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:

$y= y_0 + m(x - x_0)$

  $= f(a) + f'(a)(x - a)$

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

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

Solution

$f'(x) = 1/(2x^{1/2}),$
$f'(4) = 1/(2 \cdot 4^{1/2}) = 1/4.$

$L(x) = f(4) + (x-4)f'(4)$
       $ = 2 + (x-4)/4$
       $ = 0.25x + 1.$

Example 2 Linear Approximation of the Logarithm

Solution

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):

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

Error Estimation

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

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

Using the delta notation, this becomes

$Δf \approx Δx f'(x).$

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

 
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

   $V= \frac{4}{3}πr^3.$

$\frac{dV}{dr} = 4πr^3$

$V= \frac{4}{3}π(1.2)^3 \approx 7.24 mm^3$

$\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$

Example 4 Little Cones

$V = \frac{1}{3}  πr^2 h.)$

You can now now go on and try the exercise set for this topic.