### Linear Approximations and Differentials

We have seen that a curve lies very close to its tangent line near the point of tangency. In fact, by zooming in toward a point on the graph of a differentiable function, we noticed that the graph looks more and more like its tangent line.

This observation is the basis for a method of finding approximate values of functions.

*The idea is that it might be easy to calculate a value f(a) of *
a function, but difficult (or even impossible) to compute

*nearby values of f.*

### Linear Approximations and Differentials

So we settle for the easily computed values of the linear
*function L whose graph is the tangent line of f at (a, f(a)). *

(See Figure 1.)

### Linear Approximations and Differentials

*In other words, we use the tangent line at (a, f(a)) as an *
*approximation to the curve y = f(x) when x is near a. An *
equation of this tangent line is

*y = f(a) + f′(a)(x – a)*
and the approximation

**is called the linear approximation or tangent line **
**approximation of f at a.**

### Linear Approximations and Differentials

The linear function whose graph is this tangent line, that is,

**is called the linearization of f at a.**

## Example 1

*Find the linearization of the function f(x) = at a = 1 *
and use it to approximate the numbers and . Are
these approximations overestimates or underestimates?

Solution:

*The derivative of f(x) = (x + 3)*^{1/2} is
*f′(x) = (x + 3)*^{–1/2}

*and so we have f(1) = 2 and f*′(1) = .

*Example 1 – Solution*

Putting these values into Equation 2, we see that the linearization is

*L(x) = f(1) + f′(1)(x – 1)*

*= 2 + (x – 1)*

The corresponding linear approximation (1) is

cont’d

*Example 1 – Solution*

In particular, we have

and

cont’d

*Example 1 – Solution*

The linear approximation is illustrated in Figure 2.

**Figure 2**

cont’d

*Example 1 – Solution*

We see that, indeed, the tangent line approximation is a
*good approximation to the given function when x is near 1.*

We also see that our approximations are overestimates because the tangent line lies above the curve.

Of course, a calculator could give us approximations for
and , but the linear approximation gives an
*approximation over an entire interval.*

cont’d

### Linear Approximations and Differentials

In the following table we compare the estimates from the linear approximation in Example 1 with the true values.

### Linear Approximations and Differentials

Notice from this table, and also from Figure 2, that the

*tangent line approximation gives good estimates when x is *
close to 1 but the accuracy of the approximation

*deteriorates when x is farther away from 1.*

**Figure 2**

### Linear Approximations and Differentials

The next example shows that by using a graphing calculator or computer we can determine an interval throughout which a linear approximation provides a specified accuracy.

## Example 2

*For what values of x is the linear approximation*

accurate to within 0.5? What about accuracy to within 0.1?

Solution:

Accuracy to within 0.5 means that the functions should differ by less than 0.5:

*Example 2 – Solution*

Equivalently, we could write

This says that the linear approximation should lie between the curves obtained by shifting the curve

upward and downward by an amount 0.5.

cont’d

*Example 2 – Solution*

*Figure 3 shows the tangent line y = (7 + x)/4 intersecting *
*the upper curve y = + 0.5 at P and Q.*

**Figure 3**

cont’d

*Example 2 – Solution*

Zooming in and using the cursor, we estimate that the

*x-coordinate of P is about –2.66 and the x-coordinate of Q*
is about 8.66.

Thus we see from the graph that the approximation

*is accurate to within 0.5 when –2.6 < x < 8.6. (We have *

cont’d

*Example 2 – Solution*

Similarly, from Figure 4 we see that the approximation is
*accurate to within 0.1 when –1.1 < x < 3.9.*

**Figure 4**

cont’d

## Applications to Physics

## Applications to Physics

Linear approximations are often used in physics. In

analyzing the consequences of an equation, a physicist sometimes needs to simplify a function by replacing it with its linear approximation.

For instance, in deriving a formula for the period of a pendulum, physics textbooks obtain the expression

*a*_{T}*= –g sin *θ for tangential acceleration and then replace
sin θ by θ with the remark that sin θ is very close to θ if θ is
not too large.

## Applications to Physics

You can verify that the linearization of the function
*f(x) = sin x at a = 0 is L(x) = x and so the linear *

approximation at 0 is

*sin x ≈ x*

So, in effect, the derivation of the formula for the period of a pendulum uses the tangent line approximation for the sine function.

## Applications to Physics

Another example occurs in the theory of optics, where light
rays that arrive at shallow angles relative to the optical axis
*are called paraxial rays.*

In paraxial (or Gaussian) optics, both sin θ and cos θ are replaced by their linearizations. In other words, the linear approximations

sin θ ≈ θ and cos θ ≈ 1 are used because θ is close to 0.

## Differentials

## Differentials

The ideas behind linear approximations are sometimes
*formulated in the terminology and notation of differentials.*

*If y = f(x), where f is a differentiable function, then the *

* differential dx is an independent variable; that is, dx can *
be given the value of any real number.

* The differential dy is then defined in terms of dx by the *
equation

*dy = f′(x) dx*

## Differentials

*So dy is a dependent variable; it depends on the values of *
*x and dx.*

*If dx is given a specific value and x is taken to be some *
*specific number in the domain of f, then the numerical *
*value of dy is determined.*

## Differentials

The geometric meaning of differentials is shown in Figure 5.

**Figure 5**

## Differentials

*Let P(x, f(x)) and Q(x + ∆x, f(x + ∆x)) be points on the *

*graph of f and let dx = ∆x. The corresponding change in y is*

*∆y = f(x + ∆x) – f(x)*

*The slope of the tangent line PR is the derivative f′(x). Thus *
*the directed distance from S to R is f′(x) dx = dy.*

*Therefore dy represents the amount that the tangent line *
*rises or falls (the change in the linearization) when x *

## Example 3

Compare the values of ∆y and dy if

*y = f(x) = x*^{3} *+ x*^{2} *– 2x + 1 and x changes (a) from 2 to 2.05 *
and (b) from 2 to 2.01.

Solution:

(a) We have

*f(2) = 2*^{3} + 2^{2} – 2(2) + 1

= 9

*f(2.05) = (2.05)*^{3} + (2.05)^{2} – 2(2.05) + 1

= 9.717625

*Example 3 – Solution*

*∆y = f(2.05) – f(2)*

= 0.717625 In general,

*dy = f′(x) dx*

*= (3x*^{2} *+ 2x – 2) dx*

*When x = 2 and dx = ∆x = 0.05, this becomes*

cont’d

*Example 3 – Solution*

*(b) f(2.01) = (2.01)*^{3} + (2.01)^{2} – 2(2.01) + 1

= 9.140701

*∆y = f(2.01) – f(2)*

= 0.140701
*When dx = ∆x = 0.01,*

*dy = [3(2)*^{2} + 2(2) – 2]0.01

= 0.14

cont’d

## Differentials

Our final example illustrates the use of differentials in estimating the errors that occur because of approximate measurements.

## Example 4

The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?

Solution:

*If the radius of the sphere is r, then its volume is V = *π*r*^{3}. If
*the error in the measured value of r is denoted by dr = ∆r, *
*then the corresponding error in the calculated value of V is *

*∆V, which can be approximated by the differential*
*dV = 4*π*r*^{2} *dr*

*Example 4 – Solution*

*When r = 21 and dr = 0.05, this becomes*
*dV = 4*π(21)^{2}0.05

≈ 277

The maximum error in the calculated volume is about
277 cm^{3}.

cont’d

## Differentials

**Note:**

Although the possible error in Example 4 may appear to be rather large, a better picture of the error is given by the

**relative error, which is computed by dividing the error by **
the total volume:

## Differentials

Thus the relative error in the volume is about three times the relative error in the radius.

In Example 4 the relative error in the radius is

*approximately dr/r = 0.05/21 ≈ 0.0024 and it produces a *
relative error of about 0.007 in the volume.

**The errors could also be expressed as percentage errors **
of 0.24% in the radius and 0.7% in the volume.