Premed · Premed · Calculus 1
Lecture 11: Linear Approximation and Differentials
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Use the tangent line to approximate function values near a point
- Write and apply the linear approximation (linearization) formula
- Define differentials dx and dy and relate them to actual changes
- Use differentials to estimate errors and changes in measurements
- Understand the geometric meaning of linear approximation
Lecture Content
I. Linear Approximation (Linearization)
Near a point x = a, the tangent line provides a good approximation to the curve y = f(x). The linearization of f at a is L(x) = f(a) + f'(a)(x - a), and the resulting linear approximation is f(x) approximately equals L(x) for x near a. This is the first-order Taylor approximation -- the best linear approximation to f in a neighborhood of a.
II. Examples of Linear Approximation
To approximate sqrt(4.02), use f(x) = sqrt(x) with a = 4. Since f(4) = 2 and f'(x) = 1/(2*sqrt(x)) gives f'(4) = 1/4, the linearization is L(x) = 2 + (1/4)(x - 4). Evaluating at x = 4.02 yields sqrt(4.02) approximately equals 2 + (1/4)(0.02) = 2.005. The actual value is 2.00499..., so the approximation is excellent.
To approximate sin(0.1), use f(x) = sin x with a = 0. Since f(0) = 0 and f'(0) = cos 0 = 1, the linearization gives sin(0.1) approximately equals 0.1. The actual value is 0.09983..., which is very close.
To approximate (1.01)^{10}, use f(x) = (1 + x)^{10} with a = 0, treating x as the small deviation from 1. Since f(0) = 1 and f'(0) = 10, the approximation is (1.01)^{10} approximately equals 1 + 10(0.01) = 1.1. The actual value is 1.10462..., a reasonable estimate.
<image>A graph showing f(x) = sqrt(x) (solid blue curve) and its linearization L(x) = 2 + (1/4)(x - 4) (dashed red line tangent at x = 4). The point (4, 2) is marked where they touch. At x = 4.02, both curves are nearly indistinguishable, with a zoomed inset showing the tiny gap between f(4.02) = 2.00499... and L(4.02) = 2.005. Title: "Linear Approximation: the tangent line approximates the curve near the point of tangency."</image>
III. The Approximation (1 + x)^n ≈ 1 + nx for Small x
An important special case of linear approximation arises from linearizing f(x) = (1 + x)^n at x = 0. Since f(0) = 1 and f'(0) = n, the approximation is (1 + x)^n approximately equals 1 + nx when |x| is small. For example, (1.02)^5 approximately equals 1 + 5(0.02) = 1.10, and 1/sqrt(1.04) = (1.04)^{-1/2} approximately equals 1 + (-1/2)(0.04) = 0.98.
IV. Differentials
Let y = f(x). The differential dx = Delta x is an independent variable representing a small change in x, and the differential dy = f'(x) * dx represents the change in y along the tangent line when x changes by dx. The actual change in y is Delta y = f(x + dx) - f(x), and the key relationship is that Delta y approximately equals dy when dx is small, with Delta y = dy + higher-order terms.
Geometrically, Delta y is the change along the curve, while dy is the change along the tangent line. The difference |Delta y - dy| is the approximation error.
<image>A graph showing the curve y = f(x) with a point P = (a, f(a)). From P, a horizontal distance dx is marked. The tangent line at P extends to the right, rising by dy = f'(a)*dx. The curve rises by Delta y = f(a + dx) - f(a), which is slightly different from dy. Both dy and Delta y are labeled with vertical arrows from the baseline at height f(a). The small gap between them (Delta y - dy) is labeled as the "approximation error." Title: "Differentials vs. Actual Change."</image>
V. Using Differentials to Estimate Errors
If a quantity Q depends on a measured value x with measurement error dx, then the propagated error in Q is approximately dQ = Q'(x) dx. The relative error is dQ/Q, and the percentage error is (dQ/Q) 100%.
For example, suppose the radius of a sphere is measured as r = 10 cm with error dr = 0.1 cm. The volume V = (4/3)pir^3 has differential dV = 4pir^2 dr = 4pi(100)(0.1) = 40pi, approximately 125.7 cm^3. The relative error in volume is dV/V = (4pir^2dr)/((4/3)pir^3) = 3(dr/r) = 3(0.1/10) = 0.03 = 3%. Notice that the relative error in the volume is three times the relative error in the radius, reflecting the cubic relationship.
VI. Differentials and Leibniz Notation
The notation dy/dx now carries a dual meaning: it represents the derivative of y with respect to x (a single symbol), and it can also be interpreted as the ratio of the differentials dy and dx, which equals f'(x). This is why Leibniz notation is so powerful: the "fraction" dy/dx truly behaves like a ratio of differentials. This interpretation justifies manipulations like dy = f'(x) dx and the chain rule in the form dy/dx = (dy/du)(du/dx).
VII. When Is the Approximation Good?
Linear approximation is most accurate when x is close to a (|x - a| is small) and the function does not curve too sharply near a (|f''| is small). The error is approximately (1/2)f''(c)(x - a)^2 for some c between a and x, which is the next term in the Taylor expansion. Functions with large second derivatives have larger approximation errors. For practical purposes, if |x - a| < 0.1 and f is "well-behaved," the linear approximation is typically quite accurate.
<image>A comparison of linear approximation quality for two functions. Panel A: f(x) = sin x near a = 0 — the curve and tangent line (y = x) agree closely over a wide range because f''(0) = 0 and the curve is gentle. Panel B: f(x) = e^x near a = 0 — the curve and tangent line (y = 1 + x) diverge more quickly because the exponential curves upward steeply (f''(0) = 1). Both panels show the error shaded between the curve and the tangent line. Title: "Approximation quality depends on curvature."</image>
