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 line tangent to the graph of a function at a point is very close to the graph of the function near that point.

This tangent line is the graph of a linear function, called the linear approximation.

Let $f$ be a function that is differentiable on some interval I that contains the point $a$ . The graph of a function $f$ and the line tangent to the curve

$y=f(x)$ at the point where $x=a$ are given in the figure below. Find the equation of the tangent line.

First, find the expression for $m$ , the slope of the tangent line to the curve $y=f(x)$ at the point $(a,f(a))$ . Select the correct choice. Since we know that the point $(a,f(a))$ lies on the tangent line, we can write an equation of the tangent line.

$y= f(a)+f'(a)(x-a).$ Now, we define a function, $L$ , by $L(x)= f(a)+f'(a)(x-a)$ . This function is linear and its graph is the line tangent to the curve $y=f(x)$ at the point where $x=a$ . This function deserves a special name.

If $f$ is a function differentiable at $x=a$ , then a linear approximation to the function $f$ at $x=a$ is given by $L(x) =f(a)+ f'(a)(x-a).$

Note that the graph of $L$ is just the tangent line to the graph of $f$ 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 the graph of $f$ sufficiently, then the graphs of $f$ and $L$ are nearly indistinguishable.

As a first example, we will see how linear approximations allow us to approximate “difficult” computations.

Let $f$ be a function defined by $f(x) =\sqrt [3]{x}.$ Approximate $\sqrt [3]{50}$ , using $L$ , a linear approximation to the function $f$ at $a=64$ .

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

Now we evaluate $L(50)= \frac {50}{48} +\frac {8}{3}$ . This was easy to compute, since $L$ is a linear function! Using a calculator we can see that $L(50)= \frac {50}{48} +\frac {8}{3}\approx 3.71$ and that $\sqrt [3]{50}\approx 3.68$ . Therefore, $L(50)$ is not only easy to evaluate, but also gives us a close approximation of $\sqrt [3]{50}$ .

What would happen if we chose $a=27$ instead?

Then we would use $L_{27}$ , the linear approximation to the function $f$ at $a=27$ . In that case, $L_{27}=f(27)+f'(27)(x-27)=3+\frac {1}{27}(x-27)$ . The graph of $L_{27}$ , together with the graphs of $f$ and $L=L_{64}$ is given in the figure below.

From the picture we can see that

$L_{27}(50)>L_{64}(50)>f(50)$ .

So, our choice, $a=64$ , was better!

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.

Use a linear approximation of $f(x) =\sin (x)$ at $a=0$ to approximate $\sin (0.3)$ .

To start, write $\ddx f(x) = \answer [given]{\cos (x)},$ so our linear approximation is $\begin{align*} L(x) &= 0+\answer[given]{\cos(0)}\cdot(x-0)\\ &= x. \end{align*}$

Hence, a linear approximation for $\sin (x)$ at $a=0$ is $L(x) = x$ , and so $L(0.3) = 0.3$ . Comparing this to $\sin (.3) \approx 0.295$ , we see that the approximation is quite good. For this reason, it is common to approximate $\sin (x)$ with its linear approximation $L(x) = x$ when $x$ is near zero.

Differentials

The graph of a function $f$ and the graph of $L$ , the linear approximation of $f$ at $a$ , are shown in the figure below. Also, two quantities, $\d x$ and $\d f$ , and a point $P$ are marked in the figure. Look carefully at the figure when answering the questions below.

Select all the correct expressions for the quantity $\d x$ .

You can see that $x=a+\d x$ .

Select all the correct expressions for the quantity $\d f$ .

You can see that $L(x)=f(a)+\d f$ .

Recall: $L(a)=f(a)$ .

Based on the figure and the expression for $L(x)$ , select all the correct expressions for $\d f$ .

Recall: $\d f=L(x)-f(a)=f(a)+f'(a)(x-a)-f(a)=f'(a)(x-a)=f'(a)\d x$ .

So, we can write $\d f=f'(a)\d x$ and call it a differential of $f$ at $a$ . Notice that we can define a differential at any point $x$ of the domain of $f$ , provided that $f'(x)$ exists. We will do that in our next definition.

Let $f$ be a differentiable function, let $x$ be a point in the domain of $f$ , and let $\d x$ be some quantity, called a differential of $x$ . We define $\d f$ , a differential of $f$ , at a point $x$ by $\d f=f'(x)\cdot \d x$

Geometrically, differentials can be interpreted via the diagram below.

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

We should not be surprised, since the slope of the tangent line in the figure is $f'(x)$ , and this slope is also given by $\frac {\d f }{\d x}$ .

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, $f'(x)h \approx f(x+h)-f(x).$ We can replace a quantity $h$ with a quantity $\d x$ 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 $f(x) + \d f\approx f(x+\d x)$ or, equivalently $f(x+\d x)\approx f(x) + \d f .$ There are contexts where the language of differentials is common. Here is the basic strategy:

We will repeat our previous examples 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 $\d x=50-64=-14$ , 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 $f(50) \approx f(64) + \frac {-7}{24} \approx 3.71$ .

Use differentials to approximate $\sin (0.3)$ .

Set $y = \sin (x)$ . We want to know $\sin (0.3)$ . Since $\sin (0) = 0$ , we will set $x = 0$ and $\d x=0.3$ . Write with me $\begin{align*} \sin(0.3) = \sin(x + \d x) &\approx \sin(x) + \d y\\ &\approx 0 + \d y. \end{align*}$ Here we see a plot of $y=\sin (x)$ with the differentials above marked:

Now we must compute $\d y$ : $\begin{align*} \d y &= \left(\ddx \sin(x)\right) \cdot \d x\\ &=\answer[given]{\cos(0)} \cdot \d x\\ &= 1 \cdot (\answer[given]{0.3})\\ &= 0.3 \end{align*}$ Hence $\sin (0.3) \approx \sin (0) + 0.3 \approx 0.3$ .

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.

A glass containing water is shown in the figure below.

Let $y$ denote the height of water (measured in centimeters). The volume of the water in the glass (in milliliters), as a function of $y$ , is given by $V(y) = \frac {2\pi y^{3/2}}{\sqrt {3}}.$

There is a mark indicating when the glass is filled to $250$ ml at $16.8$ cm from the base of the glass. If the glass is filled within $\pm 2$ millimeters of the mark, what are the bounds on the volume?

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 roughly $\pm \answer [given,tolerance=.01]{4.46062}$ 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.

$\Delta y=f(x+\Delta x)-f(x)$ ; it is the change in $y=f(x)$ associated to $\Delta x$ .
$\d y=L(x+\d x)-L(x)$ , it is the change in $y=L(x)$ associated to $\Delta x=\d x$ . $\d y =f'(x)\d x$ Note: $L(x+\d x)= f(x)+f'(x)\d x$ .
So, the change
$\begin{align*} \d y &= L(x+\d x)-L(x)\\ &= f(x)+f'(x)\d x-L(x)\\ &= f(x)+f'(x)\d x-f(x)\\ &=f'(x)\d x \end{align*}$

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 $\d y$ ?

$\Delta y=f(1+\Delta x)-f(1)=f(1.1)-f(1)$

$\d y=f'(1)\cdot \d x=f'(1)\cdot 0.1$

$\Delta y = \answer {0.21}\qquad \d y = \answer {0.2}$

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

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Differentials

Error approximation

New and old friends

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