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These notes probably belong towards the end of chapter 6 rather than at the start of chapter 7. Unfortunately, I was unable to write them in time for that. In this discussion, we will concern ourselves with linear approximations. Some of this we have already discussed in class.
Definition 1. Let f : I → R be a function on the open interval I. Let a ∈ I. If f is differentiable at a then the linear approximation of f at a is given by
(1) f (x) ≈ L(x) = f ′(a)(x − a) + f (a).
Notice that L is simply the equation of the tangent line to f at x = a. The point is that if f is differentiable at a then L is a good approximation of f for x near a. Put another way, if f is differentiable at a then under a microscope f will look very much like a straight line.
Exercise 2. Sketch a graph of a non-linear function f that illustrates this point.
Example 3. Let f (x) = x^3. At x = 2 the linear approximation of f is given by
(2) f (x) ≈ L(x) = (3 · 22 )(x − 2) + 2^3 = 12(x − 2) + 8
since f ′(x) = 3x^2.
Example 4. Let f (x) =
x + 4. Then f ′(x) = 2 √^1 x+4. The linear approximation
to f at x = 5 is
(3) f (x) ≈ L(x) =
x − 5 2
x − 5 6
As an immediate application we can approximate square roots of numbers near 10 by hand. Computing
10 is difficult to compute directly since
10 is irrational. The linear approximation gives us a simple way of estimating
Notice that
(4)
6 + 4 = f (6) ≈
It turns out that
16 ≈ 3 .16227766 (correct to 8 decimal places) so our linear approximation is only accurate to one decimal place.
Exercise 5. In the preceding example our approximation of f (6) was an overes- timate. Show that we always get an overestimate: that is, show that L(x) ≥ f (x) for x > −4. (Hint: What does the second derivative tell you about f ?)
With advanced calculators and computing software it may not appear necessary to use linear approximations. Keep in mind, however, that when computing with irrational numbers both calculators and software use approximations. As such, it can be helpful to have some idea of how the approximations happen. Although these approximations may not be linear, an advantage of studying linear approximations initially is that they are relatively simple. 1
It should also be pointed out that one can estimate how good our linear approxi- mation is without actually knowing the precise value of what we are estimating. We discuss this more when we consider approximate integration methods like Simpson’s rule. As a practical example, consider the trigonometric functions sin x and cos x. These functions come up frequently in any science where angles play a rˆole^1.
Exercise 6. Show that the linear approximation of sin x at x = 0 is x and the linear approximation of cos x at x = 0 is 1. Do not use the subsequent discussion.
We will see in math 1220 that
(5) sin x =
∑^ ∞
n=
(−1)nx^2 n+ (2n + 1)!
= x −
x^3 6
x^5 120
and
(6) cos x =
∑^ ∞
n=
(−1)nx^2 n (2n)!
x^2 2
x^4 24
Notice that if x is close to zero then sin x ≈ x and cos x ≈ 1 because the higher order terms are much smaller than x. This linearization simplifies many calculations without serious loss of accuracy and indeed makes otherwise intractable calculations feasible.
Definition 7. Let y = f (x) be a differentiable function. We define a new indepen- dent variable dx. The domain of dx is any real number. We define dy = f ′(x)dx and note that dy is a dependent variable. We say that dx and dy are differentials.
Notice that dy is a function both of x (since f ′(x) is a function of x) and of dx. Here is the reason for introducing differentials: Fix a point P (a, f (a)). Let
(7) ∆x = x − a.
Note that if x is near a then ∆x is small. Set dx = ∆x and let ∆y = f (x) − f (a). If ∆x is small then dy ≈ ∆y and dy is the linear approximation of f (x) for x near a. We illustrate the point in figure 1.
Exercise 8. Let f (x) = x^4. If a = 1 and dx = ∆x = 12 what are ∆y and dy?
Exercise 9. Let f (x) =
x. If a = 1 and dx = ∆x = 101 what are ∆y and dy?
Exercise 10. Let f (x) = sin(2x). If a = π and dx = ∆x = 100 π what are ∆y and dy?
Exercise 11. Using differentials estimate the amount of paint needed to apply a coat of paint 0.02 cm think to a sphere with diameter 40 meters. (Recall that the volume of a sphere of radius r is given by the formula V = 4 πr
3
- Notice that you are given that dr = 0.02.)
(^1) For example, optics and mechanics.
