Linear Approximation

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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 online 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:
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

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Example 1 Linear Approximation of the Square Root
Solution
SinceWe can use $L(x)$ to approximate the square root of any number close to $4$ very easily without using a calculator. For example,
_{} 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:
Substituting and simplifying gives (numerical answers should be accurate to 4 decimal places):
Before we go on... You can use $L(x) = x1$ to find approximations to the natural logarithm of any number close to 1: for instance,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 $x = a,$Using the delta notation, this becomes
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
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

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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
Thus,
_{} Example 4 Little Cones
The Little Cones Operad Co. manfactures coneshaped ornaments of various colors. All the ornaments have height $10mm$ and radius of base $2mm.$
You can now now go on and try the exercise set for this topic.
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