Calculus

Lesson 1: Limits and Continuity

Calculus begins with a deceptively simple question: what value does f(x) approach as x approaches c? The limit lim(x→c) f(x) = L captures behavior near c, regardless of f(c) itself. This distinction powers everything: instantaneous velocity is the limit of average velocities over shrinking time intervals, and the slope of a curve is the limit of secant slopes. Techniques: direct substitution, factoring and canceling (for 0/0 forms), rationalizing, and the squeeze theorem — which traps sin(x)/x between cos(x) and 1 to prove lim(x→0) sin(x)/x = 1.

A function is continuous at c if lim(x→c) f(x) = f(c): you can draw through c without lifting the pen. Discontinuities come in three flavors: removable (a hole, like (x²−1)/(x−1) at x = 1), jump (left and right limits disagree, common in piecewise functions), and infinite (vertical asymptotes, like 1/x at 0). Polynomials, exponentials, and sine/cosine are continuous everywhere; rational functions are continuous on their domains.

The Intermediate Value Theorem (IVT) turns continuity into existence proofs: if f is continuous on [a, b] and N lies between f(a) and f(b), some c in (a, b) has f(c) = N. To prove x³ − x − 2 = 0 has a real root, note f(1) = −2 and f(2) = 4; by IVT a zero lies between 1 and 2. Limits at infinity describe end behavior and horizontal asymptotes, completing the graphing toolkit from precalculus.

Try it yourself: Evaluate: (a) lim(x→−3) (x² − 9)/(x + 3), (b) lim(x→∞) (2x² + 3)/(5x² − x). Use the IVT to show f(x) = x³ + x − 5 has a zero between 1 and 2.

Lesson 2: Limits at Infinity and Asymptotes

A limit at infinity asks where a function settles as x grows without bound. For rational functions, divide by the highest power of x. For f(x) = (3x² + 2x)/(x² − 5), dividing top and bottom by x² gives (3 + 2/x)/(1 − 5/x²), which approaches 3/1 = 3 as x → ±∞. So y = 3 is a horizontal asymptote. Shortcuts: if the numerator's degree is smaller, the limit is 0; if larger, the function grows without bound (no horizontal asymptote).

End behavior models use these limits: the long-run market share of a product, the terminal velocity of a falling object, the steady-state temperature of a heated rod. The function f(x) = 1 − e⁻ₛ approaches 1 as x → ∞, modeling quantities that saturate: learning curves, charging capacitors, spread of a rumor to a fixed population.

Infinite limits describe vertical asymptotes: lim[x→2⁺] 1/(x − 2) = +∞ because the denominator is tiny and positive; from the left it is −∞. The two-sided limit does not exist, but the one-sided limits tell the full story of the graph's behavior near the asymptote.

Try it yourself: Find lim[x→∞] (4x + 1)/(x² + 3) and lim[x→∞] (x³)/(x² + 1). Answers: 0 (degree of numerator smaller); +∞ (numerator degree larger).

Lesson 3: The Definition of the Derivative

The derivative f′(x) is defined by the limit f′(x) = lim[h→0] (f(x + h) − f(x))/h. This is not a formula to memorize and forget — it is the engine that produces every differentiation rule. For f(x) = x³: (f(x + h) − f(x))/h = ((x + h)³ − x³)/h = (3x²h + 3xh² + h³)/h = 3x² + 3xh + h², which approaches 3x² as h → 0. So (x³)′ = 3x², matching the power rule.

Differentiability requires a smooth, unbroken graph with a single tangent direction. A function fails to be differentiable at corners (like |x| at 0, where left and right slopes disagree), cusps, vertical tangents, and discontinuities. Differentiable implies continuous, but continuous does not imply differentiable — |x| is continuous at 0 but has no derivative there.

The derivative has two faces: as a function, f′(x) gives the slope at every point; as a value, f′(a) is the slope at one point, hence the equation of the tangent line y − f(a) = f′(a)(x − a). Leibniz notation dy/dx and Newton's f′(x) describe the same object.

Try it yourself: Use the definition to show the derivative of f(x) = 5x is 5. Is f(x) = |x| differentiable at x = 0? Answers: (5(x+h) − 5x)/h = 5, limit 5. No — corner at 0, left slope −1, right slope +1.

Key vocabulary: Difference quotient — (f(x + h) − f(x))/h; differentiable — having a derivative at a point; corner — a point where left and right slopes disagree.

Lesson 4: Derivatives and Their Applications

The derivative f'(x) is the instantaneous rate of change of f at x — the slope of the tangent line, defined as the limit of difference quotients: f'(x) = lim(h→0) [f(x+h) − f(x)]/h. If f measures position, f' is velocity; if f is velocity, f' is acceleration. Differentiability implies continuity, but not conversely — |x| is continuous at 0 but has a corner, so no derivative there.

Computing derivatives uses rules, not limits, after the first derivation. The power rule: d/dx[x^n] = nx^(n−1). The product rule: (fg)' = f'g + fg'. The quotient rule: (f/g)' = (f'g − fg')/g². The chain rule — the most important — differentiates compositions: d/dx[f(g(x))] = f'(g(x))·g'(x). For sin(x²): outer derivative cos(x²) times inner derivative 2x. Memorize: d/dx[sin x] = cos x, d/dx[cos x] = −sin x, d/dx[e^x] = e^x, d/dx[ln x] = 1/x.

Derivatives locate extrema: f' = 0 (or undefined) marks critical points, candidates for maxima/minima. The first derivative test reads sign changes of f' (rising → falling means a max); the second derivative test uses concavity (f'' < 0 at a critical point means a max). Optimization turns word problems into extrema hunts: maximize area with fixed fencing, minimize material for a can of fixed volume. Related rates differentiate equations with respect to time — a ladder sliding down a wall, a balloon inflating.

Try it yourself: Differentiate: (a) f(x) = x³ ln x, (b) g(x) = e^(2x)/x, (c) h(x) = cos³(x). Find and classify the critical points of f(x) = x³ − 3x².

Lesson 5: The Chain Rule and Implicit Differentiation

Most real functions are layered: sin(x²), (3x + 1)⁰, e⁻²ₛ. The Chain Rule peels the layers: if y = f(g(x)), then y′ = f′(g(x)) · g′(x) — “derivative of the outside, times derivative of the inside.” For y = (3x + 1)⁰: outside power rule gives 4(3x + 1)³, inside derivative gives 3, so y′ = 12(3x + 1)³. For y = sin(x²): y′ = cos(x²) · 2x. Forgetting the “times inside derivative” step is the classic chain rule error.

Implicit differentiation handles curves that are not functions, like the circle x² + y² = 25. Differentiate both sides with respect to x, treating y as a function of x: 2x + 2y y′ = 0, so y′ = −x/y. At the point (3, 4), the slope is −3/4, and the tangent line is y − 4 = −(3/4)(x − 3). Every time you differentiate a y-term, multiply by y′ by the chain rule.

Implicit differentiation also derives the derivatives of inverse functions and logarithms: differentiating ln y = x ln ... style relations gives the derivative of logarithmic and exponential functions, and differentiating sin y = x gives the derivative of arcsine.

Try it yourself: Find y′ for (a) y = e^(3x), (b) y = (x² + 2)⁵, (c) x³ + y³ = 9 at (2, 1). Answers: (a) 3e^(3x); (b) 10x(x² + 2)⁴; (c) 3x² + 3y²y′ = 0, y′ = −x²/y² = −4.

Key vocabulary: Chain Rule — (f(g(x)))′ = f′(g(x)) · g′(x); implicit differentiation — differentiating an equation in x and y to find y′.

Lesson 6: Related Rates

Related rates problems ask how fast one quantity changes given the rate of another, with both tied to time. The method: (1) write an equation relating the quantities, (2) differentiate both sides with respect to t (chain rule!), (3) substitute the known values and solve for the unknown rate. The equation comes from geometry — Pythagoras, similar triangles, volume formulas — and the rates come from the story.

Classic: a 13-foot ladder slides with its top falling at 2 ft/s when the bottom is 5 feet from the wall. How fast is the bottom sliding? Relation: x² + y² = 169. Differentiate: 2x dx/dt + 2y dy/dt = 0. When x = 5, y = 12, and dy/dt = −2: 2(5) dx/dt + 2(12)(−2) = 0, so 10 dx/dt = 48, dx/dt = 4.8 ft/s outward. Note the sign: dy/dt is negative because y is decreasing.

Common traps: substituting the numbers before differentiating (which kills the variables you need to differentiate), and unit mismatches. Always differentiate first, substitute second.

Try it yourself: A square's side grows at 3 cm/s. How fast is its area growing when the side is 10 cm? Answers: A = s², dA/dt = 2s ds/dt = 2(10)(3) = 60 cm²/s.

Key vocabulary: Related rates — rates of change linked by an equation; differentiate with respect to time — applying d/dt to both sides of a relation.

Lesson 7: Integrals and the Fundamental Theorem

If derivatives slice change into instants, integrals accumulate instants into totals. The definite integral ∫[a,b] f(x) dx is the limit of Riemann sums — areas of thin rectangles under the curve — giving net signed area: regions below the x-axis count negative. If f is velocity, the integral is displacement; if f is a density, the integral is total mass. The integral exists for continuous functions, no exceptions needed here.

The Fundamental Theorem of Calculus (FTC) is the bridge. Part 1: d/dx[∫[a,x] f(t) dt] = f(x) — differentiation undoes integration. Part 2: ∫[a,b] f(x) dx = F(b) − F(a), where F is any antiderivative (F' = f). This turns area problems into antiderivative evaluations: ∫[0,2] 3x² dx = [x³][0,2] = 8. Antiderivatives reverse the derivative rules: ∫x^n dx = x^(n+1)/(n+1) (n ≠ −1), ∫e^x dx = e^x, ∫(1/x) dx = ln|x| — plus the constant C, since derivatives kill constants.

U-substitution reverses the chain rule: for ∫ 2x·cos(x²) dx, set u = x², du = 2x dx, giving ∫cos(u) du = sin(u) + C = sin(x²) + C. For definite integrals, either convert the limits to u-values or substitute back before evaluating. Integration by parts (∫u dv = uv − ∫v du) reverses the product rule — choose u by LIATE priority (Log, Inverse trig, Algebraic, Trig, Exponential).

Try it yourself: Evaluate: (a) ∫[0,1] (4x³ − 2x) dx, (b) ∫ x e^(x²) dx (u-substitution), (c) ∫[0,π/2] sin x dx. What does your answer to (c) mean geometrically?

Lesson 8: Substitution (u-Substitution)

u-substitution is the chain rule in reverse. To integrate ∫ 2x cos(x²) dx, notice 2x is the derivative of x²: set u = x², du = 2x dx, and the integral becomes ∫ cos u du = sin u + C = sin(x²) + C. The telltale sign of a substitution problem: the integrand contains a function and (a multiple of) its derivative.

For definite integrals, you can either change the limits to u-values or substitute back to x before evaluating. For ∫₀¹ 2x e^(x²) dx with u = x²: when x = 0, u = 0; when x = 1, u = 1, so the integral becomes ∫₀¹ e^u du = e − 1. Changing the limits avoids the round trip.

Some substitutions need a constant adjustment: ∫ x² (x³ + 1)⁵ dx with u = x³ + 1 gives du = 3x² dx, so x² dx = du/3, and the integral is (1/3)∫ u⁵ du = (1/3)(u⁶/6) + C = (x³ + 1)⁶/18 + C. Always check by differentiating your answer.

Try it yourself: Evaluate (a) ∫ 4x³ cos(x⁴) dx, (b) ∫₀¹ 2x/(x² + 1) dx. Answers: (a) sin(x⁴) + C (u = x⁴, du = 4x³ dx); (b) ln 2 (u = x² + 1, limits 1 → 2).

Key vocabulary: u-substitution — replacing an inner function with u to undo the chain rule; integrand — the function being integrated.

Lesson 9: Integration by Parts

Integration by parts comes from the product rule: ∫ u dv = uv − ∫ v du. It handles products where one factor gets simpler when differentiated and the other is easy to integrate. For ∫ x e^x dx, choose u = x (simplifies to 1) and dv = e^x dx (v = e^x): the integral becomes x e^x − ∫ e^x dx = x e^x − e^x + C = e^x(x − 1) + C.

Choosing u follows the LIATE priority: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — pick the earliest type as u. For ∫ x sin x dx, u = x (algebraic beats trigonometric): x(−cos x) − ∫ (−cos x) dx = −x cos x + sin x + C. For ∫ ln x dx, u = ln x and dv = dx (v = x): x ln x − ∫ x(1/x) dx = x ln x − x + C.

Sometimes parts must be applied twice, or the original integral reappears: ∫ e^x sin x dx leads, after two rounds, to an equation in which the original integral appears on both sides — solve for it algebraically. This “boomerang” trick is a rite of passage in calculus.

Try it yourself: Evaluate ∫ x cos x dx. Answers: u = x, dv = cos x dx: x sin x − ∫ sin x dx = x sin x + cos x + C.

Key vocabulary: Integration by parts — ∫ u dv = uv − ∫ v du; LIATE — the priority order for choosing u.

Lesson 10: Applications — Area, Volume, and Motion

Integrals compute areas between curves: for f above g on [a, b], area = ∫[a,b] (f − g) dx. Find intersection points first — they set the limits. The area enclosed by y = x² and y = x: intersections at x = 0 and x = 1 (solve x² = x), area = ∫[0,1] (x − x²) dx = [x²/2 − x³/3][0,1] = 1/6. When curves cross, split the integral at each crossing so the integrand stays non-negative.

Volumes of revolution spin a region around an axis. The disk method: rotating y = f(x) about the x-axis gives V = π∫[a,b] f(x)² dx (each cross-section is a disk of radius f(x)). With a gap (region between two curves), washers: V = π∫ (R² − r²) dx. The shell method (V = 2π∫ radius·height dx) often wins when rotating around a vertical axis. A sphere of radius r emerges as V = (4/3)πr³ from rotating a semicircle — calculus proving geometry's formula.

In motion, calculus unifies the quantities: position s(t), velocity v = ds/dt, acceleration a = dv/dt. Given acceleration, integrate upward: a(t) = −9.8 (gravity) integrates to v(t) = v_0 − 9.8t, then to s(t) = s_0 + v_0 t − 4.9t² — the projectile formula derived, not memorized. Net change theorem: ∫[a,b] f'(x) dx = f(b) − f(a) — the integral of a rate is the total change. A water tank filling at r(t) liters/min accumulates ∫ r(t) dt liters.

Try it yourself: Find the volume when y = √x (0 ≤ x ≤ 4) is rotated about the x-axis. A particle has v(t) = 3t² − 2t; find its displacement from t = 0 to t = 2. Set up (don't evaluate) the integral for the area between y = x³ and y = 2x.

Lesson 11: Differential Equations — Separation of Variables

A differential equation is an equation involving a derivative: dy/dx = ky says “the growth rate is proportional to the current amount.” Separation of variables solves equations of the form dy/dx = f(x)g(y): move all y-terms to one side, all x-terms to the other, and integrate. For dy/dx = ky: dy/y = k dx, integrate: ln|y| = kx + C, so y = Ae^(kx). This one equation models population growth, radioactive decay, continuously compounded interest, and cooling.

An initial condition pins down the constant: if y(0) = 100 and k = 0.03, then y = 100e^(0.03x). A particular solution satisfies both the equation and the condition; the general solution has the undetermined constant. Direction fields — tiny slope ticks at grid points — let you sketch solution curves without solving at all.

Logistic differential equation dP/dt = kP(1 − P/K) adds the carrying capacity: growth slows as P approaches K. Its solution is the S-curve met in precalculus, now derived rather than assumed.

Try it yourself: Solve dy/dx = x with y(0) = 1. A bacteria culture satisfies dP/dt = 0.5P with P(0) = 200; find P(4). Answers: y = x²/2 + 1; P = 200e^(2) ≈ 1,478.

Key vocabulary: Differential equation — an equation involving derivatives; separation of variables — isolating x-terms and y-terms before integrating; initial condition — a known value selecting one solution.

Lesson 12: L'Hôpital's Rule and Indeterminate Forms

Some limits produce indeterminate forms — 0/0 or ∞/∞ — where the limit cannot be read off directly. L'Hôpital's Rule says that for such forms (when f and g are differentiable near a and g′ ≠ 0), lim f/g = lim f′/g′. For lim[x→0] sin x / x, both parts approach 0, so differentiate: lim cos x / 1 = 1. The rule turns a stuck limit into a computable one.

Other indeterminate forms can be rewritten into quotients: 0 · ∞ (rewrite as a fraction), ∞ − ∞ (combine), 1^∞, 0⁰, ∞⁰ (take logarithms). For lim[x→0⁺] x ln x (0 · ∞): rewrite as ln x / (1/x), giving ∞/∞; L'Hôpital's gives (1/x)/(−1/x²) = −x → 0. So x ln x → 0 as x → 0⁺, and x^x → 1.

L'Hôpital's Rule is a scalpel, not a hammer: it applies only to 0/0 or ∞/∞ forms, and you must re-check the form after each application — sometimes the rule must be applied twice, sometimes it never resolves and another method is needed. Never apply it to a form like 1/0.

Try it yourself: Find lim[x→0] (1 − cos x)/x² using L'Hôpital's Rule twice. Answers: 0/0 form; first: sin x/(2x), still 0/0; second: cos x/2 → 1/2.

Key vocabulary: Indeterminate form — 0/0, ∞/∞, etc., needing further analysis; L'Hôpital's Rule — lim f/g = lim f′/g′ for indeterminate quotients.

Lesson 13: Integration by Trigonometric Substitution and Partial Fractions

Two more integration techniques handle the cases substitution and parts cannot. Trigonometric substitution removes a radical: for √(a² − x²), set x = a sin θ; for √(a² + x²), set x = a tan θ; for √(x² − a²), set x = a sec θ. The radical collapses by a Pythagorean identity, the integral becomes trigonometric, and a reference triangle converts the answer back to x. The substitution is chosen by the form of the radical, not by guessing.

Partial fractions split a rational integrand into simple pieces: 1/((x + 1)(x + 2)) becomes A/(x + 1) + B/(x + 2), and each piece integrates to a logarithm. The method needs a proper fraction (divide first if the numerator's degree is not smaller) and a factored denominator. Together with substitution and parts, these techniques integrate nearly every function in a first calculus course.

Real-world application: Trig substitution computes the area of an ellipse and the arc length of a parabola; partial fractions integrate the rational models of chemical kinetics and electrical circuits. Choosing the technique is the skill; the algebra is the execution.

Try it yourself: Evaluate ∫ dx/(x² + 4x + 3). Answer: Factor to (x+1)(x+3); decompose 1/((x+1)(x+3)) = (1/2)/(x+1) − (1/2)/(x+3); integral = (1/2) ln|(x+1)/(x+3)| + C.

Key vocabulary: Trigonometric substitution — replacing x to remove a radical; partial fractions — splitting a rational integrand into logarithmic pieces.

Lesson 14: Improper Integrals

An improper integral has an infinite limit of integration or an integrand that blows up (a vertical asymptote) inside the interval. It is defined as a limit: ∫_a^∞ f(x) dx = lim_b→∞ ∫_a^b f(x) dx. If the limit exists and is finite, the integral converges; otherwise it diverges. The area under 1/x² from 1 to ∞ is finite (it equals 1), even though the region is infinitely long — the curve shrinks fast enough.

The p-integral test summarizes the power functions: ∫₁^∞ 1/x^p dx converges exactly when p > 1. The integral of 1/x (p = 1) diverges — the harmonic area is infinite, though the curve itself approaches zero. The comparison test extends this: if 0 ≤ f(x) ≤ g(x) and g converges, then f converges; if f diverges and f ≤ g, then g diverges. For a vertical asymptote at c, split the integral at c and take one-sided limits.

Real-world application: Improper integrals give total probabilities (the normal curve's total area is 1), expected lifetimes, and the total energy in an infinite signal. Gabriel's horn — finite volume, infinite surface area — is the famous paradox of improper integrals.

Try it yourself: Determine whether ∫₁^∞ 1/x³ dx converges, and evaluate it if so. Answers: p = 3 > 1, converges; antiderivative −1/(2x²), value 1/2.

Key vocabulary: Improper integral — an integral with an infinite bound or unbounded integrand; converges/diverges — the defining limit exists / does not exist.

Lesson 15: Sequences and Series: Convergence Tests

An infinite series Σ a_n either converges to a finite sum or diverges. The first check is the nth-term test: if a_n does not approach 0, the series diverges (if a_n does approach 0, the test is inconclusive). The geometric series Σ arⁿ converges to a/(1 − r) exactly when |r| < 1. The integral test compares a positive decreasing series to the improper integral of the matching function.

Other tests: the comparison tests (direct and limit) compare with a known series; the ratio test computes L = lim |a_n+1/a_n| — converges if L < 1, diverges if L > 1, inconclusive if L = 1; the root test uses lim |a_n|^1/n the same way; the alternating series test says Σ (−1)ⁿ b_n converges if b_n decreases to 0. Absolute convergence (convergence of Σ |a_n|) implies convergence; a series that converges but not absolutely converges conditionally.

Real-world application: Series approximate functions in calculators and physics (a calculator's sin button uses a series); convergence tests decide whether a numerical method's infinite sum is trustworthy. The alternating harmonic series converges conditionally to ln 2.

Try it yourself: Does Σ (−1)ⁿ/n converge absolutely, conditionally, or diverge? Answer: Σ 1/n diverges (harmonic), but the alternating series test gives convergence (1/n decreases to 0), so it converges conditionally.

Key vocabulary: Convergence tests — nth-term, integral, comparison, ratio, root, alternating series; absolute/conditional convergence — convergence of Σ |a_n| / convergence without it.

Lesson 16: Taylor and Maclaurin Series

A Taylor series represents a function as an infinite polynomial: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x − a)ⁿ. A Maclaurin series is the special case a = 0. The partial sums are Taylor polynomials: better and better approximations near a. The familiar series are worth memorizing: e^x = Σ xⁿ/n!; sin x = Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)!; cos x = Σ (−1)ⁿ x²ⁿ/(2n)!; 1/(1 − x) = Σ xⁿ for |x| < 1.

Two questions matter: where does the series converge (the radius and interval of convergence), and how big is the error of a partial sum? The ratio test usually finds the radius. Taylor's inequality bounds the remainder: if |f⁽ⁿ⁺¹⁾(x)| ≤ M on the interval, then |R_n(x)| ≤ M|x − a|ⁿ⁺¹/(n+1)!. For alternating series, the alternating series estimation theorem says the error is at most the first omitted term. Calculators and computers evaluate sin, cos, and e^x with these series.

Real-world application: GPS receivers, physics engines, and calculators use Taylor polynomials to evaluate transcendental functions. Engineers approximate a nonlinear system by its Taylor polynomial near an operating point — linearization is just the first-degree Taylor polynomial.

Try it yourself: Write the first 3 nonzero terms of the Maclaurin series for sin x and use them to estimate sin 0.2. Answers: x − x³/6 + x⁵/120; sin 0.2 ≈ 0.2 − 0.001333 + 0.00000267 ≈ 0.19867 (true: 0.198669).

Key vocabulary: Taylor series — a function's infinite polynomial centered at a; Maclaurin series — centered at 0; radius of convergence — how far from the center the series converges.

Lesson 17: Parametric Equations, Polar Coordinates, and Calculus

For a curve given parametrically by x = f(t), y = g(t), the slope is dy/dx = (dy/dt)/(dx/dt), found by the chain rule — no need to eliminate the parameter. The arc length from t = a to b is ∫ √((dx/dt)² + (dy/dt)²) dt, the integral of the speed. The area under a parametric curve is ∫ y dx = ∫ g(t) f′(t) dt. These formulas reduce to the familiar ones when the curve is a function graph.

In polar coordinates, the area enclosed by r = f(θ) from θ = a to b is (1/2)∫ r² dθ, and the slope is found from x = r cos θ, y = r sin θ with the parametric formula. The cardioid r = 1 + cos θ encloses area (1/2)∫₀^2π (1 + cos θ)² dθ = 3π/2. Parametric and polar calculus describe orbits, spirals, and curves that fail the vertical line test.

Real-world application: Planetary orbits, satellite paths, and roller-coaster tracks are parametric curves; radar displays are polar. The arc-length integral tells a CNC machine how far its tool travels along a curved cut.

Try it yourself: Find the arc-length integrand for x = cos t, y = sin t. Answer: dx/dt = −sin t, dy/dt = cos t, speed = √(sin² + cos²) = 1, so arc length over [0, 2π] is 2π (the unit circle's circumference).

Key vocabulary: Parametric slope — dy/dx = (dy/dt)/(dx/dt); polar area — (1/2)∫ r² dθ; arc length — the integral of speed.

Lesson 18: Vectors and Motion in Space

A vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ describes motion in space. Its derivative r′(t) = ⟨x′(t), y′(t), z′(t)⟩ is the velocity vector, pointing tangent to the path; its magnitude |r′(t)| is the speed. The second derivative r′′(t) is the acceleration vector. Integrating acceleration with initial velocity and position recovers the motion — exactly how a navigation system reconstructs a path from accelerometer readings.

Two vector operations matter here. The dot product u · v = |u||v| cos θ tests perpendicularity (zero means perpendicular) and finds components. The cross product u × v is a vector perpendicular to both, with magnitude |u||v| sin θ — the area of the parallelogram they span. The unit tangent vector T = r′/|r′| gives direction; acceleration splits into tangential (speeding up) and normal (turning) components.

Real-world application: Spacecraft trajectories, robot-arm motion, and video-game physics all run on vector-valued functions: position, velocity, and acceleration vectors updated every frame. The cross product orients cameras and computes surface normals for lighting.

Try it yourself: r(t) = ⟨cos t, sin t, t⟩. Find r′(t) and the speed. Answers: r′(t) = ⟨−sin t, cos t, 1⟩; speed = √(sin² + cos² + 1) = √2 (constant — a helix climbed at steady speed).

Key vocabulary: Vector-valued function — r(t) giving position over time; velocity/acceleration vectors — first and second derivatives; dot/cross product — scalar / perpendicular-vector products.

Lesson 19: Applications: Work, Fluid Pressure, and Centers of Mass

Work is force times distance, but when the force varies, work is an integral: W = ∫ F(x) dx. Pumping water out of a tank requires lifting each horizontal slice: a slice at depth h with area A and thickness dy weighs 62.4 · A · dy pounds (water weighs 62.4 lb/ft³) and must be lifted h feet, so the work is ∫ 62.4 · A(y) · h(y) dy. Hooke's law for a spring, F = kx, gives W = ∫ kx dx = (1/2)kx² — the familiar spring-energy formula.

Fluid pressure at depth h is P = wh (weight density times depth), and the force on a submerged surface is F = ∫ (wh) · (width) dh — pressure times area, integrated because pressure varies with depth. The center of mass (centroid) of a region is the balance point: for a region under y = f(x) from a to b, the centroid's x-coordinate is (1/A)∫ x f(x) dx, where A is the area. The centroid of a triangle is at the average of its vertices; symmetry often gives a centroid coordinate for free.

Real-world application: Civil engineers compute the force on a dam (fluid pressure integrated over its face); mechanical engineers find centroids to balance rotating parts; petroleum engineers compute pumping work for oil wells. The integral turns a varying force into a total.

Try it yourself: A spring with k = 20 N/m is stretched from 0 to 0.5 m. How much work is done? Answer: W = ∫₀^0.5 20x dx = 10x² from 0 to 0.5 = 2.5 J.

Key vocabulary: Work — ∫ F(x) dx for a varying force; fluid pressure — P = wh at depth h; centroid — the balance point of a region.

Lesson 20: Calculus Capstone: Review and the Big Theorems

The capstone unifies the course around its big theorems. The Mean Value Theorem: if f is continuous on [a, b] and differentiable inside, some c in (a, b) has f′(c) = (f(b) − f(a))/(b − a) — the instantaneous rate equals the average rate somewhere. The Fundamental Theorem of Calculus: differentiation and integration undo each other; the integral of a rate is the total change. L'Hôpital's Rule resolves 0/0 and ∞/∞ limits with derivatives. Taylor's theorem says smooth functions are, locally, polynomials.

Capstone problem: a car's position is s(t) = t³ − 6t² + 9t (meters, seconds). Velocity v = 3t² − 12t + 9 = 3(t − 1)(t − 3); acceleration a = 6t − 12. The car moves forward when v > 0 (t < 1 or t > 3), backward between t = 1 and 3. Total distance over [0, 4] is ∫ |v| dt = [s(1) − s(0)] + [s(1) − s(3)] + [s(4) − s(3)] = 4 + 4 + 4 = 12 m, while displacement is s(4) − s(0) = 4 m. Distance needs absolute value; displacement does not.

Real-world application: The Mean Value Theorem justifies average-speed cameras (if your average speed exceeded the limit, your instantaneous speed did too, somewhere). The Fundamental Theorem powers every accumulation computation in science and engineering.

Try it yourself: A particle's position is s(t) = t² − 4t. Find its velocity, when it is at rest, and its displacement from t = 0 to t = 5. Answers: v = 2t − 4; at rest at t = 2; displacement s(5) − s(0) = 5 − 0 = 5.

Key vocabulary: Mean Value Theorem — instantaneous rate equals average rate somewhere; displacement vs. distance — net change vs. total path length.