WEBVTT
Kind: captions
Language: en
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There are many ways in science in which exponential
increase or decrease, exponential growth,
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occurs. And it's very useful, therefore, to
be familiar with some of the terminology and
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reasoning associated with that phenomenon.
Here's a first example. It's of a word problem
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type. As you can see, not just an equation
you have to solve, but a situation described
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in words. I love these kind of problems, personally,
but some students find them a little tricky.
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So we have a sheet of material. And the thickness
of this sheet is 1/10 of a millimeter. You're
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going to fold it a certain number of times
over on itself. How many times must you fold
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it so that the thickness is at least one meter?
Quite an increase. Well, to analyze the problem,
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you see that when you fold it once, you will
have doubled the thickness. So the thickness
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will be 2 times 1 over 10 in millimeters.
When you fold it over again, you will double
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that thickness. So you will get 2 squared
over 10 millimeters, and so on. You'll see--
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could be proved by induction-- that after
n folds, the thickness of your material will
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be 2 to the power n over 10 millimeters. Now
you're aiming for a thickness of at least
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one meter. That's 1000 millimeters. So you
want the number of folds n to be such that
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2 to the power n over 10 will be greater than
1000, which is 10 cubed.
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In other words, 2 to the n has to be greater
than 10 to the fourth. You could solve this
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inequality explicitly. That means that n has
to be greater than 4 log to base 2 of 10,
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which you could look up on a calculator. However,
this problem is sufficiently simple that you
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can find the n by hand. It's not very hard
to calculate 2 to the power 13. You already
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have 2 to the power 6 is 64. And you find
that 2 to the 13th is over 8000. And 2 to
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the 14th is over 16000. And your answer, therefore,
is n equals 14 folds. Notice how exponential
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growth is awfully fast. Just 14 folds got
you up to a meter.
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Now more generally, problems of exponential
growth and decay occur frequently in science,
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in physics and biology and other places. And
there's a definition that goes with this,
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a useful definition that you should retain.
If you have a quantity y that depends on time,
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so y of t, it is said to have an exponential
law, either of growth or decay, if the y of
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t is described as a function of the following
type-- y naught, or y0, times the exponential
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of k t. Here, y0 is taken always to be a positive
constant. Now we say that it's growth if k
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is positive and decay if k is negative. Why?
Because if k is positive, the function y is
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an increasing function of time. And it's decreasing
if k is negative. Incidentally, we note that
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the constant y0 is simply the value of y at
time 0. It's the amount of y you have at the
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beginning of the discussion. Here's an example
of how this kind of concept occurs in, again,
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a word problem. You have a certain bacterial
population that experiences exponential growth.
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And you observe-- you can't count these bacteria,
but by looking at the mass or the weight or
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whatever, you see that its initial size has
doubled in 3 hours. And the question is, how
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many times will the initial population have
been multiplied by after 9 hours?
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Well, here's how to solve this problem. You
have to carefully use all the given information.
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First of all, it's said to exhibit exponential
growth, this population. That means, by definition,
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that y of t is of the form y0 e to the k t.
t here is in hours. Now you're also given
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that y of 3, that's the population after 3
hours, is twice the initial population. So
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that tells you that 2 y0 is equal to y0 e
to the 3 k. You'll notice that the y0 will
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cancel on both sides of this equation, and
you'll be able to solve for k, which we do.
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We find k is ln 2 over 3, a positive number.
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Once we know the value of k, then we know
that the population after 9 hours is equal
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to y0 times e to the 9 k. We put in the k
that we found, we do a little simplification
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using the laws of logarithm that we're familiar
with, and we find 8 y0. In other words, after
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9 hours, the population is multiplied by 8.
And that's the answer to the question that's
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been posed. Let's look now at an example in
which things decrease rather than increase--
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decay. We have a radioactive substance. And
these substances, in fact, behave as exponential
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decay. We observe for this particular substance
that the 200 grams we had initially reduces
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to 50 grams after one hour.
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The question is, how much of the substance
will be left after 3 hours? We have to carefully
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use all the information in these short sentences.
Here's how it works. First of all, we know
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that y of t is of the form y0 e to the k t.
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We're also told that 200 grams reduces to
50 grams in one hour. That means that the
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200 grams we had initially, the value of y0,
when multiplied by e to the k times 1, will
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give you 50. We find e to the k equals a quarter.
We easily solve that simple equation, as we
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solved earlier ones, to find that k is minus
ln 4-- a negative number, because here, it's
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a matter of decay rather than growth. Once
we know the value of k, we can plug it in
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to find y of 3. After a little simplification,
we find 50 over 16 grams. Now let me ask you
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a question.
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For the same radioactive substance, at which
point will there be none left at all? When
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will the substance be completely gone? The
answer is never, because a function like exponential
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of k t, although it goes down to 0, it decays
towards 0 asymptotically, it never attains
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0. So a radioactive substance is never completely
gone. It's eternal. It doesn't have a finite
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life. And yet oddly enough, it has a half
life. Now that sounds like a paradox. Let
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me explain. A well-known definition in physics
is the half life of an exponentially decreasing
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quantity-- that means the time required for
the quantity to become half of what it was
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initially. That turns out not to depend on
the actual quantity.
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It's an intrinsic constant linked to the substance
itself. Let's look at the definition more
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carefully. You have a y of t, quantity y depending
on t, that is decaying exponentially. That
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means, if you look at the graph of y, it's
going to be the graph of an exponential function
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with a negative exponent. And we know that
it's going to go down towards 0 asymptotically.
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And its initial value is y0 at time t. Now
if you look at y0 over 2, that's a strictly
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positive quantity. And there will be, at some
point, some time at which you hit that value.
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I'm going to call that time tau.
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You know, tau is another beautiful Greek letter
that we like to use, especially for time--
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tau. Why is there such a tau? Well, there
is such a tau, because the function y goes
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down to 0, and eventually, the graph must
cross the value y0 over 2. And furthermore,
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there's only one such point, because the function
is strictly decreasing. And what defines the
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point tau? Well, what defines it is that the
value of y at tau-- which, on the one hand,
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is equal to y0 e to the tau k-- is 1/2 the
initial value, 1/2 y0. So the y0 cancels from
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both sides, and you find a simple equation
for tau, which you proceed to solve.
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And you find that tau is equal to minus k over
ln 2. That's a positive number, by the way.
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And that's called the half life of the radioactive
substance. It's the usual measure by which
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you measure the persistence of a radioactive
element.