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Mom's Basement University

Linear Approximations and Differentials

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to use the tangent line to approximate function values near a known point, and how differentials give you a quick way to estimate small changes.

The Concept

Section titled “The Concept”

Linear Approximation

Section titled “Linear Approximation”

Near a point x = a, the tangent line is a good stand-in for the curve. The tangent line equation at x = a is:

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

This is called the linearization of f at a. When x is close to a, L(x) is close to f(x). The further you go from a, the worse the approximation gets.

The blue curve is f(x) = the square root of x. The green line is the tangent at x = 9, which gives L(x) = 3 + (x - 9)/6. Near x = 9, the green line hugs the blue curve closely. Further away, they diverge. That’s the whole idea: the tangent line approximates the function locally.

Differentials

Section titled “Differentials”

The differential dy is a related concept:

dy=f′(x) dxdy = f'(x) \, dx

Here dx is a small change in x, and dy is the approximate change in y. This is useful for estimating how much a function’s output changes when the input changes by a small amount.

Worked Example

Section titled “Worked Example”

Example 1: Approximating a square root

Approximate the square root of 9.1 using linearization.

Let f(x) = the square root of x, a = 9, x = 9.1.

f(9) = 3. The derivative is f’(x) = 1/(2 times the square root of x), so f’(9) = 1/6.

L(9.1)=3+16(9.1−9)=3+0.16≈3.0167L(9.1) = 3 + \frac{1}{6}(9.1 - 9) = 3 + \frac{0.1}{6} \approx 3.0167

The actual value is about 3.01662. The approximation is off by less than 0.0001.

Example 2: Differentials for volume change

A spherical balloon has radius 5 inches. If the radius increases by 0.1 inches, approximately how much does the volume increase?

V = (4/3) pi r³, so dV = 4 pi r² dr.

dV=4π(5)2(0.1)=4π(25)(0.1)=10π≈31.4 cubic inchesdV = 4\pi(5)^2(0.1) = 4\pi(25)(0.1) = 10\pi \approx 31.4 \text{ cubic inches}

A tiny change in radius (0.1 inches) causes about 31.4 cubic inches of volume change. That’s because volume scales with the cube of the radius, so even small radius changes produce large volume changes.

The blue circle is the balloon at r = 5. The thin green ring is the added volume when the radius grows by dr = 0.1. That ring’s volume is dV = 4 pi r² dr, which is the differential. The ring looks thin, but it wraps around the entire surface of the sphere, so it adds up to about 31.4 cubic inches.

Example 3: Approximating a fourth power

Approximate (1.02)⁴ using linearization.

Let f(x) = x⁴, a = 1, x = 1.02.

f(1) = 1. f’(x) = 4x³, so f’(1) = 4.

L(1.02)=1+4(1.02−1)=1+4(0.02)=1.08L(1.02) = 1 + 4(1.02 - 1) = 1 + 4(0.02) = 1.08

The actual value is about 1.0824. The approximation is close because 1.02 is near 1.

Real-World Application

Section titled “Real-World Application”

Linear approximation and differentials are practical tools:

  • Engineering: estimating how manufacturing tolerances affect a final product. If a part’s diameter is off by 0.01 mm, how much does the volume change?
  • Physics: approximating small oscillations. For small angles, sin(theta) is approximately theta, which is a linear approximation.
  • Finance: estimating how a small change in interest rate affects the present value of a bond.
  • Medicine: estimating how a small dosage change affects drug concentration.

Quiz

Section titled “Quiz”
The linearization of f at x = a is:
Linear approximations work best when:
The differential dy equals:
Using linearization at a = 9, the square root of 9.1 is approximately:
Differentials are especially useful for: