Mathematics for Physics · Chapter 22

Calculus Essentials

Calculus is the mathematics of change and accumulation. Every equation of motion, every field law, every thermodynamic relation in this curriculum is written in the language of calculus.

PrerequisitesAlgebraicmanipulation⋅Functionsandgraphs⋅BasictrigonometryAlgebraic manipulation \cdot Functions and graphs \cdot Basic trigonometry
Learning Goals
  • Interpret derivatives as instantaneous rates of change and connect position, velocity, and acceleration.
  • Apply standard derivative rules including the product rule and chain rule.
  • Interpret definite integrals as accumulated quantities and use the Fundamental Theorem of Calculus.
  • Choose substitution or integration by parts for common physics integrals.
  • Read partial derivative notation and explain what variables are held fixed.

22.1 The Derivative

The derivative of a function f(x) at a point x measures the instantaneous rate of change — the slope of the tangent line. It is defined as the limit of the difference quotient:

f′(x)=df/dx=limh→0[f(x+h)−f(x)]/hf'(x) = df/dx = lim_{h\to0} [f(x+h) - f(x)] / h(22.1)

In physics, the most common derivative is with respect to time. If x(t) is position, then ẋ = dx/dt is velocity, and ẍ = d²x/dt² is acceleration. Newton's second law F = ma is F = m d²x/dt² — a differential equation.

Theorem 22.1 — Standard Derivatives
d/dx[xnd/dx [x^{n} =nxn−1nx^{n-1}d/dx [eˣ] =eˣd/dx [ln x] =1/xd/dx [sin x] =cos xd/dx [cos x] =−sinx-sin xd/dx [cf(x)] =c f'(x)d/dx[f\cdotgd/dx [f\cdotg =f'g + fg' (product rule)d/dx [f(g(x))] =f′(g)\cdotg′(chainrulef'(g)\cdotg' \quad (chain rule
Example 22.1 — Velocity and Acceleration from Position

Aparticle′spositionisx(t)=3t3−12t2+9t+2.Finditsvelocityandacceleration,aA particle's position is x(t) = 3t^{3} - 12t^{2} + 9t + 2. Find its velocity and acceleration, and determine when it is momentarily at rest.

Velocity:v(t)=dx/dt=9t2−24t+9v(t) = dx/dt = 9t^{2} - 24t + 9
Acceleration:a(t)=dv/dt=18t−24a(t) = dv/dt = 18t - 24
At rest (v=0):9t2−24t+9=0→t2−(8/3)t+1=0→t=[8/3±(64/9−4)]/29t^{2} - 24t + 9 = 0 \to t^{2} - (8/3)t + 1 = 0 \to t = [8/3 \pm \sqrt(64/9 - 4)]/2
Solve:t=[8/3±(28/9)]/2=(8±28)/(6)→t≈0.45sandt≈2.22st = [8/3 \pm \sqrt(28/9)]/2 = (8 \pm \sqrt28)/(6) \to t \approx 0.45 s and t \approx 2.22 s

22.2 The Integral

The integral is the inverse of the derivative and measures accumulation. The definite integral ∫_a^b f(x) dx gives the area under f(x) from a to b, and by theFundamental Theorem of Calculus:

∫abf(x)dx=F(b)−F(a)whereF′(x)=f(x)\int_a^b f(x) dx = F(b) - F(a) \qquad where F'(x) = f(x)(22.2)

F is the antiderivative. The indefinite integral ∫f(x)dx = F(x) + C finds the family of antiderivatives. In physics, integrating velocity gives displacement: x = ∫v dt; integrating force gives work: W = ∫F·dx; integrating power gives energy.

Theorem 22.2 — Standard Antiderivatives
∫xndx(n≠−1\int x^{n} dx (n≠-1 =xn+1/(n+1)+Cx^{n+1}/(n+1) + C∫exdx\int eˣ dx =eˣ + C∫1/xdx\int 1/x dx =ln|x| + C∫sinxdx\int sin x dx =−cosx+C-cos x + C∫cosxdx\int cos x dx =sin x + C∫eaxdx\int eᵃˣ dx =eᵃˣ/a + C∫1/(1−x2)dx\int 1/\sqrt(1-x^{2}) dx =arcsin x + C∫1/(1+x2)dx\int 1/(1+x^{2}) dx =arctan x + C
Example 22.2 — Work Done by a Variable Force

AspringexertsforceF(x)=−kx(Hooke′slaw).Findtheworkdonecompressingitfromx=A spring exerts force F(x) = -kx (Hooke's law). Find the work done compressing it from x = 0tox=A(amplitude0 to x = A (amplitude

Work integral:W=∫0AFdx=∫0Akxdx(wepushwith+kxagainstthespringW = \int_{0}^A F dx = \int_{0}^A kx dx \qquad (we push with +kx against the spring
Integrate:W=k[x2/2]0A=kA2/2W = k [x^{2}/2]_{0}^A = kA^{2}/2
Result:W=12kA2—thisisexactlytheelasticpotentialenergystoredinthespring.✓W = \frac{1}{2}kA^{2} — this is exactly the elastic potential energy stored in the spring. ✓

22.3 Techniques of Integration

Three techniques cover most physics integrals:

Definition 22.1 — Substitution (u-substitution)
Iftheintegrandcontainsf(g(x))\cdotg′(x),letu=g(x),du=g′(x)dxIf the integrand contains f(g(x))\cdotg'(x), let u = g(x), du = g'(x)dx:∫f(g(x))g′(x)dx=∫f(u)du\int f(g(x)) g'(x) dx = \int f(u) duExample:∫2x\cdotsin(x2)dx→letu=x2,du=2xdx→∫sinudu=−cos(x2)+CExample: \int 2x\cdotsin(x^{2}) dx \to let u = x^{2}, du = 2x dx \to \int sin u du = -cos(x^{2}) + C
Definition 22.2 — Integration by Parts
Derivedfromtheproductrule:d(uv)=udv+vdu→\intudv=uv−\intvduDerived from the product rule: d(uv) = u dv + v du \to \intu dv = uv - \intv du.∫udv=uv−∫vdu\int u dv = uv - \int v duChooseutobethefactorthatsimplifieswhendifferentiated;dv=theremainder.ExampleChoose u to be the factor that simplifies when differentiated; dv = the remainder. Example: ∫t\cdoteatdt—letu=t,dv=eatdt→teat/a−∫eat/adt=(t/a−1/a2)eat+C\int t\cdoteᵃᵗ dt — let u = t, dv = eᵃᵗ dt \to te^{at}/a - \int eᵃᵗ/a dt = (t/a - 1/a^{2})eᵃᵗ + C
Example 22.3 — Moment of Inertia by Integration

Find the moment of inertia of a uniform thin rod of mass M and length L about its center.

Setup:I=∫r2dm.Linearmassdensity:λ=M/L,sodm=λdx=(M/L)dxI = \int r^{2} dm. Linear mass density: \lambda = M/L, so dm = \lambda dx = (M/L) dx
Limits:Rodextendsfromx=−L/2tox=+L/2Rod extends from x = -L/2 to x = +L/2
Integrate:I=∫from−L/2toL/2ofx2(M/L)dx=(M/L)[x3/3]evaluatedfrom−L/2toL/2I = \int from -L/2 to L/2 of x^{2}(M/L) dx = (M/L) [x^{3}/3] evaluated from -L/2 to L/2
Evaluate:I=(M/L)×(1/3)[(L/2)3−(−L/2)3]=(M/L)(1/3)(L3/4)=I = (M/L) \times (1/3)[(L/2)^{3} - (-L/2)^{3}] = (M/L)(1/3)(L^{3}/4) = ML2/12ML^{2}/12

22.4 Partial Derivatives and Gradients

When a function depends on multiple variables, the partial derivative∂f/∂x treats all other variables as constants. The gradient vector collects all partial derivatives and points in the direction of steepest increase:

\nablaf=(\partialf/\partialx)ı^+(\partialf/\partialy)ȷ^+(\partialf/\partialz)k^\nablaf = (\partialf/\partialx) î + (\partialf/\partialy) ĵ + (\partialf/\partialz) k̂(22.3)

In physics: electric field E = −∇V (gradient of potential); force from potential energy F = −∇U. The divergence ∇·E and curl ∇×Bappear in Maxwell's equations, connecting all of electromagnetism.

Definition 22.3 — Common Traps
  • A derivative is local rate of change: it is not always a constant slope.
  • Definite integrals accumulate signed area: negative regions subtract.
  • Use the chain rule with nested functions: missing inner derivatives is a common source of errors.
  • Partial derivatives hold other variables fixed: the constraint matters in physics formulas.
Exercises — 22.1–22.4 Calculus
1.
Findthederivativeoff(x)=3x3−12x2+9x+2andevaluateitatx=1Find the derivative of f(x) = 3x^{3} - 12x^{2} + 9x + 2 and evaluate it at x = 1.
Straightforward
2.
Findtheworkdonecompressingaspring(F=kx)fromx=0tox=A.ExpressasamultiplFind the work done compressing a spring (F = kx) from x = 0 to x = A. Express as a multipleofkA2e of kA^{2}.
kA²
Straightforward
3.
A uniform rod (mass M, length L) rotates about its center. Find its moment of inertia I.
ML²
Intermediate
4.Evaluate∫0πxsinxdxusingintegrationbypartsEvaluate \int_{0}^\pi x sin x dx using integration by parts.
Intermediate
5.Derive the gravitational field g(r) both inside and outside a uniform solid sphere of mass M and radius R. At what radius is g maximum?
Challenging
Key Takeaways
  • Derivativef′(x)=lim[f(x+h)−f(x)]/h—instantaneousrateofchange;slopeoftangentDerivative f'(x) = lim[f(x+h)-f(x)]/h — instantaneous rate of change; slope of tangent.
  • Fundamentaltheorem:∫abfdx=F(b)−F(a),linkingdifferentiationandintegrationFundamental theorem: \int_a^b f dx = F(b) - F(a), linking differentiation and integration.
  • Power rule, chain rule, product rule — these three handle 90% of physics derivatives.
  • Standard techniques: u-substitution reverses chain rule; integration by parts reverses product rule.
  • Partialderivativestreatothervariablesasconstants;thegradient\nablafpointstowardsteePartial derivatives treat other variables as constants; the gradient \nablaf points toward steepest increase.
  • E=−\nablaVandF=−\nablaUconnectfieldequationstopotential—calculusisthebridgeE = -\nablaV and F = -\nablaU connect field equations to potential — calculus is the bridge.