The Timpson Lyceum · Calculus 1 (2024)

Newton’s Method & Antiderivatives - Lecture 14 (Calculus 1)

Overview

This lecture introduces two foundational tools. The first is Newton's method, a fast iterative technique for approximating where a function equals zero — useful precisely because most equations (degree-five polynomials and higher) have no general solution formula, and even when a formula exists it is often too messy to use. The second is the antiderivative and its packaging as the indefinite integral, the operation that undoes differentiation. We build Newton's iteration formula from the geometry of tangent lines, catalog the basic antiderivative rules, and close with a projectile (hot-air-balloon) problem that recovers the kinematic equations using nothing but antidifferentiation.

Key Concepts

Why we need Newton's method

Linear ( $ax+b=0$ ) and quadratic ( $ax^2+bx+c=0$ ) equations have simple closed-form solutions. Cubics and quartics still have formulas, but they grow rapidly in complexity.
For polynomials of degree five and higher there is provably no general solution formula — not for lack of cleverness, but as a proven fact. Specific cases (e.g. $x^5-32=0$ ) can still be solved by inspection.
Newton's method sidesteps this: it approximates a root quickly, even when solving exactly is impossible or simply inconvenient.

The idea behind Newton's method

Start with an initial guess $x_0$ near a root.
Draw the tangent line to the curve at that point and find where the tangent crosses the $x$ -axis. Because a line's root is trivial to compute, this gives a new, usually better, guess.
Repeat: each guess feeds the next, and the sequence converges toward the true root.

The iteration formula

Using the point-slope tangent line $y - f(x_n) = f'(x_n)\,(x - x_n)$ and setting $y=0$ , solve for the next $x$ -intercept to get: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
Requirement: $f'(x_n)\neq 0$ . Geometrically, a horizontal tangent never crosses the $x$ -axis; algebraically, $f'(x_n)=0$ forces division by zero.
Failure modes (rare in practice): landing on a point where $f'(x_n)=0$ , or hitting two points whose tangents bounce back and forth, trapping the iteration in a cycle. The method is reliable but not foolproof.

Antiderivatives and the indefinite integral

A function $F(x)$ is an antiderivative of $f(x)$ if $F'(x)=f(x)$ — it "undoes" differentiation.
The collection of all antiderivatives of $f$ is the indefinite integral, written $\int f(x)\,dx$ . It is "indefinite" because it represents a whole family of functions, not a single one.
Every antiderivative carries an added constant of integration $+C$ , since the derivative of any constant is zero.
The $dx$ specifies the variable of integration; any other symbol is treated as a constant.

Basic antiderivative rules

Power rule (add one to the exponent, then divide by the new exponent): $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \qquad n \neq -1$ The restriction $n=-1$ is excluded because it would divide by zero.
Constant rule: $\displaystyle\int k\,dx = kx + C$ .
Trig antiderivatives (each verified by differentiating the right side and using the chain rule on the inner $kx$ ):

\begin{aligned} \int \sin(kx)\,dx &= -\tfrac{1}{k}\cos(kx) + C \\ \int \cos(kx)\,dx &= \tfrac{1}{k}\sin(kx) + C \\ \int \sec^2(kx)\,dx &= \tfrac{1}{k}\tan(kx) + C \\ \int \csc^2(kx)\,dx &= -\tfrac{1}{k}\cot(kx) + C \\ \int \sec(kx)\tan(kx)\,dx &= \tfrac{1}{k}\sec(kx) + C \\ \int \csc(kx)\cot(kx)\,dx &= -\tfrac{1}{k}\csc(kx) + C \end{aligned}

Connection to motion (kinematics)

Velocity is the derivative of position, and acceleration is the derivative of velocity. Reversing this, velocity is the antiderivative of acceleration, and position is the antiderivative of velocity.
Starting from a constant acceleration $a$ , antidifferentiating twice reproduces the familiar kinematic equation $s(t) = \tfrac{1}{2}a t^2 + v_0 t + s_0$ , where the integration constants are exactly the initial velocity $v_0$ and initial position $s_0$ .

Worked Example

Approximate $\sqrt{2}$ using Newton's method by finding the root of $f(x)=x^2-2$ , with $f'(x)=2x$ , starting from $x_0=1$ .

Apply $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$ :

\begin{aligned} x_1 &= x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{1^2 - 2}{2(1)} = 1 - \frac{-1}{2} = \frac{3}{2} \\ x_2 &= x_1 - \frac{f(x_1)}{f'(x_1)} = \frac{3}{2} - \frac{\left(\tfrac{3}{2}\right)^2 - 2}{2\left(\tfrac{3}{2}\right)} = \frac{3}{2} - \frac{\tfrac{1}{4}}{3} = \frac{3}{2} - \frac{1}{12} = \frac{17}{12} \approx 1.4167 \end{aligned}

After just two iterations the estimate $1.4167$ already matches the true value $\sqrt{2}\approx 1.4142$ to three decimal places; another iteration tightens it further.

Projectile from a balloon. A bag is released from $80$ ft up while rising at $12$ ft/s; with $a(t)=-32$ ft/s $^2$ , antidifferentiate twice (the constants are $v_0=12$ and $s_0=80$ ) to get $s(t) = -16t^2 + 12t + 80$ . Setting $s(t)=0$ and applying the quadratic formula gives $t = \dfrac{3+\sqrt{329}}{8} \approx 2.64$ seconds until it lands.

Summary

Newton's method approximates roots by repeatedly following the tangent line to its $x$ -intercept; the core formula is $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$ .
It exists because polynomials of degree $\ge 5$ have no general solution formula, and because approximation is often easier than exact solving.
The iteration requires $f'(x_n)\neq 0$ and can rarely fail by cycling, but in practice it converges quickly.
An antiderivative $F$ satisfies $F'=f$ ; the indefinite integral $\int f(x)\,dx$ collects all of them and always includes $+C$ .
Key rules: the power rule $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$ (for $n\neq-1$ ), the constant rule, and the standard trig antiderivatives.
Antidifferentiating acceleration gives velocity and then position, which is how the kinematic equations of motion arise.