4.2 Linear approximations and differentials (Page 3/12)

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Computing differentials

For each of the following functions, find dy and evaluate when $x = 3$ and $d x = 0.1 .$

$y = x^{2} + 2 x$
$y = cos x$

The key step is calculating the derivative. When we have that, we can obtain dy directly.

Since $f (x) = x^{2} + 2 x,$ we know $f' (x) = 2 x + 2,$ and therefore

$d y = (2 x + 2) d x .$

When $x = 3$ and $d x = 0.1,$

$d y = (2 \cdot 3 + 2) (0.1) = 0.8 .$
Since $f (x) = cos x,$ $f' (x) = − sin (x) .$ This gives us

$d y = − sin x d x .$

When $x = 3$ and $d x = 0.1,$

$d y = − sin (3) (0.1) = −0.1 sin (3) .$

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For $y = e^{x^{2}},$ find $d y .$

$d y = 2 x e^{x^{2}} d x$

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We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function $f$ that is differentiable at point $a .$ Suppose the input $x$ changes by a small amount. We are interested in how much the output $y$ changes. If $x$ changes from $a$ to $a + d x,$ then the change in $x$ is $d x$ (also denoted $Δ x),$ and the change in $y$ is given by

Δ y = f (a + d x) - f (a) .

Instead of calculating the exact change in $y,$ however, it is often easier to approximate the change in $y$ by using a linear approximation. For $x$ near $a,$ $f (x)$ can be approximated by the linear approximation

L (x) = f (a) + f' (a) (x - a) .

Therefore, if $d x$ is small,

f (a + d x) \approx L (a + d x) = f (a) + f' (a) (a + d x - a) .

That is,

f (a + d x) - f (a) \approx L (a + d x) - f (a) = f' (a) d x .

In other words, the actual change in the function $f$ if $x$ increases from $a$ to $a + d x$ is approximately the difference between $L (a + d x)$ and $f (a),$ where $L (x)$ is the linear approximation of $f$ at $a .$ By definition of $L (x),$ this difference is equal to $f' (a) d x .$ In summary,

Δ y = f (a + d x) - f (a) \approx L (a + d x) - f (a) = f' (a) d x = d y .

Therefore, we can use the differential $d y = f' (a) d x$ to approximate the change in $y$ if $x$ increases from $x = a$ to $x = a + d x .$ We can see this in the following graph.

A function y = f(x) is shown along with its tangent line at (a, f(a)). The tangent line is denoted L(x). The x axis is marked with a and a + dx, with a dashed line showing the distance between a and a + dx as dx. The points (a + dx, f(a + dx)) and (a + dx, L(a + dx)) are marked on the curves for y = f(x) and y = L(x), respectively. The distance between f(a) and L(a + dx) is marked as dy = f’(a) dx, and the distance between f(a) and f(a + dx) is marked as Δy = f(a + dx) – f(a). — The differential $d y = f' (a) d x$ is used to approximate the actual change in $y$ if $x$ increases from $a$ to $a + d x .$

We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.

Approximating change with differentials

Let $y = x^{2} + 2 x .$ Compute $Δ y$ and dy at $x = 3$ if $d x = 0.1 .$

The actual change in $y$ if $x$ changes from $x = 3$ to $x = 3.1$ is given by

Δ y = f (3.1) - f (3) = [{(3.1)}^{2} + 2 (3.1)] - [3^{2} + 2 (3)] = 0.81 .

The approximate change in $y$ is given by $d y = f' (3) d x .$ Since $f' (x) = 2 x + 2,$ we have

d y = f' (3) d x = (2 (3) + 2) (0.1) = 0.8 .

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For $y = x^{2} + 2 x,$ find $Δ y$ and $d y$ at $x = 3$ if $d x = 0.2 .$

$d y = 1.6,$ $Δ y = 1.64$

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Calculating the amount of error

Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error.

Consider a function $f$ with an input that is a measured quantity. Suppose the exact value of the measured quantity is $a,$ but the measured value is $a + d x .$ We say the measurement error is dx (or $Δ x) .$ As a result, an error occurs in the calculated quantity $f (x) .$ This type of error is known as a propagated error and is given by

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Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

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