Re: How Many Of You Had To Teach Yourself The Math?

Jonathan Kirwan wrote:
On Sun, 10 Sep 2006 02:07:20 GMT, "Homer J Simpson"
<nobody@xxxxxxxxxxx> wrote:

"Bill Bowden" <wrongaddress@xxxxxxx> wrote in message
news:1157852385.930640.208130@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Yes, but it's all very confusing. A real world example would be
helpful.

Is there a real world example of using 2X, (the derivative of X^2) to
obtain some useful information from the function y=x^2 ?

Not exactly, but try to figure out the maximum size of package that can be
mailed, considering that the maximum size is 108 inches in combined length
and girth (distance around the thickest part). Note that I did not say the
shape has to be rectangular or cylindrical.

To expand on the above comment....

When a curve rises and then falls, or falls and then rises, it must
pass through a point where the slope is zero. This fact is one of
those things taken advantage of, using derivatives. By no means is
this the only use. Just one of many.

An example is finding the case where you minimize the amount of ***
metal used to make a can for soup or tuna. If you know the volume you
need to enclose, that is.

A can is a right circular cylinder and has two ends (top lid and
bottom lid) and body cylinder itself that must be made out of the
*** metal. Let's call the height H, the cylinder's diameter D, the
volume V, and the total area of *** metal required A. For these
purposes, lets ignore the bit of metal used to seal the edges or form
flaps or lips and just focus on the simplified surface area and make
that what we want to minimize (for cost reasons, let's say.)

The area is:

A = top lid area + bottom lid area + side cylinder area
= PI*(D/2)^2 + PI*(D/2)^2 + PI*D*H

and after a little algebraic manipulation,

A = PI * D * [ H + D/2 ]

Hopefully, Bill, you can get this far without too much difficulty. At
this point, we have two variables, H and D, and there are an infinite
number of possible combinations that could be used to achieve any
particular A. Worse than that, we aren't even given the area, but
instead given some volume. So let's look at the equation for the
volume:

V = PI*(D/2)^2*H

Well, that's not too bad. Except it also has those two variables, H
and D, and for any given volume we can still have an infinite number
of possible H's and D's to achieve any particular V.

In thinking just a moment about this before going further, you should
be able to think about and realize that there are two incredible
extreme cases. One where the area of the lids is about zero and the
height must be very, very high in order to achieve some volume V. The
other where the height is about zero and the area of the lids must be
very, very large to achieve some volume V. In this two extremes, it
should be fairly easy to see that a lot of *** metal would be needed
and that it would be a lot more than the optimum. So you should be
able to guess that the amount of metal goes from a very large number
when the height is near zero down to some smaller value where the
height is reasonable and then back up to a very large amount as the
height nears infinity. We know the answer must be in-between.

Imagine plotting such a curve of area versus, for example, just the
height of the can. It would go from infinity at H=0, down to some
minimum value at some H we don't yet know, then back up to infinity
when H=infinity. All this with the volume V remaining constant, of
course. Remember that when a curve rises and then falls, or falls and
then rises, it must pass through a point where the slope is zero. This
must apply here, just by thinking about it.

So what we want to do is find the point where the area is a minimum.
That says that the A= equation mentioned above needs to be minimized
and suggests that to do so we just need to find where the slope is
zero.

But we are still stuck with those two pesky variables, H and D. So
let's get rid of one, right now.

To do that, take the volume equation and solve for H, instead (or you
could do that for D as that's another path that will get you to the
same place.) It is:

H = (4*V) / (PI*D^2)

Now we can substitute that into the A= equation, thus:

A = PI * D * [ (4*V) / (PI*D^2) + D/2 ]

Again, some re-adjusting with typical algebra skills yields:

A = 4*V/D + PI*D^2/2

Now, there is just the one variable, D. At this point we can use
calculus rules to produce the derivative:

dA = (-4*V/D^2 + PI*D) dD, or
dA/dD = (-4*V/D^2 + PI*D)

This represents the slope. Just set it to zero and solve or inspect
it.

0 = (-4*V/D^2 + PI*D)
4*V/D^2 = PI*D
4*V = PI*D^3
D^3 = 4*V/PI
D = CUBEROOT( 4*V/PI )

There it is. For any volume, this will tell you what the diameter of
the lids should be.

So let's look at the height. If D is the value above, what would the
height then be. Well, we can plug in the D= equation we just
generated back into the volume equation:

V = PI*(CUBEROOT(4*V/PI)/2)^2*H
= (PI/4) * CUBEROOT(4*V/PI)^2 * H
= (PI/4) * CUBEROOT(16*V^2/PI^2) * H
= CUBEROOT(PI^3/4^3 * 16*V^2/PI^2) * H
= CUBEROOT((PI/4)*V^2) * H

Solving this for H gives us:

H = V / CUBEROOT((PI/4)*V^2)
= CUBEROOT( V^3 / [(PI/4)*V^2] )
= CUBEROOT( V / (PI/4) )
= CUBEROOT( 4*V/PI )

Note that this H value is the same as the D value we also worked out.
In other words, the optimum is found when both D and H are the same. A
kind of "square can" so to speak. At least in profile.

This could have been solved differently, as well. We could have set
up a ratio, H/D, as a new variable and then worked out the equations
in terms of that H/D variable. In the end, we would have also set
that derivative equation to zero and solved and, I predict, would have
found a value of 1 as the answer.

In fact, this is the differential equation in that case (I'm skipping
all the drudgery here.) I'll use X = H/D:

dA = PI*CUBEROOT([4*V/(PI*X)]^2)*(X-1)/(3*X) dX

Setting this to zero, is:

0 = PI * CUBEROOT([4*V/(PI*X)]^2) * (X-1) / (3*X)

Upon inspection, you can see that this is zero if X = 1 simply because
the term (X-1) there will be zero forcing the rest to go to a zero
value. Also, if V=0, then the CUBEROOT will go to zero and the
equation will be zero then, too. But I suppose that is kind of an
obvious possibility. But the main point is that you can see that
(X-1) term in there and that tells us that when X=1 the area equation
will be at a minimum. Confirmation of what we discovered otherwise.

Similar reasoning using partial differentials and a basic description
of an error term allows one to develop the standard "least squares
fit" method used to fit a line to a set of data points taken in the
face of measurement error. There are many such applications using
this idea. And this is only one of many ideas using the derivative,
each of them with many interesting applications.

So it is important.

Jon

Jon,

Thanks for the explanation of sizing a can for maximum volume and
minimum area. It was very helpful. I follow most of it, but forget the
rules for derivatives. It was a good example of using derivatives to
find some optimum value.

I checked a tuna can on my shelf that measures 3.25 diameter by 1.75
inch high, which is not optimum and apparently wastes about 1.6 sq.
inch of area. But tuna cans are probably designed for convenience of
use and not optimum volume.

Thanks again for the math lesson.

-Bill

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