Unit 4, lesson 44

Linear Approximations and Differentials

How the tangent line gives a simple approximation of a function near a point, why its error shrinks faster than the step, and how to bound that error.

  • Theory part 87
  • Practice part 88
  • Limits and Approximation

Theory, part 87

Practice, part 88

Zoom in far enough on a smooth curve and it looks straight. That straight line is the tangent line, and near the point of tangency it is a very good stand in for the function. This is how you can estimate 4.1\sqrt{4.1} in your head, and how engineers judge how much a small measurement error affects a computed result.

Key terms

Linearization

Definition Let ff be differentiable at aa. The linearization of ff at aa is the function

L(x)=f(a)+f′(a)(x−a).L(x) = f(a) + f'(a)(x - a).

Its graph is the tangent line to the graph of ff at (a,f(a))(a, f(a)).

In plain words The linearization is the tangent line, written as a function you can evaluate.

Example Take f(x)=xf(x) = \sqrt{x} at a=4a = 4. Then f(4)=2f(4) = 2 and

f′(x)=12x,f′(4)=14,f'(x) = \frac{1}{2\sqrt{x}}, \qquad f'(4) = \frac{1}{4},

so

L(x)=2+14(x−4).L(x) = 2 + \frac{1}{4}(x - 4).

Tangent line approximation

Definition The tangent line approximation, or linear approximation, of ff near aa is

f(x)≈L(x)=f(a)+f′(a)(x−a),f(x) \approx L(x) = f(a) + f'(a)(x - a),

used for xx close to aa. The error of the approximation is f(x)−L(x)f(x) - L(x).

In plain words Close to aa, read values off the tangent line instead of the curve. Pick aa so that f(a)f(a) and f′(a)f'(a) are easy to compute.

Example The linearization above gives

4.1≈2+0.14=2.025.\sqrt{4.1} \approx 2 + \frac{0.1}{4} = 2.025.

The true value is 2.024846…2.024846\ldots, so the error is only about 0.000150.00015.

Differentials and actual change

Definition Let y=f(x)y = f(x) with ff differentiable at xx. The differential dxdx is an independent variable, any real number, and the differential dydy is defined by

dy=f′(x) dx.dy = f'(x)\,dx.

For an increment Δx=dx\Delta x = dx, the actual change in yy is

Δy=f(x+Δx)−f(x).\Delta y = f(x + \Delta x) - f(x).

The differential dydy is the change along the tangent line, and Δy≈dy\Delta y \approx dy when Δx\Delta x is small.

In plain words The increment Δy\Delta y is how much the function really changes. The differential dydy is how much the tangent line predicts it changes. They are close for small steps but not equal.

Example For y=x3y = x^3 at x=2x = 2 with dx=Δx=0.1dx = \Delta x = 0.1,

dy=3⋅22⋅0.1=1.2,Δy=2.13−23=9.261−8=1.261.dy = 3 \cdot 2^2 \cdot 0.1 = 1.2, \qquad \Delta y = 2.1^3 - 2^3 = 9.261 - 8 = 1.261.

The differential misses the actual change by 0.0610.061.

The error shrinks faster than the step

Definition A function ff defined on an open interval containing aa is differentiable at aa if and only if there is a number mm such that the error E(h)=f(a+h)−f(a)−mhE(h) = f(a + h) - f(a) - mh satisfies

lim⁡h→0E(h)h=0,\lim_{h \to 0} \frac{E(h)}{h} = 0,

and in that case m=f′(a)m = f'(a). So for a differentiable function the error of the linear approximation goes to 00 faster than the increment hh itself.

In plain words It is not enough that the error gets small, since any continuous function does that. The error gets small even compared with the size of the step, and among all lines through (a,f(a))(a, f(a)) only the tangent line has this property.

Example For f(x)=x3f(x) = x^3 at a=2a = 2 we have f′(2)=12f'(2) = 12 and

E(h)=(2+h)3−8−12h=6h2+h3.E(h) = (2 + h)^3 - 8 - 12h = 6h^2 + h^3.
hhE(h)E(h)E(h)/hE(h)/h
0.10.10.0610.0610.610.61
0.010.010.0006010.0006010.06010.0601
0.0010.0010.0000060010.0000060010.0060010.006001

The ratio

E(h)h=6h+h2\frac{E(h)}{h} = 6h + h^2

tends to 00.

Big-O and little-o notation

Definition Let EE and φ\varphi be defined for hh near 00, with h≠0h \neq 0. As h→0h \to 0, write

E(h)=o(φ(h))iflim⁡h→0E(h)φ(h)=0,E(h) = o(\varphi(h)) \quad \text{if} \quad \lim_{h \to 0} \frac{E(h)}{\varphi(h)} = 0,

where φ(h)≠0\varphi(h) \neq 0 for small h≠0h \neq 0, and write E(h)=O(φ(h))E(h) = O(\varphi(h)) if there are constants C>0C > 0 and δ>0\delta > 0 with ∣E(h)∣≤C∣φ(h)∣\lvert E(h) \rvert \le C\lvert \varphi(h) \rvert whenever 0<∣h∣<δ0 < \lvert h \rvert < \delta. In this notation, ff is differentiable at aa exactly when

f(a+h)=f(a)+f′(a)h+o(h).f(a + h) = f(a) + f'(a)h + o(h).

In plain words Little-o of hh means negligible compared with hh. Big-O of h2h^2 means at most a fixed multiple of h2h^2.

Example Take E(h)=6h2+h3E(h) = 6h^2 + h^3. The limit

lim⁡h→0E(h)h=lim⁡h→0(6h+h2)=0\lim_{h \to 0} \frac{E(h)}{h} = \lim_{h \to 0} (6h + h^2) = 0

shows E(h)=o(h)E(h) = o(h). And for 0<∣h∣<10 < \lvert h \rvert < 1,

∣6h2+h3∣≤6h2+∣h∣h2≤7h2,\lvert 6h^2 + h^3 \rvert \le 6h^2 + \lvert h \rvert h^2 \le 7h^2,

so E(h)=O(h2)E(h) = O(h^2) with C=7C = 7 and δ=1\delta = 1.

Error bounds from the mean value theorem

Definition Let ff be twice differentiable on an open interval containing aa and xx.

  1. If ∣f′(t)∣≤M1\lvert f'(t) \rvert \le M_1 for every tt between aa and xx, then ∣f(x)−f(a)∣≤M1∣x−a∣\lvert f(x) - f(a) \rvert \le M_1 \lvert x - a \rvert. This follows from the mean value theorem.
  2. If ∣f′′(t)∣≤M2\lvert f''(t) \rvert \le M_2 for every tt between aa and xx, then the error of the linear approximation satisfies
∣f(x)−L(x)∣≤M22(x−a)2.\lvert f(x) - L(x) \rvert \le \frac{M_2}{2}(x - a)^2.

The second bound comes from Taylor’s theorem, which says that

f(x)−L(x)=f′′(c)2(x−a)2f(x) - L(x) = \frac{f''(c)}{2}(x - a)^2

for some cc between aa and xx. Applying the plain mean value theorem twice only gives the weaker bound M2(x−a)2M_2 (x - a)^2, without the factor 1/21/2.

In plain words The tangent line goes wrong only because the slope changes, and f′′f'' measures how fast the slope changes. A small f′′f'' and a small step mean a small error, and the error shrinks like the square of the step.

Example For f(x)=xf(x) = \sqrt{x} near 44 the second derivative is f′′(t)=−14t−3/2f''(t) = -\tfrac{1}{4}t^{-3/2}, whose size decreases as tt grows. For tt between 44 and 4.14.1,

∣f′′(t)∣≤14⋅4−3/2=132.\lvert f''(t) \rvert \le \frac{1}{4} \cdot 4^{-3/2} = \frac{1}{32}.

So

∣4.1−2.025∣≤164(0.1)2=0.00015625,\lvert \sqrt{4.1} - 2.025 \rvert \le \frac{1}{64}(0.1)^2 = 0.00015625,

and the actual error 0.0001540.000154 fits under this bound.

Check yourself

  1. Use a linearization to estimate 8.063\sqrt[3]{8.06}.
Show the answer

With f(x)=x1/3f(x) = x^{1/3} and a=8a = 8 we get f(8)=2f(8) = 2 and

f′(8)=13⋅8−2/3=112.f'(8) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{12}.

So

8.063≈2+0.0612=2.005.\sqrt[3]{8.06} \approx 2 + \frac{0.06}{12} = 2.005.

The true value is about 2.004992.00499.

  1. With xx in radians, the linearization of sin⁡x\sin x at 00 is L(x)=xL(x) = x. Bound the error of sin⁡0.1≈0.1\sin 0.1 \approx 0.1.
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Here ∣f′′(t)∣=∣sin⁡t∣≤1\lvert f''(t) \rvert = \lvert \sin t \rvert \le 1, so the error is at most

12(0.1)2=0.005.\frac{1}{2}(0.1)^2 = 0.005.

The actual error is about 0.0001670.000167, well within the bound.

  1. A sphere’s radius is measured as 1010 centimeters with a possible error of 0.050.05 centimeters. Estimate the possible error in its computed volume.
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The volume is V=43πr3V = \tfrac{4}{3}\pi r^3, so

dV=4πr2 dr=4π⋅100⋅0.05=20π≈62.8.dV = 4\pi r^2\,dr = 4\pi \cdot 100 \cdot 0.05 = 20\pi \approx 62.8.

The possible error in the volume is about 62.862.8 cubic centimeters.

How this lesson connects

It also uses linear functions.

Resources for this lesson