# Lecture 11: Linear Approximation and Differentials ## Calculus I — Differential Calculus --- ## Learning Objectives By the end of this lecture, students will be able to: 1. Use the tangent line to approximate function values near a point 2. Write and apply the linear approximation (linearization) formula 3. Define differentials dx and dy and relate them to actual changes 4. Use differentials to estimate errors and changes in measurements 5. 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. 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." ### 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. 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." ### 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)*pi*r^3 has differential dV = 4*pi*r^2 * dr = 4*pi*(100)*(0.1) = 40*pi, approximately 125.7 cm^3. The relative error in volume is dV/V = (4*pi*r^2*dr)/((4/3)*pi*r^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. 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."