[Fig. 3.1]
Chapter 2 concentrated on computing derivatives. This chapter concentrates on using them. Our computations produced dy/dx for functions built from xn and sin x and cos x. Knowing the slope, and if necessary also the second derivative, we can answer the questions about y = ƒ(x) that this subject was created for:
The information in dy/dx is entirely local. It tells what is happening close to the point and nowhere else. In Chapter 2, Δx and Δy went to zero. Now we want to get them back. The local information explains the larger picture, because Δy is approximately dy/dx times Δx.
The problem is to connect the finite to the infinitesimal—the average slope to the instantaneous slope. Those slopes are close, and occasionally they are equal. Points of equality are assured by the Mean Value Theorem—which is the local-global connection at the center of differential calculus. But we cannot predict where dy/dx equals Δy/Δx. Therefore we now find other ways to recover a function from its derivatives—or to estimate distance from velocity and acceleration.
It may seem surprising that we learn about y from dyldx. All our work has been going the other way! We struggled with y to squeeze out dy/dx. Now we use dy/dx to study y. That's life. Perhaps it really is life, to understand one generation from later generations.
The book started with a straight line ƒ = vt. The distance is linear when the velocity is constant. As soon as v begins to change, ƒ = vt falls apart. Which velocity do we choose, when v(t) is not constant? The solution is to take very short time intervals, in which v is nearly constant:
ƒ = vt is completely false
Δƒ = vΔt is nearly true
dƒ = vdt is exactly true.
For a brief moment the function ƒ(t) is linear—and stays near its tangent line.
In Section 2.3 we found the tangent line to y = ƒ(x). At x = a, the slope of the curve and the slope of the line are ƒ'(a). For points on the line, start at y = ƒ(a). Add the slope times the "increment" x − a:
Y = ƒ(a) + ƒ'(a)(x − a). (1)
We write a capital Y for the line and a small y for the curve. The whole point of tangents is that they are close (provided we don't move too far from a):
y ≈ Y or ƒ(x) ≈ ƒ(a) + ƒ'(a)(x − a). (1)
That is the all-purpose linear approximation. Figure 3.1 shows the square root function y = √x and its tangent line at x = a = 100. At the point y = √100 = 10, the slope is 1/2√x = 1/20. The table beside the figure compares y(x) with Y(x).
[Fig. 3.1]
The accuracy gets worse as x departs from 100. The tangent line leaves the curve. The arrow points to a good approximation at 102, and at 101 it would be even better. In this example Y is larger than y—the straight line is above the curve. The slope of the line stays constant, and the slope of the curve is decreasing. Such a curve will soon be called "concave downward," and its tangent lines are above it.
Look again at x = 102, where the approximation is good. In Chapter 2, when we were approaching dyldx, we started with Δy/Δx:
slope ≈ (√102 − √100) ⁄ (102 − 100). (3)
Now that is turned around! The slope is 1/20. What we don't know is √102:
√102 ≈ √100 + (slope)(102 − 100). (4)
You work with what you have. Earlier we didn't know dy/dx, so we used (3). Now we are experts at dy/dx, and we use (4). After computing y' = 1/20 once and for all, the tangent line stays near √x for every number near 100. When that nearby number is 100 + Δx, notice the error as the approximation is squared:
(√100 + Δx/20)² = 100 + Δx + (Δx)²/400.
The desired answer is 100 + Δx, and we are off by the last term involving (Δx)². The whole point of linear approximation is to ignore every term after Δx.
There is nothing magic about x = 100, except that it has a nice square root. Other points and other functions allow y ≈ Y. I would like to express this same idea in different symbols. Instead of starting from a and going to x, we start from x and go a distance Δx to x + Δx. The letters are different but the mathematics is identical.
3A At any point x, and for any smooth betion y = ƒ(x)
slope at x ≈ ƒ(x+Δx) − ƒ(x) ⁄ Δx. (5)
For the approximation to ƒ(x+Δx), multiply both sides by Δx and add ƒ(x):
ƒ(x+Δx) ≈ ƒ(x) + (slope at x)(Δx). (6)
EXAMPLE 1 An important linear approximation: (1 + x)n ≈ 1 + nx for x near zero.
EXAMPLE 2 A second important approximation: 1/(1 + x)n ≈ 1 − nx for x near zero.
Discussion Those are really the same. By changing n to -n in Example 1, it becomes Example 2. These are linear approximations using the slopes n and -n at x = 0:
(1 + x)n ≈ 1 + (slope at zero) times (x − 0) = 1 + nx.
Here is the same thing with ƒ(x) = xn. The basepoint in equation (6) is now 1 or x:
(1 + x)n ≈ 1 + nΔx (x + Δx)n ≈ xn + nxn-1Δx.
Better than that, here are numbers. For n = 3 and -1 and 100, take Δx = .01:
(1.01)³ ≈ 1/1.01 ≈ .99 (1 + 1/100)100 ≈ 2
Actually that last number is no good. The 100th power is too much. Linear approximation gives 1 + 100Ax = 2, but a calculator gives (1.01)100 = 2.7…. This is close to e, the all-important number in Chapter 6. The binomial formula shows why the approximation failed:
(1 + Δx)100 ≈ 1 + 100Δx + (100·99) ⁄ (2·1) (Δx)² + ….
Linear approximation forgets the (Δx)² term. For Δx = 1/100 that error is nearly ½. It is too big to overlook. The exact error is ƒ"(c), where the Mean Value Theorem in Section 3.8 places c between x and x + Δx. You already see the point:
y − Y is of order (Δx)². Linear approximation, quadratic error.
There is one more notation for this linear approximation. It has to be presented, because it is often used. The notation is suggestive and confusing at the same time—it keeps the same symbols dx and dy that appear in the derivative. Earlier we took great pains to emphasize that dyldx is not an ordinary fraction.* Until this paragraph, dx and dy have had no independent meaning. Now they become separate variables, like x and y but with their own names. These quantities dx and dy are called differentials.
The symbols dx and dy measure changes along the tangent line. They do for the approximation Y(x) exactly what Δx and Δy did for y(x). Thus dx and Δx both measure distance across.
Figure 3.2 has Δx = dx. But the change in y does not equal the change in Y. One is Δy (exact for the function). The other is dy (exact for the tangent line). The differential dy is equal to ΔY, the change along the tangent line. Where Δy is the true change, dy is its linear approximation (dy/dx)dx.
You often see dy written as ƒ'(x)dx.
[Fig. 3.2]
EXAMPLE 3 y = x² has dy/dx = 2x so dy = 2x dx. The table has basepoint x = 2. The prediction dy differs from the true Ay by exactly Δx)² = .01 and .04 and .09.
| dx | dy | Δx | Δy | ||
| y = x² | .1 | 0.4 | .1 | 0.41 | Δy = (2+Δx)² − 2² |
| dy = 4dx | .2 | 0.8 | .2 | 0.84 | Δy = 4Δx + (Δx)² |
| .3 | 1.2 | .3 | 1.29 |
The differential dy = ƒ'(x)dx is consistent with the derivative dy/dx = ƒ'(x). We finally have dy =(dy/dx)dx, but this is not as obvious as it seems! It looks like cancellation—it is really a definition. Entirely new symbols could be used, but dx and dy have two advantages: They suggest small steps and they satisfy dy = ƒ'(x)dx. Here are three examples and three rules:
d(xn) = nxn-1 dx d(ƒ + g) = dƒ + dg
d(sin x) = cos x dx d(cƒ) = cdƒ
d(tan x) = sec² x dx d(ƒg) = ƒdg + gdƒ
Science and engineering and virtually all applications of mathematics depend on linear approximation. The true function is "linearized,"using its slope v:
Increasing the time by Δt increases the distance by ≈ vΔt
Increasing the force by Δƒ increases the deflection by ≈ vΔƒ
Increasing the production by Δp increases its value by ≈ vΔp.
The goal of dynamics or statics or economics is to predict this multiplier v—the derivative that equals the slope of the tangent line. The multiplier gives a local prediction of the change in the function. The exact law is nonlinear—but Ohm's law and Hooke's law and Newton's law are linear approximations.
The change Δy or Δƒ can be measured in three ways. So can Δx:
Absolute change Δƒ Δx
Relative change Δƒ ⁄ Δx Δx ⁄ x
Percentage change (Δƒ ⁄ Δx) × 100 (Δx ⁄ x) × 100
3 miles ⁄ 300,000 miles < 1 inchi ⁄ 70 inches or .001% < 1.4%.
EXAMPLE 4 The radius of the Earth is within 80 miles of r = 4000 miles.
(a) Find the variation dV in the volume V = (4/3)πr³, using linear approximation.
(b) Compute the relative variations dr/r and dV/V and ΔV/V
Solution The job of calculus is to produce the derivative. After dV/dr = 4πr², its work is done. The variation in volume is dV = 4π(4000)²(80) cubic miles. A 2% relative variation in r gives a 6% relative variation in V:
dr/r = 80/4000 = 2% dV/V = 4πr(4000)²(80) ⁄ 4πr(4000)³/3 = 6%.
Without calculus we need the exact volume at r = 4000 + 80 (also at r = 3920):
ΔV/V = 4π(4080)³/3 − 4π(4000)³/3 ⁄ 4π(4000)³/3 ≈ 6%
One comment on dV = 4πr²dr. This is (area of sphere) times (change in radius). It is the volume of a thin shell around the sphere. The shell is added when the radius grows by dr. The exact ΔV/V is 3917312/640000%, but calculus just calls it 6%.
