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N2O5 is a reagent that was used synthesize
explosives like TNT. Unfortunately, N2O5 decomposes
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relatively quickly at or above room temperature,
so if you accidentally leave a bottle of it
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out on the counter, you could be in trouble!
In this video, we'll approximate a solution
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to the decomposition rate equation to figure
out whether a batch of N2O5 that you left
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at room temperature can still be used. We'll
also determine under what conditions our approximation
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is valid.
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This video is part of the Linearity video
series.
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Many complex systems are modeled or approximated
linearly because of the mathematical advantages.
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Hi, my name is Ben Brubaker, and I'm a professor
in the Department of Mathematics at MIT.
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Today we'll be talking about linear approximations.
In mathematical terms, this is just another
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name for the tangent line to a function. But
the name suggests more. Linear approximations
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can be used to simplify mathematical models
that are not analytically solvable. The approximated
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model will have a solution that is only acceptable
under suitable conditions. However, it can
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still illuminate the behavior of the system
within a certain acceptable range.
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Before watching this video, you should know
the definition of the derivative, and how
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to write the equation of a line with a given
slope that passes through a given point.
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After watching this video, you should be able
to recognize the linear approximation of a
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function as the tangent line to the function,
apply linear approximations to solve simple
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differential equations, and explain the limitations
of linear approximations both mathematically
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and graphically.
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Let's begin by defining the linear approximation.
Recall that if a function is differentiable
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at a point c, then when we zoom in on the
point c, the function begins to look more
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and more like a line. This only works when
the function is "smooth"â€”it doesn't have
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any kinks, corners, or discontinuities. Given
a function f(x), which is differentiable at
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the point c, we define the linear approximation
to be the tangent line to the function at
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c. This is a line whose slope equals f'(c).
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As an example, let's look at the following
cubic equation: f(x) = one thirtieth x times
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x minus 2 times x plus 5. This function is
differentiable everywhere, and in particular
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is differentiable at the point x=3. Find the
equation of the tangent line to this cubic
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equation at x=3 for yourself.
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We've also found the equation for the tangent
line. To do this, we found the value of the
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function at 3, which we found to be 4/5. And
then we computed the derivative and evaluated
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it at x=3, which we found to be 7/6. We'll
write the tangent line as T sub 3 of (x) to
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remind us that we are finding the tangent
to our function at the point 3. Then the equation
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for this line can be written as T sub 3 of
x equals 4/5 + 7/6 times (x-3).
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Let's take a look at the graphs of f(x) and
T sub 3 of x. The function f(x) is graphed
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in red. The further we zoom into the graph
at x=3, the more it begins to resemble a straight
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line. The slope of the line approaches the
value of the slope of the tangent line at
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3, which is drawn in blue.
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The tangent line certainly seems to be a good
approximation to our function when we are
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close to x=3. But what about when we zoom
out? Is it still a good approximation? Let's
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give a mathematical justification that the
tangent line is a good approximation. Our
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function is differentiable at a point, say
x=3. This is equivalent to the statement that
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when x is near 3, the slope of the secant
line is approximately equal to the slope of
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the tangent line. That is, for x near 3 f(x)-f(3)
over x-3, is approximately equal to the value
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of the derivative at 3.
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In order to see why these statements are equivalent,
we need explain what we mean by "approximately
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equal to". This means that we can make the
difference between these two sides as small
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as any error bound we choose, provided that
we choose x close enough to 3.
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And this is exactly the definition of differentiability
at the point 3.
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To relate this to linear approximations, we
use a bit of algebra. First we multiply both
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sides by (x-3), and we get f(x) -- f(3) approximately
equal to f'(3)(x-3). Adding f(3) to both sides,
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we get that f(x) is approximately equal to
f(3)+f'(3)(x-3). And this right hand side
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is the equation for the tangent line at 3.
So indeed the tangent line closely approximates
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the function as long as x is close to 3.
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Here we see the tangent line drawn in blue,
the function drawn in red. The graphs demonstrate
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our mathematical proof that the tangent line
is a good approximation near x=3, but the
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further we get from x=3 the worse the approximation
seems.
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Now you might see that the function and tangent
line intersect at x=-9. Is the tangent line
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a good approximation for the function near
x=-9?
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No, it isn't. If we move a small distance
along the tangent line away from x=-9, this
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does not model the behavior of the function
near x=-9. This property is extremely important
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in applicationsâ€”no measurement or observed
quantity is ever given exactly.
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A1 The linearization of a function that is
differentiable at x=c is just another name
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for the tangent line through c.
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The A2 key properties of this tangent line
are that it shares the same value as the function
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and the same first derivative of the function
at x=c. And A3 we've seen through both graphical
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intuition and the definition of the derivative
that this linearization closely approximates
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the function for points x sufficiently close
to c.
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Let's see how we can apply a linear approximation
to simplify a problem. Let's suppose that
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you are participating in undergraduate research
in a chemistry lab. You have been using a
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.01 molar solution of N2O5. You bring it out
of the refrigerator at 9:00 Monday morning,
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and promptly forget about it, leaving it on
the counter in the lab, which is 25 degrees
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Celsius. When you realize you left the solution
out, 1 hour has passed. You panic.
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The problem is that N205 decomposes into NO2
and O2 at room temperature. If the molarity
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of the solution has changed significantly,
it might ruin your experiments. Experiments
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have shown that the rate of decomposition
follows first order kinetics. This means that
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the instantaneous rate of change in concentration
of N205 is proportional to the concentration
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of N205. The constant of proportionality,
k, has been found experimentally to be 1.72
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times ten to the negative 5 inverse seconds.
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Because you are panicking, you are having
a hard time solving this differential equation.
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Instead, find the _approximate_ decomposition
in the N205 solution after 1 hour using a
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linear approximation at time t=0. Note that
because the reaction constant has units of
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inverse seconds, the variable t must have
units of seconds.
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In order to produce a linear approximation
at t=0, we need two ingredients: the concentration
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at time 0, and the value of the derivative
at time 0.
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The initial concentration was .01 molar. To
find the value of the derivative, we use our
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differential equation, and we find that the
value of the derivative at time t=0 is just
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minus k times the initial concentration at
t=0, or negative k times .01 molar. Remember
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that k has units of inverse seconds.
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Putting this together into the formula for
the tangent line at zero, we see that the
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linear approximation, is given by T0(t)= (.01
molar) times 1 minus $kt$.
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Now we are interested in approximating the
value at 1 hour, or 3600 seconds. Plugging
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in the value for k, we find that to two significant
figures T0(3600)=.0094 Molar. So the molarity
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is not so different. Phew!!
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Now you are feeling relieved, so when you
look back at the reaction rate, you realize
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that in fact, a first order reaction rate
is telling you that the concentration is some
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function of time whose derivative is proportional
to itself.
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The function with this property is the exponential
function! So the concentration function is
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equal to some constant times e to the negative
kt.
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The constant must be the initial concentration,
which is .01 Molar, so [N2O5]=(.01)e^{-kt}.
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Evaluating this expression at time t=3600,
we find that the exact concentration is also
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.0094 Molar to two significant figures. To
be more precise, we can look at the error,
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which is the absolute value of the exact solution
minus the approximate solution all divided
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by the exact solution.
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The error at time t=3600 seconds is .002 or
.2%.
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This is really quite good. Generally speaking,
we might consider an error of less than 5%
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to be acceptable.
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So our linear approximation was definitely
within the acceptable range 1 hour later.
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But now it's Tuesday. Today you leave the
.01 Molar N2O5 on the counter for a full 10
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hours. Oops.
We know that the exact solution is an exponential
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decay.
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So do you think that the linear approximation
will be an over estimate, an under estimate,
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or equal to the exact solution after 10 hours?
[pause]
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The linear approximation of the concentration
after 10 hours at room temperature is T0(36000)=
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.0038 Molar.
The exact solution of the concentration after
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10 hours is .0054 Molar. Comparing these two
solutions, we see that linear approximation
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is an under estimate as we would expect, with
error is .29 or 29%. So a linear approximation
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is definitely not acceptable for this time
range.
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But now you begin to wonder, for what times
is the linear approximation at t=0 within
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5% error of the exact solution? [pause] Because
we are only interested in positive time values,
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you should have found that the approximation
is acceptable for times 0 less than t less
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than 16, 700 seconds, which is about 4.5 hours.
Now that you've carefully examined this problem
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using linear approximation, it is probably
time to tell your graduate student adviser
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that you need to order a new batch of N_2
O_5 solution.
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Let's review. A1 The linear approximation
to a function at a point c is the tangent
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line of a function at c. A2 This linear approximation
only accurately models the function for points
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sufficiently close to c. A3 It is easy to
read off the derivative at a point from a
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differential equation, and thus give a linear
approximation to the solution. In our example
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with the concentration of N205, we could also
solve the differential equation exactly and
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demonstrate that the approximation was acceptable
near the point of approximation t=0, but not
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so for larger time intervals.
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Notice that we used this simple example to
illustrate linear approximation because we
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could also solve it exactly and use the exact
solution to determine the error. This is somewhat
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misleading, since we often want to use linear
approximation precisely when the differential
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equation that describes our physical situation
does not have an analytic solution.
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So you may wonder, is there a way to estimate
the error in a linear approximation without
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knowing the exact solution. The answer is
yes! However, it involves Taylor's Remainder
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theorem, which finds an acceptable range in
terms of higher order derivatives, and is
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beyond the scope this video.