# Linear Approximations and Differentials

## Full text

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### 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.

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### 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.)

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### 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.

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### Linear Approximations and Differentials

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

is called the linearization of f at a.

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## 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) = .

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## 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

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## Example 1 – Solution

In particular, we have

and

cont’d

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## Example 1 – Solution

The linear approximation is illustrated in Figure 2.

Figure 2

cont’d

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## 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

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### Linear Approximations and Differentials

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

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### 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

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### 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.

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## 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:

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## 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

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## 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

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## 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

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## 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

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## 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

aT = –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.

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## 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.

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## 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.

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## 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

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## 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.

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## Differentials

The geometric meaning of differentials is shown in Figure 5.

Figure 5

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## 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

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## Example 3

Compare the values of ∆y and dy if

y = f(x) = x3 + x2 – 2x + 1 and x changes (a) from 2 to 2.05 and (b) from 2 to 2.01.

Solution:

(a) We have

f(2) = 23 + 22 – 2(2) + 1

= 9

f(2.05) = (2.05)3 + (2.05)2 – 2(2.05) + 1

= 9.717625

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## Example 3 – Solution

∆y = f(2.05) – f(2)

= 0.717625 In general,

dy = f′(x) dx

= (3x2 + 2x – 2) dx

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

cont’d

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## 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

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## Differentials

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

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## 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 = πr3. 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 = 4πr2 dr

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## Example 4 – Solution

When r = 21 and dr = 0.05, this becomes dV = 4π(21)20.05

≈ 277

The maximum error in the calculated volume is about 277 cm3.

cont’d

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## 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:

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## 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.

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