Linear approximation, sometimes called linearization, is a process that we can use to find a really
good estimate about the value of a function at a certain point that we don't know, by
using some information that we have about the value of the function at a point that
we already know.
Basically, we're using the fact that at a given point on a function, let's call it (x_1,y_1),
the tangent line at that point runs really close to parallel to the graph of the function
for x-values that are close to x_1.
This is really useful in cases where the function we are dealing with is especially complicated
and finding the y-value of a given x-value proves to be a difficult process.
For example it can take this equation and turn it into this equation.
The second equation may look harder, but without a calculator, the second equation is definitely
easier to solve than the first one.
Although the answer to the second equation won't be exactly equal to the answer to the
first, the approximation made by the second equation could be good enough in a lot of
circumstances.
If the y-output of the second equation is easier to find than the first, and the two
answers are close enough, why wouldn't we just use the second equation?
For example, linear approximation can be really helpful when we're trying to estimate square
roots, since we all already know lots of other square roots, like the square root of 4, 9,
16, and so on, but we don't really know how to take the square roots of numbers that aren't
perfectly square, like 3 or 9.2.
Let's go ahead and look at how we can use linear approximation on the square root function.
Let's say we want to find the square root of 16.3.
That sounds pretty tricky, but we already know that the square root of 16 is 4, so we
could say with at least some confidence that the square root of 16.3 is going to be just
a little bit more than 4.
Let's try using linear approximation to get an even better idea of what this square root
might be.
We'll use the information that we do know - that the square root of 16 is 4 - in order
to estimate what we don't know.
If we take a look at the graph of the function f(x)=sqrt(x), we can mark the point here with
coordinates (16,4).
We know the square root of 16.3 is going to be nearby, so what we can do is draw a tangent
line to the graph at the point we do know and then use the equation of that line in
order to find a y-value for x=16.3.
Finding the equation of that line is just a matter of recalling the point-slope formula
and plugging in the information we have.
The point-slope formula, remember, gives us y-y_1=m(x-x_1).
Using the point (16,4), we have that y-4=m(x-16).
The one thing we're missing in this equation is the slope of the line.
But we're using the tangent line to the function at the point (16,4), and how do you find the
slope of the tangent line at a point?
You evaluate the derivative of the function at that point.
So the slope of this line is going to be equal to f'(x_1), or in this case, f'(16).
The derivative of the square root function is equal to (1/2)x^(-1/2), so at x=16, the
slope of the tangent line is 1/8.
The equation for the line we're using for our linear approximation, then, is y-4=(1/8)(x-16).
After a bit of algebra, we get the simple equation y=(1/8)x+2.
Now all we need to do is plug in 16.3 for x, and that's going to give us (16.3/8)+2,
or just 4.0375.
This method isn't going to give you the exact value of y on the original function, but since
the square root of 16 is so close, the linear approximation will give you a pretty good
estimate about the square root of 16.3.
Looking at the graph of the function, plus the tangent line that runs through the point
(16,4), we can see that we get really similar values when we plug 16.3 into the original
function, and into the tangent line.
In fact, the actual square root of 16.3 is 4.0373, so the answer we found seems like
a pretty good estimate.
So if you think about linear approximation you can say that, in general, you can find
the linear approximation at a point x_2 when x_2 is close to x_1, and when you already
know information about x_1.
You just plug the information you have into this linear approximation formula, which is
really just a generalized version of what we used in the previous example.
The linear approximation of f(x_2)=f(x_1)+f'(x_1)(x_2-x_1).
And again, this formula comes directly from the point-slope formula for the equation of
a line, so if we can remember that formula, this one should be pretty easy to remember,
too.
Let's take a look at another example where we can use linear approximation.
Given the function f(x)=cos(x), we want to find a linear approximation for x=1.05, which
happens to be pretty close to pi/3.
f(pi/3) is actually much simpler to find, so we can use that knowledge to find an appropriate
estimate for f(1.05).
From the unit circle, we know that f(pi/3)=cos(pi/3)=1/2.
We also need to know the derivative of the function
at x=pi/3.
The derivative of cosine is negative sine, so f'(pi/3)=-sqrt(3)/2.
Now we can simply plug what we found into the formula for linear approximation, and
we get the linear approximation f(1.05)=(1/2)-(sqrt(3)/2)(1.05-(pi/3)).
And this gives us approximately 0.4976.
This estimate seems to make sense since f(pi/3)=1/2 and 1.05 is so close to pi/3.
In fact, there's actually a simple way to check the error of a linear approximation,
just by rearranging our equation a little bit.
So if we subtract f(x_1) from both sides, the left hand side will give us the difference
in y, and this is the error in our estimation.
So the error can just be calculated using everything that's left on the right-hand side
of the equation, meaning the product of the derivative and the difference in the x-values.
In this case, the error is equal to (-sqrt(3)/2)(1.05-(pi/3)), which gives us a value of about -0.0024.
This is a pretty small margin of error, so I'd say we got a pretty good estimation of
x=1.05 thanks to linear approximation.
Keep in mind that you can get both positive and negative values for the error of a linear
approximation, but since we're looking at error, which is just the distance between
the actual value and the approximation value, you can always give a positive value for error.
So in this case we'd take the absolute value of 0.0024, and say that the error of the linear
approximation is positive 0.0024.
To review, linear approximation just uses the y-value and the derivative at a given
x-value in order to estimate values of a function close to the x-value we're interested in.
The formula for linear approximation is given by f(x_2)=f(x_1)+f'(x_1)(x_2-x_1).
If we want to calculate the error in our linear approximation, all we're looking for is the
change in y, which we can find by multiplying the derivative at x_1, by the change in x.
In general, as long as we're trying to estimate the value of a function that's close to a
value that we already know, linear approximation usually gives us a pretty good estimate.
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