning that your tank isn’t lopsided or
uneven. I’m also going to assume that each new, thin layer is
perfectly level. If you’re storing boulders or old office chairs
that pile up in uneven heaps, then all these computations are
garbage! But if you’re storing something liquid, read on!
Since each layer is going to be like a perfect, flat, circle, we can
use the formula for the area of a circle to approximate the volume of a
thin
layer:
Volumelayer
~ π (radius)2 *
thickness of the layer.
The “radius” here is just the width of the tank, measured from the
center dipstick. That is, it’s just the width (w) we used
above. So we’ll substitute in the relationships between width and
height that we wrote on the previous page.
For a hemisphere, the volume of a layer is
Volumesph-layer
~ π (wsph)2
* thickness of the layer.
= π
[1002-(h-100)2] * thickness of the layer.
For a parabolic tank, the volume of a layer is
Volumepar-layer
~ π (wpar)2 *
thickness of the layer.
= π [100h] *
thickness of the layer.
If we use a thickness of 1 foot, we can use the formulas above to get a
rough estimate for how much is in the tank up to 10 feet high, one
layer at a time. In the table below, I calculated the approximate
volume for the bottom 10 layers, using both formulas. I used the
middle height in each layer, so I could get a closer approximation.
height of the 1-ft layer
|
volume of the layer
|
volume of the layer
|
| h=? |
in a hemisphere
|
in a parabola
|
| h=? |
π*[1002-(h-100)2]*1 |
π*100h*1 |
| 0.5 |
313 |
157 |
| 1.5 |
935 |
471 |
| 2.5 |
1551 |
785 |
| 3.5 |
2161 |
1100 |
| 4.5 |
2764 |
1414 |
| 5.5 |
3361 |
1728 |
| 6.5 |
3951 |
2042 |
| 7.5 |
4536 |
2356 |
| 8.5 |
5114 |
2670 |
| 9.5 |
5685 |
2985 |
|
| subtotal |
30371 |
15708 |
Table:
Using flat disks with a thickness of
1 foot to estimate
the volume at the bottom of a hemisphere and a parabola.
Of course, this is just an estimate. To get a better answer, we’d
take incredibly thin layers (but also an incredibly large number of
those layers) and add them up. In calculus, doing that is called
“integration”, and the symbol for that looks like the “S” of a
sum. There is also a technique to find a simple formula
that adds these many skinny layers up.
I’ll show you the notation that mathematicians use, just in case you
want to impress your friends, but all you really need to know is the
simple formula that comes from that fancy notation. For a
hemisphere, the volume (in cubic feet) for material up to H feet high
is shown symbolically as
Volumesph(H) = §0H π [1002-(h-100)2]
dh,
and the simplification symbolic stuff is just
Volumesph(H) = 314*H2-1.047 H3.
So more precisely, you’ll see that the bottom 10 feet of a hemisphere
have about
Volumesph(10) =
31,400-1,047 =30,353 cubic feet of material.
That’s not far off from the estimate we had above, but it’s very far
from what you measured yourself! That’s how I know Eve was lying
to you!
So let’s do the same thing for the real tank: the parabola.
The symbol for summing up the volume of all those thin layers up to a
height of H feet is
Volumepar(H) = §0H π [100h] dh,
And the simplification that you can actually use to compute volumes is
Volumepar(H) = 50*π*H2,
or even more usefully,
Volumepar(H) = 157.08*H2.
And we can use this to give us the markings on your dipstick for any
height H. Here is a table with markings every 10 feet.
Height
|
10
|
20
|
30
|
40
|
50
|
60
|
70
|
80
|
90
|
100
|
Volume
|
15,708 |
62,832 |
141,372
|
251,327 |
392.699 |
565,487
|
769,690
|
1,005,310
|
1,272,345
|
1,570,796 |
Table: The markings of volume for the dipstick,
if you mark it
every 10 feet, using the formula Volumepar(H)
= 157.08*H2.
That’s all you need to figure out the volume of material, or how to
mark your dipstick. Get back to working with your granddad, and
leave that witchy woman alone!
Sincerely,
A concerned calculus student
• back to the
problem • back to the
writing
page •
look at the Guide
to Writing in Math Classes