Linear approximation

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We use a method called ‘‘linear approximation’’ to estimate the value of a (complicated) function at a given point.

Given a function, a linear approximation is a fancy phrase for something you already know:

The derivative is the slope of the tangent line.

Except in this section, the emphasis is on the line.

If $f$ is a differentiable function at $x=a$ , then a linear approximation for $f$ at $x=a$ is given by $\l (x) = f'(a)(x-a) +f(a).$

Note that $\l (x)$ is just the tangent line to $f(x)$ at $x=a$ .

A linear approximation of $f$ is a ‘‘good’’ approximation as long as $x$ is ‘‘not too far’’ from $a$ . If one ‘‘zooms in’’ on $f$ sufficiently, then $f$ and the linear approximation are nearly indistinguishable. As a first example, we will see how linear approximations allow us to make approximate ‘‘difficult’’ computations.

Use a linear approximation of $f(x) =\sqrt [3]{x}$ at $x=64$ to approximate $\sqrt [3]{50}$ .

To start, write $\ddx f(x) = \ddx x^{1/3} = \frac {1}{\answer [given]{3x^{2/3}}}.$ So our linear approximation is $\begin{align*} \l(x) &= \frac{1}{3\cdot 64^{2/3}} (x-64) + \answer[given]{4} \\ &= \frac{1}{\answer[given]{48}} (x-64) + 4\\ &= \frac{x}{48} +\frac{8}{3}. \end{align*}$

Now we evaluate $\l (50) \approx 3.71$ and compare it to $\sqrt [3]{50}\approx 3.68$ . From this we see that the linear approximation, while perhaps inexact, is computationally easier than computing the cube root.

With modern calculators and computing software it may not appear necessary to use linear approximations. In fact they are quite useful. In cases requiring an explicit numerical approximation, they allow us to get a quick rough estimate which can be used as a ‘‘reality check’’ on a more complex calculation. In some complex calculations involving functions, the linear approximation makes an otherwise intractable calculation possible, without serious loss of accuracy.

Differentials

The notion of a differential goes back to the origins of calculus, though our modern conceptualization of a differential is somewhat different than how they were initially understood.

Let $f$ be a differentiable function. We define a new independent variable $\d x$ , and a new dependent variable $\d f=f'(x)\cdot \d x.$ The variables $\d x$ and $\d f$ are called differentials. Geometrically, differentials can be interpreted via the diagram below

Note, it is now the case (by definition!) that $\frac {\d f}{\d x} = f'(x).$

The differential $\d x$ is:

$d$ times $x$ . A single variable.

The differential $\d f$ is:

$d$ times $f$ . A single variable that is dependent on $\d x$ .

Essentially, differentials allow us to solve the problems presented in the previous examples from a slightly different point of view. Recall, when $h$ is near but not equal zero, $f'(x) \approx \frac {f(x+h)-f(x)}{h}$ hence, $f'(x)h \approx f(x+h)-f(x).$ Since $h$ is simply a variable, and $\d x$ is simply a variable, we can replace $h$ with $\d x$ to write

$\begin{align*} f'(x)\cdot \d x &\approx f(x+\d x)-f(x)\\ \d f &\approx f(x+\d x)-f(x). \end{align*}$

Adding $f(x)$ to both sides we see $f(x+\d x)\approx f(x) + \d f.$ While this is something of a ‘‘sleight of hand’’ with variables, there are contexts where the language of differentials is common. Here is the basic strategy:

We will repeat our previous example using differentials.

Use differentials to approximate $\sqrt [3]{50}$ .

Set $f(x) = \sqrt [3]{x}$ . We want to know $\sqrt [3]{50}$ . Since $4^3 = 64$ , we set $x=64$ . Setting $\d x=-14$ , we have $\begin{align*} \sqrt[3]{50} = f(x + \d x) &\approx f(x) + \d f\\ &\approx \sqrt[3]{64} + \d f. \end{align*}$

Here we see a plot of $y=\sqrt [3]{x}$ with the differentials above marked:

Now we must compute $\d f$ : $\begin{align*} \d f &= f'(x) \cdot \d x\\ &= \answer[given]{\frac{1}{3x^{2/3}}} \cdot \d x\\ &= \frac{1}{3\cdot64^{2/3}} \cdot(\answer[given]{-14})\\ &= \frac{1}{3\cdot64^{2/3}} \cdot(-14)\\ &= \frac{-7}{24} \end{align*}$

Hence $f(50) \approx f(64) + \frac {-7}{24} \approx 3.71$ .

The upshot is that linear approximations and differentials are simply two slightly different ways of doing the exact same thing.

Error approximation

Differentials also help us estimate error in real life settings.

The cross-section of a $250$ ml glass can be modeled by the function $r(x) = \frac {x^4}{3}$ :

At $16.8$ cm from the base of the glass, there is a mark indicating when the glass is filled to $250$ ml. If the glass is filled within $\pm 2$ millimeters of the mark, what are the bounds on the volume? As a gesture of friendship, we will tell you that the volume in milliliters, as a function of the height of water in centimeters, $y$ , is given by $V(y) = \frac {2\pi y^{3/2}}{\sqrt {3}}.$ Note: If you persist in your quest to learn calculus, you will be able to derive the formula above like it’s no-big-deal.

We want to know what a small change in the height, $y$ does to the volume $V$ . These small changes can be modeled by the differentials $\d V$ and $\d y$ . Since $\d V = V'(y) \d y$ and $V'(y) = \answer [given]{\pi \sqrt {3 y}}$ we use the fact that $\d y = \pm 0.2$ with $y=\answer [given]{16.8}$ to see $\d V = \answer [given]{\pi \sqrt {3 \cdot 16.8}} \cdot 0.2.$ Hence the volume will vary by at most $\pm \answer [given,tolerance=.01]{4.46062}$ milliliters.

New and old friends

You might be wondering, given a plot $y=f(x)$ ,

What’s the difference between $\Delta x$ and $\d x$ ? What about $\Delta y$ and $\d y$ ?

Regardless, it is now a pressing question. Here’s the deal: $\frac {\Delta y}{\Delta x}$ is the average rate of change of $y=f(x)$ with respect to $x$ . On the other hand: $\frac {\d y}{\d x}$ is the instantaneous rate of change of $y=f(x)$ with respect to $x$ . Essentially, $\Delta x$ and $\d x$ are the same type of thing, they are (usually small) changes in $x$ . However, $\Delta y$ and $\d y$ are very different things.

$\Delta y$ is the change of $y$ associated to $\Delta x$ .
$\d y$ is the change in $y$ needed to make the following relation true: $\d y = f'(x)\d x$

Suppose $f(x) = x^2$ . If we are at the point $x=1$ and $\Delta x =\d x = 0.1$ , what is $\Delta y$ ? What is $\d y$ ? $\Delta y = \answer {0.21}\qquad \d y = \answer {0.2}$

Differentials can be confusing at first. However, when you master them, you will have a powerful tool at your disposal.