The Timpson Lyceum · Calculus 1 (2024)

Differentials, Linearization, and Related Rates - Lecture 11 (Calculus 1)

Overview

This lecture closes out the chapter with three closely related applications of the derivative: related rates, linearization, and differentials. Related-rates problems take a story problem relating two (or more) changing quantities, differentiate the relating equation with respect to time, and solve for an unknown rate. Linearization replaces a function near a point with its tangent line, giving a simple linear approximation that is accurate close to that point. Differentials, $dy = f'(x)\,dx$ , formalize the "small change" idea and let us approximate small changes in a function's value. Together these tools turn the derivative into a practical instrument for modeling motion, estimating values, and bounding measurement error.

Key Concepts

Related rates — the strategy

There is no fixed procedure: the hard part is translating the story into a mathematical equation relating the variables.
Draw a picture and label what changes. Write down every given rate and the quantity you are solving for.
Find an equation relating the variables (geometry/trigonometry/Pythagoras).
Differentiate both sides with respect to time $t$ , applying the chain rule to every variable — each variable's derivative produces its rate of change.
Plug in the known values only after differentiating, then solve for the unknown rate.
Watch the sign of each rate: a decreasing quantity has a negative rate (e.g. a distance shrinking over time).

Useful relations that recur in related rates

Cone volume: $V = \frac{1}{3}\pi r^2 h$ (one-third the cylinder $\pi r^2 h$ ). Use similar triangles to eliminate one variable, e.g. for a cone with height $=$ twice the radius, $r = \tfrac{1}{2}y$ .
Right-triangle trig: $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$ , used to relate an angle to a height.
Pythagorean theorem: $s = \sqrt{x^2 + y^2}$ for a distance between two legs.

Linearization

The linearization of $f$ at a point $a$ is just the tangent line to $f$ at $a$ , written as the function $L$ : $L(x) = f(a) + f'(a)(x - a)$
Requires $f$ to be differentiable at $a$ (it must have a slope/tangent there).
For $x$ near $a$ , $f(x) \approx L(x)$ . Zoom in far enough on a differentiable function and its graph is indistinguishable from its tangent line.
Why it matters: ugly functions become easy to work with near a point of interest (e.g. small-angle pendulum approximations in physics). It is the first step toward later polynomial-approximation ideas.
A common standard result: the linearization of $(1+x)^k$ at $x = 0$ is $L(x) = 1 + kx$ . For example, $\sqrt{1+x} = (1+x)^{1/2} \approx 1 + \tfrac{1}{2}x$ near $0$ .

Differentials

For a differentiable $y = f(x)$ , the differential is $dy = f'(x)\,dx$ where $dx$ is an independent variable (any chosen small change) and $dy$ is the output. So $dy$ depends on both $x$ and $dx$ .
Dividing gives $\dfrac{dy}{dx} = f'(x)$ . The Leibniz symbol $\frac{dy}{dx}$ for the derivative is not literally a quotient, but differentials are defined so that their quotient equals the derivative — which is why everyone treats them like a fraction.
Geometric meaning: $dy$ is the change in height of the tangent line (the linearization), $\Delta L$ , for a step $dx$ — not the exact change in $f$ . It is close to the true change $\Delta y = f(a+dx) - f(a)$ when $dx$ is small.
Differential rules mirror derivative rules: $d(u+v) = du + dv$ , $d(uv) = u\,dv + v\,du$ , $d\!\left(\frac{u}{v}\right) = \dfrac{v\,du - u\,dv}{v^2}$ .

Approximation and error

Rearranging the differential gives the linear approximation formula: $f(a + \Delta x) \approx f(a) + dy = f(a) + f'(a)\,\Delta x$ Choose $a$ to be a "nice" point near the value you actually want (e.g. approximate $\sqrt[3]{7.97}$ using $a = 8$ , $dx = -0.03$ ).
The error of the approximation is the true change minus the approximate change: $\text{error} = \Delta y - dy$ . As $dx \to 0$ the approximation becomes exact.

Worked Example

Related rates — hot-air balloon (angle to height). A balloon rises straight up; a rangefinder $150\text{ m}$ away tracks it. When the elevation angle is $\theta = \tfrac{\pi}{4}$ , the angle increases at $\frac{d\theta}{dt} = 0.14$ rad/min. Find the rate the balloon rises, $\frac{dy}{dt}$ .

Relate the height $y$ to the angle by $\tan\theta = \dfrac{y}{150}$ , so $y = 150\tan\theta$ . Differentiate with respect to $t$ (chain rule), then substitute:

\begin{aligned} y &= 150\tan\theta \\ \frac{dy}{dt} &= 150\sec^2\theta \cdot \frac{d\theta}{dt} \\ \frac{dy}{dt} &= 150\left(\sec\tfrac{\pi}{4}\right)^2 (0.14) = 150\,(\sqrt{2})^2 (0.14) = 150 \cdot 2 \cdot 0.14 = 42 \text{ m/min} \end{aligned}

Differentials — cube-root approximation. Approximate $\sqrt[3]{7.97}$ . Use $f(x) = x^{1/3}$ at the nice point $a = 8$ with $dx = -0.03$ , via $f(a+dx) \approx f(a) + dy$ where $dy = f'(x)\,dx$ :

\begin{aligned} dy &= \frac{1}{3}x^{-2/3}\,dx = \frac{dx}{3x^{2/3}} \\ dy &= \frac{-0.03}{3 \cdot 8^{2/3}} = \frac{-0.03}{3 \cdot 4} = \frac{-0.03}{12} = -0.0025 \\ \sqrt[3]{7.97} &\approx f(8) + dy = 2 + (-0.0025) = 1.9975 \end{aligned}

(A calculator gives $\approx 1.9975$ as well — very close.)

Summary

Related rates: model the relationship between variables, differentiate both sides with respect to $t$ using the chain rule, then substitute knowns and solve for the unknown rate; mind the signs.
Eliminate extra variables using geometry (similar triangles for the cone, trig for angles, the Pythagorean relation for distances) before or after differentiating, but substitute numbers only after differentiating.
Linearization at $a$ is the tangent line $L(x) = f(a) + f'(a)(x-a)$ ; it approximates $f$ well near $a$ , and $(1+x)^k \approx 1 + kx$ near $0$ .
A differential is $dy = f'(x)\,dx$ with $dx$ a freely chosen small change; it measures the change in height of the tangent line, and its rules match the derivative rules.
Use $f(a+\Delta x) \approx f(a) + f'(a)\,\Delta x$ to approximate values near a convenient point; the error $\Delta y - dy$ shrinks to $0$ as $dx \to 0$ .