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Topic: Integral as Accumulator...revisted
Replies: 10   Last Post: Apr 27, 1998 12:36 PM

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Chris & Sheila King

Posts: 484
Registered: 12/6/04
Re: Integral as Accumulator...revisted
Posted: Mar 22, 1998 3:27 PM
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On Sun, 22 Mar 1998 12:29:31 -0500, Richard Sisley
<keckcalc@earthlink.net> wrote:

:Sheila King asked the following questions about applications of the
:integral concept:
:
:1. Consider a sphere which is expanding as time passes. The Area is a
:function of time. A'(t) is the rate of change of the area with respect
:to time. Explain what the following integral represents:
:fnInt(A'(t), t, 3, 5)
:
:2. Considering the same sphere as in the previous problem, B(r) is the
:area of the sphere as a function of the radius. Explain what the
:following integral represents:
:fnInt(B(r), r, 3, 5)
:
:I have tried to introduce integals as measures of change from the very
:start this year. I want to help the students connect the idea of an
:integral with a process. I have recently shared a diagram with my
:students about integrals which I will try to describe here:
:
:I had an oval drawn around the space where an expression for the
:integrand function would normally be written in the standard notation
:for a (definite) integral. This oval was labeled "an expression from
:which the rates at which something is happening can be computed." Then
:around the entire integral notation a second, larger oval was drawn and
:that oval was tagged with the label "a measure of what happened."

Yes, this is one of the ideas I was getting at with my questions.
:
:Sheila's questions can be answered in this context.
:
:For number 1: Her integrand expression is A'(t) and this expression
:could be used to compute rates at which the area of the surface of the
:expanding sphere is increasing. The units for these rates would be
:square units per unit of time. For example, it could be square
:centimeters per second, or square inches per minute, etc. According to
:her limits of integration, 3 and 5, the process continued for 2 units of
:time (2 seconds, or two minutes, or, etc) The integral would then
:measure what happened during that time span--the change in the area over
:that time interval.

The above sounds good to me.
:
:For number 2: Her integrand expression is A'(r)

Ooops. Sorry. I think you've misinterpreted my question. I did not
indicate that the integrand was a rate of change of the area with
respect to the radius. In fact it WAS the area, itself, computed as a
function of the radius.

: and this expression
:could also be used to compute rates at which the area of the surface of
:the expanding sphere is increasing. However, in this second case, the
:units for these rates would be square units per unit of change in the
:radius. For example, it could be square centimeters per centimeter, or
:square inches per inch. The integral would then measure what happened
:as the radius increased from 3 to 5 units--the change of the surface
:area when the radius changed by 2 units.
:
:Here is an addition to this which some might find interesting. Imagine
:modeling points with the following description (the radius of the
:expanding sphere, the volume of the region in the interior of that
:sphere). Suppose this model function is named V.
:
: Then fnInt(V'(r),r,0,t) would represent the volume of a sphere with
:radius t. In my courses, we demonstrate and use the fact that when area
:and volume are being accumulated in a sweep by an elastic figure through
:a region, the rate of accumulation at any stage of the sweep is the size
:of the sweeping element at that stage. In this case, the sizes of the
:sweeping element would be the surface areas of the exspanding sphere.
:If it is established in some other way that the volume of the sphere
:when the radius is r is (4/3)*pi*r^3, then the rate of accumulation of
:the volume is 4*pi*r^2 which must then be a formula for the surface area
:of the sphere.

Here is another question:
Do we have to establish that the Area =B(r) is actually 4*pi*r^2 and
show that this comes out to Vol = (4/3)*pi*r^3 to use the concept that I
was employing in my second question? I wasn't sure when I wrote the
questions, to I stuck to spheres, which seemed safe to me. I also
thought it would be easier for my students to visualize in a testing
situation.

But, What about another shape of a solid...say a right cone. If it's
area is changing as a function of, say, it's slant height
say x = slant height of the cone
A(x) is the area of the cone as a function of its slant height
(because slant height is changing, it is not certain whether radius or
height of the cone are changing...possibly both or just one or the
other)
Doesn't the integral
S A(x) dx have to represent volume?
(This is where I get a bit confused...sometimes I'm thinking yes, and
sometimes I'm thinking no.)

Well, anyway, let me limit the area A(x) to the Curved outer surface of
the cone, and not including the circle on the bottom. Then I feel pretty
confident to assert that
A(x), measured in square units, multiplied with x measured in linear
units is a product in cubic units and the integral must be change in
Volume.

Is this still valid if we include the circle at the bottom as part of
A(x)? I keep wavering on that one. First I think "no" then I think "yes"
then I go back again to "no"...

I think that to only consider the definite integral as accumulating
rates of change is not interesting enough. It accumulates products where
one of the factors is very small!

Sheila King

:
:Sincerely,
:
:Richard Sisley
:keckcalc@earthlink.net

------
http://www.wenet.net/~cking/sheila/




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