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### Section 4-11 : Linear Approximations

In this section we’re going to take a look at an application not of derivatives but of the tangent line to a function. Of course, to get the tangent line we do need to take derivatives, so in some way this is an application of derivatives as well.

Given a function, $$f\left( x \right)$$, we can find its tangent at $$x = a$$. The equation of the tangent line, which we’ll call $$L\left( x \right)$$ for this discussion, is,

$L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)$

Take a look at the following graph of a function and its tangent line.

From this graph we can see that near $$x = a$$ the tangent line and the function have nearly the same graph. On occasion we will use the tangent line, $$L\left( x \right)$$, as an approximation to the function, $$f\left( x \right)$$, near $$x = a$$. In these cases we call the tangent line the linear approximation to the function at $$x = a$$.

So, why would we do this? Let’s take a look at an example.

Example 1 Determine the linear approximation for $$f\left( x \right) = \sqrt[3]{x}$$ at $$x = 8$$. Use the linear approximation to approximate the value of $$\sqrt[3]{{8.05}}$$ and $$\sqrt[3]{{25}}$$.
Show Solution

Since this is just the tangent line there really isn’t a whole lot to finding the linear approximation.

$f'\left( x \right) = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\,\sqrt[3]{{{x^2}}}}}\hspace{0.5in}f\left( 8 \right) = 2\hspace{0.25in}f'\left( 8 \right) = \frac{1}{{12}}$

The linear approximation is then,

$L\left( x \right) = 2 + \frac{1}{{12}}\left( {x - 8} \right) = \frac{1}{{12}}x + \frac{4}{3}$

Now, the approximations are nothing more than plugging the given values of $$x$$ into the linear approximation. For comparison purposes we’ll also compute the exact values.

\begin{align*}L\left( {8.05} \right) & = 2.00416667 & \hspace{0.75in} \sqrt[3]{{8.05}} & = 2.00415802\\ L\left( {25} \right) & = 3.41666667 & \hspace{0.75in} \sqrt[3]{{25}} & = 2.92401774\end{align*}

So, at $$x = 8.05$$ this linear approximation does a very good job of approximating the actual value. However, at $$x = 25$$ it doesn’t do such a good job.

This shouldn’t be too surprising if you think about it. Near $$x = 8$$ both the function and the linear approximation have nearly the same slope and since they both pass through the point $$\left( {8,2} \right)$$ they should have nearly the same value as long as we stay close to $$x = 8$$. However, as we move away from $$x = 8$$ the linear approximation is a line and so will always have the same slope while the function’s slope will change as $$x$$ changes and so the function will, in all likelihood, move away from the linear approximation.

Here’s a quick sketch of the function and its linear approximation at $$x = 8$$.

As noted above, the farther from $$x = 8$$ we get the more distance separates the function itself and its linear approximation.

Linear approximations do a very good job of approximating values of $$f\left( x \right)$$ as long as we stay “near” $$x = a$$. However, the farther away from $$x = a$$ we get the worse the approximation is liable to be. The main problem here is that how near we need to stay to $$x = a$$ in order to get a good approximation will depend upon both the function we’re using and the value of $$x = a$$ that we’re using. Also, there will often be no easy way of predicting how far away from $$x = a$$ we can get and still have a “good” approximation.

Let’s take a look at another example that is actually used fairly heavily in some places.

Example 2 Determine the linear approximation for $$\sin \theta$$ at $$\theta = 0$$.
Show Solution

Again, there really isn’t a whole lot to this example. All that we need to do is compute the tangent line to $$\sin \theta$$ at $$\theta = 0$$.

\begin{align*}f\left( \theta \right) & = \sin \theta & \hspace{0.75in} f'\left( \theta \right) & = \cos \theta \\ f\left( 0 \right) & = 0 & \hspace{0.75in}f'\left( 0 \right) & = 1\end{align*}

The linear approximation is,

\begin{align*}L\left( \theta \right) & = f\left( 0 \right) + f'\left( 0 \right)\left( {\theta - a} \right)\\ & = 0 + \left( 1 \right)\left( {\theta - 0} \right)\\ & = \theta \end{align*}

So, as long as $$\theta$$ stays small we can say that $$\sin \theta \approx \theta$$.

This is actually a somewhat important linear approximation. In optics this linear approximation is often used to simplify formulas. This linear approximation is also used to help describe the motion of a pendulum and vibrations in a string.