 ## Linear Approximation & Error Estimation Miscellaneous on-line topics for Calculus Applied to the Real World

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Español Note To understand this topic, you will need to be familiar with derivatives, as discussed in Chapter 3 of Calculus Applied to the Real World. If you like, you can review the topic summary material on techniques of differentiation or, for a more detailed study, the on-line tutorials on derivatives of powers, sums, and constant multipes.

We start with the observation that if you zoom in to a portion of a smooth curve near a specified point, it becomes indistinguishable from the tangent line at that point. In other words:

The values of the function are close to the values of the linear function whose graph is the tangent line.

For this reason, the linear function whose graph is the tangent line to $y = f(x)$ at a specified point $(a, f(a))$ is called the linear approximation of $f(x)$ near $x = a.$

Q What is the formula for the linear approximation?
A All we need is the equation of the tangent line at a specified point $(a, f(a)).$ Since the tangent line at $(a, f(a))$ has slope $f'(a),$ we can write down its equation using the point-slope formula:
$y= y_0 + m(x - x_0)$
$= f(a) + f'(a)(x - a)$
Thus, the the linear approximation to $f(x)$ near $x = a$ is given by

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

Q The above argument is based on geometry: the fact that the tangent line is close to the original graph near the point of tangency. Is there an algebriac way of seeing why this is true?
A Yes. This links to an algebraic derivation of the linear approximation.

 Linear Approximation of $f(x)$ Near $x = a$ If $x$ is close to a, then $f(x) \approx f(a) + (x-a)f'(a).$ The right-hand side, $L(x) = f(a) + (x-a)f'(a),$ which is a linear function of $x,$ is called the linear approximation of $f(x)$ near $x = a.$  Example 1 Linear Approximation of the Square Root

Let $f(x) = x^{1/2}.$ Find the linear approximation of $f$ near $x = 4$ (at the point $(4, f(4)) = (4, 2)$ on the graph), and use it to approximate $\sqrt{4.1.}$

Solution

Since

$f'(x) = 1/(2x^{1/2}),$
$f'(4) = 1/(2 \cdot 4^{1/2}) = 1/4.$
so the linear approximation is
$L(x) = f(4) + (x-4)f'(4)$
$= 2 + (x-4)/4$
$= 0.25x + 1.$ We can use $L(x)$ to approximate the square root of any number close to $4$ very easily without using a calculator. For example,

 $\sqrt{4.1}$ $\approx$ $L(4.1) = 0.25(4.1) + 1 = 2.025$ Q $\sqrt{3.82}$ $\approx$ Q The Linear approximation of the same function, $f(x) = x^{1/2},$ near $x = 9$ is given by  $L(x) =$  Example 2 Linear Approximation of the Logarithm

Use linear approximation to approximate $\ln(1.134).$

Solution

Here, we are not given a value for $a.$ The key is to use a value close to $1.134$ whose natural logarithm we know. Since we know that $\ln(1) = 0,$ we take a to be $1.$

Now use the formula for linear approximation:

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

Substituting and simplifying gives (numerical answers should be accurate to 4 decimal places):

 Q $L(x)$ $=$ $\ln(1.134)$ $\approx$ The actual value is $\ln(1.134)$ is The error in the approximation is Select one betrween 0 and 0.005 between 0.0051 and 0.01 between 0.011 and 0.5 does not exist undefined The approximation is accurate to Select one 0 decimal places 1 decimal place 2 decimal places 3 decimal places None of the above Before we go on... You can use $L(x) = x-1$ to find approximations to the natural logarithm of any number close to 1: for instance,

$\ln(0.843) \approx 0.843 - 1 = -0.157,$
$\ln(0.999) \approx 0.999 - 1 = -0.001.$

Error Estimation

When a physical measurement is made, there is always some uncertainty about it accuracy. For instance, if you are measuring the radius of a ball bearing, you might measure it repeatedly and obtain slightly differing results. Rather than concluding, say, that the radius of the ball bearing is exactly $1.2mm,$ you may instead conclude that the radius is $1.2mm ± 0.1mm.$ (The actual calculation of the range $± 0.1mm$ is often given by a statistical formula based on the standard deviation of all the separate measurements.)

Once you have an error estimate for the radius, you might wonder how this error might effect the calculation of the volume of the ball bearing. In other words, if the radius is off by $0.1 mm,$ by how much is the volume off? To answer the question, think of the error of the radius as a change, $Δr,$ in $r,$ and then compute the associated change, $ΔV,$ in the volume $V.$ The general question is therefore:

Q If $x$ changes by $Δx,$ and $y$ is a function of $x,$ what is the associated change $Δy$ in $y$?

To answer this question, let us go back to our linear approximation formula: We saw above that, near $x = a,$

$f(x) \approx f(a) + (x-a)f'(a),$     or
$f(x) - f(a) \approx (x-a)f'(a)$

The quantity $f(x) - f(a)$ represents the change in $f$ corresponding to a change in the independent variable from a to $x.$ In other words,

Change in $f \approx$ Change in $x \times f'(a).$

Using the delta notation, this becomes

$Δf \approx Δx f'(x).$

If $y = f(x),$ we can write this formula as

$Δy \approx Δx \frac{dy}{dx}$     Notice how the $dx$ and $Δx$ appear to cancel

Estimating the error of $y = f(x)$

If $x = a,$ with a possible error of $Δx,$ and $y = f(x),$ then $y = f(a),$ with a possible error of $Δy,$ given by

$Δy \approx Δx \left$\frac{dy}{dx} \right$_{x=a}$

Example

Suppose $y = x^2 + 3x,$ and $x = 2,$ accurate to $±0.2,$ then the associated value of $y$ is $2^2 + 3(2) = 10,$ and is accurate to within $±Δy,$ where

 $Δy \approx Δx \left$\frac{dy}{dx} \right$_{x=2}$ $= (0.2)(7)$ Since $dy/dx = 2x+3,$ which is $7$ when $x = 2$ $= 1.4.$

Therefore, even though the error in $x$ is only $±0.2,$ the error in $y$ is much larger; approximately $±1.4.$ Example 3 Measurement Error

Precision Corp. manufactures ball bearings with a radius of 1.2 millimeter, varying by ±0.1 millimeters. What is the volume of the ball bearings, and by how much can it vary?

Solution

The volume of a sphere and its derivative are given by

$V= \frac{4}{3}πr^3.$
$\frac{dV}{dr} = 4πr^3$

Evaluating these quantities at $r = 1.2$ gives

$V= \frac{4}{3}π(1.2)^3 \approx 7.24 mm^3$
$\left$\frac{dV}{dr} \right$_{r=1.2} = 4π(1.2)^3 \approx 18.10$

Thus,

$ΔV \approx Δr \left$\frac{dV}{dr}\right$_{r=1.2} = (0.1)(18.1) = 1.81$

Thus, the volume of the ball bearings is $7.24 ± 1.81mm^3$ Example 4 Little Cones

The Little Cones Operad Co. manfactures cone-shaped ornaments of various colors. All the ornaments have height $10mm$ and radius of base $2mm.$ The radius of the base of the cones is known to be accurate to within $0.15mm.$

(Note: The volume of a cone of height $h$ and radius of base $r$ is
$V = \frac{1}{3} πr^2 h.)$

 Q The volume of the cones is $mm^3$ Q The volume of the cones is accurate to within $mm^3$ You can now now go on and try the exercise set for this topic.  Last Updated:February, 2000